For any interested latecomers who somehow discover this question in the future, I've found a very low-tech answer, bootstrapping from the low-tech answer to Is a Lie group equivariantly formal under conjugation by a maximal torus?.
Anyway, once you believe that the conjugation action upon a compact Lie group $G$ of its maximal torus $T$ is equivariantly formal, it follows that the action of $G$ by conjugation is as well. This argument will actually work for any reasonably good space $M$ on which $G$ acts and the restricted $T$-action is equivariantly formal, because one has $$H_G(M) = H_T(M)^W = (H(M) otimes H(BT))^W = H(M) otimes H(BT)^W = H(M) otimes H(BG)$$ as $H(BT)$-modules, where $W = N_G(T)/T$ is the Weyl group of $G$.
If, say, you don't buy that the action of $W$ on $H(M)$ is trivial, a longer proof goes like this.
$require{AMScd}$
The homotopy quotient $M_G$ is a further quotient of $M_T$, and the projection $EG times M to EG$ then induces a commutative diagram
begin{CD}
M @= M\
@VVV @VVV\
M_T @>>> M_G\
@VVV @VVV\
BT @>>> BG
end{CD}
where the upper vertical maps are fiber inclusions.
The projection $BT = EG/T to EG/G = BG$ induces an inclusion $H(BG) cong H(BT)^W hookrightarrow H(BT)$ in cohomology, and there are induced maps both in cohomology and on the Serre spectral sequences for the equivariant cohomologies, starting with this $E_2$ page:
begin{CD}
H(M) @= H(M)\
@AAA @AAA\
H(M) otimes H(BT) @<<< H(M) otimes H(BG)\
@AAA @AAA\
H(BT) @<<< H(BG)
end{CD}
Because the top and bottom horizontal maps are injective, so is the middle one, so the differentials for the spectral sequence converging to $H_G(M)$ are restrictions of those for $H_T(M)$. But the differentials for $H_T(M)$ are all zero, by equivariant formality, so the spectral sequence for $H_G(M)$ collapses as well.
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