Pullback along the Frobenius morphism

582 Views Asked by At

Let $X$ be a scheme over a finite field $\mathbb{F}_q$ and let $F : X \to X$ be the absolute Frobenius morphism. If $\mathcal{L}$ is an invertible $\mathcal{O}_X$-module, then $F^*(\mathcal{L}) \cong \mathcal{L}^{\otimes q}$ (naturally in $\mathcal{L}$).

Is there also a general formula for $F^*(\mathcal{M})$ if $\mathcal{M}$ is a locally free $\mathcal{O}_X$-module of given rank $d$?

In other words, I ask for a bundle representative of the Adams operation $\psi^q$ on the $K$-theory of $X$. For example, when $q=2$, we have $\psi^2(x)=x^2-2 \lambda^2(x)$, and one could hope for a representation of $F^*(\mathcal{M})$ as the cokernel of some monomorphism $\Lambda^2(\mathcal{M}) \oplus \Lambda^2(\mathcal{M}) \to \mathcal{M} \otimes \mathcal{M}$, or as the kernel of some epimorphism $\mathcal{M} \otimes \mathcal{M} \to \Lambda^2(\mathcal{M}) \oplus \Lambda^2(\mathcal{M})$.