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If we have an extension of bundles $0 \to E \to F \to G \to 0$ on $X$, then to show that this is the zero element in $Ext^1_X(G,E)$, we need to show that this sequence splits. To produce a splitting is a concrete question in homological algebra. I understand that in general if we have an extension $ 0 \to E \to E_1 \to \cdots \to E_k \to G \to 0$, short of showing $Ext^k_X(G,E)=0$, there is no simple recipe for showing that this particular extension is zero. But are there any known examples where such things have been shown? I am basically interested only in the case $k=2$. Is there a simple recipe in this case?

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up vote 9 down vote accepted

There is a simple recipe to show that a product of two $Ext^1$ is zero (and it is clear that any $Ext^2$ can be represented as such a product). Namely, let $0 \to E_0 \to E_{01} \to E_1 \to 0$ and $0 \to E_1 \to E_{12} \to E_2 \to 0$ are two exact triples. The product of the corresponding elements $e_{01} \in Ext^1(E_1,E_0)$ and $e_{12} \in Ext^1(E_2,E_1)$, that is $e_{01}\circ e_{12} \in Ext^2(E_2,E_0)$, is zero, if and only if there is an object $E$ with a filtration of length 3 on it $F_0E \subset F_1E \subset F_2E = E$ such that the first triple is isomorphic to $0 \to F_0E \to F_1E \to F_1E/F_0E \to 0$ and the second triple is isomorphic to $0 \to F_1E/F_0E \to F_2E/F_0E \to F_2E/F_1E \to 0$.

Here is a sketch of a proof. Consider the long exact sequence obtained by applying $Hom(E_2,-)$ to $0 \to E_0 \to E_{01} \to E_1 \to 0$: $$ \dots \to Ext^1(E_2,E_{01}) \to Ext^1(E_2,E_1) \to Ext^2(E_2,E_0) \to \dots $$ It is clear that the class $e_{12}$ is mapped by the second arrow to the class $e_{01}\circ e_{12} = 0$, hence there exists a class $e \in Ext^1(E_2,E_{01})$ which is mapped to $e_{12}$ by the first arrow. Let $E$ be the corresponding extension. Then we have an exact sequence $0 \to E_{01} \to E \to E_2 \to 0$ and a morphism from this to the sequence $0 \to E_1 \to E_{12} \to E_2 \to 0$ such that the map $E_{01} \to E_1$ is the given one and the map $E_2 \to E_2$ is the identity (you can write down a commutative diagram here). It follows that the induced map $E \to E_{12}$ is a surjection and its kernel is $E_0$. This is precisely what was claimed ($F_1E$ is the image of $E_{01}$ in $E$ and $F_0E$ is the image of $E_0 \subset E_{01}$).

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I want to know the proof of this statement, could you help me, Sasha? –  TmobiusX Jun 7 '10 at 1:09
Dear TmobiusX, I added a sketch of proof to the answer. –  Sasha Jun 7 '10 at 19:30
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