# How are gauge transformations of a $G$-bundle related to the adelic points of $G$?

In a very interesting blog discussion at the $n$-category cafe, an anonymous poster made the following remark: "... using the dictionary between number fields and function fields, Weil suggested that G-with-adelic-entries is analogous to the group of gauge transformations of a principal G-bundle over a Riemann surface."

I would like to know how this particular piece of the number field / function field analogy is made precise. As a preliminary question, does anyone know a reference to where Weil (or someone else) explains it? I have a suspicion that it might be discussed in the notes from Weil's 1959-1960 lectures on "Adeles and algebraic groups", but the copy in my local library is checked out so I'm not sure.

My next question is for an explanation of the analogy in what I assume is the simplest case: holomorphic vector bundles on a compact Riemann surface. There are two obstacles here. First, I don't know what a gauge transformation is, and the "Mathematical Formalism" section of the Wikipedia page seems nonsensical to me. What is being transformed, and what is the transformation itself? Can anyone provide a physics-free definition, purely in the language of geometry?

Now given a curve $X$ over the complex numbers (or any field), we can form a topological ring $\mathbf{A}_X$ as the restricted product of the completed local rings at the closed points of $X$. The second obstacle is that when the ground field is infinite, these completions are not locally compact, so it seems unlikely that $\mathbf{A}_X$ is a good thing to consider.

Nonetheless, can one put the constructions of the previous two paragraphs together to produce, for any rank $n$ holomorphic vector bundle $\mathcal{E}$ on $X$, a bijection as follows?

$${\text{gauge transformations of }\mathcal{E}}\stackrel{?}{\leftrightarrow} \mathrm{GL}_n(\mathbf{A}_X)$$

My final question is, what is the correct version of this when the complex numbers are replaced by a finite field? Now we can consider algebraic vector bundles, and even better the adeles of $X$ are now a good thing to consider. So in trying to extend the putative bijection above to this case, the content seems to be in algebraizing the notion of gauge transform. Thus, I'm asking for a (second?) definition of gauge transformation which is native to algebraic geometry, if such exists.

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I wasn't sure whether to post this here or at mathoverflow. Do any experts have an opinion? –  Sam Lichtenstein Aug 27 '10 at 16:31
This seems like a fine MO question to me. –  Qiaochu Yuan Aug 27 '10 at 16:38
Agreed -- push this one over to MO. People will be interested. –  Pete L. Clark Aug 28 '10 at 22:21
I would... but in fact this has been answered to my satisfaction. Ramras's Atiyah-Bott reference seems like a good place for me to learn what the deal with gauge groups is (thanks! I've only glanced at it so far...) and Matt E gave a nice explanation of the link between G-bundles and $G(\mathbf{A})$. I see no need to repost on MO. –  Sam Lichtenstein Aug 29 '10 at 4:48

If $X$ is a smooth projective curve over a finite field $k$, with field of rational functions $F$, then the quotient $GL_n(F)\backslash GL_n(\mathbf A_F)/GL_n(\mathcal O_F)$ (where $\mathcal O_F$ denotes the subring of integral adeles in $\mathbf A_F$) is in natural bijection with the set of isomorphism classes of rank $n$ vector bundles on $X$, i.e. the set $Bun_n(k)$ of $k$-valued points of the moduli stack of rank $n$ bundles on $X$. (And the same should be true if we replace $GL_n$ by another reductive group $G$, and replace "rank $n$ vector bundles" by principal $G$-bundles; recall that principal $GL_n$-bundles are "the same" as rank $n$ vector bundles; one goes from the latter to the former by passing to the associate frame bundle.)

The passage from a bundle to an a point in the double coset space is made as follows: given the vector bundle, one chooses an affine open subset $U = X \setminus {x_1,\ldots,x_r}$ over which the bundle is trivialized, and also trivialiations in a n.h. $U_i$ of each of the points $x_i$. The gluing data comparing the two trivializations on $U_i \cap U$ gives an element $g_{x_i} \in GL_n(F_{x_i})$, where $F_{x_i}$ is the completion of $F$ at $x_i$. If $x$ is a point distinct from the $x_i$, set $g_{x} = 1$. We can then put all the $g_x$ together into an adele, and its class in the double coset space $GL_n(F)\backslash GL_n(\mathbf A_F)/GL_n(\mathcal O_F)$ is independent of all choices.

To see that this map is a bijection, one first observes that any element in the double coset space can be represented as $(g_x)$ with $g_x = 1$ for almost all $x$, and then one interprets these $g_x$ as formal gluing data around the points $x_i$ at which $g_{x_i} \neq 1$ and uses them to extend the trivial bundle on $X\setminus {x_1,\ldots,x_r}$ to a bundle on all of $X$.

The relationship with gauge transformations is that the $g_x$ are change of basis matrices (gauge tranformations) in formal punctured n.h.s of the points $x$.

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Thanks! Someone told me offline about the double coset space parametrizing bundles. If anyone is curious about the ground field $k$, apparently either $k$ finite or $k$ algebraically closed ensures that this bijection works. –  Sam Lichtenstein Aug 28 '10 at 16:18

I'd like a see an answer to this question, and I hope someone here or on MO can provide a good one.

I just want to address your question about gauge transformations. Usually a gauge transformation on a principal G-bundle $P\to X$ is simply a G-equivariant map $P\to P$ which sends fibers to fibers (said another way, a gauge transformation is simply an automorphism of the principal G-bundle P). A great place to read about the basics of gauge transformations is Atiyah and Bott's paper "The Yang-Mills Equations over Riemann Surfaces" (section 2 in particular).

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Hmmm... apparently LaTex doesn't work here in the same way as on MO. –  Dan Ramras Aug 28 '10 at 6:16
The "backtick" trick doesn't seem to work here as a fix when LaTex is misbehaving, but straight up LaTex usually seems to work: $P\to X$. –  Sam Lichtenstein Aug 28 '10 at 16:11
Okay, fixed it, thanks. –  Dan Ramras Aug 28 '10 at 22:18