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In Chriss and Ginzburg's "Representation Theory and Complex Geometry", they describe a geometric construction of representations of the affine Hecke algebra, using the Borel-Moore homology of generalized Springer fibers.

Briefly, let $G$ be a sufficiently nice algebraic group, and choose a semisimple $s \in G$ and some nilpotent $x \in \mathfrak{g} = $ Lie$(G)$, such that $x$ is an eigenvector for the adjoint action of $s$. Consider the variety $B_{x}^{s} $ of Borel subgroups of $G$ containing $s$ and exp$(x)$. Then $C_{G}(s,x)$ acts on this variety by conjugation, and thus acts on its Borel-Moore homology.

Now, the claim is (p.415 of C & G) that this action descends to the component group because the identity component of $C_{G}(s,x)$ acts trivially. Normally, I would expect this to be true since a path in the group from an element $g$ to the identity gives a homotopy between the action of $g$ and the identity map. But Borel-Moore homology is not a homotopy invariant, so I don't understand why an element in the identity component should necessarily act trivially on the homology.

They make a similar claim earlier in the book (p. 170) without any justification there either. If anyone can point out what I'm missing, I'd really appreciate it.

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Borel-Moore homology isn't a homotopy invariant, but it is an isotopy invariant; if two homeomorphisms are homotopic through a homotopy which is a homeomorphism over any point in [0,1], they induce the same map on Borel-Moore homology.

One can see this instantly by writing Borel-Moore homology as homology of the 1-point compactification relative to the new point. There's no reason a homotopy between arbitrary maps will extend to this space, but one which stays in homeomorphisms will by the functoriality of 1-point compactification.

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The statement follows from the following general claim: let $G$ be a connected group acting on a variety $X$ and let $F$ be an $G$-equivariant constructible sheaf on $X$ (or a complex of sheaves). Then $G$ acts trivially on $H^*(X,F)$. Borel-Moore homology is (by definition) the cohomology of the dualizing sheaf, so the statement about Borel-Moore homology is a special case of the above statement.

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  • $\begingroup$ Which raises the question of why that statement is true.... $\endgroup$
    – Ben Webster ♦
    Commented Sep 20, 2011 at 23:00
  • $\begingroup$ This is obvious - it is enough to prove this when $X$ consists of one point $\endgroup$
    – Alexander Braverman
    Commented Sep 21, 2011 at 0:49

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