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We work over $\mathbb{C}$. Let $X$ be a normal projective irreducible variety, and let $\mathbb{C}^*$ act nontrivially on $X$. The fixed point locus of $X$, namely $X^{\mathbb{C}^*}$, can be decomposed into a disjoint union of connected fixed point components, let us call them $F_1,\ldots,F_s$. Moreover, for any $k=1,\ldots,s$, we can define the plus cell as

$$X^+(F_k)=\{p\in X\mid \lim_{t\to 0} t\cdot p \in F_k\}$$

and similarly also the minus cell, by considering the limit at $\infty$.

If $X$ is smooth, then the celebrated theorem of Bialynicki-Birula tell us that \begin{equation} X=\bigsqcup_{i=1\ldots k} X^+(F_k)\tag{*}\label{star} \end{equation} (and similarly for the minus cell), and moreover this is an affince cell decomposition.

Question: If $X$ is only normal, do we still have a decomposition as in \eqref{star} (but in this case the cell will not be afine, as explained in Białynicki-Birula decomposition for singular projective variety)?

I've tried to search on the literature, but what I've found is usually much more deep of what I'm asking and I have a hard time trying to translate that result in my specific case. Any help, or reference, would be much appreciated. Thanks in advance!

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  • $\begingroup$ In your question, did you intend to write, "If $X$ is only normal, ..."? $\endgroup$
    – Jason Starr
    Commented Jun 7, 2022 at 10:11
  • $\begingroup$ Dear @JasonStarr, sure, thanks for spotting the typo! $\endgroup$
    – YetAnotherPhDStudent
    Commented Jun 7, 2022 at 10:29
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    $\begingroup$ Have you looked at Hausel-Hitchin arXiv:2101.08583, Section 2.1? $\endgroup$
    – Balazs
    Commented Jun 7, 2022 at 10:35
  • $\begingroup$ Such a generalization is I think due to Drinfeld and employed in several papers by Jelisiejew. $\endgroup$
    – Piotr Achinger
    Commented Jun 7, 2022 at 10:46
  • $\begingroup$ The references: arXiv:1308.2604 (Drinfeld), arXiv:1710.06124 (Jelisiejew, see Section 3), arXiv:1805.11558 (generalization to reductive groups, Jelisiejew-Sienkiewicz) $\endgroup$
    – Piotr Achinger
    Commented Jun 7, 2022 at 10:50

1 Answer 1

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Yes, \eqref{star} is always a disjoint union (that's obvious). Moreover, each set $X^+(F_k)$ is locally closed and the map $x\mapsto\lim_{t\to0}t\cdot x$ induces an affine morphism $\pi_k:X^+(F_k)\to F_k$. In general, the morphism $\pi_k$ is not a fiber bundle anymore.

These assertions can be easily reduced to the smooth case by using a theorem of Sumihiro (Sumihiro, Hideyasu: Equivariant completion. J. Math. Kyoto Univ. 14 (1974), 1–28) according to which $X$ can be embedded equivariantly into a projective space.

The assertions are more generally valid for any normal complete variety. The paper "Konarski, Jerzy: Decompositions of normal algebraic varieties determined by an action of a one-dimensional torus. Bull. Acad. Polon. Sci. Sér. Sci. Math. Astronom. Phys. 26 (1978), 295–300" seems to be a good reference.

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    $\begingroup$ Dear Prof. Knop, thank you very much for the answer and the references. Before accepting it though, I have few questions. You say at the beginning "it's obvious": are you referring to the fact that the union is disjoint (and I agree with you on this), or are you referring to the fact that $X$ coincides with the union? I was precisely asking about the second point, but with your suggestion I think I know how to prove the second question (see the comment below): $\endgroup$
    – YetAnotherPhDStudent
    Commented Jun 8, 2022 at 6:27
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    $\begingroup$ Using Sumihiro's theorem, we embed $X$ $\mathbb{C}^*$-equivariantly into a $\mathbb{P}^n$. Now we use the BB-decomposition of the projective space, obtaining $$\mathbb{P}^n=\bigsqcup_{i=1}^k (\mathbb{P}^n)^+(F_i),$$ where $F_i$ are the connected fixed point components of $\mathbb{P}^n$ under the $\mathbb{C}^*$-action. Then since $X$ is $\mathbb{C}^*$-invariant we consider the intersection of the BB-decomposition with $X$, obtaining $$\mathbb{P}^n\cap X=X=\bigsqcup_{i=1}^k X^+(F_i\cap X),$$ where maybe some intersection will be empty, but we still obtain a $+$-decomposition of $X$. $\endgroup$
    – YetAnotherPhDStudent
    Commented Jun 8, 2022 at 6:28
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    $\begingroup$ @Mago: That's correct. That $(*)$ is a disjoint union follows from the fact that for every $x$ the limit $\lim_{t\to0} t\cdot x$ exists (since $X$ is complete) and is unique (since $X$ is separated) and that the limit point is a $\mathbb C^*$-fixed point. But all this follows also from the Sumihiro argument once you accept the assertions for the projective space. $\endgroup$
    – Friedrich Knop
    Commented Jun 8, 2022 at 14:44
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    $\begingroup$ @Mako: One more thing: Observe that $F_i\cap X$ might have several connected components. Then $X^+(F_i\cap X)$ decoposes accordingly. $\endgroup$
    – Friedrich Knop
    Commented Jun 8, 2022 at 14:54
  • $\begingroup$ Dear Prof. Knop, I see. Thank you very much for the clear explanation! $\endgroup$
    – YetAnotherPhDStudent
    Commented Jun 9, 2022 at 6:58

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