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.

  1. 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 $Ext(O_{Y},O_{Z})$?
  2. 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.

  • 4
    $\begingroup$ A minor remark : the exact sequences you've written are classified by $Ext(B,A)$ (extensions of $B$-quotient by $A$-kernel) not $Ext(A,B)$. $\endgroup$
    – BS.
    Sep 30, 2012 at 10:15
  • $\begingroup$ @BS Thank you for pointing out the mistake. I corrected the typo. $\endgroup$
    – user2013
    Sep 30, 2012 at 18:48

3 Answers 3


In general, unlike say singular homology, constructions in homological algebra are usually not all that geometric. Nevertheless they do sometimes translate into geometry. Let's consider the simplest case, where both subvarieties are $X$ itself. Then you are asking what does $Ext^1(O_X,O_X)\cong H^1(X,O_X)$ mean geometrically? One answer is that it is the tangent space to the Picard variety (or more correctly scheme) at some given line bundle $L$. In fact, the $Ext$ interpretation gives an nice way to see this. Let $k$ denote the ground field. Then a tangent vector to $Pic(X)$ at $L$ is just a first order deformation of $L$, i.e. a line bundle $\mathcal{L}$ on $\mathcal{X}= X\times Spec\ k[\epsilon]/(\epsilon^2)$ which restricts to $L$ on $X$ viewed as subscheme of $\mathcal{X}$. It follows that there is an exact sequence $$0\to \epsilon O_{\mathcal{X}}\otimes\mathcal{L}\to \mathcal{L}\to L\to 0$$ which can be identified with $$0\to L\to \mathcal{L}\to L\to 0$$ This in turns gives an extension $$0\to O_X\to \mathcal{L}\otimes L^{-1}\to O_X\to 0\in Ext^1(O_X,O_X)$$

Perhaps I can do one more case, which may be more typical. Say $Y$ and $Z$ are curves on a smooth surface $X$ with no common components. Then relevant $Ext$ group is easy compute using some standard tools from homological algebra. The so called local to global spectral sequence implies that $$Ext^1(O_Y, O_Z) \cong \bigoplus_{p\in Y\cap Z} Ext^1_{O_{X,p}}(O_{Y,p}, O_{Z,p})$$ The latter is just the sum $$\bigoplus_{p\in Y\cap Z}O_{X,p}/(f_p,g_p)$$ where $f_p$ and $g_p$ are the local equations for these curves. I admit however that I haven't thought about what this means in terms of extensions.

  • $\begingroup$ I really like this answer. Do you think you could add some insight for the case of Ext between actual _sub_varieties? If I understood it correctly, what you did here was computing the tangent space of the space Pic(X) at L, and then you remembered that there is an automorphism of Pic(X) given by tensoring by $L^{-1}$ which gives you the tangent space at the identity $O_X$. Is there any heuristic for the more general case of subvarieties? $\endgroup$
    – Jacob Bell
    Sep 30, 2012 at 12:18
  • $\begingroup$ Right, I'm just translating to the origin. I added another case which might help. $\endgroup$ Sep 30, 2012 at 13:50
  • $\begingroup$ Thank you for the enlightening answer. $Ext^1(L,L)\cong Ext^1(O,O)\cong H^1(X,O)$ for any line bundle $L$, so the extension represent the tangent space of the Picard group scheme at $L$. That makes sense. I don't know how to understand $\sum_{p}O_{X,p}/(f_{p},g_{p})$ either. $\endgroup$
    – user2013
    Sep 30, 2012 at 21:08

[Edit: this answer mistakenly addresses the Ext complex rather than the first Ext group, which is what is being asked about..]

If you replace Ext by Tor, you are defining functions on the derived intersection of $Y$ and $Z$, let's call it $W=Y\times_X Z$ (this is the fiber product in the world of derived schemes, which by definition corresponds to this Tor).

One can then describe the (first) Ext group you asked about as almost functions on the derived intersection of Y and Z (for closed subvarieties). Let's write $i,j$ for the inclusions of $Y$ and $Z$ (yes I know $j$ is usually reserved for open immersions, but anyway..). Then the derived Hom (Ext complex) $Hom(i_*O_Y, j_* O_Z)$ can be calculated by Grothendieck duality for the proper map $i$ as $Hom(O_Y, i^!j_* O_Z)$. Let $p:W\to Y,q:W\to Z$ denote the two projections. Then by (derived) base change we can identify $i^!j_*O_Z$ with $p_*q^! O_Z$. So finally we summarize:

$$Hom(i_*O_Y, j_*O_Z)= Hom(O_Y, i^!j_*O_Z)= \Gamma_Y(i^!j_*O_Z)=\Gamma_Y(p_*q^! O_Z)=\Gamma_W(q^!O_Z).$$

So we find global sections (again I mean the derived version, i.e., cohomology) of the restriction with supports of $O_Z$ to the (derived) intersection -- i.e., local cohomology of the intersection with coefficients in functions on $Z$. Maybe there's a nice way to say this more intuitively. Note for yet another variant that if you consider $Ext(O_Y,\omega_Z)$ (replace structure sheaf of $Z$ by its dualizing sheaf) then the same calculation yields $\Gamma_W(\omega_W)$, global "top-forms" on the derived intersection. (I guess the Tor version is again an Ext with $\omega_Y$ instead of $O_Y$.)

  • $\begingroup$ @David Thank you for the answer from the viewpoint of derived AG. I need to read your answer carefully as I am not familiar with derived AG. Also, sorry for confusing you by simply writing $Ext$ for $Ext^1$. $\endgroup$
    – user2013
    Sep 30, 2012 at 21:14

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.