Is there an easy proof of the fact that the intermediate image functor respects weights? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T12:23:58Z http://mathoverflow.net/feeds/question/54516 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/54516/is-there-an-easy-proof-of-the-fact-that-the-intermediate-image-functor-respects-w Is there an easy proof of the fact that the intermediate image functor respects weights? Mikhail Bondarko 2011-02-06T13:54:17Z 2011-02-08T16:46:47Z <p>It was proven in BBD (see Corollary 5.3.2) that for an open immersion $j$ the functor $j_{!*}$ preserves weights of mixed sheaves. The proof relies on several previous results; it is especially complicated in the case when $j$ is not affine. Does an easier proof (or a plan of it:)) exist? I would like to have a proof that (mostly) relies on the properties of $j_{!*}$ (and on the 'formal' properties of weights).</p> http://mathoverflow.net/questions/54516/is-there-an-easy-proof-of-the-fact-that-the-intermediate-image-functor-respects-w/54542#54542 Answer by Alex for Is there an easy proof of the fact that the intermediate image functor respects weights? Alex 2011-02-06T18:19:45Z 2011-02-08T16:46:47Z <p>The proof in BBD is not that complicated, and it doesn't matter much whether $j$ is affine or not. It uses the three following facts :</p> <ul> <li><p>If $f$ is a morphism of schemes, then $f_*$ sends a complex of weight $\geq a$ to a complex of weight $\geq a$, and $f_!$ sends a complex of weight $\leq a$ to a complex of weight $\leq a$ (a very natural property of weights; of course that's not so easy to prove for weights of $\ell$-adic complexes, and it is the main result of Deligne's Weil II). Cf BBD 5.1.14.</p></li> <li><p>If $K$ is an $\ell$-adic complex, then $K$ is of weight $\leq a$ (resp. $\geq a$) if and only if, for every $k\in\mathbb{Z}$, the $k$th perverse cohomology sheaf of $K$ (call it <code>${}^pH^k K$</code>) is of weight $\leq a+k$ (resp. $\geq a+k$). Cf BBD 5.4.1. Again, hard to prove, but a natural enough property of weights, and a reason in my opinion why perverse sheaves are so much more natural than constructible sheaves (one out of many).</p></li> <li><p>If $j$ is a locally closed immersion (more generally, a quasi-finite map), then <code>$j_{!*}$</code> is the image of <code>${}^pH^0j_!$</code> in <code>${}^pH^0j_*$</code>. This is the definition of the intermediate extension.</p></li> </ul> <p>Now the result you want is obvious : Take <code>$j:X\rightarrow Y$</code> a quasi-finite morphism. If the perverse sheaf $K$ on $X$ is of weight $\leq a$, then <code>$j_!K$</code> is of weight $\leq a$ (as a complex), so the perverse sheaf <code>${}^pH^0j_!K$</code> is of weight $\leq a$, and so is its quotient <code>$j_{!*}K$</code>. Likewise for weights $\geq a$, using this time $j_*$. </p> <p>Note that you could also define $j_{!*}K$ (for $K$ pure of weight $a$) as the weight $\leq a$ part of $j_*K$, or as the weight $\geq a$ part of $j_!K$. I think it's not too hard to recover the usual properties of $j_{!*}K$ from that definition, but I would have to think more to see how to make it work for mixed (but not pure) perverse sheaves.</p> <p>Edited to add two remarks :</p> <p>(1) I don't think that it is so hard to go from the affine case to the general case. Consider an open embedding $j:U\rightarrow X$, let $i:Y\rightarrow X$ be the complement. Let $\pi:X'\rightarrow X$ be the blowup of $Y$ in $X$, and $j':U\rightarrow X'$ be the inclusion. Then $j'$ is affine, and, for every perverse sheaf $K$ on $U$, <code>$j_{!*}K$</code> is a direct factor of <code>${}^pH^0\pi_*j'_{!*}K$</code>, so the result for <code>$j_{!*}K$</code> follows if you know it for <code>$j'_{!*}K$</code>, without any need of BBD 5.3.1. (You don't need the decomposition theorem to prove my claim. It is an exercise in perverse sheaves to prove that the map <code>${}^pH^0\pi_*j'_!K={}^p H^0j_!K\rightarrow j_{!*}K$</code> factors through a map <code>${}^pH^0\pi_*j'_{!*}K\rightarrow j_{!*}K$</code>. Likewise, or by duality, there is a natural map <code>$j_{!*}K\rightarrow{}^pH^0\pi_*j_{!*}K$</code>. The composition <code>$j_{!*}K\rightarrow{}^pH^0\pi_*j'_{!*}K\rightarrow j_{!*}K$</code> is the identity when restricted to $U$, so it is the identity.)</p> <p>(2) If $K$ is pure, there is a slightly different way to prove what you want (you might be able to do something if $K$ is mixed too, but I didn't try to work it out). Notation : $j$ is an open immersion from $U$ to $X$. First, the problem is local in $X$, so you can assume that $X$ is affine. Then $Y:=X-U$ is defined by a finite number of functions on $X$. By induction over the number of functions necessary to define $Y$, you can reduce to the case where there exists a function <code>$f:X\rightarrow\mathbb{A}^1$</code> such that <code>$Y=f^{-1}(0)$</code>. Now you can use the result of Beilinson-Bernstein (cf "A proof of Jantzen conjectures") that the Jantzen filtration on <code>$j_!K$</code> coincides with (a shift of) the weight filtration if $K$ is pure. The Jantzen filtration on <code>$j_!K$</code> is induced by the monodromy filtration on the maximal extension <code>$\Xi_f K$</code>, and it is an exercise to identify the quotient <code>$j_{!*}K$</code> of <code>$j_!K$</code> with one of the graded pieces of this filtration and to conclude that it has the expected weight. This proof avoids BBD 5.2, but it relies on the article of Beilinson-Bernstein instead; as fat as I can tell, the methods Beilinson-Bernstein use to prove the result that you need are natural extensions of the methods of Weil II, and you have to assume Weil II anyway, so maybe this is slightly more natural.</p>