show/hide this revision's text 5 added 552 characters in body

Nick Shepherd-Barron once asked me this question, and I think I can remember what was eventually concluded.

The short answer to the original question is negative for the additive group $\mathbb{G_a}$. Map $SL_2(\mathbb{C})$ to $X=\mathbb{C}^2-(0,0)$ by letting a matrix act on the vector $(1,0)^T$. This realizes the group as a torsor over $X$ for the group of unipotent matrices \[ \begin{pmatrix} 1 & a \\ 0& 1 \end{pmatrix}. \] We can trivialize it over the vectors $(v_1,v_2)^T$ with $v_1\neq 0$ with the section \[ \begin{pmatrix} v_1 & 0 \\ v_2& v_1^{-1} \end{pmatrix},\] while it trivializes on the set $v_2\neq 0$ via the section \[\begin{pmatrix} v_1 & -v_2^{-1} \\ v_2& 0 \end{pmatrix}.\] The transition function on the overlap is easliy computed to be \[\begin{pmatrix} 1 & (v_1v_2)^{-1}\\ 0& 1 \end{pmatrix}.\] This represents the standard non-trivial generator of $H^1(O_X)$, and hence, the bundle is non-trivial. On the other hand, if you could extend it to $\mathbb{C}^2$, it would trivialize, since there are no non-trivial $\mathbb{G}_a$-bundles on an affine variety.

There is indeed a correspondence between principal bundles and tensor functors from representations to vector bundles. But if I recall correctly, the functor is required to be exact. In the case at hand, the extension of vector bundles is the direct image with respect to the inclusion $X\hookrightarrow \mathbb{C}^2$, which fails this.

Added: OK, I see this was just an elaborate way to say: take any principal $\mathbb{G}_a$-bundle corresponding to a non-zero element of $H^1(O_X)$. Note, anyways, that the derived group is trivial in this example. Certainly the statement is false for general connected groups, contrary to some of the comments.

Added, 25, November, 2011:

This question came back to me today while I was thinking about something unrelated. It occurred to me then to point out that for the example above, if we work in the analytic category, we have $$H^1(X, \mathbb{G}_a)\simeq H^1(X, \mathbb{G}_m),$$ via the exponential sequence. On the other hand, $$H^1(\mathbb{C}^2, \mathbb{G}_a)=H^1(\mathbb{C}^2, \mathbb{G}_m)=0.$$

So the desired extension property is false on analytic spaces even for reductive structure groups.
show/hide this revision's text 4 added 352 characters in body

Nick Shepherd-Barron once asked me this question, and I think I can remember what was eventually concluded. I believe the

The short answer to the original question is negative for the additive group $\mathbb{G_a}$. Map $SL_2(\mathbb{C})$ to $X=\mathbb{C}^2-(0,0)$ by letting a matrix act on the vector $(1,0)^T$. This realizes the group as a torsor over $X$ for the group of unipotent matrices \[ \begin{pmatrix} 1 & a \\ 0& 1 \end{pmatrix}. \] We can trivialize it over the vectors $(v_1,v_2)^T$ with $v_1\neq 0$ with the section \[ \begin{pmatrix} v_1 & 0 \\ v_2& v_1^{-1} \end{pmatrix},\] while it trivializes on the set $v_2\neq 0$ via the section \[\begin{pmatrix} v_1 & -v_2^{-1} \\ v_2& 0 \end{pmatrix}.\] The transition function on the overlap is easliy computed to be \[\begin{pmatrix} 1 & (v_1v_2)^{-1}\\ 0& 1 \end{pmatrix}.\] This represents the standard non-trivial generator of $H^1(O_X)$, and hence, the bundle is non-trivial. On the other hand, if you could extend it to $\mathbb{C}^2$, it would trivialize, since there are no non-trivial $\mathbb{G}_a$-bundles on an affine variety.

There is indeed a correspondence between principal bundles and tensor functors from representations to vector bundles. But if I recall correctly, the functor is required to be exact. In the case at hand, the extension of vector bundles is the direct image with respect to the inclusion $X\hookrightarrow \mathbb{C}^2$, which fails this.

Added: OK, I see this was just an elaborate way to say: take any principal $\mathbb{G}_a$-bundle corresponding to a non-zero element of $H^1(O_X)$. Note, anyways, that the derived group is trivial in this example. Certainly the statement is false for general connected groups, contrary to some of the comments.
show/hide this revision's text 3 added 131 characters in body

Nick Shepperd-Barron Shepherd-Barron once asked me this question, and I think I can remember what was eventually concluded. I believe the short answer to the original question is negative for the additive group $G_a$. Consider \mathbb{G_a}$. Map $SL_2(C)$ acting SL_2(\mathbb{C})$ to $X=\mathbb{C}^2-(0,0)$ by letting a matrix act on the vector $X=C^2-(0,0)$, realizing it (1,0)^T$. This realizes the group as a torsor over $X$ for the group of unipotent matrices \[ \begin{pmatrix} 1 & a \\ 0& 1 \end{pmatrix} end{pmatrix}. \] We can trivialize it over the vectors $(v_1,v_2)^T$ with $v_1\neq 0$ with the section \[ \begin{pmatrix} v_1 & 0 \\ v_2& v_1^{-1} \end{pmatrix},\] while it trivializes on the set $v_2\neq 0$ via the section \[\begin{pmatrix} v_1 & -v_2^{-1} \\ v_2& 0 \end{pmatrix}.\] The transition function on the overlap is easliy computed to be \[\begin{pmatrix} 1 & (v_1v_2)^{-1}\\ 0& 1 \end{pmatrix}.\] This represents the standard non-trivial element generator of $H^1(O_X)$, and hence, the bundle is non-trivial. On the other hand, if you could extend it to $C^2$, \mathbb{C}^2$, it would trivialize, since there are no non-trivial $G_a$-bundles \mathbb{G}_a$-bundles on an affine variety, it would trivialize.

It is correct that there

There is indeed a correspondence between principal bundles and tensor functors from representations to vector bundles. But if I recall correctly, the functor is required to be exact. In this the case at hand, the extension of vector bundles is the direct image via with respect to the inclusion $X\hookrightarrow C^2$, \mathbb{C}^2$, which fails this.
show/hide this revision's text 2 Fixed the LaTeX
show/hide this revision's text 1