Consider the Grassmannian $Gr(n, N)$ of $n$-dimensional subspaces of $\mathbf C^N$. Its cohomology ring is isomorphic to $\mathbf C[x_1, \ldots, x_n, \bar x_1, \ldots \bar x_{N-n}]/I_{n,N}$, wherethe ideal is generated by homogeneous elements of

$$ (1+x_1t+\ldots+ x_nt^n)(1+\bar x_1 t+\ldots +\bar x_{N-n}t^{N-n}) -1$$

This ring is isomorphic to the ring

$$\frac{\mathbf C[\varepsilon_1, \ldots, \varepsilon_n]}{(\text{$h_i=0$ for $i>N-n$})}$$

of elementary symmetric polynomials $\varepsilon_i$ in the $x_i$, modulo the complete homogeneous symmetric polynomials $h_i$ for $i>N-n$.

Ellis/Khovanov/Lauda now show similar results for odd polynomials $\mathbf C\langle x_1, \ldots, x_n\rangle/(\text{$x_ix_j=-x_jx_i$ for $i\neq j$})$, see [2]. One can define odd analogues of the elementary and complete homogeneous symmetric polynomials. Then one can recover the odd analogue of the above quotient as some odd cohomology, see [1].

The even case can be generalized for $n$-step flag varieties $\{0=F_0\subseteq F_1 \subseteq\ldots\subseteq F_n = \mathbf C^N\}$, see [3, §5]. The respective ideal to be quotiented out then is generated by homogeneous parts of

$$ (1+x_{11}t+x_{12}t^2+\ldots)(1+x_{21}t+x_{22}t^2+\ldots)\cdots -1,$$

and the “dual” elements $\bar x_{ij}$ are given by homogeneous parts of that ideal, leaving out the $i$-th factor.

Question: Can the results for odd symmetric polynomials and their relations to cohomology of Grassmannians be generalized to some odd cohomology of $n$-step flag varieties? I am struggling with the problem that I cannot pull out factors of the above ideal easily anymore in the non-commutative case.

[1] https://sites.math.washington.edu/~julia/AMS_SFSU_2014/Lauda.pdf

[2] https://arxiv.org/abs/1111.1320

[3] https://arxiv.org/abs/0807.3250


1 Answer 1


Well I'm a bit late, but I think what you are asking is in Lauda-Russell paper : https://arxiv.org/pdf/1203.0797.pdf

It's written OH(X) in the reference.

  • 3
    $\begingroup$ You should probably expand on your answer given that it has been a while. $\endgroup$ Commented Nov 27, 2017 at 22:28

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