# A question on $Isom(p_1^*E,p_2^*E) \rightrightarrows X$

I am learning the moduli stacks of vector bundles and have trouble understanding some definitions. Let $E$ be a rank $n$ vector bundle over the scheme $X$. We denote by $p_i$ the $i$th projection $p_i:X\times X\rightarrow X$ for $i=1,2$. I would like to understand the following statement.

The scheme of isomorphisms $Isom(p_1^*E,p_2^*E)$ is a principal $GL(n)$-bundle over $X\times X$. We define the symmetry groupoid of $E\rightarrow X$ to be $Isom(p_1^*E,p_2^*E) \rightrightarrows X$.

As a trivial example, take $X=Spec(k)$ for a field $k$. Then the vector bundle $E$ is a simply $k$-vector space and $Isom(p_1^*E,p_2^*E) \cong GL(n,k)$. So it makes sense to make a such definition. For a general $X$, the definition above is hard for me to digest. My questions are:

1. Why do we need $Isom(p_1^*E,p_2^*E)$, instead of $Isom(E,E)$?
2. What is the two maps $Isom(p_1^*E,p_2^*E) \rightrightarrows X$?
3. Why is it reasonable to call $Isom(p_1^*E,p_2^*E) \rightrightarrows X$ the symmetry groupoid of $E\rightarrow X$?

I would appreciate it if some experts could help me understanding these things.

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For question number 2, $\operatorname{Isom}(p_1^\ast E,p_2^\ast E)$ is equipped with a canonical map to $X \times X$. The two arrows are the composite of this map followed by the two projections to $X$. –  S. Carnahan Dec 23 '12 at 15:25
Thank you for the comment, S.Carnahan. OK, that's the only reasonable map, but I still don't see why such maps are important in this context. –  Pooya Dec 24 '12 at 1:00
Perhaps, to see why such a groupoid might arise, one should look at the infinitesimal version. It is simply $TE / GL(n)$ and is known as the Atiyah bundle of $E$ (it is a Lie algebroid as opposed to the finite version being a groupoid). It appears, for example, when one defines connections on $E$. –  Pavel Safronov Dec 24 '12 at 5:44
Thanks for the comments, Pavel. I am not familiar with the Atiyah bundle and would appreciate it if you could briefly explain what it is (I don't quite understand your comment in the bracket). –  Pooya Dec 25 '12 at 0:45
The Atiyah bundle $A_E$ is the bundle of $G$-invariant vector fields on $E$. These integrate to isomorphisms of torsors between different fibers, i.e. your $Isom$ groupoid. The Atiyah bundle has an anchor map $A_E\rightarrow T_X$ given by the differential of the projection $E\rightarrow X$. It is a map of Lie algebras, which is an infinitesimal version of the groupoid multiplication. The analogy is Lie algebroids are to groupoids, as Lie algebras are to groups. Finally, a connection is a splitting $T_X\rightarrow A_E$. –  Pavel Safronov Dec 30 '12 at 8:36

I'm not an algebraic geometer so let's call $X$ a real manifold. I don't think that really matters it could be a topological space or even a set for what I am about to say. Assume that $E \to X$ is a rank $n$ real vector bundle. For any $x \in X$ let $E_x$ be the fibre of $E$ over $x$. The natural structure you have in this situation is that if $f \colon E_{x} \to E_{y}$ and $g \colon E_{y} \to E_{z}$ are isomorphisms then you can compose to get $g \circ f \colon E_{x} \to E_{z}$ also an isomorphism. From this follows the fact that you have a groupoid whose objects are all $x \in X$ and whose morphisms from $x$ to $y$ are $Isom(E_x, E_y)$ (or $Isom(E_y, E_x)$ depending on how you like to compose morphisms.)

(1) $Isom(E, E)$ is a perfectly reasonable object, it's a bundle of groups over $X$. But it doesn't capture all the information such as isomorphisms from $E_x$ to $E_y$ where $x \neq y$.

(2) $Isom(p_1^*E, p_2^*E)$ is the union over all $x, y \in X$ of $Isom(E_x, E_y)$ and if $f \in Isom(E_x, E_y)$ then the two maps are $f \mapsto x$ and $f \mapsto y)$.

(3) I am not sure of the answer to this but it seems reasonable to me that this groupoid captures information about the symmetries of $E \to X$.

I don't think that $Isom(p_1^*E, p_2^*E)$ being a principal $GL(n, \mathbb{R})$ bundle is correct. I don't see any reason why $Isom(E_x, E_y)$, for example, is acted on by $GL(n, \mathbb{R})$.

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Regarding your last paragraph, any local trivialization of $E$ (hence its pullbacks) induces a local isomorphism from the Isom sheaf to the trivial $GL_n$ principal bundle. In general the action is by conjugation. –  S. Carnahan Dec 24 '12 at 3:59
But to be a principal bundle you need the action to be defined without choosing a trivialisation surely? In any case conjugation won't be a free action. –  Michael Murray Dec 24 '12 at 4:08
Conjugation won't be transitive either. You can make $Isom(E_x, E_x)$ and $Isom(E_y, E_y)$ act freely and transitively on the left and right of $Isom(E_x, E_y)$, and they are groups isomorphic to $GL(n, \mathbb{R})$ but there is no canonical isomorphism. –  Michael Murray Dec 24 '12 at 4:11
Thanks for the answer, Michael. I am now happy with your answer to (1) and (2), but still confused with (3). –  Pooya Dec 25 '12 at 0:48