14
$\begingroup$

Given a smooth algebraic variety $X$, and an $\mathcal{M}\in \text{Mod}(D_X)$, there is the characteristic variety of $\mathcal{M}$ defined as $$ \text{Char}(\mathcal{M}):= V\left(\sqrt{Ann(\mathcal{M})}\right) \subset T^*X $$ These varieties have a number of nice properties

  1. Their dimension is equal to the dimension of the underlying $D_X$-module
  2. Their dimension is greater than or equal to the dimension of $X$
  3. Behaves well with restriction to open subsets of $X$
  4. They behave well with respect to exact sequences of coherent $D_X$-modules
  5. They are coisotropic subvarieties of $T^*X$
  6. They are lagrangian iff the underlying D-module is holonomic

Unfortunately, it's not clear why these varieties are useful and what their motivation for construction is.

$\endgroup$
  • 1
    $\begingroup$ I don't recall the details (I am not an analyst), but the motivation comes primarily from distribution theory; the characteristic variety of a holonomic D-module (which as you know is cyclic, generated by a distribution) is related to the singular spectrum of the distribution. I would go have a look at the original work of Kashiwara and Saito, it might be enlightening. $\endgroup$ – Ketil Tveiten Sep 8 '16 at 8:15
  • 4
    $\begingroup$ You may also think of it as an invariant of the PDE, for example the classical distinction between elliptic parabolic or hyperbolic PDE can be read from the characteristic variety. $\endgroup$ – Michael Bächtold Sep 8 '16 at 18:12
6
$\begingroup$

Here’s one way to think about them: they tell you how far a $D$-module is from being an integrable connection (i.e. finitely generated over $\mathcal O$). Here’s what I mean: let $M$ be a $D$-module on $X$. Then $M$ is an integrable connection in a neighborhood of a point $x\in X$ if and only if $\operatorname{Char}(M)\cap T^*_x X$ is zero (i.e. is contained in the zero section).

I also want to correct your point number 1. The dimension of a $D$-module is by definition the dimension of its characteristic variety.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.