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In nearly every reference on the classical springer correspondence (for example Chriss/Ginzburg's book on Complex Geometry) it is stated that the action of the Weyl Group on the homology of the springer fiber is not induced by an action of the Weyl group on the fiber itself. But I couldn't find any reference or general statement that would make this precise (and I'm just learning about geometric representation theory so I don't have any good intuition/working knowledge).

For concreteness (and I think this should be a simpler case), consider the springer fiber over zero -- then $\mathcal{B}_0 = \mathcal{B} = G/B$ is the flag varitey (where $B$ is a Borel subalgebra). Now there is an action of the Weyl group $W$ on the quotient $G/T$ (where $T\subseteq B$ is a maximal torus, which allows to write $W = N_G(T)/T$). Now in Yun's notes, (1.5.4), he writes that the map $G/T\to G/B$ is an "affine space bundle", that this implies that the their homologies are isomorphic and that under this isomorphism, the action of $W$ on $H(G/T)$ gives the springer action of $W$ on $H(\mathcal{B})$ (I couldn't find a reference for that though). But I don't see why there should be no way to use this projection to define an action of $W$ on $\mathcal{B}$. Also, in "Schubert cells and cohomology of the spaces $G/P$" (1973), Bernstein Gelfand and Gelfand construct in chapter 5 a correspondence that gives rise to the springer action. I was hoping that this very explicit construction in the paper would shed some more light on what's going on, but I haven't been able to do that myself.

Any help, and also references, are really appreciated.

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    $\begingroup$ It appears to me that the most simple example explaining the situation is the subregular Springer fibre (i.e. Springer fibre at a subregular unipotent element) of SL_3: In this case the Springer fibre is two lines intersected at a single point and the two lines form the two irreducible component, so, if you let the Weyl group S_3 act (continuously) on the Springer fibre, then it either preserves or exchanges the two components, and as a result, its induced action on the cohomology is either a multiple of the trivial rep or a sum of the trivial rep and the sign rep; however...... $\endgroup$
    – user148212
    Commented Jun 6, 2021 at 8:37
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    $\begingroup$ ......, according to the Springer correspondence, the rep afforded by the cohom should be the 2-dim irrep of S_3. $\endgroup$
    – user148212
    Commented Jun 6, 2021 at 8:37

1 Answer 1

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Here is a general fact: Given a group $G$ and a subgroup $H$, there is an isomorphism between

  • The space $\text{End}(G/H)$, consisting of $G$-equivariant maps $G/H\to G/H$.
  • The group $N/H$, where $N$ denoted the normalizer of $H$ in $G$.

This isomorphism is induced by the surjective homomorphism $N\to \text{End}(G/H)$, taking $n\in N$ to $gH \mapsto gnH$, which kernel is $H<N$.


Consider now a reductive group $G$, a Borel subgroup $B<G$ and a maximal torus $T<B$ and observe that $\text{End}(G/T)\simeq W$, while $\text{End}(G/B)$ is trivial, as $B$ is its own normalizer.

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  • $\begingroup$ The above is written with complex groups in mind, as considered in the OP. Over other fields, one considers a maximal split torus $T$ and then $W\simeq \text{End}(G/Z(T))$, where $Z(T)$ denotes the centralizers of $T$. For this, note that $T$ and $Z(T)$ have the same normalizer in $G$. $\endgroup$
    – Uri Bader
    Commented Jun 6, 2021 at 15:35
  • $\begingroup$ Thank you for your answer! But this doesn't really explain what happens in user148212's comment under the question, right? $\endgroup$
    – Aaron Wild
    Commented Jun 8, 2021 at 9:25
  • $\begingroup$ No, I was referring to the fiber over the identity element. I thought this is what you asked for when you wrote "For concreteness". $\endgroup$
    – Uri Bader
    Commented Jun 8, 2021 at 15:14

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