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Let $X$ be the Grassmannian variety $\operatorname{Gr}(k,n)$ of $k$-planes in $\mathbb{C}^n$. I'm aware of two ways to describe its $T$-equivariant cohomology:

  1. (Quotient ring) $H_T^*(X)=\Lambda[e_1(x|t),\dotsc,e_k(x|t)]/(h_{n-k+1}(x|\dotsc,h_{n}(x|t))$ where $\Lambda=\mathbb{C}[t_1,\dotsc,t_n]$, and $e_i(x|t)$ (resp. $h_i(x|t)$) are the elementary (resp. complete homogeneous) factorial symmetric polynomials.

  2. (GKM) The fixed points $X^T$ are indexed by Young diagrams $\lambda$ inside the $k\times (n-k)$ box, and $H_T^*(X)$ is a certain subring of the ring $\bigoplus_{\lambda \in X^T}\Lambda$. Specifically, a tuple $(f_\lambda)$ in this subring must satisfy certain congruences of the form $t_i-t_j \mid f_{\mu} - f_{\nu}$.

What is the (ring) isomorphism from 1. to 2.?

This is definitely somewhere in the literature, but I haven't been able to find a reference stating the explicit map. A more general reference for e.g. partial flag varieties would also be greatly appreciated.

Edit: I believe the right map is the evaluation given by Hunter Spiker's comment, I'm looking for a reference stating that fact.

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    $\begingroup$ I don’t have an answer, but didn’t Knutson and Tao construct a nice basis for the GKM realization? Since the cohomology is a free $\Lambda$ module you can define a map by just choosing where the basis goes $\endgroup$
    – Exit path
    Jun 18, 2022 at 14:59
  • $\begingroup$ @leibnewtz I need the map to be a ring map (or rather a map of $\Lambda$-algebras), so any $\Lambda$-module map won't cut it. I'll edit the post to specify this, thanks $\endgroup$ Jun 18, 2022 at 15:16
  • $\begingroup$ I see. Maybe the same approach works. Unless I’m mistaken both the Knutson-Tao basis and the evident one in your first description correspond to the equivariant Schubert classes $\endgroup$
    – Exit path
    Jun 18, 2022 at 15:53
  • $\begingroup$ @leibnewtz that makes sense. I was hoping to be able to define the map on the $x_i$ (and then restrict it to this subquotient get the desired map). I know how to do this for $\mathbb{P}^n$ but not in general, and maybe someone points to a reference that deals with this $\endgroup$ Jun 18, 2022 at 16:07
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    $\begingroup$ The fixed points aren't indexed by Tableux, but instead by subsets A of {1,...,n} of size k - they correspond to the kxn matrices M such that M restricted to the columns indexed by A is the identity matrix and M restricted to the columns outside of A is the zero matrix. Then the specialization to the fixed point corresponding to A is specializing $x_1,\ldots,x_k$ to $\{t_i\}_{i\in A}$ in no particular order. $\endgroup$ Jun 18, 2022 at 17:10

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$\def\Fl{\mathcal{F}\ell}$I'd been hoping someone else would answer this, since I don't have a good reference to cite and I tend to make minor errors in things like this, but some answer is better than none. As in your question, $\Lambda := H_T^{\ast}(\text{point}) \cong \mathbb{Z}[t_1, t_2, \ldots, t_n]$.

Let's do the full flag manifold first. The corresponding descriptions are

(1) $H_T^{\ast}(\Fl) \cong \Lambda[x_1, x_2, \ldots, x_n]/\langle e_j(x) - e_j(t) \rangle$. In other words, we take the symmetric functions in the $x$-variables and set them equal to the symmetric functions in the $t$-variables.

(2) The fixed points $\Fl^T$ are indexed by the symmetric group $S_n$. The ring $H_T^{\ast}(\Fl)$ is a subring of $\bigoplus_{w \in S_n} \Lambda$. Namely, if $f$ is a function from $S_n$ to $\Lambda$ then we must have $f(u) \equiv f((ij) \cdot u) \bmod t_i - t_j$. The subring $H_T^{\ast}(\text{point})$ is identified with the functions which take the same value at every permutation.

Given $g(x_1, x_2, \ldots, x_n, t_1, t_2, \ldots, t_n)$ in the first ring, we define an element of the second ring by $$f(w) = g(t_{w(1)}, t_{w(2)}, \ldots, t_{w(n)}, t_1, t_2, \ldots, t_n).$$ So the generator $x_j$ turns into the function $w \mapsto t_{w(j)}$ and $t_j$ is the function which is $t_j$ on every permutation.

The projection $\Fl_n \to G(k,n)$ induces an injection $H_T^{\ast}(G(k,n)) \to H_T^{\ast}(\Fl_n)$. We can describe $H_T^{\ast}(G(k,n))$ as a subring of both (1) and (2).

In (1), we take the ring of invariants for $S_k \times S_{n-k}$, permuting the $x$-variables and leaving the $t$-variables fixed. This should match your description (1).

In (2), we take the ring of functions which obey $f(u) = f(up)$ for $p \in S_k \times S_{n-k}$. In other words, we can think of $f$ as a function on cosets $S_n/(S_k \times S_{n-k})$, which is in easy bijection with $k$-element subsets of $\{ 1,2,\ldots, n \}$, these can be identified with Young diagrams in the $k \times (n-k)$ box in a standard way, giving your (2).

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  • $\begingroup$ This is perfect, thank you!! $\endgroup$ Jun 20, 2022 at 17:40

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