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Let us have a (possibly singular) irreducible projective variety $X$ over $\mathbb{C}$, with an algebraic $\mathbb{C}^*$-action that has finitely many fixed points $\{x_1,\dotsc,x_n\}$. One can define the attracting sets $$U_k = \{x \in X \mathrel| \lim_{t\rightarrow \infty}t\cdot x =x_k\}$$ that decompose $X$ into a disjoint union. When $X$ is smooth, Białynicki-Birula's theorem is that these are affine bundles over $x_k$, hence affine spaces. That gives us an affine cell decomposition of $X$.

Is this still true when $X$ is singular?

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    $\begingroup$ Maybe you want to add the hypothesis that $X$ is irreducible ? (else there are easy counterexamples). $\endgroup$ Apr 9 '20 at 13:47
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    $\begingroup$ No, take a toric variety and the action of a randomly chosen rank one subtorus. You will get toric singularities in the cells. $\endgroup$ Apr 9 '20 at 17:01
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    $\begingroup$ @Filip92 : You can take 2 $\Bbb P^1$ touching at a point, and take the torus action so that the commun vertex is attracting for both action. $\endgroup$ Apr 9 '20 at 17:06
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    $\begingroup$ There are some results, just not as nice, e.g., decomposition is not into affine spaces; but maybe it is still useful for you? For instance, arxiv.org/abs/1308.2604 is for $G_m$ action on algebraic spaces of finite type; doi.org/10.1016/j.matpur.2019.04.006 generalizes it to actions of reductive groups. $\endgroup$ Apr 10 '20 at 5:00
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    $\begingroup$ @Filip92 : it's still wrong in the non-irreducible case (you can take $3$ $\Bbb P^1$ forming a triangle that will give you a non-trivial $H^1$) $\endgroup$ Apr 10 '20 at 20:16
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No, not at all. Take any Schubert variety, $X_w \subset G/B$. Then for one choice of $\mathbb{C}^* \subset T$ the BB decomposition is a cell decomposition, but for others (e.g., when the Schubert variety is singular and the torus is chosen to be attractive at the "base point" $B/B$) it will not be.

The simplest example is probably given by a singular quadric cone in $\mathbb{P}^3$ with unique singular point $x$. This has many $\mathbb{C}^*$-actions which are attractive near $x$ with finitely many fixed points, and hence the attractive set is a singular affine quadric.

My guess is that it is a miracle when it is, and studying singular examples shows how amazing it is that it holds in the smooth case.

PS: I just did the classic "read the comments after drafting an answer" and I see that Piotr (1) and Nicolas (1 2 3) have already answered the question. Posting anyway in case it helps someone someday ….

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