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?
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> 
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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](https://mathoverflow.net/questions/20671/what-is-the-equivariant-cohomology-of-a-group-acting-on-itself-by-conjugation) 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.



Is a Lie group equivariantly formal under conjugation by a maximal torus?
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 $\mathbb{Q}$).


