Question regarding a statement in A proof of Jantzen conjectures' - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T15:49:59Z http://mathoverflow.net/feeds/question/106299 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/106299/question-regarding-a-statement-in-a-proof-of-jantzen-conjectures Question regarding a statement in A proof of Jantzen conjectures' Reladenine Vakalwe 2012-09-04T04:00:38Z 2012-09-05T12:38:19Z <p>So I am trying to understand a statement in the proof of Corollary 5.2.3 in A proof of Jantzen conjectures' (a copy of the paper can be found at <a href="http://www.math.harvard.edu/~gaitsgde/grad_2009/" rel="nofollow">http://www.math.harvard.edu/~gaitsgde/grad_2009/</a>).</p> <p>The starting assumptions of the Corollary are:</p> <p>Let $M_1, M_2$ be pure perverse sheaves of weights $w_1, w_2$ that are both $*$- and $!$-pointwise pure. Suppose that $Ext^1_{mixed}(M_1, M_2)\neq 0$. Then ...'</p> <p>The first line of the proof says:</p> <p>Clearly either $Y_1 \subset Y_2$ or $Y_2 \subset Y_1$ (otherwise $Ext^1 = 0$)'</p> <p>Here $Y_i = Supp(M_i)$. This statement confuses me. The ordinary ( = non-mixed) $Ext^1$ group should be the extensions between the restrictions of $M_1$ and $M_2$ to the intersection $Y_1 \cap Y_2$. I don't see why this should vanish if either $Y_1$ isn't contained in $Y_2$ or vice versa. Anyway, even if the unmixed group does vanish, without vanishing of the unmixed $Hom$ group I don't see how I would get $Ext^1_{mixed}$ vanishes. </p> <p>Presumably I am making a stupid error here and both of the unmixed groups above do vanish? Any comments would be appreciated.</p> http://mathoverflow.net/questions/106299/question-regarding-a-statement-in-a-proof-of-jantzen-conjectures/106301#106301 Answer by Ben Webster for Question regarding a statement in A proof of Jantzen conjectures' Ben Webster 2012-09-04T05:05:23Z 2012-09-05T12:38:19Z <p>The problem is your statement: "The ordinary ( = non-mixed) $Ext^1$ group should be the extensions between the restrictions of $M_1$ and $M_2$ to the intersection $Y_1\cap Y_2$." This is absolutely not true in any generality I can think of. (<strong>EDIT:</strong> I should have said this doesn't work if you use <code>*</code>-restriction in both cases. The spectral sequence mentioned below that it does if you use <code>*</code>-restriction for one, and <code>!</code>-restriction for the other. The spectral sequence below applied to $X\supset Y_1\cup Y_2 \supset Y_1\cap Y_2$ shows this.) </p> <p>The way one actually can calculate Ext groups using the geometry of the intersections is the spectral sequence given at the start of section 3.4 of <a href="http://www.ams.org/journals/jams/1996-9-02/S0894-0347-96-00192-0/" rel="nofollow">Koszul duality patterns...</a>. </p> <p><strong>EDIT</strong>: Perhaps it's better to think of it this way: assume $Y_2\not\subset Y_1$ and let $j$ be the inclusion of $Y_2\setminus Y_1\cap Y_2$. Then any non-trivial extention $M_1 \to M \to M_2$ has a map $j_!j^!M_2\to M$ induced by the isomorphism $j^!M_2\cong j^!M$ which factors through the perverse truncation $H^p_0(j_!j^!M_2)$. As a map of perverse sheaves $H^p_0(j_!j^!M_2)\to M$ must be surjective since otherwise its image would split the exact sequence. Thus, $M_1$ must be a composition factor of $H^p_0(j_!j^!M_2)$ and so $Y_1\subset Y_2$.</p> <p><strong>MORE EDIT</strong>: If $M$ is a nontrivial extension of $M_1 \to M \to M_2$, it cannot have a subobject isomorphic to $M_2$. If it did, then then we would have an isomorphism $M\cong M_1\oplus M_2$ using the inclusion of $M_1$ we had before, and the inclusion of $M_2$ we just assumed existed.</p> http://mathoverflow.net/questions/106299/question-regarding-a-statement-in-a-proof-of-jantzen-conjectures/106304#106304 Answer by Reladenine Vakalwe for Question regarding a statement in `A proof of Jantzen conjectures' Reladenine Vakalwe 2012-09-04T06:18:10Z 2012-09-04T06:18:10Z <p>Too long to leave as a comment.</p> <p>Ben: Let $i_k$, $k=1,2$ be the closed inclusions $Y_i \to X$ (where $X$ is my ambient space). Then $M_k = i_{k*}i_k^*M_k$. So</p> <p>$Ext^1(M_1, M_2) = Ext^1(i_{1*}i_{1}^*M_1, i_{2*}i_2^*M_2) = Ext^1(i_2^*i_{1*}i_1^*M_1, i_2^*M_2)$</p> <p>Now let $r\colon Y_1\cap Y_2 \to X$ and $s\colon Y_1\cap Y_2 \to Y_2$ be the closed inclusions and we get:</p> <p>$Ext^1(i_2^*i_{1*}i_1^*M_1, i_2^*M_2) = Ext^1(s_*r^*M_1, i_2^*M_2) = Ext^1(r^*M_1, s^!i_2^*M_2)$</p> <p>Ah, so my initial error was to magically convert the $s^!$ to $s^*$, but it still reduces the computation of the Ext group to the intersection (or did I do something screwy again?). On the other hand, I don't see how to sanely deal with the $s^!i_2^*$. Regardless, I still don't see why the mixed Ext group in the original question is vanishing.</p>