## Schubert Classes and Cellular Cohomology

So, as of the last post in the series, we defined Schubert cells.  We’re going to use them to discuss the Cohomology of the Grassmannian, and to write down an explicit basis.  With an eye looking forward, next time, we’ll work out the cup product in this cohomology ring, and then finally we’ll use it to solve some problems.  So today, we’ll discuss Cellular Homology, Poincaré Duality, and the cohomology of the Grassmannian.

The first thing we need is that the Grassmannian is the union of the Schubert cells, and that any two of them are disjoint.  Let $\lambda,\mu$ be two Young diagrams fitting into an $n\times r$ box.  Then, for each, there is a unique form of a reduced row echelon matrix $M_\lambda,M_\mu$, as described last time. (That is, all entries are arbitrary, except has a 1 in the $(i,n+1-\lambda_i)$ place and zeros in that column and to the right.)

Now, assume that $V\in\Omega_\lambda^\circ\cap \Omega_\mu^\circ$.  Then $V$ is the row span of a matrix of the form $M_\lambda$ and one of the form $M_\mu$, and the matrix in each is uniquely determined, as the matrices are in row eschelon form.  Look at the first $i$ such that $\lambda_i\neq \mu_i$, and, without loss of generality, assume that $\lambda_i>\mu_i$.  Then the condition from $M_\lambda$ says that for all $k>n+1-\lambda_i$, we have the $(i,k)$ entry is zero.  One such entry is $(i,n+1-\mu_i)$, and so this entry must be both 1 and 0.  This is a contradiction, so $\Omega_\lambda^\circ\cap\Omega_\mu^\circ=\emptyset$, and so the Schubert cells are disjoint.  It is straightforward that they cover $Gr^n(E)$.

Now, as the notation suggests, $\overline{\Omega_\lambda^\circ}=\Omega_\lambda$, though we’re not going to bother proving it.  Instead, we’re going to move on to the algebraic topology we need.

We’ll need to talk about CW complexes for a bit first.  An $n$-cell is a topological space homeomorphic to the closed $n$-disc.  A CW complex is any space that can be built up in the following way:

Start with a discrete set, called $X^0$, the 0-skeleton.  Then inductively define the $n$-skeleton $X^n$ from $X^{n-1}$ by attaching $n$-cells by gluing it to $X^{n-1}$ along the boundary $(n-1)$-sphere.  There’s a third condition, but it’s irrelevant to us, because we only care about complexes with finitely many cells.

Now, to work, we’ll need to use homology rather than cohomology.  The big lemma about CW complexes is as follows:

Lemma If $X$ is a CW complex, then

1. $H_k(X^n,X^{n-1})$ is zero for $k\neq n$ and is free abelian for $k-n$, with a basis in one-to-one correspondence with the $n$-cells of $X$.
2. $H_k(X^n)=0$ for $k>n$.
3. The inclusion $i:X^n\to X$ induces an isomorphism $i_*:H_k(X^n)\to H_k(X)$ if $k.

This, along with the relative homology exact sequence, $\ldots H_n(A)\stackrel{i_*}{\to} H_n(X)\stackrel{j_*}{\to} H_n(X,A)\stackrel{\partial_n}{\to} H_{n-1}(A)\to\ldots$ when $A\subset X$

Next, we can use the long exact sequences (as above) of the pairs $(X^{n+1},X^n), (X^n,X^{n-1}), (X^{n-1},X^{n-2})$ to construct maps $d_{n+1}:H_{n+1}(X^{n+1},X^n)\to H_n(X^n,X^{n-1})$ and $d_n:H_n(X^n,X^{n-1})\to H_{n-1}(X^{n-1},X^{n-2})$ by taking $d_{n+1}=j_n\partial_{n+1}$ where $j_n$ is the $j_*$ map in the long exact sequence of $(X^n,X^{n-1})$.

This gives us a chain complex called the cellular chain complex, and its homology groups are the cellular homology of $X$, denoted $H_n^{CW}(X)$.  The big theorem is that this is isomorphic to the usual singular homology $H_n(X)$, and this gives the following results about CW complexes:

1. $H_n(X)=0$ if $X$ has no $n$-cells.
2. If $X$ has $k$ $n$-cells, tehn $H_n(X)$ is generated by at most $k$ elements.
3. If $X$ is a CW complex with no two cells in adjacent dimension, then $H_n(X)$ is free abelian with basis in correspondence to the $n$-cells.

That last one is the most important for us.  Using Schubert cells, we have the Grassmannian as a CW complex with no cells in odd dimension, so 3 holds.  That means that $H_k(Gr^n(E))$ is the free ablian group of the Schubert cells of dimension $k$.

The final ingredient is Poincaré duality.  A Grassmannian is a compact manifold (it’s a smooth projective variety over $\mathbb{C}$) and the complex structure gives it a natural orientation, so it must be orientable.  Thus, it satisfies Poincaré duality.  This says that there’s an isomorphism $H^k(G,\mathbb{Z})\to H_{n-k}(G,\mathbb{Z})$.  This tells us that $H^k(Gr^n(E),\mathbb{Z})$ consists of Schubert cells of codimension $k=|\lambda|$.

So, finally, we have that the cohomology ring $H^*(Gr^n(E),\mathbb{Z})$ of a Grassmannian has basis the classes of Schubert varieties (because when we use Poincaré duality, we want to replace the represenatives of the classes with their closures).

There are two more facts that we want to mention today:

First is that the cohomology classes $[\Omega_\lambda]=\sigma_\lambda$ don’t depend on the choice of flag, so for any two flags, we get different $\Omega_\lambda(F_*)$, but they represent the same cohomology class.  This is because the group $GL(E)$ acts transitively on the space of complete flags in $E$.

The last thing to mention now is that the basis $\{\sigma_\lambda\}$ is self dual.  That is, for each $\lambda$, there exists $\mu$ such that $\sigma_\mu\cup \sigma_\lambda=\sigma_{(r,\ldots,r)}$.  This relationship will hold if and only if $\lambda_i+\mu_{r+1-i}=n$ for all $1\leq i\leq r$.  More geometrically, this says that two Schubert varieties intersect in a unique point if and only if their Young diagrams fit together to form the rectangle.  That is, if the Young diagrams are $\lambda,\mu$, then if we rotate $\mu$ by 180 degrees and put the two diagrams together, they form the whole rectangle that diagrams are allowed to fit into.

That’ll be it for today.  Next time: the Pieri and Giambelli formulas that will allow us to compute arbitrary intersections. (For those following who know more, I will NOT be discussing the Littlewood-Richardson rule, partly because I don’t need to prove that things are positive anywhere, and the LR rule is complicated and combinatorial.) Charles Siegel is currently a postdoc at Kavli IPMU in Japan. He works on the geometry of the moduli space of curves.
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### 6 Responses to Schubert Classes and Cellular Cohomology

1. Alexander Ellis says:

A minor typo in the second to last paragraph:”form the whole square” should be “form the whole rectangle.”

Great series so far, I’m looking forward to the rest! Will you get to talk about the cohomology of more general flag varieties? Or about converting between the Schubert and other other generators for the cohomology (e.g., Chern class for the tautological bundle)?

2. Charles Siegel says:

Thanks for the correction! I had been working out the example of $Gr^2(\mathbb{C}^4)$ to make sure my proofs were ok, and that has a square.

3. Bryan says:

“because when we use Poincaré duality, we want to replace the represenatives of the classes with their closures”

4. Charles Siegel says: