# Is a Lie group equivariantly formal under conjugation by a maximal torus?

Given an action of a group $G$ on a topological space $X$, the associated homotopy quotient is $$X_G := (EG \times X)/G,$$ where $EG$ is the total space of a universal principal $G$-bundle $EG \to BG$ and the quotient is by the (free) diagonal action of $G$ on $EG \times X$. The $G$-equivariant cohomology of $X$ is defined to be $$H^\ast_G(X) := H^\ast(X_G),$$ the cohomology of $X_G$ (say singular with $\mathbb{Q}$ coefficients).

The homotopy quotient $X_G$ is the total space of a $X$-bundle over $BG$, given by $[(e,x)] \mapsto eG$, so the fiber inclusion $X \to X_G$ induces a restriction homomorphism $$H_G^\ast(X) \to H^\ast(X).$$ One says the $G$-space $X$ is equivariantly formal if this homomorphism is surjective.

A compact Lie group $G$ has a maximal torus $T$, which acts on $G$ by conjugation. I would like to find a class in $H^{\dim G}_T(G)$ that restricts to a generator for the top-dimensional cohomology $H^{\dim G}(G)$. Is that always possible? More generally, I am curious:

Is $G$ an equivariantly formal $T$-space?

So far, I only know this for the abelian case $G = T$, where $G_T = T_T = (ET \times T)/T = BT \times T$ and a Künneth formula applies.

Edit:

I see from this question that $H^\ast_G(G) = H^\ast(G) \otimes H^\ast(BG)$, but don't understand the proof there well enough to see if or how it could carry over. Anyway, thanks.

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Let's see, $H^*_{T_{Ad}}(G) = H^*_{T_{Ad}\times G}(G\times G) = H^*_{G_{Ad}}(G\times^T G) \from H^*_{G_{Ad}}(G)$ where the last is pullback along the multiplication map $G\times^T G \to G$. So I suspect you could use the representative from that last cohomology group. –  Allen Knutson May 21 '13 at 1:51
@jdc, as you saw, $G$ is equivariantly formal for the adjoint action, so $H_G^*(G) \to H^*(G)$ is surjective. But this factors through the pullback $H_G^*(G) \to H_T^*(G)$, so the map $H_T^*(G) \to H^*(G)$ is surjective, as well. (@Allen, is the map you describe different from this change-of-groups homomorphism?) –  Dave Anderson May 21 '13 at 19:02
No, I think it's the same -- in both cases the only interesting map is a certain $G/T$-bundle. I like your description a lot more, though! –  Allen Knutson May 22 '13 at 11:48

The answer is yes, if $G$ is connected.

This follows from the following theorem that I believe is due to Borel and can be found in his Seminar on Transformation Groups.

Theorem: Let $M$ be a compact $T$-manifold (this can be weakened) with fixed point set $M^T$. The sum of rational Betti numbers of $M$ is greater than or equal to the sum of rational Betti numbers of $M^T$, with equality if and only if $M$ is equivariantly formal.

In the case of a compact connected Lie group $G$ acted on by the maximal torus $T$, we have $G^T = T$. The cohomology rings $H^*(G)$ and $H^*(T)$ are both exterior algebras on rank$(G)$ generators so they have the same dimension. It follows from Borel's theorem that the action is equivariantly formal (over $\mathbb{Q}$).

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Thank you so much. My next step is actually trying to find the equivariant extensions. Do you know anything about the accepted answer to the related question linked to above? I tried to ask about it, but it seems to have flown under the radar. –  jdc Jul 7 '13 at 2:52
Do you perchance know where in the Seminar this result can be found? –  jdc Aug 1 '14 at 7:09

I found an arguably simpler answer (statement 4.3 in this paper):

then again, it's arguable it just moves the difficulty elsewhere (to other results of Borel and Hopf).

The trick is that the fiber restriction homomorphism is surjective if and only if the Serre spectral sequence $M \to M_T \to BT$ collapses at $E_2$, and then and only then does one have $\dim H^*(M) = \mathrm{rank}_{H^*(BT)} (H^*_T(M))$.

Since $G^T = T$, one gets $H^*_T(T) = H^*(BT) \otimes H^*(T)$, and by the Borel localization theorem applied to conjugation of $G$ by $T$, the rank over $H^*(BT)$ of $H^*_T(G)$ is also $\dim H^*(T)$; but $\dim H^*(T) = \dim H^*(G)$.

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