If $X$ is a smooth $G$-variety with trivial canonical bundle, then does $X^G$ also have trivial canonical bundle?

Let $$G$$ be a reductive group and $$X$$ a smooth $$G$$-variety. Then the fixed point subvariety $$X^G$$ is also smooth (this is theorem 13.1 of Milne's book on algebraic groups). Suppose in addition that the canonical bundle $$\omega_X$$ is trivial. Then is $$\omega_{X^G}$$ also trivial?

One idea I had was to use the adjunction formula, which gives that $$\omega_{X^G} = \det(I/I^2)^\vee$$, where $$I$$ is the ideal sheaf of $$X^G\subset X$$. When $$X = \text{Spec}(A)$$ is affine, we have $$I=(g\cdot f - f)_{g\in G, f\in A}$$, but in the general case I'm having trouble giving a sufficiently effective description of $$I/I^2$$ to calculate its determinant.

No. Let $$C \subset \mathbb{P}^2$$ be a smooth sextic curve, let $$X$$ be the double covering of $$\mathbb{P}^2$$ ramified over $$C$$, and let $$G = \mathbb{Z}/2$$ acting on $$X$$ be the involution of the double covering. Then $$X$$ is a K3 surface, hence its canonical bundle is trivial. But $$X^G = C$$ is a curve of genus $$10$$, its canonical class is not trivial.

• Usually (but probably not always in the literature) "reductive group" is by definition connected. Thus, when one uses this more restrictive notion, the (beautiful!) counterexamples given in the two answers of Sasha and Francesco Polizzi do not work anymore. So for (connected) reductive groups the question still remains open. Nov 12, 2021 at 8:42
• @AlexIvanov: The case of connected groups is tricky. The point is that smooth projective varieties with trivial canonical class rarely admit faithful actions by connected groups (most of examples of such actions come from abelian varieties, where the action is free). Nov 12, 2021 at 13:14
• Oh, I see. Thank you. This means that if one hopes to construct a counterexample, it should better be non-projective (which still might be possible). Nov 12, 2021 at 13:56
• @AlexIvanov If you pick your favorite smooth variety, then I believe the total space of the cotangent bundle will have trivial canonical bundle. You can let $\mathbb{G}_m$ act by scaling on the fibers, and then the fixed points are the original variety (zero section).
– afh
Nov 13, 2021 at 2:07
• In other words, there are/should be plenty of counterexamples. Thank you! Nov 13, 2021 at 5:23

We can generalize Sasha's counterexample in all dimensions as follows.

Let $$f \colon X \to \mathbb{P}^n$$ be a double cover branched over a smooth hypersurface $$Y=Y_{2n+2}$$ of degree $$2n+2$$. Then $$\omega_X = f^*(\omega_{\mathbb{P^n}} \otimes \mathcal{O}_{\mathbb{P}^n}(n+1)) = \mathcal{O}_X$$ is trivial. On the other hand, the double cover involution gives an action of $$G=\mathbb{Z}/2$$ on $$X$$ such that $$X^G$$ is isomorphic to $$Y$$ and, by adjunction, we find that $$\omega_Y = \mathcal{O}_Y(n+1)$$ is non-trivial.

For a non-projective example, you could take your favorite smooth variety $$Y$$ with nontrivial canonical bundle, and let $$X$$ be the total space of the cotangent bundle of $$Y$$. Then I believe $$X$$ will have trivial canonical bundle. You can let $$\mathbb{G}_m$$ act by scaling on the fibers of $$X$$. The $$\mathbb{G}_m$$-fixed points will be the original variety $$Y$$ (sitting as the zero section inside the cotangent bundle $$X$$).