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Suppose we have a smooth algebraic variety $X$ with an action of $\mathbb{C}^*$ with finitely many fixed points. Suppose $X$ can be covered by invariant quasi-affine open sets and suppose for each $x\in X$ the limit $\lambda x$ when $\lambda\to 0$ exists. Then Bialynicki-Birula proves that $X$ is a union of locally closed sets $W_i$ isomorphic to affine spaces.

I find often in the literature it is concluded that $X$ has a paving by affine spaces, but "paving" requires more: that $W_i$ are ordered in such a way that $\cup_{j \leq i} W_i$ is closed. But this issue is often ignored. Is there something I'm missing, for instance some general statement that would guarantee existence of such an ordering?

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  • $\begingroup$ To get a paving, it is enough to assume that X admits a $\mathbb{C}^*$-equivariant embedding inside some projective space $\mathbb{P}(V)$ with linear $\mathbb{C}^*$-action. In my experience this is often enough to get what you want. If you want to be more fancy: it is enough for $X$ to admit an ample and $\mathbb{C}^*$-equivariant line bundle. $\endgroup$
    – Geordie Williamson
    Commented Nov 2, 2021 at 19:23
  • $\begingroup$ I agree that for projective varieties it is clear. I was wondering if there's some more general argument. $\endgroup$
    – Anton Mellit
    Commented Nov 2, 2021 at 21:13
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    $\begingroup$ ahh OK. I have vague recollections of looking at the the relatively easy case of non-projective complete toric varieties (e.g. discussed in Fulton) and thinking it was a bit tricky. Other than that I have nothing to add. $\endgroup$
    – Geordie Williamson
    Commented Nov 3, 2021 at 0:29
  • $\begingroup$ I'd be very interested to hear if you do find something though... $\endgroup$
    – Geordie Williamson
    Commented Nov 3, 2021 at 0:29
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    $\begingroup$ There is an article "Filtrations of meromorphic $\mathbb{C}^*$-actions on complex manifolds" by Carrell and Sommese that is essentially about this question, but I am not sure if you would find their criteria useful (corollaries 1-4). $\endgroup$
    – Gjergji Zaimi
    Commented Nov 10, 2021 at 21:40

1 Answer 1

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Consider the partial ordering defined by $$ i \le j \qquad\text{if $\dim(W_i) \le \dim(W_j)$}. $$ Since $\overline{W_i} \setminus W_i$ has dimension less than $W_i$ and is $\mathbb{C}^\times$-invariant, it is the union of some $W_j$ with $j < i$. Therefore, $$ \bigcup_{j \le i} W_j = \bigcup_{j \le i} \overline{W_j} $$ is closed.

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    $\begingroup$ @AntonMellit: where did I write $W_i \subset \overline{W_{i+1}}$? $\endgroup$
    – Sasha
    Commented Nov 2, 2021 at 18:06
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    $\begingroup$ @AntonMellit: I do not mean it is the union of all $W_j$ with $j < i$; I mean it is the union of some $W_j$ with $j < i$. $\endgroup$
    – Sasha
    Commented Nov 3, 2021 at 4:45
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    $\begingroup$ I see. I edited your answer to include "some". So everything would follow from this statement that it is the union of some $W_j$. Can you prove this statement? $\endgroup$
    – Anton Mellit
    Commented Nov 3, 2021 at 8:51
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    $\begingroup$ Your claim is false, see "Białynicki-Birula, Some properties of the decompositions of algebraic varieties determined by actions of a torus. Bull. Acad. Polon. Sci. Sér. Sci. Math. Astronom. Phys. 24 (1976), no. 9, 667–674", example 1. This is a blow up of $P^2$ with action with weights $(1,2,3)$ in the hyperbolic fixed point. The closure of a cell doesn't have to be a union of other cells. Your answer and its apparent popularity kind of confirms my complaint about the issue being ignored. $\endgroup$
    – Anton Mellit
    Commented Nov 3, 2021 at 20:48

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