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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 there is a natural isomorphism $F^*(\mathcal{L}) \cong \mathcal{L}^{\otimes q}$.

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

Of course, we have $F^*(\mathcal{M}) = \mathcal{M} \otimes_{\mathcal{O}_X} F_*(\mathcal{O}_X)$ (since $F$ is affine), but I would like to have a formula which is independent of the quasi-coherent algebra $F_*(\mathcal{O}_X)$ (unless you can describe this algebra via generators and relations), but only depends on $\mathcal{M}$ and uses the usual operations on quasi-coherent modules ($\otimes$, $\ker$, $\mathrm{coker}$, $\oplus$, $\mathrm{Sym}^n$, $\Lambda^n$, $\dotsc$).

For $q=2$ I have found that $F^*(\mathcal{M})$ is the kernel of the homomorphism $\mathrm{Sym}^2(\mathcal{M}) \twoheadrightarrow \Lambda^2(\mathcal{M})$, $m \cdot n \mapsto m \wedge n$. In general, there is an embedding$$F^*(\mathcal{M}) \to \mathrm{Sym}^q(\mathcal{M})$$ which maps $m \otimes 1 \mapsto m^q$.

My question may be also phrased as follows: On the $K$-theory of $X$, $F^*$ induces the Adams operation $\psi^q$. Hence, I ask for a bundle representative of the Adams operation $\psi^q$. (Not just a formal difference of bundles.)

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Consider the $p$-th tensor power $\mathcal{M}^{\otimes p}$. The group $\mathbb{Z}/p\mathbb{Z}$ acts by cyclic permutations on it. Denote its generator by $\sigma$. There is a map from coinvariants to invariants $$(\mathcal{M}^{\otimes p})_{\sigma}\xrightarrow{1+\sigma+\dots+ \sigma^{p-1}}(\mathcal{M}^{\otimes p})^{\sigma}$$ Its kernel is canonically isomorphic to $F^*\mathcal{M}$ via the map $s\mapsto s^{\otimes p}$. This is proven in Lemma 6.9 here:http://arxiv.org/pdf/1509.08784v1.pdf for the affine case and the general case follows by functoriality.

Edit: By the way, its cokernel is also isomorphic to $F^*\mathcal{M}$.

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  • $\begingroup$ Interesting, but this only holds when $M$ is flat. (See the comments before Lemma 6.9, and the proof.) $\endgroup$ – Martin Brandenburg Oct 21 '16 at 13:55
  • $\begingroup$ @MartinBrandenburg Sure, but I thought you are asking about locally free sheaf which is automatically flat, aren't you? $\endgroup$ – SashaP Oct 21 '16 at 17:27
  • $\begingroup$ Oh, you are right. I forgot that. $\endgroup$ – Martin Brandenburg Oct 23 '16 at 10:51
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Using the notation from my paper on Adams operations for the Schur functors of hook type, there ought to be a long exact sequence

$$0 \to F^*(\mathcal{M}) \to L_1^q(\mathcal M) \to \dots \to L_{q-1}^q(\mathcal M) \to 0$$

that generalizes the short exact sequence that you wrote down for $q=2$; the corresponding equation in the Grothendieck group holds, as explained in the paper, and that motivates the guess, which I remember wondering about long ago.

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  • $\begingroup$ Thank you! Could you please give a self-contained definition of $L^q_k(M)$? $\endgroup$ – Martin Brandenburg Dec 3 '15 at 18:04
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In topological K-Theory, there is such an expression:

For $E=\bigoplus L_i$, $\psi^n E = \bigoplus L_i^n$. Now $\Lambda_k E$ is the $k$-th elementary symmetric polynomial in the $L_i$. Since any symmetric polynomial is expressible in terms of the elementary ones, a description follows. Note that it involves subtraction, so this is genuinely a stable thing.

Of course, to prove that this expression agrees with $\psi^n$ for stable vector bundles that don't split into line bundles, we use the splitting principle. I have no idea whether that is applicable to your setting.

(from Hatcher's VBKT, below Thm 2.20: https://www.math.cornell.edu/~hatcher/VBKT/VB.pdf)

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  • $\begingroup$ Sorry, but this doesn't answer my question; this is just the usual description of the Adams operation. I don't work in the K-theory of $X$. I work in the category of quasi-coherent (or locally free) sheaves on $X$. Therefore, the minus signs have to be replaced by cokernels or kernels, I guess. But I cannot make it work so far. $\endgroup$ – Martin Brandenburg Nov 20 '15 at 11:50

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