# Białynicki-Birula decomposition for singular projective variety

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?

• Maybe you want to add the hypothesis that $X$ is irreducible ? (else there are easy counterexamples). Apr 9 '20 at 13:47
• No, take a toric variety and the action of a randomly chosen rank one subtorus. You will get toric singularities in the cells. Apr 9 '20 at 17:01
• @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. Apr 9 '20 at 17:06
• 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. Apr 10 '20 at 5:00
• @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 $H^1$) Apr 10 '20 at 20:16

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.