Presentation of the cohomology of generalized flag varieties as graded ranks of rings of symmetric polynomials

Question

Hello!

Let $n,m\geq 0$ be integers. If I understand it correctly, there is the following description of the cohomology of the complex Grassmannian $\text{Gr}(m+n;m)$: denote by $\text{Sym}(n,m)$ the subring of the polynomial ring in $n+m$ variables consisting of polynomials which are symmetric in the first $n$ and the last $m$ variables, and by $\text{Sym}(n+m)$ the ring of symmetric polynomials in $n+m$ variables. Then $\text{Sym}(n,m)$ is a free $\text{Sym}(n+m)$ module, and its graded rank (written as a $q$-polynomial) equals the Poincare polynomial of the Grassmannian $\text{Gr}(m+n;m)$.

I'd like to know if this is true for more general flag varieties: Is it true that the graded rank of $\text{Sym}(m_1,...,m_k)$ over $\text{Sym}(m_1+...+m_k)$ is the Poincare polynomial of the variety of flags $(\{0\}=U_0,U_1,...,U_k=V)$ in $V := {\mathbb C}^{m_1+...+m_k}$ such that $\text{dim}(U_{i}) - \text{dim}(U_{i-1}) = m_i$ for $i=1,...,k$? If yes: how to prove it? :-)

On the other hand, there is a presentation of the cohomology of the full flag variety of ${\mathbb C}^N$ as the quotient of ${\mathbb C}[X_1,...,X_N]$ by the ideal generated by the symmetric polynomials of positive degree. How do these two descriptions of the cohomology of flag varieties relate?

Thank you!

Hanno

Answer 1

Yes, this is all true. The cohomology of the partial flag variety $F(m_1,\dots,m_k)$ is the quotient of the polynomials that are symmetric under $S_{m_1}\times \cdots \times S_{m_k}$ by the positive degree ones which are fully symmetric (a special case of this is the presentation you mention for the full flag variety). Since partially symmetric polynomials are a free module over full symmetric ones, the q-rank of the free module is the same as the q-dimension of the quotient by all positive degree fully symmetric polynomials.

The "reason" this presentation works is that $F(m_1,\dots,m_k)$ carries tautological vector bundles $U_i/U_{i-1}$ (these are subquotients of the trivial bundles of rank $m_1+\cdots m_k$). The Chern classes of these bundles are the elementary symmetric polynomials in the variables corresponding to $S_{m_i}$; thus, the fully symmetric polynomials, written in terms of these are the Chern classes of the the trivial bundle of rank $m_1+\cdots m_k$ by the Whitney sum formula (which says that if I think of $c_i(V)$ as elementary symmetric functions in one set of variables, and $c_j(W)$ as elementary symmetric functions in another, the Chern classes $c_k(V\oplus W)$ are the elementary symmetric functions in the union of the variables) and the splitting principle (which says that when I have a filtration $ V_1\subset V_2\subset \cdots\subset V_n=V$, then $c_i(V)=c_i(V/V_{n-1}\oplus \cdots \oplus V_2/V_1\oplus V_1)$) are thus 0. Admittedly, you then have to do some actual geometry to see that these Chern classes generate and that these are all the relations, but to me, this is the heart of the argument.

Answer 2

One way is to use the Eilenberg-Moore spectral sequence applied to the fibration $$ \mathrm{Fl}_{m_1,\ldots,m_k} \rightarrow BU_{m_1}\times\cdots\times BU_{m_k}\rightarrow BU_{m_1+\cdots+m_k} $$ which is a spectral sequence $$ \mathrm{Tor}_*^{\mathrm{Sym}(m_1+\ldots+m_k)}(\mathrm{Sym}(m_1,\ldots,m_k),\mathbb Z) \implies H^*(\mathrm{Fl}_{m_1,\ldots,m_k}). $$ However, as $\mathrm{Sym}(m_1,\ldots,m_k)$ is projective as module over $\mathrm{Sym}(m_1+\ldots+m_k)$ (by for instance the fact that is a finite Cohen-Macaulay module over a regular ring of the same dimension) the sequence collapses and gives $$ H^*(\mathrm{Fl}_{m_1,\ldots,m_k}) = \mathrm{Sym}(m_1,\ldots,m_k)\bigotimes{}_{\mathrm{Sym}(m_1+\ldots+m_k)}\mathbb Z $$ which gives what you want.