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Search options questions only not deleted user 919
7 votes
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Ample divisors on $T$-varieties

Also, $X$ admits a $T$-equivariant closed embedding into $\mathbb{P}(V)$, where $V$ is a multiplicity free $T$-module. In particular, $T$ has finitely many fixed points on $X$. …
Geordie Williamson's user avatar
21 votes
3 answers
2k views

Formality of classifying spaces

*) for example because $H(BG, \mathbb{Q})$ is a poynomial algebra, and $\mathcal{A}$ admits a graded commutative model using the de Rham complex -- see Bernstein-Lunts "Equivariant sheaves and functors …
Geordie Williamson's user avatar