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I am reading about perverse sheaves from the notes of Cataldo and Migliorini http://www.ams.org/journals/bull/2009-46-04/S0273-0979-09-01260-9/S0273-0979-09-01260-9.pdf

In page 553 example 2.2.2 they say: If $Y$ is the projective cone over a nonsingular curve $C$ of genus $G$ then the cohomology groups are $\mathbb{Q}, 0, \mathbb{Q}, \mathbb{Q}^{2g}, \mathbb{Q}$, while the intersection cohomology groups are $\mathbb{Q}, \mathbb{Q}^{2g}, \mathbb{Q}, \mathbb{Q}^{2g}, \mathbb{Q}.$

Can someone explicitly explain these two computations, and maybe other basic computations of intersection cohomology? For some reason I can't seem to understand why these are so.

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  • $\begingroup$ de Cataldo and Migliorini Example 1.3.3 gives a reference to Example 2.2.1 in Borel's intersection cohomology book for this computation. $\endgroup$
    – Jonathan Frink
    Commented Nov 24, 2018 at 20:19
  • $\begingroup$ A cheap way to compute cohomology of the projective cone is to use that in the Grothendieck ring of varieties $[Y] = 1 + \mathbf{L}[C]$ which gives the correct Poincare polynomial. $\endgroup$
    – Evgeny Shinder
    Commented Aug 23, 2021 at 22:30
  • $\begingroup$ For the intersection homology one can use the definition using cycles (this is a bit cumbersome), the open-closed exact sequence (removing the vertex of the cone) or the exact sequence involving the resolution of $Y$ (which is a $\mathbb{P}^1$-bundle over $C$). $\endgroup$
    – Evgeny Shinder
    Commented Aug 23, 2021 at 22:34

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It helps to know the hard Lefschetz theorem, and in particular the concept of primitive homology. Consider the statement: The intersection of a projective cone on a projective variety $X$ has the same primitive homology as the base $X$.

This statement immediately implies the result you quote from Cataldo and Migliorini, once you recognize the primitive homology of the curve $C$.

I'm not a technical expert in intersection homology, and don't recall seeing a proof of it. So I can't give a reference, or even with second-hand authority state that I know it to be true. But true or not, I hope it helps you understand the result you quoted.

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