I'm not a DD-module person. I'm hoping someone else can give a slightly more insightful explanation. (It looks like YBL gives a clear picture of the holonomic case.)
By definition Ch(M)Ch(M) is the support of the associated graded of MM, for a suitable filtration, so it should lie in the preimage of the support of MM. However, in many interesting cases the inclusion would be strict. If MM is holonomic, Ch(M)Ch(M) is Lagrangian, so it wouldn't coincide with the preimage of supp(M)supp(M). The simplest case where this
happens is when MM is a flat connection, then Ch(M)Ch(M) is the zero section of T∗XT∗X.
Continuation: Perhaps it's worthwhile making this a little more explicit.
The simplest example is X=mathbbAnX=mathbbAn. Then (the global sections of) DXDX
is the Weyl algebra with generators x1,ldots,xn,partial1,ldots,partialnx1,ldots,xn,partial1,ldots,partialn with
commutation relations [xi,partialj]=deltaij[xi,partialj]=deltaij.
If we force these to commute, by passing to the associated graded with respect to
the filtration by order of operators, we obtain the polynomial ring in 2n2n variables
or in other words the coordinate ring of T∗mathbbAnT∗mathbbAn. Any finitely generated DXDX-module MM, carries a (noncanonical) compatible filtration, so can define Ch(M)=Supp(GrM)subsetT∗mathbbAnCh(M)=Supp(GrM)subsetT∗mathbbAn (it is independent of the filtration). Now let
M=D/sumDpartialiM=D/sumDpartiali, which corresponds to mathcalOXmathcalOX. I'll omit the
details, but one can see that
Ch(M)=V(partial1,ldotspartialn)Ch(M)=V(partial1,ldotspartialn) (the zero section), and pi∗supp(M)=T∗mathbbAnpi∗supp(M)=T∗mathbbAn where
pi:T∗mathbbAntomathbbAnpi:T∗mathbbAntomathbbAn.
For a nonholonomic example, take M=DXM=DX. Then Ch(M)=pi−1supp(M)=T∗mathbbAnCh(M)=pi−1supp(M)=T∗mathbbAn.
No comments:
Post a Comment