# Cohomology of the Complex Grassmanian – Part I: A CW structure

This series of post will primarily be an exposition of my project for the Directed Reading Program here at the University of Chicago. The goal of this project is, primarily, to understand the cohomology ring associated with the Grassmanian.  Thus, this first post will seek to find a CW decomposition of the Grassmanian and use this to compute not only the cohomology groups but also the fundamental group and the homology groups.

Definition: The Grassmanian of dimension ${n}$ over a vector space ${V}$ is defined to be the set of all subspaces of dimension ${n}$.  We will denote this by ${G_n(V)}$.

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 ${V_n}$ denote the set of all ${n}$-tuples of linearly independent vectors in ${V}$.  We then define an equivalence relation on this space where ${a\sim b}$ if and only if ${a}$ and ${b}$ 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 ${G_n(\mathbb{C}^{m+n})}$.  The choice of notation here may seem a bit odd at first: why not just write ${G_n(\mathbb{C}^m)}$?  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 ${\lambda}$ such that ${m \geq \lambda_1 \geq \lambda_2 \geq \ldots \geq \lambda_n \geq 0}$.  We will then define the set ${e(\lambda)}$ to be the set of all ${n}$-planes ${W}$ in ${\mathbb{C}^{m+n}}$ such that ${\dim(W \cap \mathbb{C}^{m + i - \lambda_i}) = i}$ and ${\dim(W \cap \mathbb{C}^{m + i - \lambda_i - 1}) = i - 1}$.  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 ${\mathbb{C}^{m+ i - \lambda_i}}$ is the lowest dimension with the property that its intersection with ${W}$ has dimension ${i}$.  We can most easily represent this as a matrix.  Consider the example of ${G_3(\mathbb{C}^5)}$.  Let ${\lambda = (2,1,1)}$.  Then any subspace in ${e(\lambda)}$ has a basis that can be expressed in the following way:

${\left(\begin{array}{ccccc}a_1 & 0 & 0 & 0 & 0 \\ b_1&b_2&b_3&0&0\\ c_1&c_2&c_3&c_4&0\end{array}\right)}$

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

${\left(\begin{array}{ccccc}1 & 0 & 0 & 0 & 0 \\ 0&b'_2&1&0&0\\ 0&0&c'_3&1&0\end{array}\right)}$

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 ${e(\lambda)}$ has dimension precisely equal to ${2(mn - \sum \lambda_i)}$.

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

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 ${X \in G_n(\mathbb{C}^{m+n})}$, we know that ${X \in e(\lambda)}$ for exactly one ${\lambda}$.

Proof: To verify this, we will need to simply consider the sequence given by ${\dim(X \cap \mathbb{C}^k)}$. This will uniquely determine the sequence ${\lambda}$ associated with the subspace, so the claim holds. ${\square}$

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 ${\bar{e(\lambda)}}$ has the property that ${\bar{e(\lambda)} - e(\lambda)}$ 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 ${\lambda_1 \leq \lambda_2}$ if and only if ${\lambda_{1,i} \geq \lambda_{2,i}}$ for all ${i}$.  We claim then that the set defined by ${g(\lambda) = \cup_{\sigma \leq \lambda}e(\sigma)}$ is closed.  If this is the case, then our proof will be complete, as we will have found a closed set containing ${e(\lambda)}$ with the desired property, and ${\overline{e(\lambda)} \subset g(\lambda)}$.  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 ${n \times (m+n)}$ 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:

${\left(\begin{array}{ccccc}a_1 & a_2 & a_3 & a_4 & a_5 \\ b_1&b_2&b_3&b_4&b_5\\ c_1&c_2&c_3&c_4&c_5 \end{array}\right)}$

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

${\left(\begin{array}{ccccc}a'_1 & 0 & 0 & 0 & 0 \\ b'_1&b'_2&b'_3&0&0\\ c'_1&c'_2&c'_3&c'_4&0\end{array}\right)}$

We first note that we must have ${a_5=b_5=c_5=0}$Thus, we can consider, for example, the submatrix given by:

${\left(\begin{array}{c}a_4 \\ b_4\\ c_4\end{array}\right)}$

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

${\left(\begin{array}{cc} a_2&a_3 \\ b_2&b_3 \end{array}\right)}$

must have rank 1.  This means that ${\frac{a_2}{a_3} = \frac{b_2}{b_3}}$.  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 ${e(\lambda)}$ 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 ${\sigma \leq \lambda}$, 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. ${\square}$

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

Corollary: ${\pi_1(G_n(\mathbb{C}^{m+n}))}$ 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. ${\square}$

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 ${d_n: X^n \mapsto X^{n+1}}$ associated with the cochain complex of the skeletons.  If ${n}$ is odd, then ${X^n}$ is empty, so ${\ker d_n = \varnothing}$ and the cohomology group is trivial.  If ${n}$ is even, then ${\text{im } d_{n-1} = \varnothing}$ and the cohomology group is the free group generated by the ${n}$-cells. ${\square}$

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.