# Does S-equivalence imply a deformation relation?

Let $V$ and $V'$ be two semi-stable vector bundles, on some algebraic curve, that are S-equivalent. My question is whether it follows that $V$ and $V'$ are related by a deformation, i.e. whether they sit in a single extension group (hopefully this is the correct way to phrase this).

In particular, I'm interested in the case where $V$ and $V'$ are vector bundles on an elliptic curve $E$.

I suspect this is true because, on $E$, the direct sum $\mathcal{O}_E\oplus\mathcal{O}_E$ is S-equivalent to the unique non-trivial extension of $\mathcal{O}_E$ by itself, and I believe this is meant to be a very representative case. However I don't have an intuition for the general case, and I'm not sure how to go about a proof.

I expect the answer to your question on an elliptic curve to be yes. Let us consider the case of bundles of rank two and degree zero. A direct sum of two line bundles $\mathscr{L}$ and $\mathscr{M}$ will be semistable if both have degree $0$. For one such sum to be $S$-equivalent to another, the line bundles must be the same, possible with order interchanged. So in this case there is nothing to prove. Now suppose that $V$ is a nonsplit extension of $\mathscr{L}$ by $\mathscr{M}$. For this to happen, the $\text{Ext}$ group with $\mathscr{L}$ and $\mathscr{M}$ must be nonzero, and this can happen only when $\mathscr{L}$ is isomorphic to $\mathscr{M}$. (Think of $H^1(\mathscr{L} \otimes \mathscr{M}^*)$.)
So now, the family of extensions of $\mathscr{L}$ by $\mathscr{M}$ will contain $V$ and in the limit the direct sum. This answers your question in the case of rank two, degree zero. I think you should be able to generalize this argument to show that your question is affirmative for all ranks and degrees, at least in the sense of linking the two bundles by a sequence of extensions. If $V_1$ and $V_2$ are semistable, the first being a direct sum of two indecomposable bundles of rank $3$, and the second a sum of indecomposable bundles of ranks $2$ and $4$, I don't know if you can get them both in a single extension family. Perhaps you can figure this out.