Adelic description of moduli of $G$-bundles on a curve - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-18T20:59:55Z http://mathoverflow.net/feeds/question/112593 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/112593/adelic-description-of-moduli-of-g-bundles-on-a-curve Adelic description of moduli of $G$-bundles on a curve Justin Campbell 2012-11-16T17:07:44Z 2012-11-17T10:43:32Z <p>Let $X$ be a smooth, projective, geometrically connected curve over a field $k$ and $G$ an an affine algebraic group group over $k$ (we can put more hypotheses on $G$ if necessary). If $K$ denotes the function field of $X$ and $\mathbb{A}$ the corresponding ring of adeles with integral adeles $\mathcal{O}$, I expect that there is a bijection from $G(\mathcal{O}) \backslash G(\mathbb{A}) / G(K)$ to the set of isomorphism classes of $G$-bundles on $X$ (although I have only seen this fact stated when $G = \text{GL}_n$). Also, I think there should be an interpretation of the "class group" $G(\mathbb{A}) / G(K)$ as $G$-bundles with level structure, but I don't even know how to make this into a precise statement.</p> <p>Can someone explain these things to me? I would also be happy with a reference.</p> http://mathoverflow.net/questions/112593/adelic-description-of-moduli-of-g-bundles-on-a-curve/112598#112598 Answer by Will Sawin for Adelic description of moduli of $G$-bundles on a curve Will Sawin 2012-11-16T17:44:31Z 2012-11-16T19:06:18Z <p>First let's remove the condition that $X$ is projective. Then we will replace the adeles with a product over the points actually in $X$, and do the same for the integral ideles.</p> <p>Then let's make this set into a category. A map from $x \in G(\mathbb A_X)$ to $y \in G(\mathbb A_X)$ is a pair $a \in G(\mathcal O_{\mathbb A_X})$, $b\in G(K)$ such that $axb=y$.</p> <p>Then we can define a section of $x$ as a map from the trivial adele, $1$, to $x$, or a pair $a \in G(\mathcal O_{\mathbb A_X})$, $b\in G(K)$ such that $ab=x$.</p> <p>Then the sections of the trivial bundle are $G(\mathcal O_X)$, as desired.</p> <p>Then every element of $G(\mathbb A_X)$ has a set of sections on each open set - in fact a set with a free action of $G(\mathcal O_U)$. (If $(a,b)$ is a section, and $c \in G(\mathcal O_U)$, then $(ac,c^{-1}b)$ is a section.) Moreover there are natural restriction maps, and it is easy to check that this satisfies the sheaf condition - just glue together $a$ and $b$ separately.</p> <p>Then I claim we're done because you can find an open neighborhood of each point that's trivial - the key point being that an element of $G(\mathbb A_X)$ is in $G(\mathcal O_{\mathbb A_X})$ except at finitely many primes, and that you can cancel it at any single prime with an element of $G(K)$.</p> <p>This is probably a pretty silly way of looking at it.</p> <p>To turn a $G$-bundle back into an adele we just need to know how to glue an adele on $U$ and an adele on $V$ together to form an adele on $U \cup V$ given an isomorphism between them on $U \cap V$. If $x \in G(\mathbb A_U)$ and $y \in G(\mathbb A_V)$ satisfy $axb=y$ on $U \cap V$ for $a \in G(\mathcal O_{\mathbb A_{U \cap V}})$, $b \in G(K)$, then the adele that looks like $axb$ on $U$ and $y$ on $V$, for $a$ pulled up to $G(\mathcal O_{\mathbb A_U})$ by adding a bunch of trivial factors, is an appropriate gluing-together.</p> <p>Does that make sense?</p> http://mathoverflow.net/questions/112593/adelic-description-of-moduli-of-g-bundles-on-a-curve/112614#112614 Answer by Alexander Chervov for Adelic description of moduli of $G$-bundles on a curve Alexander Chervov 2012-11-16T21:05:09Z 2012-11-16T21:16:34Z <p>Concerning the bijection G-Bun &lt;-> G(K) \ G(A) / G(O) - it has rather simple intuitive explanation:</p> <p>1) G-Bun by definition can be described like this - choose covering U_i and consider $G(U_i) \backslash G(U_i \cap U_j) /G(U_j)$ with condition on triple intersection which we do not care for the moment.</p> <p>2) The main point is that <strong>adelic description is particular case of the one above for specific choice of covering</strong>. Indeed, choose the following covering: for each point "x" consider its infinitesimal neighbourhood U_x, so regular functions on it are O_x - power series regular at "x" and hence G(O) can be seen as direct product of G(U_x) for all "x". And there is one more chart - "generic point" - regular function on it is "K" and so G(K) is the set of G-regular functions on generic point. Now the intersection of U_x and U_y is empty so we only care about the intersection between U_x and generic point - hence we see G(A) arising as a product over "x" of G(U_x intersect generic point).</p> <p>So we get that G-Bun = G(K) \ G(A) / G(O). The triple intersection is empty - so we do not have any condition.</p> <hr> <p>Now about the level structure. My point of view might not be conventional, but it is close to Serre's "Algebraic groups and class fields".</p> <p>The point is that we can make sense of this level structures via considering the curves with n-cusp singularities.</p> <p>Let me do only locally. Consider the curve as Spec of f(x): f'(0)=f''(0)=...f'' ''(0) = 0. (for only one derivative we get cusp curve). We might be interested how to describe the bundles on it. The point is that instead of coset G(K) \ G(A) / G(O) : we will have coset G(K) \ G(A) / G(O_{n,0}) i.e. at point zero regular functions are not all power series but such that f'(0)=f''(0)=...f'' ''(0) = 0 .</p> <p>Now if change "n" and number of points we will get similar cosets and for bundles on singular curves, which might be considered as "level" structures on the original curve.</p> <p>PS</p> <p>Sometime ago we played with such curves and bundles on them: <a href="http://arxiv.org/abs/hep-th/0309059" rel="nofollow">http://arxiv.org/abs/hep-th/0309059</a></p> http://mathoverflow.net/questions/112593/adelic-description-of-moduli-of-g-bundles-on-a-curve/112652#112652 Answer by nosr for Adelic description of moduli of $G$-bundles on a curve nosr 2012-11-17T06:03:31Z 2012-11-17T10:43:32Z <p>The one-sentence answer to this question is: use fpqc descent theory (and an "answer" which doesn't address the role of fpqc descent -- sometimes presented in the form of a reference to a paper of Beauville and Laszlo -- is missing the key technical issue in the rigorous proof when working with general $G$, as far as I know).</p> <p>We make two "necessary" hypotheses (as noted in the comments to Sawin's answer) for our smooth connected affine $k$-group $G$: ${\rm{H}}^1(K,G) = 1$ for $K = k(X)$ and ${\rm{H}}^1(k',G) = 1$ for all finite extensions $k'/k$. For example, if $k$ is finite then this holds for any simply connected semisimple (connected) $G$ by the theorems of Harder and Lang respectively. If instead $k$ is algebraically closed of char. 0 then it holds for any $G$ by theorems of Tsen and Springer. If $k$ is algebraically closed of positive characteristic then it holds for any connected reductive $k$-group $G$, but Springer's theorem doesn't literally apply; see Remark 2(b) of the Drinfeld-Simpson paper mentioned in the comments to Sawin's answer.</p> <p>We shall allow $X$ to be any 1-dimensional reduced and irreducible $k$-scheme of finite type, not assumed to be proper or even normal. This way we incorporate Chervov's observations about using singular curves to build in more level structure. </p> <p>We construct the desired bijection as follows. Consider a left $G$-torsor $E \rightarrow X$ (local triviality equivalent for the fpqc and etale topologies due to the smoothness of $G$; the equivalence will be crucial later on and is false in general if we try using the Zariski topology, and officially we work with the etale topology in the definition as is traditionally done). Since ${\rm{H}}^1(K,G) = 1$, the generic fiber $E_{\eta}$ has a $K$-point and this spreads out over a dense open $U$ in $X$. Fix such a $U$ and trivialization $\xi \in E(U)$. </p> <p>Let $X^0$ be the set of closed points of $X$. Consider the pullback of $E$ over the completion $O^{\wedge}_x$ at $x \in X^0$. This pullback is a smooth $O^{\wedge}_x$-scheme whose special fiber is a $G$-torsor over the finite extension $k(x)$ of $k$ and so has a $k(x)$-point due to the other vanishing hypothesis. By smoothness (!) of $G$ (and the henselian property of $O^{\wedge}_x$) this lifts to a point $\xi_x \in E(O^{\wedge}_x)$.</p> <p>For each $x \in X^0$ consider the pullbacks of $\xi$ and $\xi_x$ over $U \times_X {\rm{Spec}}(O^{\wedge}_x) = {\rm{Spec}}(K_x)$ where $K_x$ is the total ring of fractions (product of finitely many fields) of the 1-dimensional reduced complete local ring $O^{\wedge}_x$. These two $K_x$-points of a common $G$-torsor are related through the action of a unique $g_x \in G(K_x)$; to be precise, $\xi = g_x \xi_x$ in $E(K_x)$. If $x \in U^0$ then clearly $g_x \in G(O^{\wedge}_x)$, so $(g_x) \in G(\mathbf{A}_X)$.</p> <p>If we change $\xi$ then we multiply every $g_x$ on the left by some common $g \in G(U)$, and if we change the various $\xi_x$'s then we multiply each $g_x$ on the right by an element of <code>$G(O^{\wedge}_x)$</code>. Finally, taking into account that we may shrink $U$ (and thereby enlarge $X - U$), we obtain an element <code>$$(g_x) \in G(K)\backslash G(\mathbf{A}_X)/G(O^{\wedge})$$</code> (where $O^{\wedge} = \prod_{x \in X^0} O^{\wedge}_x$) that depends only on the isomorphism class of $E$ over $X$.</p> <p>Our problem is to show that (i) this adelic double coset determines the isomorphism class of $E$ and (ii) all double cosets arise in this way. </p> <p>The assertion (i) is proved as follows. Assume $E$ and $E'$ give rise to the same double coset, so for a Zariski-dense open $U$ in $X$ trivializing $E$ and $E'$ we have $(g'_x) = \gamma (g_x) h$ for some $\gamma \in G(K)$ and $h \in G(O^{\wedge})$ with $g'_x, g_x \in G(O^{\wedge}_x)$ for all closed points $x$ of $U$. By shrinking $U$ we may assume $\gamma \in G(U)$. We may replace $\xi_x$ with $h_x\xi_x$ for all $x \in X^0$ and replace $\xi$ with $\gamma \xi$ so that $g'_x = g_x$ for all $x$. In other words, $E_U$ and $E'_U$ are each identified with the trivial $G_U$-torsor and has corresponding trivial $K_x$-fiber identified with the generic fiber of $G_{O^{\wedge}_x}$ via $g_x$-translation. Working one point of $X - U$ at a time, we just have to check:</p> <p>${\mathbf{Claim}}$: The category of $G$-torsors over $O_x$ is equivalent to the category of $G$-torsors over $O^{\wedge}_x$ equipped with a $K$-descent on its generic fiber over $K_x$. </p> <p>The categorical aspect of this Claim is essential (i.e., we do not just consider sets of isomorphism classes).</p> <p>Proof: By fpqc descent theory (the "Beauville-Laszlo step", though for us all we need was provided by Grothendieck) applied to the fpqc cover $${\rm{Spec}}(K) \coprod {\rm{Spec}}(O^{\wedge}_x) \rightarrow {\rm{Spec}}(O_x)$$ whose fiber square is ${\rm{Spec}}(K_x)$, the category of affine $O_x$-schemes is equivalent to the category of affine $O_x^{\wedge}$-schemes equipped with a $K$-structure on the generic fiber over $K_x$. Since the notion of $G$-torsor is well-behaved for the fpqc topology (and recall that fpqc $G$-torsors are automatically etale-topology torsors, due to the smoothness of $G$!!!), this equivalence specializes to the case of $G$-torsors. QED</p> <p>Now we run the game in reverse. Pick a class in $G(K)\backslash G(\mathbf{A}_X)/G(O^{\wedge})$ represented by some $(g_x) \in G(\mathbf{A}_X)$. Since $g_x \in G(O^{\wedge}_x)$ for all but finitely many $x \in X^0$, we can pick a Zariski-dense open $U$ in $X$ such that $g_x \in G(O^{\wedge}_x)$ for all $x \in U^0$, so we may change our representative to satisfy $g_x = 1$ for all $x \in U^0$. Applying the above Claim then enables us to extend the trivial $G_U$-torsor to a $G$-torsor $E$ over $X$ by fpqc-gluing using the elements $g_x$ for each $x \in X - U$ one at a time, and by design this $E$ gives rise to the chosen adelic double coset (if we are careful not to mix up $g_x$ and $g_x^{-1}$ for $x \in X - U$).</p>