Question regarding a statement in `A proof of Jantzen conjectures' 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.
 A: 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.
A: 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.
