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3 added 219 characters in body

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.

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 $axb=y$.

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 $ab=x$.

Then the sections of the trivial bundle are $G(\mathcal O(X))$O_X)$, as desired. Then every element of$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$separately. Then I claim we're done because you can find an open neighborhood of each point that's trivial - the key point being that an element of$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)$. Then on each open neighborhood we have the set of sections, a$G$-torsor, and on each intersection we have a natural isomorphism between them, producing a principle$G$-bundle. This is probably a pretty silly way of looking at it. To turn a$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})$by adding a bunch of trivial factors, is an appropriate gluing-together. Does that make sense? 2 deleted 3 characters in body First let's remove the condition that$X$is projective. Then we will replace the adeles with a product over the points actually in$X$, and the integral ideles. Then let's make this set into a category. A map from$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)$such that$akx=y$.axb=y$.

Then we can define a section of $x$ as a map from the trivial adele, $1$, to a given adele, $x$, or a pair $a \in G(\mathcal O_{\mathbb A_X})$, $b\in G(K)$ such that $ax=y$.ab=x$. Then the sections of the trivial bundle are$G(\mathcal O(X))$, as desired. Then I claim we're done because you can find an open neighborhood of each point that's trivial - the key point being that an element of$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)$. Then on each open neighborhood we have the set of sections, a$G$-torsor, and on each intersection we have a natural isomorphism between them, producing a principle$G$-bundle. This is probably a pretty silly way of looking at it. To turn a$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})$by adding a bunch of trivial factors, is an appropriate gluing-together. Does that make sense? 1 First let's remove the condition that$X$is projective. Then we will replace the adeles with a product over the points actually in$X$, and the integral ideles. Then let's make this set into a category. A map from$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)$such that$akx=y$. Then we can define a section as a map from the trivial adele,$1$, to a given adele, or a pair$a \in G(\mathcal O_{\mathbb A_X})$,$b\in G(K)$such that$ax=y$. Then the sections of the trivial bundle are$G(\mathcal O(X))$, as desired. 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$G(K)$. Then on each open neighborhood we have the set of sections, a$G$-torsor, and on each intersection we have a natural isomorphism between them, producing a principle$G$-bundle. This is probably a pretty silly way of looking at it. 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.

Does that make sense?