TITLE: How to understand $Ext(\mathcal{O}_{Y},\mathcal{O}_{Z})$ for subvarieties $Y,Z\subset X$? 
QUESTION [10 upvotes]: By standard homological algebra we know that $Ext(A,B)$ of $R$-modules classifies certain equivalence classes of short exact sequences $0\rightarrow B\rightarrow C \rightarrow A \rightarrow 0$ of $R$-modules, where $R$ is a commutative ring. I now would like to understand this fact in geometry. 

Let $X$ be a variety (or a scheme if you want), how should I understand $Ext(O_{Y},O_{Z})$ for subvarieties $Y,Z\subset X$? Of course it classifies extensions of $O_{Z}$ by $O_{Y}$, but are there any geometric or intuitive way to understand $Ext(O_{Y},O_{Z})$? 
More generally are there any geometric way to understand $Ext(\mathcal{E},\mathcal{F})$ for coherent $O_X$-modules $\mathcal{E},\mathcal{F}$? 

I would appreciate any idea about "seeing" these extensions.

REPLY [4 votes]: Here is some special example.
If j : Y --> X is a closed embedding of smooth varieties over a field then 
$\mathcal{Ext}^{i}_{X}(j_{*}O_{Y}, j_{*}O_{Y}) = \wedge^{i} N_{Y|X}$
where$ N_{Y|X}$ is the normal bundle of Y in X.
In particular, il the local global spectral sequence degenerates, for example if all is affine, 
then
$ Ext^{i}_{X}(j_{*}O_{Y}, j_{*}O_{Y}) = H^{0}(\wedge^{i} N_{Y|X})$
The case i = 1 express the intuition that $ Ext^{1}_{X}(j_{*}O_{Y}, j_{*}O_{Y}) $
is a (infinitesimal) measure of the ability to deform Y in X.
By taking X the product of Y by itself and j the diagonal embedding, 
one obtains a interpretation of $ Ext^{i}_{X}(j_{*}O_{Y}, j_{*}O_{Y}) $ in terms
of (dual of) De Rham cohomology.