First I would like to apologize to ask many questions in a few days.

I would like to check that the perverse $t$-structure $(^pD^{\leq 0},\ ^pD^{\geq 0})$ is indeed a $t$-structure.

The second axiom is that $Hom(F,G) = 0$ if $F \in \ ^pD^{\leq 0}$ and $G \in \ ^pD^{\geq 1}$. However, in "D-modules, perverse sheaves and representation theory" the authors instead just show that for $j<0$, $H^jR\mathscr Hom(F',F'') = 0$ for $F' \in \ ^pD^{\leq 0}$ and $F'' \in \ ^pD^{\geq 0}$. I am not sure why it's enough.

I understand that to see that a complex is zero iff its cohomology vanishes, but I don't understand why it's enough to check it for $j<0$. $F \in \ ^pD^{\leq 0}$ implies that $H^j(F) = 0$ for $j >0$ and if I'm not mistaken $G \in \ ^pD^{\geq 0 }$ implies that $H^j(G) = 0$ for $j < - n = \dim X$.

If what I wrote is correct, it seems to implies that $R \mathscr Hom(F,G[k]) = 0$ if $k < -\dim X$. This gives that $R\mathscr Hom(F,G)$ is concentrated in $[-\dim X, +\infty]$. Is it correct ? If yes why $R \mathscr Hom(F,G)$ is concentrated in negative degrees ?


Note that $Hom(F,G)=H^0R\Gamma(R\mathscr{H}om(F,G))$. Now your proposition says that if $F\in {}^pD^{\leq 0}$ and $G\in{}^pD^{\geq 0}$ then $R\mathscr{H}om(F,G)$ is concentrated in degree $\geq 0$. But because $\Gamma$ is left exact, $R\Gamma(R\mathscr{H}om(F,G))$ is also concentrated in degree $\geq 0$.

Now if $G\in{}^pD^{\geq 1}$ instead, apply the proposition with $G[1]$, which gives that $R\Gamma(R\mathscr{H}om(F,G[1]))$ is concentrated in degree $\geq 0$, so $R\Gamma(R\mathscr{H}om(F,G))$ is concentrated in degree $\geq 1$. It follows that $Hom(F,G)=H^0R\Gamma(R\mathscr{H}om(F,G))=0$.

  • $\begingroup$ Many thanks for your very clear explanation. $\endgroup$ – student May 22 '18 at 21:28

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