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:

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 $0$.

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.

- 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.