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Let $X\subset \mathbb{P}^n$ be a smooth complex projective variety, and consider a non-trivial action of $\mathbb{C}^*$ on $X$. For any connected fixed component $Y$ of the fixed point locus, we may define the plus and minus Białynicki-Birula cells $X^{\pm}(Y)$ as $$X^{\pm}(Y)=\{x\in X\mid \lim_{t\to 0} t^{\pm 1} t\cdot x \in Y \}.$$ It is a famous result by Białynicki-Birula that $X^{\pm}(Y)$ are locally closed, and they provide two decompositions of $X$.

My questions is: can it happen that $X^+(Y)=X^+(M_1)\sqcup X^+(M_2)$, with $M_i$ two subvarieties of $Y$?

I've always thought of those cells as "tubes" joining two connected fixed components $Y$ and $Y_1$, but maybe they could join also another component $Y_2$. Intuitively, can the Białynicki-Birula cells appear as "disjoint pants"?

The disjoint union is forced since we are considering orbits, but I cannot come up with a proof or a counterexample.

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    $\begingroup$ A quick attempt at an answer (standard health warnings apply): I guess the fixed point loci are smooth (this is explained e.g. in Chriss-Ginzburg) and then on knows (see BB's first paper) that the attracting spaces are affine bundles over Y. Because you take Y to be a connected component, X^+(Y) is also connected. If you didn't assume X is smooth things would be much more complicated... $\endgroup$
    – Geordie Williamson
    Commented Jun 3 at 4:23
  • $\begingroup$ You're right! I was indeed aware of these results, but I didn't put them together. Thank you! $\endgroup$
    – YetAnotherPhDStudent
    Commented Jun 3 at 4:49

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