According to Chapter 4 of Beilinson, Bernstein, and Deligne's "Faisceaux Pervers" (Asterisque 100, 1980) the inverse image $Rf^*$ with respect to a finite morphism $f$ is right t-exact with respect to the perverse t-structure. Is there an example that shows that it is not necessarily t-exact? Is there a class of finite morphisms other than the etale morphisms where it is t-exact?
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2 Answers
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By Noether normalisation any affine variety admits a finite map $f$ to affine space. Hence it is enough to find an example of an affine variety for which the constant sheaf is not perverse. (In this case $f^* \mathbb{Q}_{\mathbb{A}^n}[n]$ will provide a counter-example.)
Now it is easy to find examples where the constant sheaf is not perverse. For example any affine cone $C_X$ over a smooth projective variety $X$ provides a counter-example as long as the cohomology of $C_X \setminus \{ 0 \}$ doesn't look like that of a sphere. So I guess the cone over an elliptic curve would do it.
Edit (...not quite:) see Daniel's comment below.
answered Dec 12, 2014 at 12:07
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4$\begingroup$ "as long as the cohomology of $C_X \setminus \{0\}$ doesn't look like that of a sphere": that's not true (here you are negating being IC, rather than being perverse). The cone over an elliptic curve does not work, the constant sheaf is a perverse $!$ extension. Actually the constant is perverse for any complete intersection. However the strategy is nice. A correct example would be the cone over the closure of the minimal nilpotent orbit in $SL_3$. Then the constant sheaf has perverse cohomology in degree $-1$. $\endgroup$ Commented Jul 15, 2015 at 9:17
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1$\begingroup$ Of course you are right... I won't change the above as your example is correct. $\endgroup$ Commented Nov 3, 2015 at 16:50
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My idea was to consider $f \colon X \to Y$ of dimension $d$ such that $a_Y^\ast \mathbf Q[d] \cong IC_Y$ but $a_X^\ast \mathbf Q[d] \not\cong IC_X$, where $a$ denotes the map to a point and all functors are derived. Then $a_X^\ast \mathbf Q[d] = f^\ast IC_Y$ can not be perverse, so $f^\ast$ is not t-exact.
(Incorrect example removed)
For a concrete example let $X$ be two copies of $\mathbb A^2$ glued at a point, let $f\colon X \to Y \cong \mathbb A^2$ be the quotient by the involution switching the two copies. If it were true that $IC_X \cong a_X^\ast \mathbf Q[2]$ then the same should be also true for $\mathbf P^2$ with two points identified, as perversity is a local property. But the cohomology of $\mathbf P^2$ with two points identified does not have Poincaré duality.
answered Dec 11, 2014 at 8:07
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$\begingroup$ Thank you very much! I am confused because I thought that the shifted constant sheaf on a locally complete intersection is perverse. If two lines glued at a point means the union of coordinate axes in the coordinate plane then it is a locally complete intersection, so I am getting a contradiction. Do you mean something else by two lines glued at a point? $\endgroup$ Commented Dec 11, 2014 at 15:03
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$\begingroup$ Sorry, I just wanted to say that I found the fact that the shifted constant sheaf on a locally compete intersection is perverse here: mathoverflow.net/questions/76186/… $\endgroup$ Commented Dec 11, 2014 at 15:10
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$\begingroup$ You're absolutely right, I managed to confuse myself when writing the answer. I changed the example, now it should be OK. $\endgroup$ Commented Dec 11, 2014 at 17:47