# When is equivariant cohomology generated by equivariant Euler classes?

Let $X$ be a smooth complex projective variety acted upon by algebraically by a complex torus $T$. Let $F_1,\ldots,F_n$ be the connected components of $X^T$ and assume that the restriction map $$H_T^*(X)\rightarrow H_T^*(X^T)=\bigoplus_{i=1}^nH_T^*(F_i)$$ is injective. (For me, all equivariant cohomology is over $\mathbb{Q}$.) For each $i$, let $N_i$ denote the normal bundle of $F_i$ in $X$, and let $e_T(N_i)\in H_T^*(F_i)$ denote its $T$-equivariant Euler class.

$\textbf{Question}:$ Viewing $H_T^*(X)$ as a subalgebra of $\bigoplus_{i=1}^nH_T^*(F_i)$, when is $H_T^*(X)$ generated as an algebra by the equivariant Euler classes $e_T(F_i)$?

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A simple necessary condition is for at least one component $F_i$ to have (complex) codimension one, since $H_T^2(X)$ is nonzero. This rules out lots of the common examples, e.g., ${\Bbb P}^n$ with the standard $n$-dimensional torus action. (On the other hand, if you invert those Euler classes, both sides become isomorphic.) – Dave Anderson Jan 24 '14 at 19:16
Even requiring all $F_i$ to have codimension $1$ is not sufficient, though: take a positive-genus curve $C$ and consider ${\Bbb P}^1 \times C$ with the standard ${\Bbb C}^*$ action on the first factor. – Dave Anderson Jan 24 '14 at 19:25
I'm of the opinion that Dave's necessary condition above shows that the situation asked for is pretty weird. Did you have any examples to motivate it? – Allen Knutson Jan 25 '14 at 9:08
The only explicit example that occurs to me is the usual action of $\mathbb{C}^*$ on $\mathbb{P}^1$. I ask because I recall having heard a speaker suggest there was a large class of examples in equivariant symplectic geometry in which this occurs. I will try to ask the speaker for clarification, and possibly the examples themselves. – Peter Crooks Jan 25 '14 at 9:52
Are you sure it was the classes of the actual fixed point sets, rather than the closures of their unstable manifolds? – Allen Knutson Jan 26 '14 at 20:03