# ${\rm Ext}^1$ and extensions of line bundles on a curve

I am confused about the following. I know that for two line bundles $L_1, L_2$ on an algebraic curve $C$ the vector space ${\rm Ext}^1(L_1,L_2)$ classifies isomorphism classes of rank two vector bundles on $C$ which are extensions of $L_2$ by $L_1$. My question is, does this mean that there is a "universal" rank two vector bundle $E$ on ${\rm Ext}^1(L_1,L_2) \times C$ ?

I.e. if I let $p_2 : {\rm Ext}^1(L_1,L_2) \times C \to C$ be the projection then does there exist a rank two vector bundle $E$ on ${\rm Ext}^1(L_1,L_2) \times C$ such that $0 \to p_2^*L_1 \to E \to p_2^*L_2 \to 0$ is a short exact sequence and for all $x \in {\rm Ext}^1(L_1,L_2)$ if we restrict the above short exact sequence to $x \times C$ we get the exact sequence $0 \to L_1 \to E_x \to L_2 \to 0$ of vector bundles on $C$ which corresponds to the extension class $x$ ?

• Yes. See chapter 7 of Le Potier's "Lectures on Vector Bundles" for details. Commented Nov 9, 2013 at 2:00
• @JackHuizenga: I don't have the book with me, are there any problems if the bundles are not stable (or simple). A universal extension certainly exists on the stack parameterising Ext, but do you still have it on the coarse space? Commented Nov 9, 2013 at 2:08
• I don't believe there are any such problems. The book certainly doesn't require any other hypotheses. Commented Nov 9, 2013 at 3:25

I wanted to work this out for myself anyways, so here's a summary of the argument in Le Potier.

Suppose $X$ is a projective variety, and $E,G$ are locally free sheaves on $X$. Put $S = {\rm{Ext}}^1(G,E)$, and let ${\bf E},{\bf G}$ be the constant families on $S\times X$, namely ${\bf E} = p_2^\ast E$ and ${\bf G} = p_2^\ast G$. We want to construct a universal extension

$$0\to {\bf E} \to {\bf F} \to {\bf G}\to 0.$$

We can specify $\bf F$ by giving an element of ${\rm Ext}^1({\bf G},{\bf E})$.

What is ${\rm Ext}^1({\bf G},{\bf E})$? Intuitively, we can specify an extension of ${\bf G}$ by ${\bf E}$ by specifying for each $s\in S$ an extension of $G$ by $E$; that is, it should be the space of morphisms $S\to S$; furthermore, if $f:S\to S$ and ${\bf F}(f)$ is the sheaf corresponding to $f$, then, ${\bf F}(f)_s$ is the extension of $G$ by $E$ corresponding to the extension class $f(s)$. More formally, considering the diagram

\begin{array}{ccc} S\times X &\stackrel{p_2}{\to} &X\\ \downarrow {p_1} & & \downarrow p_1'\\ S & \stackrel{p_2'}{\to} & pt\end{array}

we find

\begin{align}{\rm Ext}^1({\bf G},{\bf E})&= H^1( {\mathcal H}om({\bf G},{\bf E})) \\&= H^0(R^1 p_{1*} {\mathcal H}om({\bf G},{\bf E})))\\ &= H^0(p_2'^* R^1 p'_{1*} {\mathcal H}om(G,E))\\ &= H^0(\mathcal O_S \otimes {\rm Ext}^1(G,E)),\end{align}

making use of various identities from Hartshorne III.6-9. The desired universal family corresponds to the distinguished identity morphism $S\to S$.

• If all you know is that $\bf F$ is an extension of $\bf G$ by $\bf E$, then the extension class corresponding to $\bf F$ cannot be determined, since any nonzero multiple of the class determines the same sheaf. Thus the points of ${\rm Ext}^1(G,E)$ are not even in bijective correspondence with the isomorphism classes of sheaves which are extensions. To get a moduli space, you'll need to remember the extensions. If your functor assigns to $S$ the family of all diagrams $0\to {\bf E} \to {\bf F} \to {\bf G} \to 0$ on $X\times S$, I believe you will have a fine moduli space, Commented Nov 9, 2013 at 21:15
• but by all means someone should correct me if I am wrong. (In your comment, I believe you meant to send $S$ to the sheaves on $X\times S$, not $X$.) Commented Nov 9, 2013 at 21:17
• Thanks a lot! The reason why I thought it was not a fine moduli space: if it were we would have that the functor of points of the space ${\rm Ext}^1(G,E)$ would be the functor that assigns to a scheme $Y$ the set of diagrams of the form $0 \to {\bf E} \to {\bf F} \to {\bf G} \to 0$ on $Y \times X$ where ${\bf E} = p_2^* E$ and ${\bf G} = p_2^* G$. But the vector space ${\rm Ext}^1(G,E)$ is an affine scheme and we know that for an affine scheme ${\rm Spec} (A)$ the functor of points looks like $Y \mapsto {\rm Hom}(A,H^0(\mathcal O_Y))$ - looks different! Help me to resolve this contradiction! Commented Nov 10, 2013 at 2:14
• @user42066: I think Jack is absolutely right. Say we are working over a ground field $C$. Let R be a C-algebra. As a functor, R sends any affine Spec A to Hom(Spec A , Spec R), which is just $C-\text{Alg}(R,A)$, ie C-algebra morphisms from R to A. I think the confusion arises when you think $\text{Ext}^1_X(G,E)$ as a vector space. Denote Ext by V. When we say that a vector space V is an affine scheme, we are really considering the scheme $\text{Sym} V^*$ given by the symmetric algebra of the dual vector space. So, the moduli functor determined by V sends (contd) Commented Nov 11, 2013 at 11:27
• sends an affine A to $C-\text{Alg}(\text{Sym}V^*,A)$, which (by the universal property of symmetric algebras) is the same as $\text{Hom}_C(V^*,A) = V \otimes_C A$. If you notice, $V \otimes_C A$ is precisely the last line in Jacks' answer when $S = Spec A$. Commented Nov 11, 2013 at 11:29