Principal bundles over smooth projective curve

Question

Let $X$ be a smooth and connected projective curve over $\mathbb{C}$ and $G$ a reductive connected group over $\mathbb{C}$. Fix a faithful representation $G \subseteq \mathrm{GL}_n$.

Given a $G$-principal bundle $P \to X$, we get an associated vector bundle bundle $\widetilde{P}=P \times_G \mathbb{C}^n$. Is it always true that if $\widetilde{P} \cong \widetilde{P'}$, then $P \cong P'$?

It seems to me that this is true for the trivial one. Namely,if $\widetilde{P}$ is trivial, then so it is $P$. We look indeed at an open covering $X=\bigcup_{i \in I}U_i$ certain cocycle $\{g_{i,j}:U_i \cap U_j \to G\}_{i,j}$ defining $P$. If $\widetilde{P}$ is trivial, we can find $\{h_i:U_i \to \mathrm{GL}_n\}_{i \in I}$ such that $$g_{i,j}=\dfrac{h_i}{h_j} .$$

In particular, the functions $h_i$ glue to a function $X \to \mathrm{GL}_n/G$. Since $X$ is projective and $\mathrm{GL}_n/G$ this function is constant and this implies that $g_{i,j}$ was the trivial cocycle.

The same reasoning however does not seem to extend to any pair of principal bundles easily.

Answer 1

This is not true. For a simple counterexample, take any two $G$-bundles that are not isomorphic. Their fiber product is then a $G \times G$ bundle in two different ways, depending on which copy of $G$ we assign to which principal bundle, which are not isomorphic. But if the faithful representation of $G \times G$ we choose is the direct sum or tensor product of two copies of the same faithul representation of $G$, the associated $GL_n$-bundles will be isomorphic.

Worse, there can be non-isomorphic $G$-bundles whose associated $GL_n$-bundles are isomorphic for every faithful representation of $G$. This can be accomplished for $G = PGL_n, n>2$.

Associated to a map $\pi_1(X) \to G$ there exists a principal $G$-bundle that can be described analytically as the universal cover of $X$ times $G$ modulo $\pi_1(X)$.acting diagonally. If the image of $\pi_1(X)$ is finite, then two $G$-bundles are isomorphic if and only if the representations of $\pi_1$ are isomorphic (since the isomorphism would lift to the covering associated to the kernel of the map on $\pi_1$ where both bundles become trivial so the only isomorphisms are constant isomorphisms, which descend if and only if they are $\pi_1$-equivariant). This is even true for compact image but we don't need this.

Now $\mathbb Z/n \times \mathbb Z/n$ embeds into $PGL_n$ where the two generators act respectively by a diagonal matrix whose $j$'th diagonal entry is $e^{2\pi i j/n}$ and a permutation matrix that sends the $j$'th unit vector to the $j+1$st unit vector (mod) $n$. We choose one homomorphism to be a surjective map $\pi_1(X) \to (\mathbb Z/n\times \mathbb Z/n) \to PGL_n$ and the other to be the same map except composed by an automorphism of $(\mathbb Z/n\times \mathbb Z/n) $ with nontrivial determinant. Then the representations are not isomorphic since this automorphism doesn't arise from conjugation by any element of $PGL_n$ so the principal $PGL_n$-bundles are not isomorphic but the representations send each element to an element in the same conjugacy class and thus their compositions with any representation of $PGL_n$ have the same character and hence their compositions with any representation of $PGL_n$ are isomorphic.