Let $X$ be a scheme. It is well known that vector bundles, or locally finite free sheaves of $\mathcal{O}_X$-modules, "correspond" to principal $GL_n$ bundles, or $GL_n$ torsors. This correspondence can be sketched via cocycles, or gluing data, yielding a bijection on isomorphism classes.

It is natural to ask whether this can be enriched to an equivalence of categories, but naively this is hopeless, as vector bundles admit plenty of non-invertible morphisms (say, the zero map), and every morphism of torsors is an isomorphism.

However, we can try and ask this question on associated grupoids. Given a vector bundle $E$ over $X$ with associated principal bundle $P$, do we have a canonical isomorphism $$ \operatorname{Aut}(E) = \operatorname{Aut}(P) $$ ?

I believe this might also not be the case, as I somehow expected $G$-torsors to have $G$ as a sheaf of automorphisms, and I have concrete examples of vector bundles with automorphism group/sheaf not being $GL_n$.

  • $\begingroup$ I should mention that the rank 1 case is easy. If $L$ is a line bundle then it is invertible, and therefore $\operatorname{End}(L) = \operatorname{End}(\mathcal{O}_X)$ and therefore the automorphisms of $L$ are $GL_1 = \mathbb{G}_m$. $\endgroup$
    – Thiago
    Commented May 28, 2023 at 8:41
  • $\begingroup$ Oh and of course if $E$ is trivial then it is also clear, As automorphisms of $E$ are just $GL_n$ and so is the automorphism sheaf of any principal bundle.... $\endgroup$
    – Thiago
    Commented May 28, 2023 at 10:43
  • $\begingroup$ I suppose you consider automorphisms over $X$. Note that there is a natural map of $ $\endgroup$ Commented May 29, 2023 at 11:11
  • 2
    $\begingroup$ I suppose you consider automorphisms over $X$. Note that there is a natural morphism of $X$-group schemes ${\rm Aut}_X(E)={\rm GL}(E)\to{\rm Aut}_X(P)$, which locally on $X$ sends a fibrewise linear automorphism $a$ of $E$ to the automorphism of $P$ sending a frame to its image under $a.$ This morphism is an $X$-isomorphism by your second comment. So the $X$-points (the sections) of ${\rm Aut}_X(E)$ and ${\rm Aut}_X(P)$ are the same and these $X$-points correspond to both your groups. $\endgroup$ Commented May 29, 2023 at 12:24

1 Answer 1


Hey Thiago:D I really like your question; I'll illustrate my thoughts in some detail so my apologies if I say lots of really obvious things.

Fix a scheme $X$, a sheaf of groups $G$ on $X$ and a $G$-torsor $P$; suppose we work in the Zariski topology for concreteness (as in your case with $\text{GL}_n$ torsors anyway, although I'm rather sure this generally works) and let $Y = \bigcup_i U_i$ be an open cover of $Y\subset X$ such that $P_{\mid U_i}$ is isomorphic to the trivial left $G$-torsor $G_{\mid U_i} \xrightarrow{\phi_i} P_{\mid U_i}$. I first think it's worth noting how $\text{Aut}_G(P)$ is a sheaf of $G$-sets: given any automorphism of $G$-torsors $\alpha : P_{\mid Y} \to P_{\mid Y}$, we have that $\phi_i\alpha_{\mid U_i}\phi_i^{-1}: G_{\mid U_i}\to G_{\mid U_i}$ is given by right multiplication by some element $a_i \in G(U_i)$ and it's rather tempting, at least for me, to define the action on $\alpha \in \text{Aut}_G(P)(Y)$ by an element $g\in G(Y)$ in imposing $\phi_i(g\cdot \alpha)_{\mid U_i}\phi_i^{-1}$ to be right multiplication by $a_i\cdot g^{-1}$, but of course there are gluing problems... we can examine them: if we set $U_{i,j}:= U_i\cap U_j$ then we have a commutative diagram $$ \require{AMScd} \begin{CD} G_{\mid U_{i,j}} @>{\cdot (a_i)_{\mid U_{i,j}}}>> G_{\mid U_{i,j}}\\ @V{(\phi_i)_{\mid U_{i,j}}}VV @VV{(\phi_i)_{\mid U_{i,j}}}V \\ P_{\mid U_{i,j}} @>{\alpha_{\mid U_{i,j}}}>> P_{\mid U_{i,j}}\\ @A{(\phi_j)_{\mid U_{i,j}}}AA @AA{(\phi_j)_{\mid U_{i,j}}}A \\ G_{\mid U_{i,j}} @>{\cdot (a_j)_{\mid U_{i,j}}}>> G_{\mid U_{i,j}} \end{CD} $$ which reads $$ a_{i} = (\phi_j^{-1}(\phi_i(1_G))\cdot a_{j}\cdot \phi_i^{-1}(\phi_j(1_G)) $$ and thus substituting $a_i,a_j$ for $a_i\cdot g^{-1},a_j\cdot g^{-1}$ might potentially hinder this equality, preventing us from getting a new automorphism $g\cdot \alpha : P_{\mid Y} \to P_{\mid Y}$. If instead we try replacing $a_i$ and $a_j$ with $a_i^g,a_j^g$ (where I use the group theorist's notation $x^y := yxy^{-1}$) we get an equality $$ a_i^g = (\phi_j^{-1}(\phi_i(1_G))^g\cdot a_{j}^g \cdot (\phi_i^{-1}(\phi_j(1_G)))^g $$ and the sweet thing is that we can replace the trivialisation $\phi_i$ with $\phi_i\circ(\cdot g)$ (which of course still provides an isomorphism between $G_{\mid U_i}$ and $P_{\mid U_i}$) thus defining an action of $G(Y)$ on $\text{Aut}_G(P)$ swapping $a_i$ for $a_i^g$, by the equation above. In particular, the automorphism sheaf of the trivial $G$-torsor $\text{Aut}_G(G) \cong G$ is endowed with the action by conjugation.

If we now consider the presheaf $$ F : X_{\text{Zar}}^{\text{op}} \to (\text{sets}) $$ given by $$ F(U) = P(U) \times G(U) / (h\cdot p,g)\sim (p,hgh^{-1}) $$ whose sheafification is the contracted product $F^\# = P \wedge^G G$ where $G$ acts on itself via inner automorphisms, we can construct a morphism $F \to \text{Aut}_G(P)$ by mapping the pair $(p,g)$ to the automorphism $$ P_{\mid U} \xrightarrow{\phi_p^{-1}} G_{\mid U} \xrightarrow{\cdot g} G_{\mid U} \xrightarrow{\phi_p} P_{\mid U} $$ where $\phi_p : G_{\mid U} \to P_{\mid U}$ is the trivialisation corresponding to the section $p \in P(U)$. Then applying $(-)^{\#}$ yields an isomorphism of sheaves $P\wedge^G G \xrightarrow{\cong} \text{Aut}_G(P)$ since for the trivial torsor, as mentioned above, we have $\text{Aut}_G(G) \cong G \cong G\wedge^G G$.

Finally, turning to your particular situation, we can actually make the same considerations for $\text{Aut}_{\mathcal{O}_X}(E)$ where $E$ is a vector bundle on $X$ (by which I mean a locally free sheaf of $\mathcal{O}_X$-modules of rank $n$): the $\text{GL}_{n,X}$-torsor $P$ corresponding to $E$ is the sheaf $$ U \mapsto \text{Isom}_{\mathcal{O}_X}(E_{\mid U}, \mathcal{O}_U^{\oplus n}) $$ and thus we can define a morphism of presheaves $$ F \to \text{Aut}_{\mathcal{O}_X}(E) $$ by mapping the pair $(p,g) \in P(U) \times \text{GL}_n(\mathcal{O}_U)$ to $$ E_{\mid U} \xrightarrow{p^{-1}} \mathcal{O}_U^{\oplus n} \xrightarrow{g} \mathcal{O}_U^{\oplus n} \xrightarrow{p} E_{\mid U} $$ which once again defines an isomorphism $F^\# = P\wedge^{\text{GL}_n}\text{GL}_n \cong \text{Aut}_{\mathcal{O}_X}(E)$ for the same reason.

So in conclusion we have $$ \text{Aut}_{\text{GL}_{n,X}}(P) \cong P\wedge^{\text{GL}_{n,X}}\text{GL}_{n,X} \cong \text{Aut}_{\mathcal{O}_X}(E).$$

I do want to mention that generally speaking I dislike working with cocycles... I'd love to see an explanation of the $G$ action on $\text{Aut}_G(P)$ without needing to fix a trivialisation :P

Hope this helps and I didn't mess something silly up (:


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