Looking around, it seems this is kind of a “folklore proof,” in that I haven’t found a fully rigorous version in any book, but everybody knows about the idea. I liked the proof a lot, so here I’ll present a short account of it.

Implicitly, this idea takes the nice correspondence of nonsingular curves with Riemann surfaces as the origin of genus of curves. Riemann surfaces, once you forget the complex structure, just become smooth orientable real surfaces, and that’s pretty much the appeal – genus for smooth orientable surfaces is very geometric (“number of handles”), and probably the first place people see “genus,” so it’s an intuitive way to correspond it to the algebraic case. Of course, in modern treatments, the first principles of genus (arithmetic and geometric) are usually taken to be cohomological, but who cares.

So let’s say we have a projective plane curve of degree – hence, defined by a homogenous polynomial of degree . The degenerate case is just a union of lines. Generically, each pair of lines intersects at a point; taking the analytification correspondences, this is the wedge sum of spheres.

The idea is that by shifting the coefficients slightly to make the curve nonsingular, the global topology stays the same besides at the singular wedge points, which open up into holes between the spheres. Then, if we take one sphere as a “central sphere,” it’s clear that each pair of the remaining spheres creates a handle, so for the Riemann surface, we get , as desired. (Actually, this first stage can be replaced by ANY exhibition of a curve with genus , so there are many ways to proceed, but this one is most in line with the spirit of the proof.)

This is the first step which is usually skipped: proving that this is indeed what happens to the topology upon perturbing the coefficients to get a nonsingular curve. To rigorize this argument, we use the Milnor fibration.

Let our curve be ; initially take it to be hyperplanes intersecting generically, with the consequent polynomial being a product of linear factors. What’s more, we can take all intersections to be away from, say , so we affinize in and to , defining an affine slice of the curve where all the singularities happen. Fix one singularity; WLOG it is at the origin . If we take a ball of sufficiently small radius about the origin of , then for every sufficiently small disc of radius about the origin of the complex plane, the map is a locally trivial fibration.

I won’t go into the general theory of Milnor fibers here; consult chapter 6 of Wall’s book for a treatment. It is clear in this case that the Milnor fiber is a cyllinder (without the ends); see lemma 6.3.3. To visualize why this is true, just picture what is happening locally as the hyperbola .

But the Milnor fiber above is precisely the curve defined by in the small ball . This same logic applies to every singularity, so if we make a sufficiently small perturbation to the constant term, all the singularities indeed get resolved into connecting tubes. For sufficiently small , we have to check that the global geometry away from the singularities does not change.

Notice on the full projective curve, what we have done is replace with . In particular, there can clearly be no new singularities on the hyperplane , so for the global geometry, we can continue restricting ourselves to the affine case. In this case, the Milnor map at infinity, the same as above where we take to be an arbitrary **large** ball, is also a locally trivial fibration for a certain class of polynomials, known as tame polynomials; see Nemethi and Zaharia. This class includes all polynomials with isolated critical points, including our . Hence *globally* the analytic surfaces corresponding to , parametrized by , are all topologically the same for in some punctured disc in the complex plane. In particular, no new singularities can appear, and the global geometry away from the singularities remains the same. Thus we get spheres pairwise joined by little tubes, and consequently the desired genus of

Now that we have our “base case,” we work inside the parameter space: the homogenous polynomials of degree in the coordinate ring , which we will denote , given the obvious topology as a -vector space – finite dimensional, too, since there are finitely many monomials of fixed degree. We have the subspace of polynomials yielding nonsingular curves .

First, we will show that genus is locally constant on this subspace. This is the other rigorizing section typically skipped. Let be the monomials of degree , so that they are a basis for . Then take the projective variety defined by the homogenous (in ) coordinate ring – here we are affine in , projective in . There is an obvious rational map (affine space over ), whose fiber over a point is precisely the degree curve with those coefficients on the monomials. Every curve is equivalent to some curve in this family because the coefficient of can always be made nonzero via a projective transformation.

Notice that it’s clear that the only singular points of are singular points on the individual curves. Further, it’s not hard to see that the image of the singular curves on affine space is Zariski closed, since it ultimately consists of polynomial conditions on the coefficients, by the theory of resultants. Hence if we take the analytification of the morphism , yielding , by local compactness we can find an (analytic/Euclidean) compact neighborhood of a point on whose fiber is a nonsingular curve. The preimage of this is compact, as a closed subset of , since every Proj construction yields a compact manifold and closed subsets of compact spaces are compact.

Then restricted to this preimage is a surjective map of a smooth manifold onto its image, which is clearly a submersion by checking algebraic tangent spaces. It is also a (analytic) proper map by construction. Hence by We will use Ehresmann’s theorem, it’s a locally trivial fibration, so every fiber is the same topologically, hence has the same genus. By identifying with -scaling orbits in , we have our desired result.

Finally, it’s a snap to show that is connected. Indeed, inside the vector space , it’s the complement of . A polynomial defines a singular curve if and only if there is a common zero of with its partial derivatives in , , and simultaneously. Since the partial derivative is just a linear transformation of the coefficients (hence of the vector space), by the theory of resultants, this corresponds to a codimension surface in considered as an affine space in the coefficients.

Thus, is indeed connected as the complement of a codimension , so genus is constant. The result follows.

One reason I like this proof a lot is that it uses proto-versions of two big ideas in algebraic geometry. First, family arguments, where we turn an easy case into a hard one while keeping something invariant, hinting towards cycle classes, flat families, and deformation theory. Indeed, the second step is trivial with the theory of flat families, just a specialization of the fact that the Hilbert polynomial is constant. Second, working with a geometric space which parametrizes the objects we’re talking about (a proto-moduli space).

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It turns out both Dirichlet’s unit theorem and the finiteness of class number, when looked at the right way, have all their “deepness” hidden in a notion of compactness in a certain lattice, and this can be made explicit using the adeles (roughly, a way to glue together all the places of a global field). This is the beginning of many different long, strange stories, none of which I really understand yet; hopefully, this will be the first of many posts detailing my exploration.

This post already assumes a basic familiarity with -adics, places, and valuation theory. A secondary aim of this post is to serve as an introduction to adeles and the adelic perspective. It is somewhat atypical, in that it will not prove or even discuss many of the elementary properties of adeles other guides will, but I think that this introduction via application and intuition is in many ways preferable. Optimally, it would be read in conjunction with a more conventional introduction to adelic number theory, which will march through all the basic properties I gloss over.

**I. Finiteness of class number and Dirichlet’s unit theorem, revisited**

Let’s reiterate the proofs of finiteness of class number and Dirichlet’s unit theorem quickly, phrasing them in a manner so as to emphasize the role of compactness. We’ll work over the (hopefully) familiar case of a number field , though will keep an eye towards the end goal of generalizing to any global field.

**Aside:** a *global field*, if you’re unfamiliar with the term, is either a number field, i.e. a finite extension of , or a function field of an algebraic curve over a finite field, i.e. finite extension of . This might seem pretty arbitrary, but there’s a reason for it beyond the (in)famous finite function field/number field analogy: the completions of such fields at their places give rise to all the local fields, which are precisely the locally compact topological fields. Working backwards, we can use this to give an abstract characterization of global fields in terms of the product formula for valuations. Indeed, I think this is probably the strongest justification for the number field/finite function field analogy in the first place.

**Theorem.** Let be a number field, with ring of numbers , and let be the associated fractional ideal group. Let the class group be defined as ; i.e. fractional ideals modulo principal ones. Then is finite.

**Aside:** this is also a familiar construction in algebraic geometry: the *divisor group* of formal linear combinations of height-1 prime ideals (so, all of them, in a Dedekind domain, but in general, codimension 1 subvarieties, geometrically speaking) by the subgroup of * principal divisors*, formal linear combinations arising from an element of (*principal divisors*), where we include in the linear combination each prime appearing in the factorization of the ideal generated by the element, multiplied by its multiplicity in the factorization. It is easy to see how this matches up with the more number theoretic definition. For “nice” schemes, including the ones we’re working with, the class group is actually naturally isomorphic to , the Picard group of “line bundles” (locally free rank one modules) over the structure sheaf of with tensor products, so the two are sometimes used interchangeably. In the global field context, “line bundles” are associated with fractional ideals, linear equivalence by multiplication by a principal ideal, and all ideals are flat modules, so tensor products are just ideal multiplication, so this correspondence is obvious.

**Proof.** There are two cleanly separated steps: to show that every ideal class contains an integral ideal with uniformly bounded norm, and to show that there are finitely many integral ideals with bounded norm. The latter is obvious for number fields, since a prime ideal lying over has norm divisible by .

The former is (generally) proven using Minkowski’s theorem, applied to the lattice created by the embedding into Minkowski space (considered as -algebras).

**Aside:** You will also see Minkowski space described more explicitly in a non-canonical way involving choosing one of each pair of complex embeddings, which is workable but less neat – the volume calculations of the fundamental parallelpiped in terms of the Lebesgue measure on the resulting space introduces an awkward power of two, which of course cancels out in the end, but is still bothersome. A canonical alternate description from Wikipedia is the pithy “fixed subspace of the embedding into under complex conjugation” (where is the degree of the extension), which is pretty misleading, since it means complex conjugation acting simultaneously on the coordinates themselves, and to permute the coordinates by acting on the associated embeddings. It is an easy exercise to see how this identifies with the tensor definition, and that the natural volume forms on the two agree.

Then it is straightforward to see that the square root of the determinant is the volume of the fundamental parallelpiped (considering as the lattice), and the lattice determined by any given ideal can be calculated in terms of its index (ideal norm). The standard way to finish from here is to take the lattice associated to the inverse of an arbitrary fractional ideal, then to use a suitably scaled convex body to find a lattice point corresponding to an element of that multiplies the ideal to become integral, yet has sufficiently small norm to uniformly bound the norm resulting representative of the ideal class. The technical details are boring and meaningless.

So what’s actually happening with this Minkowski space business? The perspective we need to take is that each coordinate corresponds to a distinct real/complex *place* of – that is, an equivalence class (equivalence under induced topology) of absolute values on it. In this case, we are only considering the *archimedean* (non--adic, with the Archimedean property) places, so by Ostrowski’s theorem there is a unique one on , corresponding to the usual absolute value. By extension, each real embedding and each Galois orbit of complex embeddings corresponds to a unique absolute value and hence place on general . Then looking at Minkowski space as a product of s and s, in general, the space should be a product of completions of with respect to its various places, with canonically embedded inside.

We will need to extend our theory of “norm” here, but the already-present absolute values are suggestive enough of the general idea that we can await for this to arise naturally in the adelic perspective later. More pressing is the need to connect this to the class group. Heuristically, the idea that we can find ideal class representatives with bounded norm will correspond to compactness of the class group, and finiteness of integral ideals with bounded norm will correspond to discreteness; the proper way to think of finiteness of , then, is as a general topological consequence of compactness and discreteness.

To precisely give the correct link between as a topological group and Minkowski space, we require adelic language and concepts, but we can give a preview here. Elements of the fractional ideal group, since we can write them as unique products of prime ideals, can be identified with a kind of “-adic lattice” of their valuations at every prime (or, their nonarchimedean places). (Equivalently, the fractional ideal group is the free abelian group generated by the primes – again, this is basically the idea of associated divisors from algebraic geometry.) So this group is already discrete! (Discrete under what topology? Well, I did say this would be an imprecise preview.) And one can clearly see the link between the discreteness of the prime lattice, and the argument we saw above for finiteness of ideals with bounded norm.

To get the class group, we just have to quotient out by the sublattice of principal ideals, induced by the embedding of . If we can prove the result is compact, we’re done – so we have isolated the compactness argument.

To do this, we find a continuous surjection from a compact space onto it: we use the subset of our lattice which corresponds to ideals with bounded norm! The role of the Minkowski’s theorem argument in our original proof is to demonstrate we can find this subset; this will be replaced by a similar adelic result later. The oddly circumspect route (finding a lattice point in Minkowski space corresponding to inverse ideal, then multiplying and seeing that the dependence on ideal norms cancels) of the classical proof will be streamlined in the adelic result, which considers the archimedean topology of Minkowski space and the non-archimedean properties of the ideals simultaneously, as we will see.

Adopting a slightly more general perspective on the ring of integers and the class group will help throw this into better relief, by way of explicitly involving the non-archimedean (-adic) places of . Let be a finite set of places of , containing all the archimedean places. Then analogously to , we define to be the ring of integers of localized at all the places of , so that the primes corresponding to non-archimedean become units. ( exactly the set of archimdean places corresponds to the usual .) We can analogously define the class group or . It is not hard to demonstrate the following exact sequence, reproduced from Neukirch’s wonderful book:

(Note: the middle term has a more natural expression adelically, which we’ll see later. Each summand is also isomorphic to , by the associated valuation; here’s our “-adic lattice”!) This sequence by itself shows that is finite given the result for the usual case. But our compactness approach generalizes easily to prove the result for all simultaneously. This is really just two slightly different algebraic manifestations of the same point of view, because can be identified with a quotient of the “-adic lattice” we talked about earlier, eliminating the dimensions corresponding to the places of . Hence in fact there is a surjection from the normal class group into the -class group, as we see in the above exact sequence, and we are given another hint of what its kernel, the middle term in the exact sequence, is adelically.

Notice this makes sense, since the -ring of integers has fewer ideals, since we turned some elements into units.

**Aside:** You may be wondering, as I did: is there a way for this to make sense if does not contain all the archimedean places? No, obviously not, stupid.

Our treatment of Dirichlet’s unit theorem will be brief by contrast, since the compactness and adelic viewpoint there are practically begging to be discovered.

**Theorem.** Given a number field with real embeddings and pairs of complex embeddings, is a finitely generated abelian group; in particular, it is the product of a finite torsion group (the roots of unity) with a free abelian group of rank .

We again have the classical embedding of and hence of into Minkowski space; take log absolute values of each coordinate (the kernel is clearly the roots of unity, which we can prove are finite in number with a separate argument). By considering the norm condition on a unit, the image lies on a codimension hyperplane. If we can prove that the fundamental domain associated to the lattice (the cokernel of the map) is compact, we have that it is a full lattice on this dim- hyperplane, so we’re done.

The usual way to prove this using Minkowski’s theorem, again. Every algebraic number theory text I’ve seen does this in a somewhat different (though, modulo convolution, basically equivalent) way, but many of them obscure what’s actually going on behind seemingly arbitrary ideas and strange constructions – except for Neukirch, another reason it’s the only ANT book worth its salt. The idea which Neukirch lays bare is that we directly show that the fundamental domain is compact by showing it lies in the image of a compact set, the same trick as our earlier proof.

To see how this works, let’s go back and look at our embedding more closely. What we did was form the chain , where the second arrow is taking logarithms of absolute values (and also where the roots of unity disappear into the kernel). Inside we have our norm-one hyperplane , whose preimage in is a codimension-one hypersurface , where the unit group is injectively embedded – indeed, we could have done the entire thing (besides throwing out roots of unity) sans log-linearizing by thinking of the unit group as a “lattice” (discrete abelian subgroup) on the abelian variety . So we construct our preimage as the compact “fundamental domain” of (modulo the unit “lattice”) inside Minkowski space, as this is the more natural setting to work in.

**Aside:** this is the reason why most books’ treatment of this material is very bad: the log-linearization is nice because it handles the roots of unity, and we get the lattice and hyperplanes, etc. to work with so we don’t need a theory of multiplicative abelian varieties and the “lattice” on it. So books just work everything out inside , often using superfluous linear algebraic ideas. But Minkowski space is the more natural setting to make the geometric/number-theoretic connections apparent.

The idea is that we simply take a “box” in where the absolute value of each coordinate is bounded by a certain number, such that the product of the bounds is large enough so we can apply Minkowski’s theorem. Then we take the union of translates of the box over a collection of algebraic integers which can correspond to every possible norm less than that product. (Again, we see the finiteness of integral ideals of a given norm in action.) The intersection of these boxes with our surface gives us our bounded covering of the fundamental domain, after a little technical legwork, and we are done.

To generalize to -units , the idea is very easy: just add on the part of the “-adic lattice” corresponding to the non-archimedean places of onto Minkowski space, and repeat an almost identical argument. Indeed, this is basically captured by the exact sequence from earlier: there is an injection , whose image is the kernel of . This latter map has a finite kernel since the class group is finite, so the kernel only affects the torsion part of . Hence the rank of the free part is just (since and account for the archimedean valuations, and the primes account for the non-archimedean).

In summary, we saw that finiteness of class number and Dirichlet’s unit theorem have proofs which can be seen in a very similar way, as the compactness of a quotient of the “-adic lattice” and the compactness of a quotient of a hypersurface in Minkowski space, respectively. Look back at our exact sequence and otice that extending these results to -rings entails “subtracting dimensions” from the former and “adding dimensions” to the latter corresponding to the non-archimedean places of , a clear complementary relationship. In the adelic formulation, which glues together the archimedean places of Minkowski space with the non-archimedean -lattice, then, we should expect these to correspond to the two ends of an exact sequence in which everything is compact, which is indeed precisely what will happen.

**II. Adeles**

Let be a global field. The **adele ring** is simply what we get when we take a product of all the completions of with respect to its various places – almost. Actually, we have to take a restricted product in the following sense: let be the completion of with respect to place/valuation ; then the restricted product is the set of all elements of the full product for which all but finitely many of the coordinates are in that coordinate’s respective discrete valuation ring (with nonnegative valuation; so e.g. the -adic integers for the place of ). So it’s somewhere in between a direct product and a direct sum.

Note this restriction only makes sense for non-archimedean places, because archimedean places don’t give rise to discrete valuations rings. But this is fine, because there are always only finitely many archimedean places: this is well-known for number fields, and for function fields of algebraic curves over finite fields, there are no archimedean places, and indeed the concept is not even really applicable.

**Aside:** Here’s a somewhat involved explanation of what’s going on with these various valuations; this is very skippable. The fact that there are no archimedean places of function fields over a finite field is an immediate consequence of the general version of Ostrowski’s theorem, which says that all such places are inherited from an embedding into , i.e. there are none when the base field has nonzero characteristic. Indeed, in general, when we have a function field over , we enforce valuations of the transcendental extension to be trivial on the latter (though this is only an extra restriction in the zero-characteristic case). What happens is that we have the typical places corresponding to the prime ideals of (generated by irreducible polynomials), and then a number of infinite places.

The easiest way to see the infinite places is when our function field is precisely . Then there is just one infinite place, which is given by the total degree (numerator minus denominator) of a rational function. This might seem out of place and similar to the exceptional archimedean places of a number field, but in fact it is no different in character from the other places! To see this, consider how the map permutes the places – one can consider the infinite place generated by the “prime” ! This is legitimate because taking amounts to just choosing a different valuation ring inside , which was an arbitrary choice to begin with. Heuristically, in the algebraically closed case, can swap the infinite place with any one of the finite places (though this doesn’t actually work in practice unless we pass to completions); the infinite place is hence associated to any of the “primes” , since they are the same asymptotically. In the case, we can actually take as the uniformizing parameter of the associated DVR.

To put this precisely, when is algebraically closed, the projective linear group acts freely and transitively on the places of . In fact, we can see this very geometrically when : the places of trivial on can be identified with plus a point at infinity – nothing more than the projective line ! This can be extended to (projective) complex algebraic curves to give an abstract nonsingular model of all such curves in terms of their discrete places, the work of Andre Weil in the pre-EGA era. We can push this to higher-dimension varieties with function fields of higher transcendence dimension, but the theory becomes significantly less nice.

It is harder to get a geometric intuition, but this applies just as well to positive characteristic. If we take , the result still applies and we obtain projective curves over . The places over a finite field, like the places over e.g. , have a more complex structure since there are places of different “degrees,” corresponding to irreducible nonlinear polynomials. Then the resulting “geometry” has not just the base projective curve, but various higher-degree points which correspond to orbits under the absolute Galois group. This is again a spectral space; in fact still the projective closure of of the function field. Notice in analogue with the classical theory of algebraic curves that this interpretation in terms of projective closure gives a geometric reason why we need to take $S$ to be nonempty even in the function field case: if $S=\emptyset$, then we just get the full divisor class group of a projective curve over a field, which is $\mathbb{Z}$ by the degree homomorphism, so our result about the finiteness of class number doesn’t actually hold.

The asymmetry of archimedean places in the two types of global fields makes the arithmetic case more difficult to work with (among other things). Intuitively, the archimedean-ness of exceptional places of means unlike the function field case, in which we can take all the non-archimedean places in the nice form of a projective closure, it isn’t easy to make a complete scheme. This precludes us from using certain tools which exist in the highly-structured, close-to-analytic, projective context. (This is a common theme in the global field analogy; e.g. there is no known arithmetic derivative with nice properties either, which is a major roadblock to e.g. the abc conjecture. I wonder if there’s actually a way to connect these ideas of lack of analyticity.) This line of thinking is the beginning of Arakelov theory.

This allows us actually to give a tensor-product definition of adeles, as , where is the discrete valuation ring associated to the completion at , and the product is split up into the archimedean and non-archimedean places. This isn’t particularly important; I just thought I’d include it to show the usefulness of tensor products and demonstrate that we can give a definition without the somewhat arbitrary-seeming restricted product.

The adeles have a natural topology as a profinite group, since before tensoring with , the non-archimedean portion of the product is naturally the profinite completion of . Then we take the product topology with the usual topology on the archimedean completions (if applicable); this gives the **integral adeles**. In the full adele ring, after tensoring with , we consider the integral adeles to be an open subring. Note this is much finer than the subspace topology of the product topology on the unrestricted product.

Another way to look at this for the categorically is to see as glued together from products where a fixed finite subset of the terms are allowed to be the whole field, which is a useful perspective because it allows us to reintroduce the concept of – finite sets of places containing the archimedean ones. Indeed, we can set the **-adeles** , with the product topology. Then there are obvious inclusion morphisms from -adeles to -adeles, where , so we get a directed system, and can set as the colimit .

**Aside:** If you’re unfamiliar with the colimit (a more general categorical term for what is referred to as a direct or inductive limit in narrower settings), just think of it as “gluing together”. All the objects in the directed system (the partial -adeles) are like pieces of the whole adele ring, and the further down the chain you go (the bigger is) the closer you are to the full ring, since we want *any* finite collection of places to be the full field, not just fixed . There is a dual categorical concept called the limit (or the projective limit, or the inverse limit) which “glues together” in a slightly different way, by a kind of approximation. The best references in this case are actually Wikipedia: 1, 2. An important element (or, really, the defining element) of these two constructions is their universal property: for an inverse limit, every object with maps into the objects of the directed system (which commute with the maps between them) sees all those maps factor through the inverse limit, and for a direct limit, every object with maps from the inverse system sees all those maps factor through the direct limit.

If you’re familiar with the colimit, you might be wondering about the strange juxtaposition between finding the adele ring to be the colimit of the -adeles, while simultaneously having the “main chunk” of it be a profinite group (and hence a limit). There’s not really a reason here; the two limit/colimit constructions are basically unrelated. In some sense, the -adele direct limit is a pretty trivial gluing which just propagates the -coordinates to whatever finite coordinates we want, while the profinite completion of structure actually constructs the heart and soul of the adeles by bringing out the non-archimedean places.

However, there is a great example of limit-colimit adjunction related to the adeles, in the identification of with the solenoid on one hand and with the character group of on the other. See Keith Conrad’s notes for an exposition of the latter identification.

In this way, the adeles become a locally compact topological group; this is not difficult to prove with a general topology argument. Multiplication is also continuous, so they are in fact a topological ring. Notice we have a natural embedding given by taken the image in each completion, which sits inside the restricted product roughly because there are only finitely many places in the denominator of an element of .

The following result is fundamental. Not strictly necessary for our purposes, so it will only be a sketch, but it’s a good illustration of the kinds of arguments the topology of the adeles give us:

**Theorem.** The image of is discrete and cocompact in . (Cocompact: the quotient is compact.)

**Proof.** For the discreteness, it suffices to find an open neighborhood isolating , since we can translate it using the group operation. This is very simple; bound the norm of each coordinate by ; strictly in the -norm case. Check that this is open, and that this isolates zero (break it into the number field and function field case).

For cocompactness, we need to find a compact set which maps surjectively to the quotient in ; that is, a compact set with a representative of each equivalence class. (How familiar!) We can take actually the same open set as above but with the -norm restricted to instead of , non-strictly, since then we have a product of compact balls coordinate-wise which is hence compact by Tychonoff’s theorem. This amounts to saying that there is an element of which matches all the valuations fairly well, which is a consequence of the well-known weak approximation theorem for Dedekind domains. See here for an exposition of this theorem which mixes a sorta-pre-adelic viewpoint with a classical perspective.

There are many other properties of adeles (self-duality with character group, realization as a solenoid, strong approximation, etc.) which are important in the theory, but we will not cover them here. Consult Pete Clark’s notes for a sample.

Now we pass to the idele group. Element-wise, it’s just the unit group , but the topology is *not* the subspace topology, since, again, inversion is not continuous. The standard way to remedy this is give it the finer induced topology as a subspace of , identified with ordered pairs with .

This seems rather arbitrary, but actually it’s just a case of poor exposition – this is actually the natural topology of the ideles when identified with .

embeds into in a very understandable way.

**Theorem. (Product formula)** Let the adelic (or idelic) norm on be given as , and let be the subgroup of given by elements of norm . Then the natural embedding of into (coordinatewise, since there are of course natural embeddings ) lands in .

(Note that this makes sense for global fields, since by construction there are only countably many places, and eventually all the norms are at most .)

(Note two: the absolute value on -adic completions is familiar for number fields, being on , and being the normalized version of this (extension of norm with “denominators cleared”) in general. For function fields, it’s similarly defined as some constant , but the constant is uniform across all primes, and basically arbitrary. It is usually taken as the size of the finite field we are working over – I’m not sure if there’s any time this choice is actually significant.)

**Proof.** This is of course just a fancy way of saying that (notice this is actually always essentially a finite product) for all , where the product is over all places. A moment’s thought on the case of will give an idea of “why” this is true – the -adic valuations will switch the powers of between the numerator and the denominator, so together they will cancel out with the infinite place.

There are two common ways to prove this: the elementary method is to use the theory of norms of field extensions to reduce the problem to simply the cases of and , which are then obvious by explicit reasoning as above. There is subtlety involved in the calulations; recall the note above on the correct normalizations of the absolute values involved.

The other method is using something called the Haar measure, which you may have encountered before if you have done any study of locally compact topological groups. It is simply a measure (in the sense of measure theory) adapted to groups in the most natural way. This is the last piece of set-up we need.

**Definition.** A **Haar measure** on a topological group is a measure on the Borel algebra of its open subsets which is invariant under translation, i.e. for all open . (Notice : we’re dealing with an abelian group under addition, which makes our life a bit easier, since right vs. left translates aren’t a thing.)

The key thing that one needs to know about Haar measure is Haar’s theorem, which states that for locally compact , there is a unique (up to multiplication) nontrivial Haar measure satisfying some nice properties – inner and outer regularity and finiteness on compact subsets. The only one really relevant to us is the last one. There are more thorough treatments of the (in my opinion, tedious) theory behind Haar measure readily available; I like this more algebraic treatment by Terry Tao more than some of the others, but you can look around if you’re interested in technical tedium.

The point is, there is a Haar measure with these nice properties on , and it is an extremely important part of adelic theory. We’re not going to really do too much with it, but the ability to integrate over the adeles allows one to do quite a bit of impressive (cf. “beautiful” – Jack) analysis in the unified adelic context, which is the the basis of Tate’s famous thesis.

Denote by the Haar measure on , scaled (since uniqueness is only up to a constant) so that the induced measure on the compact quotient (“fundamental domain”) is , so that the measure “agrees with the counting measure on ” in a nice way. (Think in analogy with a measure on with the typical lattice points with fundamental domain the unit square, if it helps.)

The following lemma is the last piece of “niceness” we need to know about adelic constructs to proceed to the main proof:

**Lemma.** For any , .

**Proof.** This almost follows from the analogous statement, of scaling by , on the natural Haar measure (think -adic metric topology in the non-archimedean case, think Euclidean topology in the archimedean case) of each factor, which is easy to prove directly – you should work this out yourself if this doesn’t seem natural to you. This is, intuitively, breaking the problem into local cases at each place. (“Local” in the same sense as “local field”!) If were a true product space so that the Haar measure could be taken as the product of the individual Haar measures, we would be done – however, recall it’s actually a restricted product, so there is subtlety involved.

Instead of naively taking the product of the measures associated to our familiar -adic and Euclidean topologies all at once, we have to build it up through finite approximations, by using our construction from earlier, each of which is a genuine product space of locally compact topological groups, where the factors simply are simply given the restricted Haar measure, where they have volume . Hence we have natural Haar measures on each of them which are genuine product measures. Then it is a theorem (which is not actually difficult to prove, given uniqueness of Haar measure!) that the pullback of the measure by the inclusion restricts it to . More conceptually, this is saying is that we can construct the Haar measure through the direct limit, because the category of measured locally compact topological spaces has all limits.

In particular, for any particular , simply consider some “big enough” so that all the places where has nonzero valuation are included – possible by definition of the adeles. In particular, we can choose so that . Then the scaling effect of in is clearly (because ), and since and is an open subgroup of the whole adele ring, this coincides with the general scaling effect.

The product formula is immediately a trivial consequence of this lemma. (This is the other, more conceptual way to prove it.)

But more importantly, with the topological and metric machinery all set up, we’re ready for the main course.

**IV. Compactness, and two finiteness theorems**

The ultimate result will be this:

**Theorem.** The following statements are equivalent, for any nonempty finite set of places containing all the archimedean ones:

1) is compact.

2) is finite and the rank of is .

This begs the question: what exactly is the topology we’re considering ? Is it inherited from the ideles or the adeles? It turns out that the subspace topologies inherited from the ideles and the adeles coincide, and this is important. But more on this later.

Let’s work backwards and put the finiteness of class number and Dirichlet’s unit theorem into adelic terms.

Let be a nonempty finite set of places, containing all the archimedean ones. (By automorphism, we can actually assume that it contains the infinite place of a function field.) To get a handle on , we identify the mythical “-adic lattice” we discussed so much earlier: it is nothing more than ! Indeed, we need a group in which each element is identified with a certain valuation on every place not in . In , the places in are indeed quotiented out, since the -adeles have in the product for . On the other hand, the places not in only have a factor in the -adeles, which becomes just the -units upon passing to the unit group. Hence the quotient on those coordinates can be identified simply by giving the valuation of the element. Finally, finiteness is assured by the definition of the adeles.

But then immediately, from our earlier discussion, we obtain the isomorphism . (Exercise: put our informal discussion rigorously by giving the explicit isomorphism.) We thus need this group to be compact and discrete.

It will help to write the group a different way. It is in fact isomorphic to ! (Here, we work *a priori* with the inherited topology from the ideles, since we don’t know that the induced topologies are the same yet.) Indeed, we need only prove that modulo , we can always find a idelic-norm- representative. This is obvious for number fields, since we can alter the archimedean places however we want to achieve overall norm . It is only slightly less obvious for function fields. Since the absolute values in the function field case are all just exponentiated valuations with the same base, all we really need is to have the valuations sum to zero, which is of course possible: since is nonempty, alter any given representative of a class to have whatever valuation you want at a place of by multiplying by an adele in with the appropriate valuation there, and units everywhere else.

**Aside:** A not entirely obvious point is that we can find a uniformizing parameter at each non-archimedean place: that is, given non-archimedean , some with . This is of course necessary to get arbitrary-valuation elements in . Think about the case of a number field for simplicity: if our place corresponds to a non-principal prime like , it’s not trivial that there is a uniformizing parameter.

To see that this is indeed the case, think about it this way: for a non-archimedean place corresponding to a prime , we have a ring of integers with respect to in , . It’s not hard to prove that this will be a Dedekind domain for any global field. Then to get a uniformizing parameter, it suffices to obtain a uniformizing parameter for the localization , since the field of fractions of the completion of this ring (under the adic topology) is . Hence we simply need to prove that this is a DVR, or equivalently a PID since it’s a local ring. But the fact that localizations of Dedekind domains are DVRs is a standard algebraic fact, e.g. see here, so we’re done.

An intuitive way to think of this is as follows: in a Dedekind domain, we have unique ideal factorization. If we have an element so that the exponent of in is , once we pass to the localization, will be a uniformizing parameter! So now it suddenly seems very plausible that we can find such a , since we just need to make principal by combining it with other primes. But given finiteness of class number of global fields, every prime is torsion with respect to becoming principal, so we can just find a bunch of other ideals not divisible by but in its ideal class, which when multiplied together, give our desired principal ideal and uniformizing parameter.

Obviously this is not a way we should actually prove it, because it is circular logic (in the context of this blog post), and more generally, uses a more powerful result to prove a less powerful one. (A specific case of a less powerful one, no less – notice this proof doesn’t work for general Dedekind domains, which can have infinite class group!) But hopefully it is a useful way to think about it.

**Aside double combo:** The group is, importantly, not necessarily compact. It is known as the idele class group (notice the analogy with the ideal class group, which is of course a quotient group), and is very important in global class field theory. Indeed, the (arguably) fundamental theorem of the theory is that finite-index closed/open (equivalent, since the topology is profinite) subgroups of the idele class group are in bijection with finite abelian extensions of (inside some separable closure of the field). The quotients by these subgroups are the Galois group of the corresponding extension. In fact, the -class groups provide examples of these; each of them gives an extension of ! What a world.

**Aside triple combo:** A little more about arithmetic surfaces: taking the number field-function field dictionary even further, the fact that we have the isomorphism correspond to the fact that *for global fields, the connected component of the identity in the Picard group is the entire group*. This connected component is generally denoted for a scheme . This connected component on traditional algebraic curves is also known as the Jacobian , which classifies degree line bundles. So what we are really seeing is that ideal classes are isomorphism classes of line bundles, and that all of them are of degree . This is, heuristically, the product formula for global fields.

Disclaimer: terminology is used loosely in this aside, and I’m not sure if everything said here is *technically* true. But they are all things which *should* be true.

So now compactness is easy to deal with: is obviously a quotient group of . So we immediately have half of one way of our implication: the latter’s compactness implies the the finiteness of class number.

One last thing on class number: can we relate this to our ad hoc Minkowski argument for number fields? Sort of; the surjection from a bounded subset of the Minkowski lattice is a bit like our isomorphism with the adelic-norm- version. Of course, the latter doesn’t actually bound the norms of the archimedean places, since the non-archimedean ones still can have arbitrarily small contribution, which is why there’s still work to do.

Before tackling Dirichlet’s unit theorem, we are finally forced to prove the technical lemma regarding ‘s inherited topology.

**Theorem.** inherits the same topology as a closed subgroup (resp. subspace) of the ideles and the adeles.

**Proof.** First we check that is a closed subgroup of the adeles – the topology of the ideles is strictly finer, so this will suffice for both closedness statements. Let . Take a finite set of places so . In particular, contains all the places of where has norm greater than in that place.

If the adelic norm of is less than , then take open balls in every place where with radius as close to the respective norm of in that place as possible. By definition of adelic norm, if we take larger and larger finite subsets of the remainder of the places where has norm less than , . The product of all these open balls with the unit balls in the remaining places (just all the rings of integers so it’s a genuine open subset of the adeles) is by construction an open neighborhood of disjoint from the unit ideles.

If the adelic norm of is greater than , it’s a little trickier. We use the convergence of the product to see that we can take a sufficiently large (with, minimally, the same stipulations as earlier) so that for any outside of this set, the exponent of the valuation (that is, the such that ) is greater than : this is clear because of the increasing exponents of number fields, and the fact that function fields only actually have finitely many places to begin with. We can similarly make large enough so that (defined in the obvious way) is in . Once again by taking sufficiently small neighborhoods in the places of of and allowing elsewhere, a simple computation shows that if any place in this neighborhood outside of has norm , the whole idele has norm less than , and otherwise of course the idele has norm greater than . So we again have an open bounding away from the unit ideles. Hence between our two cases, we have the desired closedness.

(There’s some philosophizing to be done about the dichotomy of the approaches in this last paragraph between the function field/number field case once again, but I’ll leave you to think about it, since I don’t have any particularly fresh points to make.)

Since we know that the idelic subspace topology on is *a priori* at least as fine as the adelic subspace topology, it remains to show. So let be an open subset of the unit ideles in the idelic topology; we need to show that it contains an open subset in the adelic topology. We can assume WLOG that contains by multiplicative translation (which is continuous even in the adelic topology), and by contraction that is a product of neighborhoods for for some finite set of places , with in every other coordinate.

is an open set in the adelic topology. If we intersect with , we don’t necessarily get something in since the coordinates outside of might not be units. But we can ensure that they are if we make the -coordinates have sufficiently small norm, so that the non--coordinates must have exactly norm (rather than just , as is *a priori* the case) to ensure it lies in the unit adeles. But this can be done by just shrinking the s suitably. The details of the computation are left for if you’re into that kind of thing.

Now Dirichlet’s unit theorem falls out easily. In place of the log embedding of just multiplicative Minkowski space, we have an “-log embedding” in general: , given by taking . We can then take the composite , where the first map is the obvious map in to the -places.

Obviously, the image of the first arrow lies in , which maps to the hyperplane with coordinates summing to zero. Then as before, we need to make a few claims to show this does what it’s supposed to. We only sketch the proofs, since they’re all straightforward:

1) The first arrow is a discrete embedding, and the composite image is discrete as well.

That it is an embedding is clear enough, since it’s an embedding in each coordinate. Hence it suffices to establish discreteness. By the previous theorem, it suffices to prove that the “additive version” is discrete (check out the subspace topologies yourself). In other words, we need a finite neighborhood of which contains finitely many elements of . It is easy to explicitly construct such a neighborhood, but also note that it follows from discreteness of in by passing to open subrings and then dropping the other factors. (Actually we also easily find that the analogous statement about cocompactness of in is true as well.)

To prove the composite image is discrete, we note that the map is proper: take a basic compact set; the product of bounded closed intervals bounded each above and below. The result is simply an annulus in (a compact closed ball minus an open ball), so the result follows. Then the image of a discrete group under a proper map of locally compact topological groups is discrete (easy result; prove this! it’s not true for general spaces), so we’re done.

2) The kernel is finite and torsion.

We saw in the number field case that it was precisely the torsion elements; the roots of unity in , which are finite in number. The kernel for the function field is the (multiplicative group of) functions in (which, as you should’ve intuited, are functions generically regular, with their only possible poles at points corresponding to places of ) which are also regular and nonzero on the places of . But then these functions can have no poles anywhere, hence no zeros by the product formula, and hence are nothing more than constants in the coefficient field . So again: finite, and torsion.

**Aside:** It feels a little weird to call the nonzero coefficients of the function field case “roots of unity,” since it’s obvious *a priori* that all nonzero coefficient field constants are roots of . As it turns out, you should get over that weirdness, because this link between the units of the coefficient field and the roots of unity turns up in plenty of places. The number of these roots of unity, with degree relatively prime to the characteristic associated to the place, in an archimedean completion is always one greater than the residue field (think about what this means for, e.g., a -adic extension, vs. the function fields – it’s almost vacuous in the latter case).

In class field theory, cyclotomic extensions and extensions of the coefficient field (for finite function fields) both give abelian extensions for their respective global fields. The celebrated Kronecker-Weber theorem tells us that *every* abelian extension is contained in a cyclotomic field, and Hilbert’s twelfth problem asks us to generalize this to arbitrary number field; essentially, we need to find nice descriptions of the maximal abelian extension . Besides special cases where nice abelian varieties (like the unit circle!) show up in the form of complex multiplication (read Kevin’s ongoing posts), the problem remains open. (Actually, there’s a nice adelic proof of Kronecker-Weber, using the idele class group machinery mentioned earlier. This even gives a description of in the general case, though the abstraction of the description is not particularly useful for computation.) But for function fields, the correct formulation has been found in the Kronecker-Weber-Hayes theorem: it’s not an exact analogue; the maximal extension of the “roots of unity” coefficient field is not the whole story, but it’s an important piece.

Finally, Pete Clark claims that counterexamples to one of Malle’s asymptotics conjectures in the number field/function field correspond to these extensions arising from “roots of unity,” but I can’t really find anything on the function field case.

3) The quotient by the composite is compact.

This is really the crux of it, since with this, we obtain that the lattice is full rank in the hyperplane, so we finally get that the image of the unit group, hence the unit group, has rank . Denote by the hyperplane in . Then it’s not hard to see we can pass to the restricted quotient map .

Dirichlet’s unit theorem is now that the target is compact. We want to prove that this is equivalent to the source being compact, to get closer to our desired equivalence into adelic language. It’s not hard to check that , like the map it’s descended from, is proper, so one direction is quick. In the other, the image of the target is compact by general topology, hence a closed subgroup in the Hausdorff target. It is a classic fact (and a good exercise) that if we have an exact sequence of locally compact topological groups, compactness is “additive in exact sequences”: what this means in this case is that is compact if and only if and the quotient

is compact. Now we really just need to get our hands dirty and show that contains linearly independent vectors. This is a bit of straightforward greasework, reminiscent of the classical proof, which I will leave to you.

Hence this last point (3) is equivalent to the statement “ is compact.” We take it just one more step and note there is an obvious quotient map with kernel the product of all those other unit groups not in , which is compact by Tychonoff’s. So we are reduced from Dirichlet’s unit theorem to compactness of .

Now, finally:

**Proof of equivalence.**

**V. Adelic Minkowski**

Let’s take a big-picture view of what we’ve accomplished.

The class group is seen to be precisely a quotient of the (discrete) -adic lattice, which has a natural adelic description as . We are able to find norm- representatives of every equivalence class, so the quotiented -adic lattice is also . Hence finiteness of class number is equivalent to compactness of this.

Geometrically, it is obvious that the unit group (once you remove torsion) embeds nicely as a lattice in a hyperplane of dimension , so the Dirichlet unit theorem is equivalent to the quotient of the hyperplane by the lattice – which can be pulled back to the quotient of the abelian variety . (Indeed, as in the classical case, everything can be done here, on the quotient of the “hypersurface,” with a little more care and theory, but the embedding provides a nice way to get rid of the torsion and put it in a linear algebraic context.) This is the “fundamental domain” of the adelic-norm-1 hypersurface under the lattice given by the unit group. Tacking on the (compact) product of the unit balls of all the places not in allows us to replace this with the more adelic (a “fattened fundamental domain”), and so we are reduced from Dirichlet to the compactness of this last group.

We are then able to glue these two groups together (in the sense of short exact sequence, -style gluing) so that their individual compactness was equivalent to the compactness of their gluing. On a technical level, it’s clear that we have the short exact sequence where the glued object is . But intuitively, why do the “fattened fundamental domain” and the “quotiented -adic lattice” fit together nicely like this? Well, the latter’s lattice structure is precisely on the places not in ; one can imagine it as a finite torsion grid-like structure along those coordinates. The former has the “fattening” – more or less extraneous product of unit balls – in the non- coordinates, to fill in the gaps of the grid, and then a simple fundamental domain along the dimensions/coordinates corresponding to the -places. It also as hypersurface of sorts in the full idelic space, quotiented by the lattice . The first arrow in the exact sequence is then just the immersion of a fundamental polygon which is extant along some coordinates (the places of ) but not along others (the complement).

At long last, it’s time to actually prove that any of these things are actually compact. And naturally, we will need a Minkowski lemma of sorts.

**Theorem. (Adelic Minkowski)** For a fixed element , consider the set of adeles with at every place . Then there exists a constant depending only on the global field so that for any with , contains a nonzero element.

**Proof.** This is basically an exercise in manipulating Haar measure, but the idea of the proof is very geometric (as one would expect, since it’s basically the same as the proof of classical Minkowski). It’s not essential to understand the technical details, but the main thread should be easy to follow.

Let be the nicely scaled Haar measure we had from earlier. Let be the product of the closed unit discs in the non-archimedean places, with the unit disc of radius (diameter ) in the archimedean places: the key is that any two elements in can’t differ by more than in valuation at any place. is compact and contains an open neighborhood, so is finite and nonzero.

Take , and with . It is fairly clear then that to prove our desired statement, it suffices to prove that contains two points with the same equivalence class modulo .

Indeed, we have that (by the lemma about scaling effect of multiplying by constants on the Haar measure from way back), and if we take the projection map , we can proceed with a simple volume argument:

where we implicitly exchange a sum with the integral, and abuse notation to let denote both the Haar measure on both the adeles and the induced one on the fundamental domain. Hence if all the are , ; contradiction.

Take a moment to note how the above proof is pretty much exactly like the proof of classical Minkowski. Really, all we’re doing is changing the lattice, if we just broaden our understanding of “lattice” to “discrete cocompact subgroup of a locally compact topological abelian group with a nice norm associated to its Haar measure”. Indeed, it is possible to generalize the statement to this context, which then includes all versions of Minkowski.

Let’s wrap things up.

**Theorem.** For a global field , is compact.

Proof. Recall that the unit ideles inherit the topology from the adeles. So we just need a compact set of which projects surjectively onto . But just choose to be the set of adeles bounded at each place by the norm at that place of , where as in the adelic Minkowski lemma. Then for any unit idele , there is some nonzero so that by the lemma, so projects onto .

Dirichlet’s unit theorem and the finiteness of class number are immediate consequences.

Let’s conclude with a sort of bird’s-eye view of what this all means. It will be helpful to have read the asides to understand this.

So hopefully we have answered Sameer’s question adequately: the importance of Minkowski’s theorem and lattice geometry is not an accident; a general notion of “lattice” is extremely important because this idea manifests itself in algebraic/arithmetic groups suited to analyzing global fields, like the adeles/ideles. The classical “geometry of numbers” and Minkowski’s theorem arise because of the special phenomenon of archimedean places of number fields, whose corresponding completions give us copies of and . When analyzing , which is the ring of integers corresponding to just the archimedean places, Minkowski space then naturally becomes an important object of study. and have special structure that can be exploited to obtain results in this manner.

On the other hand, the presence of these places is a difficulty in some ways, since unlike the function field case, these infinite places can’t be thought of “projective completion” of the spectrum of a number field, and behave very differently, which is why a lot of results are much easier for function fields than number fields. (I mean, the Riemann hypothesis for one.) The main working approach to this is the same attitude which inspired the geometry of numbers: to work in the archimedean places, take them as they are, and use their familiar analytic structure in conjunction with the nicer non-archimedean places to make progress. This is the foundation of Arakelov theory, a modern approach to number theory. We can define Arakelov divisors, which are like elements of our “-adic lattice” with archimedean real/complex components tacked on, and even the Arakelov class group, a kind of “fattened class group,” which very much resembles in spirit our “fattened fundamental domain” from the adelic discussion of Dirichlet’s unit theorem. (Remember “obviously not, stupid?” Sorry, that was a lie; that’s precisely what this is.) We can obtain analogues of a lot of algebraic geometry this way, including Riemann-Roch, sheaves, and intersection theory.

Relatedly, but with almost the opposite attitude there is a view that there should be a **field with one element**, a mythical object over which would be a curve, whose projective completion would somehow bypass, or perhaps reveal hidden depths of, the archimedean obstacle. The field with one element is a fascinating aspect of mathematical folklore which has many, many more interesting hypothetical connections and properties than the ones arising here; see the Wikipedia page and MathOverflow for more. It should be noted that quite a few pretenders exist, but none are fully satisfying.

Some of the terminology used here was nonstandard; to further get into adelic applications in class field theory and Arakelov theory, some of this should be clarified. An “-class group” is more typically described as a **ray class group** – though the latter concept is slightly more general. A set of places is a kind of **modulus**, which figures in Artin reciprocity, a fundamental result of global class field theory. Moduli can count places with multiplicity, however.

Huge credit to Brian Conrad’s notes, from which the bulk of this material was taken. Indeed, this post is basically a more fleshed out version of those notes, recast as a general philosophical overview/introduction to adeles.

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**Definition.** A *vector bundle* over a topological space is a (continuous) map satisfying the following properties:

(1) For any , the pre-image is homeomorphic to . (The pre-image of a point is called a *fiber*)

(2) (*Local Triviality*) For any , there is an open neighborhood around that point such that is homeomorphic to

is called the base space and is called the total space.

Examples of vector bundles are fairly easy to come by. We have, for example, the tangent bundle of a manifold, which can easily shown to satisfy the two conditions above. We will re-visit tangent bundles several times in future blog posts. Another important example is the Möbius bundle over . Intuitively, this assigns a line to each point on the unit circle in such a way that it “twists” once before it comes around the circle (geometrically, this would look like a Möbius strip, hence the name). This example is important because it is one of the easiest example of a non-trivial line bundle (that is, it doesn’t look globally like ).

A mild technical note: in general, we will be assuming that our base spaces are compact and Hausdorff, though a weaker assumption (such as paracompactness) is generally sufficient to get all of the properties that we want.

* * *

While there is a lot more to be said about the theory of vector bundles (which will be the case in future blog posts), instead I will focus on two related notions. The first is the theory of fibrations. This generalizes the idea of a vector bundle in a way that yields useful results in homotopy theory.

**Definition. **A *fiber bundle* is defined in the same way as a vector bundle, but instead of requiring that the fibers be homeomorphic to , we allow them to be any topological space . (Of course this also has to satisfy the local triviality condition)

While fiber bundles are useful, we want a slightly more specific homotopy theoretic property:

**Definition. **A map is said to satisfy the *homotopy lifting property* if, for any space , a homotopy lifts (not necessarily uniquely) to a homotopy satisfying . Such a map is called a *fibration*.

A reader familiar with some of the basic ideas of homotopy theory might recognize something familiar in these definitions: the idea of a covering space satisfies these properties exactly. An -sheeted covering space if precisely a -fibration. The most important property of fibrations (from the viewpoint of homotopy theory) is the following:

**Claim.** Given a fibration , there is a long exact sequence of homotopy groups:

The first two homomorphisms are the obvious ones induced by the fibration maps. The third one can be obtained through some diagram chasing (in a way analogous to the Snake Lemma).

**Example.** The *Hopf fibration* is historically an important example of a fibration. We can think of as . In an analogous way we can define as equivalence classes of pairs of complex numbers with if and only if . We can think of this equivalence relation as sending a pair of complex numbers to their quotient with an additional point at infinity – the one point compactification of . This, then, naturally induces a map sending a pair of complex numbers to their equivalence class. The pre-image of a given point is the set of complex numbers with norm 1, which is precisely . This is the Hopf fibration, and it leads to the following result on homotopy groups:

**Claim.** For , we have . In particular, .

**Proof.** We can directly apply our long exact sequence, along with the fact that al higher homotopy groups of are trivial, to get the following short exact sequence:

Because this is exact, the groups must be isomorphic. Using the fact that for all yields the desired result.

This is just one example of the usefulness of fibrations. (Ironically, from the perspective of homotopy theory, vector bundles are one of the least interesting example of fibrations because the homotopy groups of are not of much interest.)

* * *

The final area I would like to discuss is the theory of classifying spaces. First, we will explore a bit more about the theory of vector bundles:

**Definition.** Define the *infinite Grassmanian* to be the set of all -dimensional subspaces of . We can also define it as a limit of with the weak limit topology. We will denote as . There is an analogous construction in the complex case, which we will denote .

For the sake of simplicity, we will be working with complex vector bundles. The reason that this case is simpler is because every manifold has a unique complex orientation, so we don’t need to worry about issues of orientability.

**Theorem.** There is a bijection between complex line bundles and maps .

The reasoning behind this statement should be apparent: we are assigning to each point in a point in , which is precisely an -dimensional vector space. We would like to generalize this construction to fiber bundles where the fiber is any topological group. (This is a generalization because we can think of a vector bundle as having fibers in the infinite unitary group ). Unfortunately, solving this problem for any topological group is extremely difficult. However, if we assume that we are working with a discrete topological group, then there is a solution. As it turns out, is just an example of what is known as a *classifying space*:

**Definition. **Let be a discrete (topological) group. Then define a space called the *classifying space* of to be a topological space such that and all higher homotopy groups are trivial.

Of course, the task of constructing such a topological space is non-trivial. One solution is to resort to the Eilenberg-Maclane space . this certainly satisfies the definition. However, an observant algebraic topologist will notice that this definition only defines a classifying space up to homotopy equivalence, so there are many different models of classifying spaces. Thus, we will give another construction that is in some ways a “better” model – this will have the property . As the functor doesn’t preserve products, this new model can be seen as “better”.

**Definition.** Given a small category , define the *nerve* of the category to be a simplicial set based on morphisms in the category. We will give the construction here:

Our 0-simplices will simply be the objects in the category. Our 1-simplices will be morphisms between the objects. 2-simplices will be the diagrams with edges given by two composable morphisms along with their composition (that is, given two composable morphisms , we take the commutative triangle with edges ). We can continue this construction to get a simplicial set, which we will call the nerve of the category. We still need to specify the face and degeneracy maps in order to have the entire structure of a simplicial set. Define the face map by taking the simplex to the simplex by composing the morphisms into one morphism. Define the degeneracy map by inserting an identity morphism at the object .

We will use this construction to very easily construct a classifying space. Given a discrete group , we can think of it as a single-element groupoid (that is, a category with one object in which all of the morphisms are isomorphisms). Then we take the nerve of this category, and finally take the geometric realization of the nerve. This is a topological space, which will be precisely . Note that the nerve functor is right adjoint, so it commutes with limits, including products. Additionally, it is a well-known fact that geometric realization commutes with products, even though it is a left adjoint functor (see, for example, this paper for a motivation of why this is true).

That the result of this construction is a classifying space is not hard to see. The fundamental groups of a CW complex (and, thus, a simplicial complex) is obtained from the 2-skeleton by taking formal products of the 1-cells and using the 2-cells to apply relations. In this case, the 1-cells are precisely the elements of the group and the 2-cells will tell us that , so this will give us the correct fundamental group. Verifying that the higher homotopy groups are trivial is much more difficult and beyond the scope of this post.

**Example.** Consider the group with two elements. Then, in the nerve of the category, there will be exactly one non-degenerate simplex in each dimension. This will give rise to a simplicial complex with one simplex in each dimension. This is precisely analogous to infinite real projective space, which can be realized as a CW complex with one cell in each dimension. By making the proper identifications we can see that this the same space. Because our construction of classifying spaces commutes with products, we have that is the classifying space for any product of these spaces as well.

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First, a classic example which demonstrates how geometric vector bundles (equivalently locally free sheaves) on a scheme do not quite correspond to topological vector bundles. It is well-known that the only two topological real line bundles over are the trivial bundle and the Möbius bundle with one twist. Any other bundle, with a number of other twists, is homeomorphic to one of these two, since we can cancel out pairs of twists – even isotopically, if we embed in five(?) dimensions.

But if we take the scheme (as e.g. , which has underlying space homeomorphic to , it is clear we get many line bundles: specifically, the twisting sheaves for each . (Here we implicitly use the equivalence of categories between locally free sheaves and schemes over which are geometric vector bundles.) What gives?

It’s not that there’s something fundamentally geometrically different about the scheme structure: a geometric vector bundle over really does have underlying topological space a line bundle over the same space; that’s clear from typical definitions as given in, e.g. Hartshorne chapter 2. What’s different is the morphisms in the category of topological line bundles versus the category of geometric line bundles over a scheme. The latter morphisms are, of course, simply morphisms of schemes over , and so have to be given locally by regular functions. But the twist-canceling homeomorphisms can’t be given this way (intuitively, think about having to twist the lines out to infinity all the way around; these involve poles, algebraically), so they are not isomorphisms in the category of geometric vector bundles.

Intuitively, if we “allowed topological morphisms,” then the even would all be trivial whereas the odd would all be the Möbius bundle. This also explains the odd behavior of global sections (i.e. that all the nonnegative have global sections while all the negative ones don’t): the negative evens do have nonvanishing globals; it’s just that, again, they can’t be given by functions which formally satisfy the local regularity conditions of the twisting sheaves. On the other hand, the odd positive ones don’t really have nonvanishing global sections, since there are points at which the odd-degree polynomials corresponding to global sections of that twisting sheaf vanish.

**Exercise.** Someone should rigorize this line of thinking to create a gorgeous proof that every odd-degree polynomial has a real zero.

—

Second, a while back, I was trying to gain intuition for the Picard group by reading through different perspectives on the invertibility of line bundles. There’s of course the inverse transition maps, tensor-hom adjunction, and all that jazz, but I also stumbled on an amusing one: consider the classifying space for line bundles, the infinite Grassmannian . An isomorphism class of line bundles on a space is equivalently a homotopy class of maps into the classifying space, so we have reduced our problem to the much simpler and more intuitive one of putting a group structure on in the homotopy category!

A little Schubert calculus gives us the following: first, the cohomology ring is concentrated in even dimension, so it is a genuine commutative ring. Second, since the product is given by the cup product, which corresponds to intersection product, the closed points of correspond to maximal ideals generated by Schubert cycles associated to 0-dimensional subspaces, points. In fact, it turns out that is the formal affine line, , whose canonical additive structure on its closed points is induced by tensor product of the line bundles associated to the points of the Grassmannian (as a moduli space) associated to the closed points.

This structure can then be pulled back to , using the techniques in this paper. Theorem 1.1 implies that the well-known induced map in cohomology is injective. We can then compose it with the isomorphism . As a hom into an abelian scheme, this last object naturally inherits a group structure. This can then be pulled back to by naturality, giving us our group structure on isomorphism classes of line bundles.

The best part is that this doesn’t even work, since isn’t actually a strictly commutative ring if it doesn’t have vanishing odd cohomology, so it doesn’t have a spectrum. To really complete this elegant line of reasoning, I think it is necessary to develop the methods of noncommutative geometry. Alain Connes still dreams at night.

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and declaring to be the length of the shortest path between and when traveling along the edges of . On the space of finite, edge-weighted graphs, there is the equivalence relation if and are isometric. A *metric graph* is the unique (up to isometry) metric space associated with an equivalence class in , and any graph in the equivalence class is called a *model* for .

We begin with , the group (under pointwise addition) of continuous, piecewise linear functions for which every linear piece has integer slope. This is a linearized analogue of the space of meromorphic functions on a Riemann surface, over which divisor theory is classically developed. We denote by the free abelian group on , i.e.

In other words, is the group of integer-valued functions on having finite support. With this interpretation, we introduce the notation for a divisor .

There is a homomorphism given by , and is called the *degree* of . For any , we denote by the set of divisors with degree . Only when does form a subgroup of , and is the kernel of .

With each , there is an associated divisor , which records information on the “zeros” and “poles” of . Namely, for each , we set to be the sum of the slopes of in all directions of the graph emanating from . Then we define the *principal divisor* as

The principal divisors form a subgroup , and one observes for every , so that . This begs to form the quotient

which is called the *Picard group* of . More generally, two divisors , are *linearly equivalent*, denoted , if , and one can form the sets .

There is a combinatorial/dynamical interpretation of linear equivalence. Given a divisor on , one can interpret as the number of dollars person has (if , then person is in debt). For any star-shaped subset and sufficiently small , there is a function taking value 0 on , taking value outside the -neighborhood of , and with slope 1 along each outgoing direction from . Then , and this equivalence has the effect of each person on the boundary of giving one dollar to the individual situated exactly distance away. Every element of may be written as a finite -linear combination of ‘s, so two divisors are linearly equivalent precisely when connected by a sequence of these “chip-firing” moves.

In terms of the chip-firing game, is the space of inequivalent degree 0 wealth distributions on . We would like to understand the structure of this space, both algebraically and geometrically. As it turns out, one can learn quite a bit about through the *integer homology* group .

Recall that is constructed as follows. Fix a model for , fix an orientation of , fix a spanning tree of , and fix a base vertex . Then, we denote by the -vector space with basis , and by the -vector space with basis . The boundary map given by extends linearly to a map of vector spaces. One defines the *real homology* to be the kernel of . Given the choices we have made, there is a canonical basis for , consisting of simple cycles. For every , there is a unique cycle beginning and ending at that traverses no edges other than and those in . Recording each of these cycles as a sum of directed edges gives the aforementioned basis. Then is defined to be the free abelian group on this canonical basis, i.e. the associated integer lattice inside .

The Abel-Jacobi theorem states there is an isomorphism

We will not prove this fact, as it is rather difficult, but we will show how this isomorphism gives rise to interesting geometric features. One defines the *Abel-Jacobi map* via

By the isomorphism above, descends to a map from into the -dimensional torus

where is the genus of . This can be described in very geometric terms: it essentially draws a skeletonized version of on the torus .

To specify the map , we will need to equip the ambient edge space with an inner product given by

This allows us to talk about orthogonality of vectors in , and in particular, define an orthogonal projection . A line segment in can be specified by providing a vector and a base point onto which the tail of the vector should be translated. Accordingly, let denote the line segment in obtained by translating vector to . We describe the image of by constructing the map inductively. Assume in the basis the edges are ordered such that for every , the edge shares a vertex with some having . Furthermore, assume the tail of is . Begin by identifying with the segment (so that is defined on ). Now, if for shares a vertex with , where , we identify with either the segment or the segment , depending on if is the tail of or the head of , respectively. Once this process is complete, the entire image of is contained in a unique fundamental domain of the lattice , hence actually gives a map .

How much information does preserve? The only edges mapped into non-injectively are those which orthogonally project to , i.e. those edges inhabiting the orthogonal complement of . There is a nice characterization of this space when for all . Recall that a *cut* of is any disjoint partition , and any cut determines a *cut set*, consisting of the edges connecting a vertex in to a vertex in . For each cut set , there is an associated *cut vector* in encoding that set, namely . where if is oriented to flow from to and if is oriented to flow from to . We claim the *cut space*

is precisely the orthogonal complement of the cycle space .

To prove this, we first note that any cut vector is orthogonal to any cycle vector. Suppose represents a simple cycle (a cycle that traverses no edge twice) and is a cut vector. Let if traverses along its orientation, let if traverses against its orientation, and let if does not traverse . Then

But simple cycles form a basis for , so . Since has dimension , to complete the argument it will suffice to exhibit linearly independent vectors in . Fix an ordering of the vertices, and set . Then for each , the partition is a cut, with a corresponding cut set . Since always contains a vertex that , , all do not, the cut vectors corresponding to these cuts are linearly independent.

Hence, if and only if is a *bridge*, an edge whose removal would disconnect the graph. This means the only information lost by is the existence of bridges, which are collapsed into a point when the graph is drawn on ; elsewhere on , the map is injective. Because of the isomorphism between and , this geometric observation can be translated to an equivalent algebraic fact about : if , then either or there is some bridge such that , .

(Acknowledgment: the nice diagrams above are due to Samir Khan, UChicago ’19)

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We now will introduce the concept of the Betti numbers of a topological space. Intuitively, we want the -th Betti number to represent the number of “holes” in a geometric realization of the space. While this concept has an interesting history, in the end our definition ends up looking like:

**Definition:** The -th Betti number, denoted , will denote the rank of the -th homology group .

Because the homology groups (which are, conveniently, the same as the cohomology groups – this is not a coincidence at all, as we will see in a later post) are free groups, then the rank is just the number of generators, which can be found through the partitions mentioned above. There is a really deep and interesting connection between the problem of finding the Betti numbers of this space and a series of results in number theory called the Weil conjectures, which were considered to be some of the most important open problems in mathematics before the last one was proved in 1974. There are four conjectures:

**Definition: **Let be an -dimensional non-singular, projective variety defined over a **finite** field . Then we define the *zeta function* on as: , where denotes the number of points of defined over .

**Theorem:** (1) (*Rationality*) is a rational function (that is, it can be written as the quotient of two polynomials)

(2) (*Functional equation*) , where is the Euler characteristic of

(3) (*Analogue to **Riemann hypothesis*) As a rational function, we can write with and, for . When written in this form, we have that are algebraic integers and .

(4) (*Betti numbers*) If is the reduction mod of a projective variety over a field embedded in the field of complex numbers, then the degree of is the -th Betti number of .

These results have an interesting history which I definitely recommend reading about. The statements may seem a bit inaccessible, but we’ll unpack them as we go along. But first, we need to confirm that these results actually apply at all:

**Claim: **The complex Grassmanian is a smooth, projective variety embedded in a complex projective space.

**Proof:** We will use what’s called the Plücker embedding, which gives explicitly what the embedding is. Define the map as . We claim that this shows that this is a projective variety. We must verify a few points:

– *This map is well defined*: Let be another basis for the subspace . Then we have that there is an invertible transformation such that , so they are equivalent in the projective space. Thus, this map is well defined.

– *This map is injective*: Let . Then if and only if it is in the span. Thus, let . Then we have that . Thus, we can uniquely recover an element in the pre-image from its image, so this is injective.

–*This is a projective variety*: We claim that this embedding gives us a set of polynomials satisfies by the image of the Grassmanian, thus showing that this is a projective variety. To show this, we can think of the map defined above as a matrix by fixing a basis for and a basis for . Because it is injective, it must have maximal rank, which will give us a set of polynomials describing the solutions, so this is a projective variety (for more details and for a proof for general Grassmanians, see here).

Now that we have verified this, we can begin to apply the Weil conjectures to the problem of the Betti numbers. First, we will study the behavior of this projective variety over a finite field . In this special case, we can actually verify the first Weil conjecture. We will need to know how many points the Grassmanian has over a finite field, which is precisely the number of dimensional subspaces of . For notational convenience, we will denote this number by (as this notation suggests, this will have a distinctly combinatorial flavor and will be connected to the standard binomial coefficients).

**Claim: **. Furthermore, if we let , then .

Thus, the reason behind the choice of notation becomes clear. These numbers are often called the *Gaussian binomial coefficients*.

**Proof:** This proof will be a simple combinatorial argument. We want to first compute the number of sets of linearly independent vectors in . To do this, we simply note that the first vector can be any nonzero vector, so there are choices. The second vector can be any vector not in the subspace spanned by the first vector, so there are choices. Repeating for choices gives . This, however, overcounts subspaces. In fact, the number of times that each subspace is counted is equal to precisely the number of sets of linearly independent vectors in . Thus, we repeat the same calculation and divide out to get . We can factor out a to achieve the desired result.

Thus, we have an explicit computation for . However, in order to verify the rationality of the zeta function, we will need one more result about this number:

**Claim ( Generalized Pascal Identity): **.

**Proof:** We can verify this directly by computation, which will be omitted here.

**Corollary:** is a polynomial in with positive coefficients of degree .

**Proof:** This follows immediately from a simple inductive argument and the fact that .

Thus, we know that we can write . Using this, we can verify the Weil conjectures for the Grassmanian explicitly as well as compute its Betti numbers.

**Theorem 1: ** is a rational function.

**Proof:** We will use the fact that . This gives . We see that the expression inside the exponential is precisely a Taylor expansion for , so by switching the order of the summations we get that the expression is equivalent to . Thus, this is a rational function, as claimed.

First, we note immediately that all of the are positive, so this implies that all of the are trivial, which is consistent with the statement that all of the odd Betti numbers are zero. Now that we have computed the rational form of the zeta functions, the second adn third conjectures will follow fairly easily:

**Theorem 2: **The functional equation above is satisfied.

**Proof:** This is a simple computation, and so will be omitted.

**Theorem 3:** The finite analogue of the Riemann hypothesis is satisfied.

**Proof: **We will let . Then the desired result holds.

The final result we will show is that the degree of is precisely the -th Betti number. In order to verify this, we will prove a (rather suggestive) lemma:

**Lemma:** Let denote the number of partitions of into integers, each . Then .

**Proof: **We will induct on using the generalized Pascal identity discussed above. If , then the result is trivial. Now fix . By the inductive hypothesis, suppose it holds for and . Then we have .

We claim that the desired result holds immediately. To see this, suppose we have a partition of into integers . Then we have two cases: either at least one of the integers is zero, or none of them are. If one of the integers is zero, then we can remove one of the zeros and this is precisely a partition into integers . If none are zero, then we can decrease each integer by and we will have a partition of into integers . Thus, we get the identity , which allows us to combine the terms in our expression above to achieve the desired result.

Thus, we claim that the following follows immediately:

**Theorem 4:** The degree of is precisely .

**Proof:** The degree of this polynomial is precisely , which we have just shown is equal to , which we defined to be precisely the computed at the beginning of the post. Thus, the result holds.

We have now verified that all of the Weil conjectures hold for the case of the complex Grassmanian. As a final note, the process that we used here to compute the Betti numbers through the Weil conjectures can be applied to almost any projective variety. Thus, we could have computed the Betti numbers without even knowing the homology or cohomology of the Grassmanian through this process. In the next post, we will discuss the idea of Poincare duality as well as possibly the cup product and how we define it in this case.

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in displays two essentially distinct ways to write as the product of irreducible elements. Nevertheless, one can recover in a different sort of factorization: every ideal may be uniquely written as the product of prime ideals. Any integral domain in which this is possible is called a *Dedekind domain*. We will see by the end of this post that working in the world of ideals rather than numbers does not preclude us from extracting useful number-theoretic information.

Our goal for this post is to prove some of the important results on Dedekind domains (in particular, that is a Dedekind domain), and it will not hurt to work in greater generality than the number field case. In particular, the setup is as follows: let be a Noetherian integral domain, and let denote its field of fractions.

**4.1. “Modular arithmetic”**

From the module-theoretic perspective, ideals of are just finitely generated (as is Noetherian) -submodules of . This point of view is useful because it allows us to naturally generalize the notion of an ideal to include larger subsets of the entire fraction field . In particular, a *fractional ideal* is a finitely generated -submodule of . Since ideals in are also fractional ideals of , for clarity we refer to ordinary ideals in as*integral ideals*. We denote by the collection of fractional ideals in .

Given , , recall one can define their product and sum

which are themselves fractional ideals. This allows one to introduce a notion of invertibility: an ideal is *invertible* if there exists such that .

We will have to do some hard work now to demonstrate the following proposition.

**Proposition 4.1** *Let be a Noetherian integral domain. Then, the following are equivalent:*

- is integrally closed (if satisfies a monic polynomial in , then ) and every prime ideal in is maximal.
- Every integral ideal in can be written uniquely as the product of prime ideals.
- Every fractional ideal of is invertible. In particular, is an abelian group under the module product.

*If satisfies any of these conditions, one christens a **Dedekind domain**.*

Before descending into the land of commutative algebra, let us note the corollary:

**Corollary 4.2** * is a Dedekind domain.*

*Proof:* In the previous post, we showed is Noetherian and that every prime ideal in is maximal. Suppose satisfies a monic polynomial , given by

Since is a finitely generated -module, it follows from the fundamental theorem on modules over a principal ideal domain that

is finitely generated as a -module. Since is a -linear combination of , , , , it follows

is also finitely generated as a -module. Applying the fundamental theorem one last time, we have

is itself a finitely generated -module. Since , we conclude is an algebraic integer in , i.e. .

**4.2. Proof of Proposition 4.1**

As with many arguments in commutative algebra, we piece together the proof from a series of fairly sophisticated lemmas. We shall first show the direction in Prop. 4.1, so assume is Noetherian, integrally closed, and every prime ideal in is maximal.

**Lemma 4.3** *Assume . Let be an integral ideal of . Then, there exist prime ideals , , such that*

*Proof 1: (General).* Denote by the collection of integral ideals which fail to satisfy the given condition. Suppose is nonempty. Since is Noetherian, an appeal to Zorn’s lemma admits a maximal element . Of course cannot be prime, so there exist , such that but , . There are the proper inclusions and , so each of and contains a product of prime ideals. However, , which is a contradiction.

*Proof 2: (-specific).* For the axiom of choice truthers among us, one can show this result in the specific case without invoking Zorn. Use that every quotient ring is finite, and choose an ideal for which is minimal. Then argument proceeds exactly as in the first proof.

For any , define the fractional ideal

By construction, . In fact, is the only candidate for the inverse of a fractional ideal: one can easily show if is invertible, then (exercise!).

**Lemma 4.4** *Assume . Let be a prime ideal of . Then, .*

*Proof:* The argument proceeds in three steps.

*Step 1: ().* Choose a nonzero for which

such that is as small as possible. By the prime avoidance lemma, one of the (without loss of generality, assume ), is contained in . Then since is maximal. By the minimality of , one knows

so there exists such that . In particular, . However, by the choice of ,

Hence, . It follows .

*Step 2: ( for any nonzero ideal ).* Let , , be a basis for as an -module. Suppose and choose . Then there is a linear map given by , with matrix . Then is an eigenvalue of , so that . But the left-hand side is a polynomial in . By integral closure of , it follows . Thus, , contradicting step 1.

*Step 3: ().* By step 2, we see , so the inclusion is proper. Since is an -submodule of , it is an integral ideal. But is a maximal ideal, so as desired.

Using these lemmas, it is actually not so difficult to finally furnish the proof of . We demonstrate existence of the prime ideal factorization, and leave uniqueness as an (easy) exercise.

**Lemma 4.5** *Let be an integral domain and an integral ideal. If can be factored as the product of invertible prime ideals, then this factorization is unique.*

*Proof:* Exercise!

*Proof 1 of : (General).* Let denote the collection of integral ideals that do not permit a prime ideal factorization. Suppose is nonempty. As earlier, Zorn’s Lemma gives a maximal element . Since every ideal is contained in a maximal ideal, there is some prime ideal such that . Then, one may factor , where

is an integral ideal. Since is maximal in , it follows has a prime ideal factorization. But this gives a prime ideal factorization, which is a contradiction.

*Proof 2 of : (-specific).* We shall prove by induction (on ) that if an integral ideal contains a product of nonzero prime ideals, then is a product of prime ideals. By Lemma 4.3, this is sufficient to prove . For , if for some prime ideal , then by maximality. Suppose now that . Since is finite, there is a maximal ideal containing . Then the prime avoidance lemma implies, without loss of generality, that . Multiply through the inclusion by :

Induction gives a prime ideal factorization for , and hence one for .

Next on the agenda: we must show .

**Lemma 4.6** *Choose any nonzero . If for fractional ideals , then each is invertible.*

*Proof:* Simply set .

**Lemma 4.7** *Assume . Let be a nonzero prime ideal. If is invertible, then is maximal.*

*Proof:* Choose any . Suppose that . Then we may factorize

where each , is a prime ideal. Let . Then in , we have the prime ideal factorizations

so Lemma 4.6 implies each , is invertible. Equating these factorizations as

we see Lemma 4.5 tells that and each appears exactly twice in the collection . Hence is the multiset . This gives the following chain:

Thus, if , then for some and . Since and by choice, we must have . This yields another chain:

since is invertible. But by assumption.

**Corollary 4.8** *Assume . Let be a nonzero prime ideal. Then, is invertible.*

*Proof:* Choose any . We may factorize , where, by Lemma 4.6, each is an invertible prime ideal. In fact, each is also maximal, due to Lemma 4.7. The prime avoidance lemma then implies without loss of generality , so is itself invertible.

The meat of this portion is now done with, and we can give the proof of .

*Proof:* Let be a fractional ideal. Since is finitely generated, there exists such that . Factoring into prime ideals, we may write . Then is the inverse of .

Finally, it remains only to prove , which is (comparatively) not very tough at all.

*Proof:* Let be a prime ideal in , and suppose is a maximal ideal containing . Then , so the prime avoidance lemma implies either or . Suppose the former is true. Then, multiplying through by gives

which is impossible. Hence is maximal.

Suppose satisfies a monic polynomial of degree , whose coefficients are in . Consider the fractional ideal

Then , so multiplying through by proves . We conclude is integrally closed.

Let us conclude this section with a very brief exercise, which illustrates two important properties of Dedekind domains.

**Lemma 4.9** *Let be a Dedekind domain.*

- “To contain is to divide”: the ideals , satisfy iff there exists an integral ideal such that .
- Comaximal equals relatively prime: the ideals and have no common prime ideal factors iff .

*Proof:* Easy exercise.

**4.3. Introducing the ideal class group**

We have strayed a bit far from number theory with all of these maximality arguments. Let us return by introducing the *ideal class group* of a number field , whose relation to number theory is beautiful and important.

Recall denotes the collection of fractional ideals of , which gains the structure of an abelian group under the module product. There is a subgroup which consists precisely of the principal fractional ideals, those generated by exactly one element in . The class group of is then defined as the quotient

Equivalently, one may place on the integral ideals of a relation given by iff there exist , such that

The equivalence classes thus obtained gain a natural group structure under ideal multiplication, and one easily sees this is isomorphic to the structure on . In the next post, we will prove the following extremely important proposition.

**Proposition 4.10** *For a number field , the class group is finite.*

For now, let us take this fact as given. The order of is called the *class number* of , and denoted . Observe that iff is a principal ideal domain. Thus, in some sense, the class number measures the extent to which a ring of integers fails to have (or succeeds in having) unique factorization for elements.

Consequently, computation of the class number is significant in number theory, as it detects how well one can treat ideals as elements. This interpretation is scalable: finiteness of the class group implies for any integral ideal , the ideal is principal. One can use this fact to great effect in the solutions to certain Diophantine equations:

**Proposition 4.11** *Suppose is square-free, with . Let , and assume furthermore that . Then,*

- If there exists for which , then the Mordell equation has the unique solutions .
- If there exists no such , then the Mordell equation has no solutions.

The proof of this proposition is actually already well within our grasp (given temporary suspension of disbelief regarding the finiteness of ). Recall that when , the ring of integers of is exactly .

**Lemma 4.12** *Assume . Suppose there are , such that for some ideal . Furthermore, suppose and are comaximal ideals. Then and for some , and units , .*

*Proof:* Since and are comaximal, they share no prime ideal factors. Therefore, and for some ideals , with . The class of principal ideals is the identity in . By finite group theory, it follows that the order of must divide both and . As it was assumed , it follows is principal. Similarly, is a principal ideal. Writing and , the result drops out.

*Proof of Prop. 4.11:* Assume there exists a solution to Mordell’s equation . Due to the choice of , modulo considerations tell that is necessarily odd.

Additionally, we find . Otherwise, suppose there is a prime dividing both and . Then also divides , so divides , which was assumed square-free.

Now, factor the equation in and pass to ideals:

Suppose and are not comaximal, i.e. have a common prime ideal factor . Then , and hence . But also contains , thus by primality contains . Since is odd and , also , so there are integers and such that . This is a contradiction.

Therefore, and are relatively prime. By Lemma 4.12, there exists and a unit such that . The only units in are (exercise!). Setting and assuming without loss of generality, it follows

Equating coefficients gives and accordingly. A little more computation proves and as desired.

As lovely as this little characterization is, really applying it requires one to know the class number of . And in general, calculating is quite challenging. There do exist explicit formulas and algorithms designed to compute in the case where is a quadratic field, which we will discuss in the subsequent posts.

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**Definition:** The *Grassmanian* of dimension * *over a vector space is defined to be the set of all subspaces of dimension . We will denote this by .

Of course, since we are studying this structure from a topological perspective, this definition is essentially meaningless if we don’t know how it is topologized. Let denote the set of all -tuples of linearly independent vectors in . We then define an equivalence relation on this space where if and only if and span the same subspace. The Grassmanian is then the quotient space obtained by quotienting out by this equivalence relation.

Now that we have a definition of the space we’re dealing with, we can consider the specific case that this post is concerned with: the Grassmanian . The choice of notation here may seem a bit odd at first: why not just write ? The answer to this is twofold: this guarantees the non-triviality of the Grassmanian and also conveniently allows us to refer to the codimension of our subspaces, which will show up many times later on. Equipped with this notation, the main problem going forward is how we can decompose the Grassmanian into a (finite) CW complex.

Consider a weakly decreasing sequence of integers such that . We will then define the set to be the set of all -planes in such that and . This definition may seem a bit contrived at first, so let’s take a minute to dissect what everything here means. Essentially, what we’re saying is that is the lowest dimension with the property that its intersection with has dimension . We can most easily represent this as a matrix. Consider the example of . Let . Then any subspace in has a basis that can be expressed in the following way:

Applying row operations, this can be re-written as:

Thus, we can see that spaces in this cell are represented by 2 complex numbers, so it has real dimension 4.

**Claim: **More generally, we claim that has dimension precisely equal to .

**Proof: **To see this, consider the matrix representation as before. The -th row will have entries. However, we also know that, by row operations, we can cancel out entries by subtracting multiples of the rows above. We can then fix the last entry to be 1, so each row will have variable entries after row operations are performed. Summing over each row and then doubling yields the desired result.

We of course still need to verify that this is indeed a valid CW structure for our space. The first thing we will verify is:

**Claim:** Given any , we know that for exactly one .

**Proof:** To verify this, we will need to simply consider the sequence given by . This will uniquely determine the sequence associated with the subspace, so the claim holds.

The other, more difficult result to show is that the boundary of a cell contains only points in lower dimensional cells.

**Claim: **The closure of a cell has the property that is contained entirely in cells of lower dimension.

**Proof:** Essentially, what we will attempt to do is show that there is a closed set (not necessarily the closure) containing a cell that contains only that cells and cells of lower dimension. We will define a partial ordering on the sequences associated with a cell as if and only if for all . We claim then that the set defined by is closed. If this is the case, then our proof will be complete, as we will have found a closed set containing with the desired property, and . To see that this set is closed, we will refer back to the quotient map used to topologize the Grassmanian. If we can demonstrate that its pre-image is closed, then we will have shown the desired property.

Consider its pre-image in the space of matrices. Suppose that such a matrix can be row reduced to be in the form specified above. We claim that being able to be reduced into this form is equivalent to taking sub-matrices from the right and extending them downward and end up with a matrix of rank 1. To make this more precise, consider first the index of the last non-zero entry in the bottom row of the reduced matrix. Then taking all entries to the upper right of that entry must be a matrix of rank 1. This is equivalent to showing that all of those entries are constant multiples of one another. To see an example, consider the example from above. The pre-image of the quotient map is all matrices of the form:

We want to be able to tell when it can be reduced to something of the form:

We first note that we must have Thus, we can consider, for example, the submatrix given by:

This submatrix always has rank 1, so we this condition always holds. We also have that the submatrix:

must have rank 1. This means that . We can apply a similar argument to the last entry in each row of the reduced form. Thus, we know that a convergent sequence of matrices with this property should also have this property. However, we also need to take into account the fact that the pre-image of requires that they not all be 0, a property that will not necessarily hold when passing to limits. This, however, is mitigated by the inclusion of all , as having one of these submatrices be all zero simply means that we are passing to a smaller sequence.

Thus, if a sequence converges in this space, then it must converge to another entry in this space. Therefore, we have shown that this is a valid CW complex.

Now, given this CW structure, we can make several statements about the topological properties of this space.

**Corollary:** is the trivial group.

**Proof:** We will use the fact that the fundamental group depends only on the 2-skeleton of a CW complex. There is a single 0-cell, so the 2-skeleton contains disks joined at a single point. Thus, by considering the segment of a loop residing in a single disk, this will be something homotopic to a constant map on the 0-cell. Thus, all maps are nullhomotopic, as desired.

**Corollary: **The homology and cohomology groups are free groups for even numbers and trivial for odd numbers.

**Proof:** The proof of each of these is essentially identical, so we will present only the proof for cohomology (as we are much more interested in the properties of cohomology). Consider the map associated with the cochain complex of the skeletons. If is odd, then is empty, so and the cohomology group is trivial. If is even, then and the cohomology group is the free group generated by the -cells.

Hopefully in the next post we will deal with the problem of how many cells there are in each dimension, the Betti numbers, and possibly the beginning of understanding the ring structure of the cohomology.

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**I. Too many filters**

Let’s get motivation from seeing how the ultrafilter construction of from last post fails. For starters, take a simple example of a non-discrete set – in fact, take to be already compact. When we compactify, it’s clear we must get the same set back. Hence every ultrafilter needs to correspond to a point of , in the sense of the canonical embedding .

But it’s obvious that we have far, far too many ultrafilters for that: all the principal ultrafilters clearly correspond to their generator, and then we have the immeasurably more numerous nonprincipal ultrafilters leftover.

We can say more: given any nonprincipal ultrafilter , it of course converges to a single point , by compact Hausdorffness. Hence its image under any continuous into a compact Hausdorff space is an ultrafilter on which then must converge to by continuity. It is in fact true that infinitely many ultrafilters converge to each point in , so each point in then corresponds to infinitely many ultrafilters in our attempted “” which have identical behavior on every continuous function. Since behavior of continuous functions on “” is still entirely specified by behavior on , we then find that many ultrafilters cannot even be separated by continuous functions, so “” is not even completely Hausdorff, let alone compact Hausdorff.

The problem is that our “take all ultrafilters” construction never uses any data about the topology on , essentially treating it as discrete. Any function on under the discrete topology is continuous, so we are able to separate many more ultrafilters than when we have to restrict via continuity.

So we have too many ultrafilters. How do we know how to cut them down? Well, if two of them converge to the same point, we should probably make them equivalent. To see how to do this, look back at the linked proof by Qiaochu that ultrafilters on compact sets converge to at least one point. To summarize, it’s really a translation of the finite intersection property definition of compactness: the closures of the sets of an ultrafilter satisfy the finite intersection property, so their union is nonempty, and it’s not difficult to show that the ultrafilter converges to any point in that union – exactly one, since we’re working in a purely Hausdorff setting. Specifically, any neighborhood of the point must be in the ultrafilter, since its complement couldn’t be, as a closed set not containing the union of the closures of the sets in the ultrafilter.

It thus occurs to us: what if our ultrafilters were somehow defined **only** on the closed sets to begin with? Let’s suppose we had an ‘ultrafilter’ on the closed sets of , whatever this means. The closed sets still form a **lattice**, so the filter definitions still work: intersections are in the filter, the filter is upwards closed. (Consult the definition if unfamiliar; here ‘join’ is union and ‘meet’ is intersection.)

But now the ‘ultra’ condition of either or being in it doesn’t make sense, as complements of closed sets are not closed in general. Indeed, this notion of ‘ultra’ for filters only applies to complementable bounded lattices, or **Boolean algebras** (wiki), which will be important in the next post. For now, however, we just observe that while we cannot have ultrafilters, we can still have maximal filters, in the sense that there is no filter such that strictly. That such filters exist is clear from Zorn’s lemma.

Proceeding as in the original case, we observe that by the finite intersection argument, there must be at least one point contained in every set of . Every closed set containing then must be in the filter by maximality. We then have that must be unique, since points are closed, so we have our desired correspondence:

**Proposition.** Every maximal filter on the lattice of closed sets of a compact Hausdorff space is principal.

We will just refer to these as “maximal filters” from here on for brevity.

**II. From closed to zero**

So this is promising. But of course it doesn’t mean anything yet. We haven’t checked whether this construction works on non-compact spaces, what the topology is, or how functions extend.

We’ll start with the last one. Let’s drop the compactness on , and consider . How do we extend this to the set of maximal filters?

As before, a maximal filter converges to if every neighborhood of contains a filter element. Again, it is clear by the Hausdorff condition that every filter converges to at most one point. In fact, more is true.

**Proposition.** For every point , precisely one maximal filter converges to it: the principal filter generated by that point.

**Proof.** It of course suffices to prove that every maximal filter converging to must be a subfilter of the principal filter, or equivalently that every set in such a filter much contain . Indeed, suppose we had some which does not contain ; then must have a neighborhood disjoint from it. Inside this neighborhood there must be another set of disjoint from , giving a contradiction.

So we have our one-to-one correspondence between points of and filters converging to them.

The preimage of a closed set under a continuous function is closed, and hence given we can define the image filter of a maximal filter on in the normal way – if and only if . The image of a maximal filter is also a maximal filter – that is, maximal on the lattice of closed subsets of – because if we could add to the filter , then we could add to .

Let us use the notation as the set of maximal filters of . (Why are we not simply using ? Foreshadowing!) So now for any with compact Hausdorff, we can indeed extend to .

To rigorously show that this is unique, we need to find the “right” topology on – it is sufficient to find a topology that makes it a compact Hausdorff space with the image of open and dense in it such that this extension is continuous, since then certainly uniquely determines .

Taking the Stone topology on ultrafilters as a model, let’s focus on that last bit: how do we make sure nonprincipal maximal filters are limit points of principal maximal filters?

Naively, we take as a base for closed sets for closed sets . We have . In fact this still holds for as well, since maximal filters are prime on general distributive lattices, not just Boolean algebras. However, we clearly don’t have complements anymore, which makes sense, since the compactification of a non-discrete set is not going to be overflowing with clopen sets.

**Proposition.** The image of (the principal maximal filters) is dense in under this topology, and the induced topology is the topology from . (The is a homeomorphism onto a dense subspace.)

**Proof.** Certainly it is dense; finding an open set disjoint from the principal maximal filters is equivalent to finding a closed set containing all principal maximal filters. Such a set would have to be generated by a set contained every principal filter, and hence would have to be .

That the induced topology is the same is fairly obvious from the construction of the base of closed sets.

To check that it is a compactification, we just need it to be compact (or if you prefer, quasicompact) and Hausdorff.The former is easy to check using the definition of compactness by closed sets: for any collection of basis closed sets with the finite intersection property.

**Proposition.** Under this topology, is quasicompact.

**Proof.** We show that for any collection of closed basis elements satisfying the finite intersection property, their intersection is nonempty. That is, if every finite subcollection of has a maximal filter containing all of them, the entire collection has such a maximal filter. But this is obvious: every finite intersection of the is nonempty, hence they generate a filter on the closed lattice, which then extends by the usual way (something something axiom of choice; we’ll take all the theory behind lattices and filters for granted, as we have been) to a maximal filter. So just as in the original discrete case, a filter construction “translates” to topological compactness nicely.

So it all comes down to the Hausdorff condition now. But the other shoe drops: this isn’t true. (Exercise: explicitly construct a counterexample, since I can’t figure one out.)

It turns out that our construction does work for normal spaces: in sketch, for any two maximal filters , we claim that we can find closed sets with and . Indeed we can find disjoint closed sets , , by a standard argument. Then by normality, we have disjoint open sets separating ; let . It’s not hard to show that they satisfy our condition. Then for any maximal filter , , so the complements of and are disjoint open sets containing and . The last part is to check that functions to compact Hausdorff spaces extend uniquely, but this is almost trivial – we just need to check that they extend continuously at all, and we have a canonical way of doings so that we can easily show is continuous in almost perfect analogy with the discrete case. So when is normal, we have constructed our compactification.

For general Tychonoff spaces, we need a space with slightly fewer points, generated by the maximal filters on a coarser lattice. (Coarser lattices have fewer maximal filters, intuitively.) This turns out to be the lattice of **zero sets**, sets of the form for some . Every zero set is clearly a closed set, but not vice versa. The two concepts coincide for perfectly normal spaces (though seemingly there is enough similarity for normal spaces to work with the closed lattice).

It’s easy to see that zero sets form a distributive lattice, so we can apply all our constructions from above to obtain as the set of maximal filters on the zero lattice, with the same closed base for closed .

**Proposition.** , .

**Proposition.** Every maximal zero filter on a compact Hausdorff space is principal.

Defining convergence in the same way as before, we find:

**Proposition.** In any Hausdorff space, the principal maximal zero filter at a point is the unique such filter converging to it.

**Proposition.** The map using the canonical identification with principal maximal zero filters is a homeomorphism onto its image, and dense.

**Proposition.** is compact Hausdorff.

**Proposition.** satisfies the universal property of the Stone-Čech compactification.

To avoid tedium, these are left as exercises for an interested reader. A majority of these follow simply from the fact that the zero sets are a coarsification of the closed sets.

We will only prove Hausdorffness, since this is where we failed last: it suffices to show that zero sets can be separated by cozero sets, by analogy with our construction for . In a Tychonoff space, cozero sets form a base for the topology, so in fact we can simply prove that zero sets can be separated by open sets. Indeed, given zero sets and for continuous , let

This is clearly continuous, and . Then and are disjoint open sets which separate the zero sets, as desired.

**Exercise.** Find a natural way to correspond the maximal zero filter construction with the construction from last post (because I sure can’t, and it seems like there should be one).

We didn’t present much in the way of motivation or intuition for using zero sets (compared to closed sets); for more on this subject, Gillman and Jerison’s *Rings of continuous functions* is a good resource which I mean to check out soon. There, the relationship of the Stone-Čech compactification to -algebras is emphasized, something we didn’t touch on. Probably the subject of a future post.

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The separation axioms , , and are properties that a specific topological space may or may not have. By giving examples, we will see that the condition that a space be is strictly stronger than the condition that a space be . Finally, we will discuss how the properties are inhertied by the constructions on topological spaces that we know, namely taking subspaces and product spaces.

**The Axioms.** Here are the axioms. We say a space is when:

- : For any two distinct points , either there is an open set around not containing (or both).
- : For any two distinct points , there is an open set around not containing and there is an open set around not containing .
- : For any two distinct points , there are disjoint open sets and with and .

The axiom is especially important and is also called the *Hausdorff axiom*. We say a space is *Hausdorf* if it is satisfies . We know some examples of Hausdorff spaces. For example a discrete space is Hausdorff. More generally, metric spaces are Hausdorff.

**Theorem.** If is a metric space, then is Hausdorff.

*Proof.* Let be distinct points. Let . The balls and are disjoint (by the triangle inequality) open sets containing and respectively.

Clearly . Also, clearly not all spaces satisfy even the weakest axiom : an indiscrete space with more than one point does not satisfy . Let us give some examples to show that does not imply , and does not imply .

**Example 1.** A space that is but not . Consider the space whose open sets are given by the “dot-diagram” below.

There is an open set around the point on the left not containing the point on the right, but not vice-versa.

**Example 2.** A space that is but not . Let be any infinite set with the cofinite topology. Thus, a subset is open in if and only if either is empty or is finite. For distinct points , and are open sets containing but not contaning , and containing but not containing respectively, so is . On the other hand, let and be any open sets containing but not containing , and containing but not containing respectively. Then and are finite sets and so

is a finite set. Since is infinite, we must have that is not empty. Since and were arbitrary, is not .

**Example 3.** Another example of a space that is but not is with the Zariski topology. Recall that the Zariski topology has as a basis the set

where

Given and , distinct points in , the open sets

contain but not and but not respectively. This shows that this topology is . To see that the topology is not , it suffices to show that the intersection of nonempty open sets is nonempty. For arbitrary open sets and , write

Then , so it suffices to show that the intersection of two nonempty basic open sets is nonempty, i.e., that given nonzero polynomials and , there is a point such that and are both nonzero. This is clear if (since and each ahve ony finitely many zeros) and can be shown for general by induction: for all but at most finitely many values of , and are nonzero polynomials in .

Note that both examples of spaces that are not are infinite. It turns out that a finite space must be discrete. Let be a topological space whose underlying set is a finite set. For any the property implies that the intersection of all open sets containing is just the set . There are only a finite number of open sets containing (since there are only finitely many subsets of ), so eveyr one point set is open; this implies that is discrete. This observation is closely related to the following fact.

**Proposition.** A topological space is if and only if every one point subset is closed.

*Proof.* If is and , then the property implies that for each not equal to , we can find an open set containing and not containing . Then is an open set, and so is closed. Suppose on the other hand that eveyr one point set of is closed. Then for distinct points , is an open set containing but containing and is an open set containing but not containing . Hence is .

Finally, let us talk about subspaces and product spaces.

**Theorem.** Let be a topological space, and let be a subspace. If is , , or then so is .

*Proof.* Let and be distinct points in . Then and are also distinct points in , and if is , we can find an open subset of around one not containing the other. Then is an open subset of containing (the same) one of or and not the other, and so is . If is , we can find open subset and of such that , , , and . Then and are open subset of containing but not and containing but not , respectively, and so is . If is , then we can choose and to be disjoint. Then and are disjoint, and so is .

**Theorem.** Let and be topological spaces. If and are both , , or then so is .

*Proof.* Let and be disjoint points in . Then either or . We will treat the case ; the other case is entirely similar. If is , then we can find an open subset of containing one of or but not the other. Then is an open subset of containing one of or but not the other, and so is . If is , we can find open subset and of such that , , , and . Then and are open subset of containing but not and containing but not , respectively, and so is . If is , then we can choose and to be disjoint. Then and are disjoint, and so is .

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