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Yet another question of the form 'How to apply the decomposition theorem?' The example that I am considering ought to have a simple answer, but I'm getting confused and I would appreciate if someone could point out where I'm going astray. The confusing point can be stated briefly, at the end of observation 3. But I'll give an explanation of what I do understand, hoping that this will be helpful to other people and make clear what I am missing.

Let $Y$ be a quasi-projective 3-fold with a unique singular point $0 \in Y$ and suppose that the blow-up at $0$ is a resolution $p: X \rightarrow Y$ and the exceptional divisor $p^{-1}(0)=S$ is a smooth projective surface.The goal is to understand the summands of $Rp_{\ast}IC_{X} \simeq \bigoplus_{i} {}^{\mathfrak{p}}\mathcal{H}^{i}(Rp_{\ast}IC_{X})[-i]$, where the decomposition is into perverse cohomology sheaves given by the decomposition theorem.

Observations:

  1. By base change, the fact that $p$ is an isomorphism over the open set $U=Y\setminus 0$ implies that $Rp_{\ast} IC_{X}$ restricted to $U$ is just $IC_{U}$, so ${}^{\mathfrak{p}}\mathcal{H}^{0}(Rp_{\ast}IC_{X})\simeq IC_{Y} \oplus E$ for some sky-scraper $E$ at $0$. Further, the other perverse cohomology sheaves ${}^{\mathfrak{p}}\mathcal{H}^{i}(Rp_{\ast}IC_{X})[-i]$ must be supported at $0$ and so consist of shifted sky-scrapers. Thus we just have to understand the stalk of $Rp_{\ast} IC_{X}$ at $0$.

  2. By base change and the fact that $p^{-1}(0)=S$ a smooth projective surface, we have that the stalk of $Rp_{\ast} IC_{X}$ at $0$ is $H^{0}(S,\mathbb{Q})[3]\oplus H^{1}(S,\mathbb{Q})[2]\oplus H^{2}(S,\mathbb{Q})[1]\oplus H^{3}(S,\mathbb{Q})\oplus H^{4}(S,\mathbb{Q})[-1]$. Since $IC_{Y}$ is concentrated in degrees $-3,-2,-1$ with respect to the standard t-structure, $E \simeq H^{3}(S,\mathbb{Q})\otimes \mathbb{Q}_{0}$ and

${}^{\mathfrak{p}}\mathcal{H}^{1}(Rp_{\ast}IC_{X})[-1] \simeq H^{4}(S,\mathbb{Q})\otimes \mathbb{Q}_{0}[-1]$.

Further, there are no higher perverse cohomology sheaves, for degree reasons.

  1. By Verdier duality and self-duality of $Rp_{\ast}IC_{X}$, the only other perverse cohomology sheaf in the decomposition is ${}^{\mathfrak{p}}\mathcal{H}^{-1}(Rp_{\ast}IC_{X})[1]$, which must be dual to

${}^{\mathfrak{p}}\mathcal{H}^{1}(Rp_{\ast}IC_{X})[-1] \simeq H^{4}(S,\mathbb{Q})\otimes \mathbb{Q}_{0}[-1]$.

I would think that it should look like $H^{0}(S,\mathbb{Q})\otimes \mathbb{Q}_{0}[1]$, but then the degree in which $H^{0}(S)$ sits is off by $-2$ from what appears in observation 2. The only sky-scraper sitting in degree $-1$ is $H^{2}(S)$, but that isn't dual to $H^{4}(S)$ in general.

Presumably I am having a problem with applying Verdier duality, but I don't see where the problem lies. Any comments are very welcome.

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2 Answers 2

up vote 3 down vote accepted

In this example we have $p : X \to Y$ and we may assume, wlog, that $X$ is isomorphic to the total space of the normal bundle to the surface, and $p$ is the contraction of the zero section.

Then, by the Deligne construction, $IC(Y) = \tau_{\le -1} j_* \mathbb{Q}[3]$, where $j : Y^0 \hookrightarrow Y$ is the inclusion of the smooth locus (which is isomorphic to $X^0$ the complement of the zero section in $X$).

In order to work this out, we can use the Leray-Hirsch spectral sequence

$E_2^{p,q} = H^p(S) \otimes H^q(\mathbb{C}^*) \Rightarrow H^{p+q}(X^0)$

this converges at $E_3$ and we get that the degree 0, 1 and 2 parts of the cohomology of $X^0$ is given by the primitive classes in $H^i(S)$ for $i = 0, 1, 2$. Note that this is everything in degrees 0 and 1, but in degree two the primitive classes form a codimension one subspace $P_2 \subset H^2(S)$.

The Deligne construction above, gives us that $IC(Y)_0 = H^0(S)[3] \oplus H^1(S)[2] \oplus P_2[1]$.

(This is a general fact: whenever you take a cone over a smooth projective variety, the stalk of the intersection cohomology complex at 0 is given by the primitive classes with respect to the ample bundle used to embed the variety. This follows by exactly the same arguments given above.)

Then the decomposition theorem gives

$p_* \mathbb{Q} = \mathbb{Q}_0[1] \oplus ( IC(Y) \oplus H^3(S) ) \oplus H^4(S)[-1]$.

EDIT: fixed typos pointed out by Chris.

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Thanks! This was the second example that I've tried to work out and I hadn't realized that the primitive cohomology would be relevant, but now it makes sense. –  Chris Brav Mar 15 '10 at 23:18
    
Just for the record, I think that there are a few minor typos in Geordie's answer. First, we should have $IC(Y)_{0}=H^{0}(S)[3]\oplus H^{1}(S)[2]\oplus P_{2}[1]$ and second, $p_{*}\mathbb{Q}=\mathbb{Q}_{0}[1]\oplus (IC(Y)\oplus H^{3}(S))\oplus H^{4}(S)[-1]$. Also, I think $j: X^{0} \hookrightarrow X$ is meant to be $j: X^{0} \hookrightarrow Y$. –  Chris Brav Mar 16 '10 at 1:52
    
Thanks! I fixed everything you pointed out. –  Geordie Williamson Mar 16 '10 at 7:40

I'm pretty sure your intuition is just leading you astray here. The dual to H^4 in the stalk really is H^2. I think the pairing is just pullback your classes to the blowup and intersect them there. The kernel of this is the top degree of the IC sheaf, and the one bit left over is the negative perverse cohomology. Though, this is mostly irresponsible speculation based on my intuition...

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Thank you for the reply, but I don't follow. Pullback classes of what? Let me elaborate on why I'm surprised that the dual should be H^2. If I have the closed embedding of a point $i: 0 \rightarrow Y$, then $i_{\ast}=i_{!}$ and so $D_{Y}i_{*} \simeq i_{!}D_{0} s\imeq i_{*}D_{0}$, where $D$ is the Verdier duality functor. This means that if I want to compute the Verdier dual of $i_{*}H^{4}(S)[-1]$, then I can first compute the dual of $H^{4}(S)[-1]$ on the point to get $H^{0}(S)[1]$ (by usual Poincare duality) and then pushforward. What am I missing? –  Chris Brav Mar 15 '10 at 21:31

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