Confusion about D-modules and functors - MathOverflow

Confusion about D-modules and functors

2011-07-02T13:48:28Z

Given a morphism $f:X\rightarrow Y$ between smooth complex varieties, one can define functors from the bounded derived category with holonomic cohomology on $Y$ to the same category on $X$. The easiest one is $Lf^{*}$ which can be obtained by putting a $D$-module structure on the inverse image of $\mathcal O$-modules and deriving it. From this one can get two more functors:

$f^!:=D \circ Lf^{*} \circ D$, where $D$ is the duality functor and
$f^{\dagger}:=Lf^{*}[dim X-dim Y] $

Now my question is, under what conditions are these two isomorphic?

Edit: These notations are bad/confusing/wrong, since they are not compatible with the formalism of six functors. See answers below for better notations.

Answer by David Ben-Zvi for Confusion about D-modules and functors

2011-07-02T14:46:14Z

The two standard D-module pullbacks agree (up to a shift) for smooth morphisms. Bernstein, Borel or Kashiwara are standard references for this.

A couple of comments: First you should restrict to coherent D-modules to get the duality functor (on the derived level). Next I think the notation is a little off from the standard. The easy D-module pullback functor is usually the one denoted with a dagger, while its shifted version is $f^!$, not $f^\ast$ - which is its dual. These conventions are set up so as to have the "formalism of six operations" -- i.e. we have adjoint pairs $(f^\ast,f_\ast)$ and $(f_!,f^!)$ for general morphisms (on holonomic D-modules), related by duality. The two pushforwards agree for proper maps and the two pullbacks agree (after dimension shift) for smooth maps.

Answer by Sam Gunningham for Confusion about D-modules and functors

2011-07-02T15:21:44Z

Just to clear up some notational confusion (there doesn't seem to be any completely standard notation):

In Bernstein's notes:

The "easy" pullback (i.e. the one that coincides with the pullback of the underlying $\mathcal O$-module) is denoted $Lf^\Delta$.

$f^! = Lf^\Delta [dim X - dim Y]$ (right adjoint to $f_!$)

$f^\ast = \mathbb D f^! \mathbb D$ (left adjoint to $f_\ast$)

Note that $f^\ast$ and $f^!$ agree with the corresponding functors for constructable sheaves (not for the underlying $\mathcal O$-modules).

However... In Hotta, Takeuchi, Tanisaki

The easy pullback is denoted $Lf^\ast$ to agree with the $\mathcal O$-module functor.

Berstein's $f^!$ is now called $f^\dagger$.

Bernstein's $f^\ast$ is now called $f^\star$

Ok, now in David Ben-Zvi's answer above (and in many other places)

The easy pullback is $f^\dagger$, and the rest agrees with Berstein's notation.

In my opinion, the most important notational feature to be preserved is that (f^\ast , f_\ast) and $(f_! , f^!)$ form adjoint pairs.

To answer your question

When $f$ is smooth, $f^\ast = \mathbb D f^! \mathbb D = f^! [2(dim Y - dim X)]$, and the easy inverse image functor is self dual (and preserves the t-structure).

One way to think about these different pullbacks is that the easy inverse image preserves the structure sheaf $\mathcal O_Y$. This corresponds to the constant sheaf shifted in perverse degree under the RH correspondence. On the other hand $f^\ast$ preserves the "constant sheaf" whereas $f^!$ preserves the dualizing sheaf (for a smooth complex variety, these correspond to the D-modules $\mathcal O[-n]$ and $\mathcal O[n]$).