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.



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


