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I am looking for a reference for the following result:

Let $X$ be a projective variety and $\mathcal{L}$ be an ample line bundle over $X$. Suppose that there is a torus $T=(\mathbb{G}_m)^n$ which acts on $X$ and extends to a $T$-action on $\mathcal{L}$. Then $\dim(X)>0$ implies $|X^T|>1$.

I need this result for an article I am writing, but the proof doesn't fit well with the content of the paper.

Even a possible article or book where you suspect I can find this result will be much appreciated.

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  • $\begingroup$ See if this helps: chromotopy.org/torus-actions-maximal-tori-1 $\endgroup$ – T. Amdeberhan Nov 20 '16 at 20:30
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    $\begingroup$ This can fail for a non-normal variety. For a normal variety, a general orbit closure will also be normal, hence a toric variety. For a toric variety, every maximal cone comes with a torus-invariant point. If the toric variety is projective, it contains at least two maximal cones (and the projective line has precisely two maximal cones). $\endgroup$ – Jason Starr Nov 20 '16 at 20:42
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    $\begingroup$ @Jason: it seems to me that if the action is linearized, normality is not required. If a torus acts linearly on the projective space, then the closure of any non-zero-dimensional orbit contains at least two fixed points (e.g., if T=Gm, they are lim(tx) when t->0 and when t->infty). This implies the statement. $\endgroup$ – t3suji Nov 21 '16 at 21:07
  • $\begingroup$ @t3suji. You are correct. In the non-normal examples I had in mind, the torus action does not actually linearize to an ample invertible sheaf. $\endgroup$ – Jason Starr Nov 21 '16 at 21:10
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If t3suji wants to post an answer, then I will delete this answer. I am just posting this until further notice. There is a stratification of $X$ into locally closed subsets according to the dimension of the stabilizer group. To see this, apply Chevalley's result on upper semicontinuity of fiber dimension to the "inertia", i.e., the fiber product of the diagonal $\Delta:X\to X\times X$ and the morphism $\Psi:T\times X \to X\times X$, $\Psi(t,x) = (t\bullet x,x)$. So there is a dense open subset $X^o\subset X$ over which the stabilizer group is constant. Because subgroups of $T$ are discrete, there is a subgroup $T^o$ that acts trivially on $X^o$. Because $X^o$ is dense, $T^o$ acts trivially on all of $X$. Thus, without loss of generality, replace $T$ by $T/T^o$, and assume that $T$ acts freely on the dense open $X^o$.

Identify $T$ with the standard torus embedded in $\mathbb{P}^n$. The $T$-action on $\mathbb{P}^n$ has $n+1$ fixed points. For every $x\in X^o$, the $T$-action defines a $T$-equivariant rational transformation $f:\mathbb{P}^n \dashrightarrow X$ that is regular on $T$ and that sends $[1,\dots,1]$ to $x$. Denote by $\nu:Y\to \mathbb{P}^n$ the normalization of the closure of the graph of $f$ in $\mathbb{P}^n\times X$. Thus, there is an induced $T$-action on $Y$, and $\nu$ is $T$-equivariant, projective, and birational. Also denote by $g:Y\to X$ the induced morphism, so $f\circ \nu$ equals $g$ as $T$-equivariant rational transformations.

Every irreducible component of every fiber of $\nu$ over a fixed point is $T$-invariant and projective, thus it contains a fixed point. So the $T$-fixed points of $Y$ surject onto the $T$-fixed points of $\mathbb{P}^n$. The claim is that every pair of $T$-fixed points of $Y$ that are identified under $g$ are also identified under $\nu$. Assuming the claim, then $g(Y)$ contains at least $n+1$ $T$-fixed points.

The claim is proved by contradiction. The key is that, for every pair $y$ and $z$ of $T$-fixed points of $Y$ mapping to distinct $T$-fixed points of $\mathbb{P}^n$, there is a cocharacter $\lambda:\mathbb{G}_m\to T$ such that the orbit $\lambda(\mathbb{G}_m)\cdot [1,\dots,1]$ closes up to a $\mathbb{P}^1$ whose strict transform in $Y$ contains $y$ and $z$. So, if $y$ and $z$ are identified, then $g$ restricts on this rational curve to a $\lambda(\mathbb{G}_m)$-equivariant morphism that identifies precisely two points. Moreover, there is a $\lambda(\mathbb{G}_m)$-linearized ample invertible sheaf on this curve. But for a $\mathbb{P}^1$ with $0$ and $\infty$ identified, i.e., for a "nodal plane cubic", for every ample invertible sheaf, the group of automorphisms of the nodal curve together with the invertible sheaf is finite (essentially for the same reason as in the case of Abelian varieties). This contradicts that the automorphism group contains the infinite group $\lambda(\mathbb{G}_m)$. This contradiction proves that $y$ and $z$ are mapped to distinct points under $g$, even without the hypothesis that $X$ is normal.

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  • $\begingroup$ Thank you for your post. I was going to write up the proof that I had to show that normality is not needed. I think you should keep the actual proof posted for the benefit of future viewers. $\endgroup$ – Sergio Da Silva Nov 22 '16 at 22:21
  • $\begingroup$ I still need a reference for this result. The proof would seem misplaced in the paper that I am working on, so I was hoping to reference it out for the benefit of the reader. $\endgroup$ – Sergio Da Silva Nov 22 '16 at 22:23
  • $\begingroup$ I couldn't find a reference, so in the end I will include a condensed proof. Thank you Jason for your answer. Since it is the best answer provided, I will accept your answer as the most useful. $\endgroup$ – Sergio Da Silva Dec 1 '16 at 23:52

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