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 $axb=y$.

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 $G(\mathcal O(X))$, 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?

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 $G(\mathcal O_X)$, 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 $G(K)$.

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?

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?

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 $axb=y$.

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 $G(\mathcal O(X))$, 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?