# Cohomology of the Complex Grassmanian – Part II: The Betti Numbers and the Weil Conjectures

So, in the last post, we found a CW decomposition for the Grassmanian ${G_n(\mathbb{C}^{m+n})}$ and computed its CW cohomology.  We found that it was a free group generated by the number of cells in its decomposition.  The number of generators can be found by considering ${n}$-tuples ${\lambda}$ of integers ${\leq m}$.  We said that each cell corresponds to such a sequence and that the dimension of the associated cell is precisely ${2(mn - \sum \lambda_i)}$. Thus, the number of generators for a given cohomology group ${H^{2k}(G_n(\mathbb{C}^{m+n}))}$ is precisely the number of partitions of ${mn - k}$ into a sequence of ${n}$ integers ${\leq m}$.

We now will introduce the concept of the Betti numbers of a topological space.  Intuitively, we want the ${n}$-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 ${k}$-th Betti number, denoted ${b_k}$, will denote the rank of the ${k}$-th homology group ${H_k}$.

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 ${X}$ be an ${n}$-dimensional non-singular, projective variety defined over a finite field ${\mathbb{F}_q}$.  Then we define the zeta function on ${X}$ as: ${ \zeta(s) = \exp\left(\sum_{r=1}^{\infty} \frac{N_r}{r}s^r\right) }$, where ${N_r}$ denotes the number of points of ${X}$ defined over ${\mathbb{F}_{q^r}}$.

Theorem: (1) (Rationality${\zeta(s)}$ is a rational function (that is, it can be written as the quotient of two polynomials)

(2) (Functional equation${\zeta\left(\frac{1}{q^ns}\right) = \pm q^{nE/2}s^E\zeta(s)}$, where ${E}$ is the Euler characteristic of ${X}$

(3) (Analogue to Riemann hypothesis) As a rational function, we can write ${\zeta(s) = \frac{P_1(s)P_3(s)\cdots P_{2n-1}(s)}{P_0(s)P_2(s)\cdots P_{2n}(s)}}$ with ${P_0(s) = 1-s, P_{2n}(s) = 1-q^ns}$ and, for ${1 \leq i \leq 2n-1, P_i(t) = \prod_j (1- \alpha_{ij}t)}$.  When written in this form, we have that ${\alpha_{ij}}$ are algebraic integers and ${|\alpha_{ij}| = q^{\frac{i}{2}}}$.

(4) (Betti numbers) If ${X}$ is the reduction mod ${p}$ of a projective variety ${Y}$ over a field embedded in the field of complex numbers, then the degree of ${P_i}$ is the ${i}$-th Betti number of ${Y}$.

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 ${\iota: G_n(\mathbb{X}^{m+n}) \mapsto P(\wedge^n\mathbb{C}^{m+n})}$ as ${\iota(\text{span}(v_1,\ldots,v_n)) = [v_1 \wedge \ldots \wedge v_n]}$.  We claim that this shows that this is a projective variety.  We must verify a few points:

This map is well defined: Let ${\{w_1,\ldots,w_n\}}$ be another basis for the subspace ${\text{span}(v_1,\ldots,v_n)}$.  Then we have that there is an invertible transformation ${A}$ such that ${v_1 \wedge \ldots \wedge v_n = \det(A)(w_1\wedge \ldots \wedge w_n)}$, so they are equivalent in the projective space.  Thus, this map is well defined.

This map is injective: Let ${v \neq 0}$.  Then ${v \wedge (v_1 \wedge \ldots \wedge v_n) = 0}$ if and only if it is in the span.  Thus, let ${\varphi(x) = x \wedge (v_1 \wedge \ldots \wedge v_n)}$.  Then we have that ${\ker(\varphi) = \text{span}(v_1,\ldots,v_n)}$.  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 ${\varphi}$ defined above as a matrix ${M}$ by fixing a basis for ${G_n(\mathbb{C}^{m+n})}$ and a basis for ${P(\wedge^n\mathbb{C}^{m+n})}$.  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). ${\square}$

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 ${\mathbb{F}_q}$.  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 ${k}$ dimensional subspaces of ${(\mathbb{F}_q)^n}$.  For notational convenience, we will denote this number by ${\binom{n}{k}_q}$ (as this notation suggests, this will have a distinctly combinatorial flavor and will be connected to the standard binomial coefficients).

Claim: ${\binom{n}{k}_q = \frac{(q^n-1)(q^{n-1}-1)\cdots(q^{n-k+1}-1)}{(q^k-1)(q^{k-1}-1)\cdots(q-1)}}$.  Furthermore, if we let ${m!_q = (q^m-1)(q^{m-1}-1)\cdots(q-1)}$, then ${\binom{n}{k}_q = \frac{n!_q}{(n-k)!_qk!_q}}$.

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 ${k}$ linearly independent vectors in ${(\mathbb{F}_q)^n}$.  To do this, we simply note that the first vector can be any nonzero vector, so there are ${q^n-1}$ choices.  The second vector can be any vector not in the subspace spanned by the first vector, so there are ${q^n - q}$ choices.  Repeating for ${k}$ choices gives ${(q^n-1)(q^n-q)\cdots(q^n - q^{k-1})}$.  This, however, overcounts subspaces.  In fact, the number of times that each subspace is counted is equal to precisely the number of sets of ${k}$ linearly independent vectors in ${(\mathbb{F}_q)^k}$.  Thus, we repeat the same calculation and divide out to get ${\frac{(q^n-1)(q^n-q)\cdots(q^n - q^{k-1})}{(q^k-1)(q^k-q)\cdots(q^k - q^{k-1}}}$.  We can factor out a ${q^{\frac{k(k-1)}{2}}}$ to achieve the desired result. ${\square}$

Thus, we have an explicit computation for ${N_r = \binom{m+n}{n}_{q^r}}$.  However, in order to verify the rationality of the zeta function, we will need one more result about this number:

Claim (Generalized Pascal Identity): ${\binom{n}{k}_q = q^k\binom{n-1}{k}_q + \binom{n-1}{k-1}_q}$.

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

Corollary: ${\binom{n}{k}_q}$ is a polynomial in ${q}$ with positive coefficients of degree ${k(n-k)}$.

Proof: This follows immediately from a simple inductive argument and the fact that ${\binom{n}{0}_q = 1}$.

Thus, we know that we can write ${\binom{n}{k}_q = \sum a_iq^i}$.  Using this, we can verify the Weil conjectures for the Grassmanian explicitly as well as compute its Betti numbers.

Theorem 1: ${\zeta(s)}$ is a rational function.

Proof: We will use the fact that ${N_r = \binom{m+n}{n}_{q^r} = \sum a_iq^{ri}}$.  This gives ${\zeta(s) = \exp\left(\sum_{r=1}^{\infty} \frac{\sum a_i(q^is)^r}{r}\right)}$.  We see that the expression inside the exponential is precisely a Taylor expansion for ${\log(1-x)}$, so by switching the order of the summations we get that the expression is equivalent to ${\zeta(s) = \exp(-\sum a_i \log(1 - q^is)) = (1-s)^{-a_1}(1-qs)^{-a_2}\cdots (1-q^{mn}s)^{-a_{mn}}}$.  Thus, this is a rational function, as claimed. ${\square}$

First, we note immediately that all of the ${a_i}$ are positive, so this implies that all of the ${P_{2i+1}}$ 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 ${P_{2i}(s) = (1 - q^is)^{a_i}}$.  Then the desired result holds.

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

Lemma: Let ${\lambda_{m,n}(j)}$ denote the number of partitions of ${j}$ into ${n}$ integers, each ${\leq m}$.  Then ${\binom{m+n}{m}_q = \sum_{j=0}^{mn} \lambda_{m,n}(j)q^j}$.

Proof: We will induct on ${m}$ using the generalized Pascal identity discussed above.  If ${m = 0}$, then the result is trivial.  Now fix ${n}$.  By the inductive hypothesis, suppose it holds for ${(m-1,n)}$ and ${(m,n-1)}$.  Then we have ${\binom{m+n}{n}_q = q^n\binom{m+n-1}{n}_q + \binom{m+n-1}{n-1}_q = \sum_{j=0}^{(m-1)n} \lambda_{m-1,n}(j)q^{j+n} + \sum_{j=0}^{m(n-1)} \lambda_{m,n-1}(j)q^j}$.

We claim that the desired result holds immediately.  To see this, suppose we have a partition of ${j}$ into ${n}$ integers ${\leq m}$.  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 ${n-1}$ integers ${\leq m}$.  If none are zero, then we can decrease each integer by ${1}$ and we will have a partition of ${j-n}$ into ${n}$ integers ${\leq m-1}$.  Thus, we get the identity ${\lambda_{m,n}(j) = \lambda_{m-1,n}(j) + \lambda_{m,n-1}(j-n)}$, which allows us to combine the terms in our expression above to achieve the desired result. ${\square}$

Thus, we claim that the following follows immediately:

Theorem 4: The degree of ${P_{2i}}$ is precisely ${b_i}$.

Proof: The degree of this polynomial is precisely ${a_i}$, which we have just shown is equal to ${\lambda_{m,n}(i)}$, which we defined to be precisely the ${b_i}$ computed at the beginning of the post.  Thus, the result holds. ${\square}$

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.