6
$\begingroup$

Suppose that $X$ is a complex, irreducible, projective variety with at most terminal singularities. Furthermore, assume that $\mathbb{C}^*$ acts on $X$ with exactly $k$ fixed points, where $k>0$.

Question. Is it true that $k > \dim_{\mathbb{C}}(X)$?

$\endgroup$
2
  • 4
    $\begingroup$ I dunno about terminal singularities, but here's a proof in the smooth case. If there's infinitely many fixed points, done; if not, there's finitely many and by Bialynicki-Birula your space decomposes into a union of cells, where the number of these is the number of fixed points and also the Euler characteristic. In particular there is only even cohomology, the dimension of which is therefore the Euler characteristic, but is bigger than the complex dimension by Lefschetz. $\endgroup$ Mar 16, 2021 at 13:43
  • $\begingroup$ @VivekShende Lefschetz is even sort of overkill here - you just need $H^k$ is nonzero for all $k$ from $0$ to the dimension, with $H$ the hyperplane class, but that follows from the case where $k$ is the dimension, which follows from projective varieties having nonzero degree. $\endgroup$
    – Will Sawin
    Mar 23, 2021 at 0:14

1 Answer 1

11
$\begingroup$

This is true in a much more general setting. Let $X$ be any normal projective $T$-variety defines over an algebraically closed field. Then I claim that $\# X^T>\dim X$. We show this in two steps.

(1) According to a theorem of Sumihiro, there is an equivariant embedding of $X$ into a projective space ${\bf P}^n$. Normality is used only in this first step. Often this step is superfluous because $X$ is sitting already in a projective space. Then all we need for the second step is that $X$ is closed.

(2) Now let $T$ act linearly on ${\bf P}^n$ and let $X\subseteq{\bf P}^n$ be closed and $T$-stable. Then there are three substeps:

(a) Assume $X={\bf P}^n$. Then ${\bf P}^n={\bf P}(V)$ where $V$ is an $n+1$-dimensional representation of $T$. This representation is diagonalizable. Hence $V=U_1\oplus\ldots\oplus U_n$ where each $U_i$ is $1$-dimensional with $T$ acting with a character $\chi_i$. Then $T$ has $n+1$ fixed points namely ${\bf P}(U_0),\ldots,{\bf P}(U_n)$.

(b) Assume $X^T=({\bf P}^n)^T$. Then $\# X^T=\#({\bf P}^n)^T>n\ge\dim X$ by (a).

(c) Finally assume $X^T\subsetneq ({\bf P}^n)^T$. Then there is a $T$-fixed point $x$ not lying in $X$. Let $\pi:{\bf P}^n\setminus\{x\}\to{\bf P}^{n-1}$ be the projection with center $x$. Then the restriction of $\pi$ to $X$ is both affine and projective hence finite. Let $\tilde X:=\pi(X)\subseteq{\bf P}^{n-1}$. Then $\#\tilde X^T>\dim\tilde X=\dim X$ by induction on $n$. We conclude with $\# X^T\ge\#\tilde X^T$ since the fibers of $\pi|_X$ are finite and $T$ is connected.

Edit: Part (2) is Prop. IV.13.5 of A. Borel, Linear Algebraic Groups 2nd ed., Springer GTM 126.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.