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Let $G$ be an affine algebraic group over $\mathbb{C}$. It is well known that when working with principal $G$ bundles it is too restrictive to require bundles to be locally trivial in the Zariski topology. Instead $G$-bundles are defined to be locally trivial in the etale topology. The finite map $z \to z^n$ from $\mathbb{C}^\times$ to itself is a $\mu_n$ (= $n$th roots of unity) bundle that is not Zariski locally trivial. Also setting $B = Spec \mathbb{C}[s^\pm,t^\pm]$ then $\{x^2 + s y^2 + t z^2 = 0\} \subset \mathbb{P}^2_B$ is a $\mathbb{P}^1$-bundle that gives rise to a $PGL_2$-bundle that is not Zariski locally trivial.

However for some groups (dubbed special groups) being locally trivial in the etale topology implies locally trivial in the Zariski topology. If $G$ is semisimple then Grothendieck proved the only such special $G$ are products of $SL_n$ and $Sp_{2m}$.

QUESTION

I would like an example of a non Zariski locally trivial principal $G$-bundle for a simple, simply connected group $G$, e.g. $Spin(n)$ or $G_2$.

My motivation is largely out curiosity. For $G$ as in the question, I study the moduli space of $G$-bundles on a curve (and although for a curve over $\mathbb{C}$ a bundle will be Zariski locally trivial it wont be in families) and I would like to have a non Zariski locally trivial bundle in my back pocket.

A vague idea would be to mimic the $PGL_2$ example above by looking at some varying family of $G/P$. You could try this with $G = E_8$ because it is simply connected and adjoint and if you're lucky you might have $Aut(G/P) = G$. I have no idea if this works but even if it does it would be nice if there was a simpler example.

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1 Answer 1

This may not be what you are looking for, but the first example would be over the spectrum of a commutative field $K$. Then a principal $G$-bundle is a $G$-torsor over $K$; any nontrivial such torsor will answer the question. For instance, a $G_2$-torsor is given by a $K$-form of the standard octonion algebra; there is a classical such form over $\mathbb{R}$, called the split-octonion algebra, see here.

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This seems to rely on working over $\mathbb{R}$. I'd like an example over $\mathbb{C}$. –  solbap Dec 17 '13 at 22:41

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