Some questions on vanishing of Ext sheaves

Question

Let $X$ be a complex manifold of dimension $3$ and $\mathcal{E}$ be a coherent sheaf such that $\dim supp(\mathcal{E})=1$. In this situation, I would like to know why we have the Ext sheaf $\mathcal{Ext}^1(\mathcal{E},\mathcal{O})=0$.

Moreover, is it true in general that $\mathcal{Ext}^i(\mathcal{E},\mathcal{O})=0$ for $i<\dim(X)-\dim supp(\mathcal{E})$?

I have poor intuition about Ext sheaves and I would appreciate it if someone could give a geometric explanation of this kind of vanishing of Ext sheaves.

Answer 1

Here is a somewhat geometric way to think about it.

Suppose that X is an equidimensional analytic space without embedded points and E is coherent with support of dimension less than dimension of X. Then it is clear that $\mathcal{Ext}^0(\mathcal{E}, \mathcal{O}) = 0$ because this is roughly what it means not to have embedded points.

Now suppose that the support of $\mathcal{E}$ has dimension 2 less than that of X and that X is a normal analytic space. Then we can find a local nonzero function $f$ such that $f$ acts as zero on $\mathcal{E}$. Consider the short exact sequence $0 \to \mathcal{O} \to \mathcal{O} \to \mathcal{O}/f\mathcal{O} \to 0$ and apply the long exact sequence of $\mathcal{Ext}^*(\mathcal{E}, -)$. Then we see that to show that $\mathcal{Ext}^1(\mathcal{E}, \mathcal{O}) = 0$ it suffices to show that $\mathcal{Ext}^0(\mathcal{E}, \mathcal{O}/f\mathcal{O}) = 0$ (hint: use $f$ acts as zero on $\mathcal{E}$). This follows because the hypersurface defined by $f$ has no embedded points as $X$ was assumed normal and we can use the result from the previous paragraph. In fact, we could have gotten away with assuming juyst that $X$ is $(S_2)$ and equidimensional.

To do this argument in general one works with analytic equidimensional CM spaces which works well as an equidimensional hypersurface in a CM space is again CM. Of course a complex manifold is CM analytic space. Hope this helps.

Answer 2

Assume that $X$ is smooth and projective and let $L$ be an ample line bundle. Note that by Serre duality $Ext^i(E,L^N) = Ext^{\dim X-i}(L^N,E(K_X))^*$ and this is zero for $i< \dim X - \dim supp(E)$. On the other hand, if $F_i := \mathcal{Ext}^i(E,O)$, then for $N \gg 0$ we have $H^{>0}(F_i\otimes L^N) = 0$, and $F_i\otimes L^N$ is globally generated for all $i$. This means that $Ext^i(E,L^N) = H^0(\mathcal{Ext}^i(E,O)\otimes L^N))$ and it is nonzero as soon as $F_i$ is nonzero. In particular, you have the required vanishing of $\mathcal{Ext}$.