What is the geometric meaning of potential automorphy, and conceptually why is it so hard to go from potential automorphy to automorphy on the nose?
Is there an obstruction to descent from potential automorphy to automorphy and does it lie in some Galois cohomology group? Is there an example where a variety is potentially automorphic, but not automorphic?
$\DeclareMathOperator\GL{GL}$There is no obstruction to descent from potential automorphy to automorphy lying in some cohomology group. Conjecturally, potential automorphy should be equivalent to automorphy.
To see why, let us take as our primary object that can be automorphic or not, Galois representation instead of varieties or motives (to fix ideas). Specifically, we consider continuous irreducible representations $r:G_K \rightarrow \GL_n(\bar{\mathbf Q}_p)$, where $K$ is a number field, $G_K$ its absolute Galois group, and $p$ a prime number. Let us recall the following "folklore" conjecture:
Conjecture (Fontaine–Mazur + Langlands). A Galois representation $r$ of $G_K$ as above is automorphic if and only if $r$ is unramified at almost all places of $K$ and potentially semi-stable at every place of $K$ dividing $p$.
Here automorphic means that $r$ has the same $L$-factors at almost all primes as a cuspidal automorphic algebraic representation $\pi$ of $\GL_n(\bf{A}_K)$.
Now fix $K'$ a finite extension of $F$. It is almost obvious that the conditions after the "if and only if" are satisfied for $r$ if and only if they are satisfied for $r_{\vert G_{K'}}$. Thus automorphy for $r$ should be the same as automorphy for $r_{\vert G_{K'}}$. In other words, automorphy is conjecturally the same as potential automorphy.
Now why is it not possible to prove on the nose that if $r_{G_K'}$ is automorphic, that is associated with an automorphic representation $\pi'$ of $\GL_n({\bf A}_{K'})$, then $r$ itself is automorphic? Because that would mean to "descend" $\pi'$ into s suitable automorphic representation $\pi$ of $\GL_n(\bf{A}_K)$, and this question, called "automorphic descent", is very hard. It is part of the extremely hard Langlands functoriality, and has been solved so far only in the case $K'/K$ solvable (by Langlands for $n=2$, Arthur–Clozel for general $n$).
