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Let $m,n$ be nonnegative integers. Let $k$ be a field of positive characteristic. Let the vector space $V$ over the field $k$ have basis $e_1,...,e_{m+n}, f_1,..., f_n$. Let $k^*$ act on $V$ by $t \cdot e_i = t^{-2i} e_i$, $t \cdot f_i = t^{-2i} f_i$. This action of $k^*$ on $V$ induces an action on the complete flag variety $X=Fl(V)$. Let $Fr: X \to X^{(1)}$ be the Frobenius morphism. Let ${\mathcal O}(\lambda)$ be the standard line bundle on $X$ corresponding to a weight $\lambda$ of $\mathfrak{sl}_{m+2n}$. Non-equivaraintly, it is known that $Fr_*(O(\lambda)) \cong O(\lambda)^{\oplus p^{\rm dim(X)}}$. What is $Fr_*(O(\lambda))$ in the equivariant setting, where the actions of $k^*$ on $X$ and $X^{(1)}$ are induced by the above action of $k^*$ on $V$? In particular, can $Fr_*(O(\lambda))$ be expressed in terms of the line bundles ${\mathcal O}(\mu)$ on $X^{(1)}$?

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  • $\begingroup$ The formula that you write down in your sentence, "Non-equivaraintly, it is known ...", makes no sense. Let $\lambda$ be the zero weight. It is not true that the Frobenius pushforward of the structure sheaf is a direct sum of $p^{\text{dim}X}$ copies of the structure sheaf: these two sheaves have different $h^0$. (If you are concerned about absolute Frobenius versus geometric Frobenius, take $k$ to be $\mathbb{Z}/p\mathbb{Z}$.) $\endgroup$ Commented Nov 9, 2022 at 10:43
  • $\begingroup$ @JasonStarr Sorry, I meant to write instead $Fr_*({\mathcal O}(-\rho))={\mathcal O}(-\rho)^{\oplus p^{{\rm dim}(X)}}$ where $\rho$ is the half sum of positive roots $\endgroup$ Commented Nov 9, 2022 at 11:24
  • $\begingroup$ That makes a big difference. $\endgroup$ Commented Nov 9, 2022 at 11:57

1 Answer 1

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EDIT. Corrected the statement ($\sigma$ should be $p-1$ times what I wrote) and answered the question in the comment.

In general, the push-forward of a line bundle on the flag variety $G/B$ will not be the direct sum of line bundles. (This holds only on toric varieties.)

However, Haboush [1] (and independently Andersen) has proved that this is true for some special values of $\lambda$. More specifically, he proved that for $\lambda = \sigma + p\mu$ where $\sigma$ is $(p-1)/2$ times the sum of the positive roots (which is $(p-1)$ times the sum of the fundamental weights), one has an equivariant isomorphism $$ F_* \mathcal{O}(\lambda) \simeq V\otimes \mathcal{O}(\mu) $$ where $V$ is the Steinberg module, irreducible of highest weight $\sigma$. In particular, $F_*\mathcal{O}(\sigma)$ is a trivial vector bundle $V\otimes \mathcal{O}_{G/B}$ corresponding to the Steinberg representation.

Notice that the statement follows from the case $\mu=0$ by the projection formula and the fact that $F^* L = L^p$ for a line bundle $L$: $$ F_* \mathcal{O}(\sigma + p\mu) = F_* (\mathcal{O}(\sigma)\otimes F^* \mathcal{O}(\mu)) = (F_* \mathcal{O}(\sigma))\otimes \mathcal{O}(\mu) = V\otimes \mathcal{O}(\mu). $$

Now, to answer the question in the comment: in the reference, they are interested in $F_* \mathcal{O}(-\rho)$ where $\rho$ is the sum of the fundamental weights. Since $\sigma = (p-1)\rho = -\rho + p\rho$, we obtain with $\lambda = -\rho = \mu$: $$ F_* \mathcal{O}(-\rho) \simeq V\otimes \mathcal{O}(-\rho). $$

This isomorphism (saying that $\mathcal{O}(-\rho)$ is an "eigenvector" for $F_*$ with "eigenvalue" $V$) is employed in Samokhin's paper [2], which you may find useful.

[1] W. J. Haboush, A short proof of the Kempf vanishing theorem, Inventiones mathematicae volume 56, pages 109–112 (1980)

[2] https://arxiv.org/abs/1611.10320

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  • $\begingroup$ Thank you very much for your answer! Sorry, I am confused: in Lemma 6.2.6 of Bezrukavnikov, Mirkovic, and Rumynin arxiv.org/abs/math/0205144 they say that it is a well-known fact that non-equivariantly we have $Fr_*(O(-\sigma)) \cong O(-\sigma)^{\oplus p^{\rm dim(X)}}$. Does this have any relation to your answer and can this be lifted to the equivariant setting? $\endgroup$ Commented Nov 9, 2022 at 8:32

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