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?



Showing an Ext^2 element is zero
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 $E\_0 \subset E\_{01}$).


