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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.

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.

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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.

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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.

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