Let $X$ be a projective smooth curve over an algebraically closed field $\mathbb k$ of any characteristic. Let also $G$ be a reductive group over $\mathbb k$. (Probably) following Ramanathan, a $G$-bundle $\mathcal E$ on $X$ is called semistable if for all parabolic subgroups $P$ of $G$, all reductions $\mathcal E_{P}$ of $\mathcal E$ to $P$ and all dominant characters $\chi:P\rightarrow \mathbb G_m$, one has $$ \deg(\mathcal E_P\times_P \mathbb A^1_{\chi})\le 0. $$

(to be clear $\mathcal E_P\times_P \mathbb A^1_{\chi}$ here denotes the natural linear bundle on $X$ obtained from $\mathcal E_P$ by applying $\chi$ to the structure group $P$).

Now, this definition is not superintuitive for me, but looking at the case $G=GL_n$ I convinced myself that it's pretty natural (and agrees with the slopes' one).

So, my question is: is it true that the group of $\mathbb k$-points $\mathrm{Aut}(\mathcal E)(\mathbb k)$ of the automorphisms of $\mathcal E$ always obtains an emdedding in $G(\mathbb k)$ if $\mathcal E$ is semistable? Or if not, is there any sort of description of the possible automorphism groups of semistable $E$'s? Can they be bounded by some other group? (In a sense that there is a group $A_G$ over $\mathbb k$ such that any automorphism group admits an embedding to $A_G$, and $A_G$ does not depend, or depends in an easy way, on $X$)

For example in the case of $GL_n$, since stable bundles are simple, homs between two stable bundles are either $\mathbb k$ or 0, and having a polystable bundle $\mathcal E=\bigoplus_i \mathcal S_i^{\oplus k_i}$ the automorphisms are given by $\prod_{i}GL_{k_i}$, which is naturally a subgroup of $GL_n$. I'm not completely sure what happens for more complicated semistable bundles.

  • $\begingroup$ Regarding your question about boundedness: for each fixed curve and fixed group, there is a bounded family of semistable objects (in positive characteristic I believe this is due to Adrian Langer). So there is some group into which all of these groups embed. Presumably you want some a priori bound, or even a group that is independent of the curve. $\endgroup$ Jun 13, 2017 at 7:37
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    $\begingroup$ To show that the atuomorphism group embeds in $G(k)$, it suffices to show that no automorphism is trivial when restricted to a single fiber. By examining the characteristic polynomial, every automorphism that is trivial on a single fiber must be unipotent on the generic fiber. Do unipotent automorphisms canonically determine a parabolic subgroup? If so, can you find a canonical dominant cocharacter whose degree measures the order of vanishing of the automorphism? $\endgroup$
    – Will Sawin
    Jun 13, 2017 at 13:53
  • $\begingroup$ "to Jason" Yeah, I guess it is also clear from the realization of the stack of semistable bundles (if we fix the degree) as a global quotient. $\endgroup$
    – user42024
    Jun 13, 2017 at 21:10
  • $\begingroup$ "to Will" Thanks, this seems like a nice idea to start with. For now, I do not know if either of this is in fact true, but both could be $\endgroup$
    – user42024
    Jun 13, 2017 at 21:13


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