Let $S$ be a scheme and $E$ be a rank $n$ vector bundle on $S$. $E$ corresponds to a $GL_n$ torsor $P$ via the definition $$ P = Isom_S(E,\mathbf{A}_S^n) $$

Tannakian theory tells us that $P$ corresponds to a fiber functor $$ \eta : Rep(GL_n) \to Vect_S. $$ (Recall that $\eta(V)$ is defined to be the pushout of $P$ via the representation $GL_n \to GL(V))$

My questions is : is there an easy description of $\eta$ in terms of $E$ without going through $P$ ?

  • 3
    $\begingroup$ A half answer is that any finite-dimensional representation of $GL_n$ can be constructed inside suitable tensor powers of the defining representation $V$. Clearly $V^{\otimes d}$ corresponds to the vector bundle $\mathcal E^{\otimes d}$, and it might be possible (?) to somehow deduce the general case from this. But this is a bit disappointing because it doesn't give a nice formula, but rather relies on explicit knowledge of the representation theory of $GL_n$. $\endgroup$ Commented Mar 24, 2017 at 5:53
  • $\begingroup$ Unfortunately I lack knowledge, and could not find, the basic object related to this: the scheme $G:=\operatorname{Aut}_S(E)$ over $S$. It must be locally isomorphic to $GL_n$; note that $P$ is in fact a $G$-$GL_n$-bitorsor. Maybe $G$ is $GL_n$ over $S$? In that case, the answer should be $\eta(V)=E\otimes_GV$ $\endgroup$ Commented Mar 24, 2017 at 9:19
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    $\begingroup$ @მამუკაჯიბლაძე: It is not true that $G = GL_{n, S}$. For example, if $\mathcal E = \mathcal O(1) \oplus \mathcal O$ on $S = \mathbb P^1$, then $\mathscr End_S(\mathcal E) = \mathscr Hom(\mathcal O(1) \oplus \mathcal O, \mathcal O(1) \oplus \mathcal O) = \mathcal O \oplus \mathcal O(1) \oplus \mathcal O(-1) \oplus \mathcal O$, which is not the trivial bundle. Then $\mathscr Aut_S(\mathcal E)$ is some open subfunctor, and it's not hard to see that it cannot be globally isomorphic to $GL_2$ (e.g. consider global sections). $\endgroup$ Commented Mar 24, 2017 at 22:12
  • $\begingroup$ @R.vanDobbendeBruyn Thanks, that was helpful! I now realized - in fact $P$ is (among other things) the "universal isomorphiser" of $G$ with $GL_n$. So, there you get $(\text{lift of $E$})\otimes_{(\text{lift of $G$})\cong GL_n}V$, and then you must push it down back over $S$ and that is your $\eta(V)$. It becomes even more burning how to get rid of $P$, after all it is a canonical thing - the object of isomorphisms, it should be eliminable... $\endgroup$ Commented Mar 25, 2017 at 5:00
  • $\begingroup$ Thanks guys for your comments. Remy I really like your suggestion. Does it really work ? I know that the category of representations of GL_n is generated by the standard representation. So I guess this means that if we now $\eta(V)$ we should now $\eta(W)$ for most $W$ but it's not clear that it works for all no ? I mean since $V$ is irreducible we don't have sub-representations which is useful but I'm not sure we really can get everything. To be (maybe) clearer can we really get the value of $\eta$ on representations inside the tensors ? $\endgroup$
    – jojo
    Commented Mar 25, 2017 at 23:02

2 Answers 2


Write $BGL_n$ for the classifying stack of rank $n$ vector bundles; in principle it is not necessary to know what a $GL_n$ torsor is in order to say this. $E$ is represented by a map $f : S \to BGL_n$, and pullback along this map is a functor

$$f^{\ast} : QC(BGL_n) \to QC(S).$$

This is $\eta$, or rather it is a "large" version of $\eta$: you can identify $QC(BGL_n)$ with representations of $GL_n$. (If you like, this is the only non-tautologous statement I'm making: you can define $QC(BGL_n)$ to consist of natural assignments of quasicoherent sheaves on $S$ to rank $n$ vector bundles on $S$.) To pass to vector bundles, $f^{\ast}$ is symmetric monoidal, so induces a corresponding functor on dualizable objects.

  • $\begingroup$ If $QC(BGL_n)$ can be identified with $Rep(GL_n)$, what kind of representations correspond to the subcategory of $QC(BGL_n)$ consisting of vector bundles? $\endgroup$ Commented Mar 25, 2017 at 19:46
  • $\begingroup$ One more thing I don't understand (sorry), from the topological perspective. Certainly a representation $GL_n\to GL(V)$ induces $BGL_n\to BGL(V)$, hence a vector bundle on $BGL_n$. But I don't quite see how to go back. Given $BGL_n\to BGL(V)$ one produces $GL_n\to\Omega BGL(V)$ which is not quite $GL(V)$: there is a map $GL(V)\to\Omega BGL(V)$ but I believe it is not an equivalence. $\endgroup$ Commented Mar 25, 2017 at 19:48
  • $\begingroup$ Oops sorry it is an equivalence - more or less by definition. Let me still leave that comment, just in case :D $\endgroup$ Commented Mar 25, 2017 at 20:01
  • $\begingroup$ Hi, thanks for your answer. I have to admit I was hoping for something slightly more explicit. In fact I believe that Remy's comment might be the closest to what I wanted that is possible to get. $\endgroup$
    – jojo
    Commented Mar 25, 2017 at 23:00
  • 1
    $\begingroup$ @ამუკა ჯიბლაძე: vector bundles on $BGL_n$ correspond to finite-dimensional representations of $GL_n$. $\endgroup$ Commented Mar 26, 2017 at 17:54

How is your vector bundle $E$ presented? (As a locally free sheaf? As it's total space? Or...) If you like to think in terms of transition functions, i.e., Čech cocycle $\underline{\sigma} = \{\sigma_{ij}\} \in H^1(S,GL_n)$ (of non-abelian cohomology group), then the representation $(\pi,V) \in Rep_k(GL_n)$ will be sent to the vector bundle corresponding to the Čech coycle $\{ \pi(\sigma_{ij}) \} \in H^1(S,GL_V)$.

To get an idea of what is going on in the previous answers: This is essentially "resolving" $S$ using a cover $\mathfrak{U}=\{U\}$ and the Čech nerve $C^\bullet(\mathfrak{U})$ and thinking of the stacks $BGL_n$ and $BGL_V$ as simplicial schemes via the Milnor bar constrcution. We get that the bundle $\eta(\pi,V)$ is classified by the composite

$C^\bullet(\mathfrak{U}) \to BGL_n^\bullet \to BGL_V^\bullet$

where the first morphism (is the classifying map and) is precisely the data $\underline{\sigma}$ and the second is $B\pi$ (the functor $B$ applied to $\pi : GL_n \to GL_V$).

Is at least the first description closer to what you wanted?

  • $\begingroup$ Hi thank you very much for your answer. I do like the first description ! To be honest what I wanted was what was suggested by Remy in the comments. $\endgroup$
    – jojo
    Commented Apr 14, 2017 at 19:33

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