# 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 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 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 at 11:48
Thank you both. Not wanting to trouble anyone unduly, I've hesitated to ask more, but I think I really am stuck. I don't understand the equalities in Allen Knutson's comment, but I'm happy if the composition is the same as Dave Anderson's map. In David Ben-Zvi's answer to the question about $H_G^*(G)$, I understand the transgression for the bundle $G \to EG \to BG$ takes the multiplicative generators of $H^*(G)$ to those of $H^*(BG)$, but don't understand how it relates to the map $\pi_* \text{ev}^*$ from $H^*(BG)$ to $H^{*-1}(LBG)$. Could you please recommend a reference for this? –  jdc Jun 24 at 14:19

The answer is yes, if $G$ is connected.
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 $\mathbb{Q}$).