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If $i:Z\hookrightarrow X$ is a closed embedding of codimension $c$, then $$i^*k_X\ =\ k_Z , \ \ \ i^!k_X\ \stackrel{(\star)}{=}\ k_Z[2c]$$ where $(\star)$ is true when $i$ is in addition regular. Here $k$ denotes the unit D module/$\ell$-adic sheaf.

Now consider replacing $k_X$ by $\text{IC}_X$ (and assume $i$ plays nicely with the Whitney stratifications). Is there anything that can be said about $i^*\text{IC}_X$ and $i^!\text{IC}_X$, apart from them being perverse sheaves?

I'm even interested in the simple case where $Z$ and $U=X\setminus Z$ are smooth, giving a two-term stratification of $X$.

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    $\begingroup$ It is not true that they are perverse sheaves (think about i = inclusion of a point). This will be true when i is a "normally non-singular inclusion"). See Goresky-MacPherson, Kashiwara-Shapira and google. This is in some sense exactly the opposite to the situation that you are interested in. If you think through your "simple case" in the last paragraph you will see that i^*IC can have perverse cohomology in many degrees, even when Z is a point. (As always, thinking about a cone singularity is helpful.) Woof woof. $\endgroup$ Commented Nov 17, 2020 at 19:30
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    $\begingroup$ @GeordieWilliamson Thank you, I think the theorem in section 5.4 Goresky-MacPherson's ''Intersection Homology II'' is what I'm looking for! However, I don't understand what you mean by it being the opposite situation: surely $i:Z\hookrightarrow X$ being regular is the algebraic-geometry version of normally non-singular inclusion, so the aforementioned topological theorem probably also gives $i^*\text{IC}_X=\text{IC}_Z[2c],i^!\text{IC}_X=\text{IC}_Z$ in algebraic geometry? $\endgroup$
    – Pulcinella
    Commented Nov 18, 2020 at 12:44
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    $\begingroup$ Yes, section 5.4 is perfect. In the second paragraph we read: "For example, suppose $X$ is a Whitney stratified subset of some manifold $M$, and $Y=H \cap X$ where $H$ is a smooth submanifold of M which is transverse to each stratum of X". Thus you want a stratification for which $IC_X$ is constructible to be transverse to $Z$. This is the opposite of when $Z$ is actually a stratum. Again, I urge you to think about a cone: in general no $Z$ passing through the cone point will satisfy that $IC_X$ restricted to $Z$ is $IC_Z$. $\endgroup$ Commented Nov 18, 2020 at 23:57

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