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The vector space dimension of the cohomology group of the $2$-plane Grassmannian $\mathrm{Gr}_{2,n}$ is given by the number of tuples $(\lambda_1,\lambda_2)$ satisfying $$ n - 2 \geq \lambda_1 \geq \lambda_2 \geq 0. $$ Explicitly this is given by $$ \binom{n}{2}. $$ This also happens to be the dimension of $V_{\pi_2}$ the second fundamental representation of $\frak{sl}_n$. I am guessing this is not an accident, especially since the $2$-plane Grassmannian corresponds (in the usual way) to $V_{\pi_2}$.

Does this extend to the general identity $$ \mathrm{dim}(H^{*}(\mathrm{Gr}_{d,n})) = \mathrm{dim}(V_{\pi_d})? $$ If it does, then what is a conceptual explanation for this?

EDIT: Since $V_{\pi_d}$ is isomorphic to the exterior power $$ \Lambda^d(V_{\pi_1}) $$ and $V_{\pi_1}$ is of dimension of $n$, we see that the RHS of the claimed identity is the binomial coefficient $$ \binom{n}{d}. $$ It follows from the general formula given in this answer that the LHS is the same binomial coefficient. Thus the identity does indeed extend from $2$-planes to $d$-planes. So the question is if there is a conceptual reason for this . . .

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  • $\begingroup$ Yes. See en.wikipedia.org/wiki/Borel%E2%80%93Weil%E2%80%93Bott_theorem $\endgroup$ Feb 12, 2023 at 21:59
  • $\begingroup$ @Sam The Borel-Weil theorem relates the space of holomorphic sections of a line bundle to $\frak{g}$-representations. I am asking about the dimension of the Schubert calculus algebra (i.e. the cohomology group of the whole untwisted de Rham complex). $\endgroup$ Feb 12, 2023 at 22:02
  • $\begingroup$ Ah of course sorry I was mixing up the coordinate ring and the cohomology ring. Still, I think it may be a starting point... $\endgroup$ Feb 12, 2023 at 22:03
  • $\begingroup$ @Sam Ok, thanks for the suggestion! $\endgroup$ Feb 12, 2023 at 22:03
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    $\begingroup$ The best explanation I know of the relation between Grassmannian cohomology and representations of SL_n is this paper of Tamvakis: arxiv.org/abs/math/0306414 $\endgroup$
    – Oliver
    Feb 13, 2023 at 15:21

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This is very standard. For a compact complex variety admitting a cell decomposition, the (co)homology is the free Abelian group generated by the cells (over $\mathbb{C}$ there is no room for the differential in the cellular complex). In the cellular decomposition of the Grassmannian, the (Schubert) cells are indexed by possible reduced row echelon forms of a $d\times n$-matrix, that is by possible positions of pivots. To the cell having pivots in positions $i_1,\ldots,i_d$ you can assign the basis element $e_{i_1}\wedge\cdots\wedge e_{i_d}$ in $\Lambda^d(V_{\pi_1})$.

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    $\begingroup$ @SamHopkins here is a speculative idea. Coordinates of the Plücker embedding should separate different points. If we wish that the embedding is defined in a universal way, we perhaps should wish that it separates points over $\mathbb{F}_1$, which naively makes one think that the number of coordinates should be given by the number of $\mathbb{F}_1$-points, that is $\binom{n}{d}$. Someone who is well versed in geometry over $\mathbb{F}_1$ can say if this can be made rigorous, but as a heuristics I certainly find it rather useful. $\endgroup$ Feb 13, 2023 at 14:11
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    $\begingroup$ Do you know of some nice natural way of defining a structure of an irreducible $\mathfrak {sl}_n$-module on this ring? AFAICT, the standard action induced from the Grassmannian is trivial. (Not sure if this was part of the question but kinda curious.) $\endgroup$ Feb 14, 2023 at 0:50
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    $\begingroup$ @IgorMakhlin the ring is graded, and the $sl_n$ action is going to mess it up, so something non-trivial will have to happen. I don't know what it could be. $\endgroup$ Feb 14, 2023 at 11:56
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    $\begingroup$ @Vladimir One can find papers such as this or this that seem to consider actions where root vectors act as derivations of the cohomology ring which shift the grading. This is kind of what I had in mind, not that I claim to have made sense of either paper. $\endgroup$ Feb 14, 2023 at 15:48
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    $\begingroup$ @IgorMakhlin - interesting! I was not aware of these papers. Particularly arxiv.org/abs/2111.08754 might be a good candidate for a conceptual answer to the OP's question. $\endgroup$ Feb 14, 2023 at 16:32
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Harry Tamvakis gives an explanation of the relation between Grassmannian cohomology and representations of SL_n in The connection between representation theory and Schubert calculus.

One fundamental obstacle for explaining this phenomenon is that there is not an analogous relationship between the cohomology rings of other flag varieties (Grassmannian-analogues for other Lie groups) and the representations of those Lie groups. For Tamvakis, the key fact that makes his explanation go is that GL_n is dense in its own Lie algebra, which is not generally true for other Lie groups. (At least this is what I remember; it's been a while since I read the paper carefully.)

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This is a special case of geometric Satake. I have to admit, I'm really struggling to find a place where this theorem is stated in an elementary-ish statement that makes this clear and am totally failing. By Mirkovic-Vilonen, Thm. 7.3 (and the discussion below), the irreps $V_{\lambda}$ of any complex algebraic group can be written as the intersection cohomology of the closure $\overline{\check{G}[[t]] \cdot t^{\lambda}\cdot \check{G}[[t]]}/G[[t]]\subset G((t))/G[[t]]$ in the affine Grassmannian. If $\lambda$ is minuscule, then this orbit is already closed and smooth, so the intersection cohomology is usual cohomology. Furthermore, it is of the form $\check{G}/\check{P}$ for $\check{P}$ the parabolic corresponding to the stabilizer of the minuscule weight in the Weyl group.

For $G=\check{G}=GL_n$, these are all Grassmannians; more generally, every minuscule representation is isomorphic to the cohomology of the corresponding cominuscule flag variety of the dual group.

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    $\begingroup$ In your last paragraph, what category is that isomorphism happening in? $\endgroup$
    – Oliver
    Feb 17, 2023 at 1:24
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    $\begingroup$ @Oliver They are isomorphic as vector spaces. $\endgroup$
    – Ben Webster
    Feb 17, 2023 at 22:18
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I accidentally (looking for something else) came across another paper where a very elegant explanation is given:

Dan Laksov, Anders Thorup: Schubert Calculus on Grassmannians and Exterior Powers

Indiana University Mathematics Journal, Vol. 58, No. 1 (2009), pp. 283-300

They introduce the "$d$-th factorization algebra" that "controls" the ways to factor a polynomial $p(x)$ of degree $n$ into a product of factors of degrees $d$ and $n-d$, relate the $d$-th exterior power of the quotient by $p(x)$ to the $d$-th factorization algebra (Main Theorem, see Sec.0.6), and also relate the $d$-th factorization algebra to the Schubert calculus (Sections 4 and 5).

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