0
$\begingroup$

Let $M$ be a toric manifold. I'm not sure what conditions on $M$ are required, but one can assume, if needed, that it is compact, smooth, etc. We consider $M$ as a quotient given by the momentum map $P$ of the action of a torus $\mathbb{K} \subset (S^1)^n$ of dimension $k$: $$M = P^{-1}(p) \ / \ \mathbb{K}, \ \ p \text{ regular value of } P : \mathbb{C}^n \to \text{Lie}(\mathbb{K})^*.$$

I am interested in the cohomology $$H^*(M,\mathbb{C})$$ of $M$ with complex coefficients. More precisely, one can show that it is naturally ismorphic to the following quotient $$H^*(M,\mathbb{C}) \simeq \mathbb{C}[u_1,...,u_n] \ / \ (I + J),$$ where:

  • $I$ is the ideal of polynomials which vanish on $\mathbb{C}^k = \text{Lie}(\mathbb{K}) \otimes \mathbb{C} \subset \mathbb{C}^n = \mathbb{R}^n \otimes \mathbb{C}$;
  • J is the ideal generated by monomials $u_1^{m_1}...u_n^{m_n}$ such that $m = (m_1,...,m_n)$, considered as linear functions on $\mathbb{R}^{n*}$, assume strictly positive values on the polyhedron image of the momentum map of the action of $(S^1)^n \ / \ \mathbb{K}$ on $M$.

My question is: is this algebra a finite dimensional vector space over $\mathbb{C}$ ? If yes, how to prove this ?

Improve this question
$\endgroup$
2
  • 3
    $\begingroup$ The singular cohomology of any smooth projective variety (not necessarily toric) is finite-dimensional. The explicit formula for the cohomology ring is not needed here. $\endgroup$
    – Sasha
    Commented Sep 17, 2018 at 19:18
  • $\begingroup$ Thanks for your comment @Sasha. However, I'm looking for a proof using properties of the ideals $I$ and $J$. Let me explain why: I'll use the notation $u = (u_1,...,u_n)$. My question comes from the fact that, I'm trying to understand why a certain other quotient is finite dimensional, mainly $$\mathbb{C}[u] \ / \ (I + J_r \cap \mathbb{C}[u]),$$ where $J_r$ is the submodule in $\mathbb{C}[u,u^{-1}]$ generated by monomials $u^m$ for which $m \in \mathbb{Z}^k$ and $p(m) \geq r$. $\endgroup$
    – BrianT
    Commented Sep 18, 2018 at 14:14

0

Reset to default

You must log in to answer this question.