## Vanishing of Self-Ext groups of vector bundles

Let $E$ be a rank two vector bundle on $\mathbb{P}^n$. Assume that $\text{Ext}^1(E, E)=0$. Will $\text{Ext}^2(E, E)$ be zero? Why? Any geometric explanation (in terms of deformation theory?)?

Edit: As pointing out by Angelo, in the case $n=2$, the answer is no. However, I really want to know when $n\geq 4$.

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To complement Angelo's answer (which you should accept, as it answers your original question):

If $\mathrm H^1(E^\vee \otimes E) = 0$ then $E$ must be homogeneous, see for instance Theorem 3 this paper by Mohan Kumar.

It is well-known that homogeneous vector bundles on $\mathbb P^n$ of rank less than $n$ splits as a direct sum of line bundles (Theorem 3.2.3 of Okonek-Schneider-Spindler). So if $n>2$ the answer to your question is yes.

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 This is interesting: I was not aware of Kumar's result. – Angelo Aug 14 2010 at 5:47 @Angelo: it was pure luck that I knew, professor Kumar told me about this after one of his talks. – Hailong Dao Aug 15 2010 at 1:57

No. When $E$ is a sum of two line bundles on $\mathbb P^2$, then $\mathrm{Ext}^1(E, E) = \mathrm H^1(E^\vee \otimes E) = 0$, but $\mathrm{Ext}^2(E, E) = \mathrm H^2(E^\vee \otimes E)$ is not necessarily 0 (for example, take $E = \mathcal O \oplus \mathcal O(3)$).

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 Thanks for pointing that. I am really interested in the case $n\geq 4$. – Fei YE Aug 13 2010 at 14:45