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One definition of regular D-modules on affine space is that a D-module is regular if it has a filtration compatible with the order filtration on differential operators whose associated graded is reduced as a coherent sheaf, that is, its annihilator is a radical ideal.

But, differential operators on affine space have a lot of filtrations. For example, if I look at $\mathbb{A}^n$ and choose $n$ positive integers $a_i$, I can define a filtration of the following form: we let $x_i$ lie in $D^{-a_i}$ and $\partial_i$ in $D^{1+a_i}$ and extend the filtration to the rest of the differential operators in the obvious way. Not all D-modules have good filtrations with respect to this filtration, but some do (for example, any homogeneous with respect to the action of $\mathbb{C}^*$ with weights $a_i$).

Does the condition of having a filtration compatible with this one whose associated graded is reduced as a coherent sheaf, that is, its annihilator is a radical ideal have an interesting consequences? Most importantly, does it imply regularity in the usual sense given above?

For that matter, has this filtration been studied by anyone? Is there any literature on it?

I'll note, if I let the $a_i$'s be negative, regularity definitely isn't preserved by this change of filtration: there are regular D-modules whose Fourier transforms aren't regular.

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Ben, I am confused about your definition of regularity. If you take the D-module generated by $e^x$ on ${\mathbb A}^1$ it obviously has a filtration such that the associated graded is the structure sheaf of the zero section in the cotangent bundle and it is not regular. What am I missing? –  Alexander Braverman Sep 20 '11 at 22:49
    
Do you mean that the $D$-module is regular on ${\mathbb A}^n$ without infinity? In this case I would be interested to see the proof of your original statement. In any case if you take your $D$-module to be $exp(1/x)$ on ${\mathbb A}^1$ and set $deg(x)=-1$ and $deg(d)=2$ then you can define a filtration $M_i$ to be generated over functions by all $p(x)exp(1/x)$ where $p(x)$ is a polynomial in $x^{-1}$ of degree $\leq i$ and I think it gives a counterexample. –  Alexander Braverman Sep 20 '11 at 23:23
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