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Let $X$ be a smooth complex projective variety and $V$ be a holomorphic bundle on $X$. Suppose we have an algebraic $\mathbb C^*$-action on $X$. Is it true that the bundle $V$ can always be deformed to a holomorphic bundle $V'$ so that the $\mathbb C^*$-action can be lifted to an action on $V'$?

UPD. It is not quite clear to me now if there is no some type of topological obstruction for such a lift, i.e., that the $\mathbb C^*$-action can not be lifted even to an action on $V$ as a topological complex vector bundle. I would of course be interested to see such an example if it exists, but then would be definitely interested to know what happen in the situation when the action can be lifted topologically.

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  • $\begingroup$ Certainly the answer is no if the group is nonabelian: consider $PSL_2$ acting on $\mathbb P^1$, an action that doesn't extend to $\mathcal O(1)$, since $PSL_2$ couldn't act on the $2$-d space of sections. I'm a little confused about extending that argument to the maximal torus, though. $\endgroup$
    – Allen Knutson
    Commented May 11, 2017 at 1:26
  • $\begingroup$ Allen thank you for this comment, I am a bit worried indeed of this type of counter-examples. Though I see them as of "topological nature". If by any chance there is such a topological obstruction for lifting $\mathbb C^*$ actions, I would also ask if one can lift instead a $\mathbb C^*$-action induced by a finite cover $\mu_n: \mathbb C^*\to \mathbb C^*$ for some $n$. $\endgroup$
    – aglearner
    Commented May 11, 2017 at 3:53
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    $\begingroup$ If I were trying to find a counterexample, I would begin with a threefold $X$ and an elliptic curve $E$ embedded in $X$. If $X$ is simply connected, then by Serre's correspondence, $E$ is the zero scheme of a regular section of a rank $2$ locally free sheaf $V$. If $h^1(X,V)$ is zero and if $E$ is rigid, then $V$ is rigid. If $h^0(X,V)$ equals $1$, then $V$ has no compatible $\mathbb{C}^*$-action. $\endgroup$
    – Jason Starr
    Commented May 12, 2017 at 15:37
  • $\begingroup$ Jason, thank you for the idea. In order to model at least some bits of what you suggest, could one take $\mathbb CP^3$ with a $\mathbb C^*$-action that fixes a $\mathbb CP^2$ and a point and blow $\mathbb CP^3$ along a smooth cubic $E\subset \mathbb CP^2$? Then there is a rigid $E$ in the blown up variety. But I don't know how to calculate the cohomology groups. Also I have a question on rigidity. Is the bundle $O(-1)\oplus O(-1)$ on $\mathbb CP^1$ considered to be rigid? (I know it can be deformed to $O\oplus O(-2)$) $\endgroup$
    – aglearner
    Commented May 12, 2017 at 18:42
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    $\begingroup$ This is from long ago, but I just remembered now. I did try to find such a counterexample at the time, but with no success. Now I believe (vaguely) that there are no such counterexamples. $\endgroup$
    – Jason Starr
    Commented Aug 3, 2017 at 12:34

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