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 http://www.math.harvard.edu/~gaitsgde/grad_2009/).

The starting assumptions of the Corollary are:

`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 ...'

The first line of the proof says:

`Clearly either $Y_1 \subset Y_2$ or $Y_2 \subset Y_1$ (otherwise $Ext^1 = 0$)'

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.

Presumably I am making a stupid error here and both of the unmixed groups above do vanish? Any comments would be appreciated.

  • $\begingroup$ Maybe one should also assume that $M_i$ are simple? $\endgroup$ Sep 4, 2012 at 18:01
  • $\begingroup$ Mikhail: Beilinson and Bernstein certainly don't do so. But to start with I am happy assuming that both $M_1, M_2$ are of the form $IC(U,constant sheaf)$, with the natural mixed structure. Heck, as a simplification even assume that $M_1$ and $M_2$ are constant sheaves (shifted ofcourse). However, ideally, the answer to my question should be some general nonsense that applies to both the $\ell$-adic and mixed Hodge settings (I mean if it is `clearly', then it really should be some basic observation that I am missing). However, the paper is written with mixed being as in the $\ell$-adic setting. $\endgroup$ Sep 4, 2012 at 19:25

2 Answers 2


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. (EDIT: I should have said this doesn't work if you use *-restriction in both cases. The spectral sequence mentioned below that it does if you use *-restriction for one, and !-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.)

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 Koszul duality patterns....

EDIT: 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 $Y_1\subset Y_2$.

MORE EDIT: 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.

  • $\begingroup$ You are right about my mistake (hopefully I didn't make a second mistake below), but I still don't see how to get vanishing of $Ext^1_{mixed}$. $\endgroup$ Sep 4, 2012 at 6:20
  • $\begingroup$ I don't fully understand. I get up to the part that the map must factor through the perverse truncation, but don't get the part about why the image must be surjective? I have an exact sequence: $0\to H^{-1}(i_*i^*M)\to H^0(j_!j^*M_2)\to M\to H^0(i_*i^*M)\to 0$ where $i$ is the complementary inclusion to $j$ and everything is perverse cohomology. Now you seem to be claiming that $H^0(i_*i^*M)=0$? Sorry, I just don't understand, could you elaborate on your argument? $\endgroup$ Sep 4, 2012 at 15:21
  • $\begingroup$ You're making this too hard; this is just a basic observation about abelian categories. If the map from $M$ isn't surjective, then $M$ is just the direct sum of $M_1$ and its image, which I assumed wasn't the case. $\endgroup$
    – Ben Webster
    Sep 5, 2012 at 12:40
  • $\begingroup$ In response to `More Edit'. By your argument there would be no such thing as a non-trivial local system on $\mathbb{C}^*$ with unipotent monodromy. In other words, take the abelian category $\mathbb{C}[t, t^{-1}]-mod$ and consider the representation $M$ given by the Jordan block $\begin{matrix} 1 & 1 \\ 0 & 1\end{matrix}$. It fits into a sequence of the form $M_1 \to M \to M_1$. It has a subobject isomorphic to $M_1$, but $M$ is certainly not $M_1\oplus M_1$. I believe this also addresses your comment. Or did I just misunderstand everything? $\endgroup$ Sep 6, 2012 at 7:18
  • $\begingroup$ You used the wrong hypotheses: we already assumed that $M_1$ and $M_2$ don't have the same support, so your example isn't an example. $\endgroup$
    – Ben Webster
    Sep 6, 2012 at 14:40

Too long to leave as a comment.

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

$Ext^1(M_1, M_2) = Ext^1(i_{1*}i_{1}^*M_1, i_{2*}i_2^*M_2) = Ext^1(i_2^*i_{1*}i_1^*M_1, i_2^*M_2)$

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:

$Ext^1(i_2^*i_{1*}i_1^*M_1, i_2^*M_2) = Ext^1(s_*r^*M_1, i_2^*M_2) = Ext^1(r^*M_1, s^!i_2^*M_2)$

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.

  • $\begingroup$ Ok, I can `sanely' deal with $s^!i_2^*$, since $i_2^*M_2 = i_2^!M_2$, but none of this helps with the original question. $\endgroup$ Sep 4, 2012 at 6:57
  • $\begingroup$ Of course it helps. You're done. Just apply what the vanishing conditions on stalks of IC sheaves. $\endgroup$
    – Ben Webster
    Sep 4, 2012 at 13:37
  • $\begingroup$ Ben, I dont understand this comment either. Care to elaborate? Sorry I am being slow. $\endgroup$ Sep 4, 2012 at 15:24

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