How to understand $Ext(\mathcal{O}_{Y},\mathcal{O}_{Z})$ for subvarieties $Y,Z\subset X$? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-21T21:29:00Z http://mathoverflow.net/feeds/question/108456 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/108456/how-to-understand-ext-mathcalo-y-mathcalo-z-for-subvarieties-y-z-su How to understand $Ext(\mathcal{O}_{Y},\mathcal{O}_{Z})$ for subvarieties $Y,Z\subset X$? Muon 2012-09-30T09:26:51Z 2012-10-01T10:13:27Z <p>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. </p> <ol> <li>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 $Ext(O_{Y},O_{Z})$? </li> <li>More generally are there any geometric way to understand $Ext(\mathcal{E},\mathcal{F})$ for coherent $O_X$-modules $\mathcal{E},\mathcal{F}$? </li> </ol> <p>I would appreciate any idea about "seeing" these extensions. </p> http://mathoverflow.net/questions/108456/how-to-understand-ext-mathcalo-y-mathcalo-z-for-subvarieties-y-z-su/108459#108459 Answer by Donu Arapura for How to understand $Ext(\mathcal{O}_{Y},\mathcal{O}_{Z})$ for subvarieties $Y,Z\subset X$? Donu Arapura 2012-09-30T11:46:19Z 2012-09-30T13:49:56Z <p>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 $$0\to O_X\to \mathcal{L}\otimes L^{-1}\to O_X\to 0\in Ext^1(O_X,O_X)$$</p> <hr> <p>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.</p> http://mathoverflow.net/questions/108456/how-to-understand-ext-mathcalo-y-mathcalo-z-for-subvarieties-y-z-su/108487#108487 Answer by David Ben-Zvi for How to understand $Ext(\mathcal{O}_{Y},\mathcal{O}_{Z})$ for subvarieties $Y,Z\subset X$? David Ben-Zvi 2012-09-30T18:39:53Z 2012-09-30T19:00:24Z <p>[Edit: this answer mistakenly addresses the Ext complex rather than the first Ext group, which is what is being asked about..]</p> <p>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).</p> <p>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:</p> <p>$$Hom(i_*O_Y, j_*O_Z)= Hom(O_Y, i^!j_*O_Z)= \Gamma_Y(i^!j_*O_Z)=\Gamma_Y(p_*q^! O_Z)=\Gamma_W(q^!O_Z).$$ </p> <p>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 $O_Y$.)</p> http://mathoverflow.net/questions/108456/how-to-understand-ext-mathcalo-y-mathcalo-z-for-subvarieties-y-z-su/108537#108537 Answer by unknown (google) for How to understand $Ext(\mathcal{O}_{Y},\mathcal{O}_{Z})$ for subvarieties $Y,Z\subset X$? unknown (google) 2012-10-01T10:13:27Z 2012-10-01T10:13:27Z <p>Here is some special example.</p> <p>If j : Y --> X is a closed embedding of smooth varieties over a field then <code>$\mathcal{Ext}^{i}_{X}(j_{*}O_{Y}, j_{*}O_{Y}) = \wedge^{i} N_{Y|X}$</code> where<code>$N_{Y|X}$</code> is the normal bundle of Y in X. In particular, il the local global spectral sequence degenerates, for example if all is affine, then <code>$Ext^{i}_{X}(j_{*}O_{Y}, j_{*}O_{Y}) = H^{0}(\wedge^{i} N_{Y|X})$</code> The case i = 1 express the intuition that <code>$Ext^{1}_{X}(j_{*}O_{Y}, j_{*}O_{Y})$</code> is a (infinitesimal) measure of the ability to deform Y in X.</p> <p>By taking X the product of Y by itself and j the diagonal embedding, one obtains a interpretation of <code>$Ext^{i}_{X}(j_{*}O_{Y}, j_{*}O_{Y})$</code> in terms of (dual of) De Rham cohomology.</p>