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Let $X$ be a complex projective and smooth toric variety of complex dimension $n$. It is acted by the real torus $T=(S^1)^n$.

Is it true that the $T$-equivariant cohomology $H^*_T(X,\mathbb{Z})$ of $X$ is isomorphic as a graded algebra to $$H^*(X,\mathbb{Z})\otimes H^*_T(pt,\mathbb{Z})\simeq H^*(X,\mathbb{Z})\otimes \mathbb{Z}[x_1,\dots,x_n],$$ where $x_i$ have degree 2? (to prove such an isomoprhism with coefficients in a field would be fine too.)

Remark. I think I have a proof of such an isomorphism (with $\mathbb{Q}$-coefficients) in the category of graded $H_T^*(pt, \mathbb{Q})$-modules rather than graded algebras (which seems to be well known).

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    $\begingroup$ Isn't this false for $\Bbb P^1$? There we have $H^*_{S^1}(\Bbb P^1) = H^*(\Bbb{CP^\infty} \vee \Bbb{CP}^\infty)$ (as that wedge is the Borel construction on $\Bbb P^1$) which has cohomology $\Bbb Z[x,y]/(xy)$. In particular, it has no nilpotent elements, unlike your tensor product. $\endgroup$
    – mme
    Commented Dec 25, 2019 at 16:19
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    $\begingroup$ As graded $H^\ast_T(pt,\mathbb Q)$-modules it is just the fact, that the action of $T$ on $X$ is Hamiltonian with respect to the invariant Kähler structure. This implies that the action is equivariant formal, which means your isomorphism is an isomorphism of $H^\ast_T(pt,\mathbb Q)$-modules. $\endgroup$
    – Panagiotis Konstantis
    Commented Dec 25, 2019 at 17:14
  • $\begingroup$ @MikeMiller: Why not make your comment into an answer? I'd be interested to know how you see the homotopy equivalence you mentioned. $\endgroup$
    – Mark Grant
    Commented Jan 6, 2020 at 12:01

1 Answer 1

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As per Mark Grant's suggestion, here are some additional details for my comment.

We may write $\Bbb P^1$ with its circle action as the adjunction space $$* \cup_{S^1 \times 0} S^1 \times [0,1] \cup_{S^1 \times 1} *,$$

with the circle acting in the obvious way on $S^1 \times [0,1]$. Passing to the Borel construction, we find that $$\Bbb P^1 \times_{S^1} ES^1 = BS^1 \cup_{ES^1 \times 0} ES^1 \times [0,1] \cup_{ES^1 \times 1} BS^1.$$

The picture is that we are connecting two copies of $BS^1$ by a contractible bit.

To make this precise, collapse the closed contractible subspace $ES^1 \times 1/2$ to a point. As this is contractible, the projection map is a homotopy equivalence; thus $\Bbb P_{hS^1}$ is homotopy equivalent to the wedge of two mapping cones $$\text{Cone}(ES^1 \to BS^1) \vee \text{Cone}(ES^1 \to BS^1).$$ Because $ES^1$ is contractible, each of these cones are homotopy equivalent to $BS^1$ itself.

Thus we see that $$\Bbb{CP}^\infty \vee \Bbb{CP}^\infty \simeq \Bbb P^1_{hS^1};$$ further the projection map $\Bbb P^1_{hS^1} \to \Bbb{CP}^\infty$ sends the two wedge summands identically onto $\Bbb{CP}^\infty$.

Thus $$H^*_{S^1}(\Bbb P^1;\Bbb Z) \cong \Bbb Z[x,y]/(xy),$$ with action of $u \in H^2_{S^1}(pt)$ given by $u \cdot x = x^2, u \cdot y = y^2.$ This is isomorphic as a graded module to $$H^*(\Bbb P^1;\Bbb Z) \otimes H^*_{S^1}(pt;\Bbb Z),$$ but not as an algebra: the equivariant cohomology has no nilpotent elements, whereas the tensor-product algebra does, given by the generator of $H^2(\Bbb P^1) \otimes H^0_{S^1}(pt)$.

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  • $\begingroup$ A previous version of this answer just wrote down a map $$\Bbb{CP}^\infty \vee [0,1] \vee \Bbb{CP}^\infty \hookrightarrow \Bbb P^1_{hS^1}$$ which Mayer-Vietoris and van Kampen show is an iso on $\pi_1$ and homology, so is a homotopy equivalence by Hurewicz and Whitehead. This seemed cleaner, though. $\endgroup$
    – mme
    Commented Jan 6, 2020 at 16:21

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