7

I am trying to develop an intuition about holonomic D-modules and find the literature formidable (I study physics). My question is, given a linear differential operator in n-variables, $x=(x_1,...,x_n)$ (using multi-index notation), $ P(x,D)=\sum_{\alpha}^m P_\alpha(x)D^\alpha$ with polynomial coefficients, $P_\alpha(x)$, what are the holonomic modules that can be associated to this operator?

I know that for $n=1$ every such operator defines a holonomic D-module, so is there a simple algorithm, for example, in $n=2$ that would determine immediately if the operator is holonomic or not?

alphanzo
  • 193

1 Answers1

3

A single PDE in $n > 1$ variables cannot be holonomic. The characteristic variety of a D-module defined by one PDE is just the vanishing locus of its principal symbol, considered as a function on the cotangent bundle. A D-module is holonomic when its characteristic variety is Lagrangian, or equivalently has dimension equal to its codimension.

In this case, the cotangent bundle is $\mathbb{R}^{2n}$ and the principal symbol is $$P(x,y) = \sum_{\alpha}^m P_{\alpha}(x)y^{\alpha},$$ where $y_1,\cdots,y_n$ are new variables. The vanishing locus of $P(x,y)$ has codimension $1$, so for $n > 1$ cannot have (co)dimension $n$, which is the holonomicity condition.

Justin Campbell
  • 7,169
  • Thanks, that answers my question. Follow up: does this mean that the set of $n$ equations $P_i(x,D),\ i=1,...,n$, in $n$-dimensions is always holonomic for $P_i(x,D)$ distinct? I apologize if it's a silly question, it's still very new to me. – alphanzo Feb 11 '14 at 04:49
  • 1
    That's a very natural question, but I think the answer is no. Generally speaking, $n$ distinct equations will have a vanishing locus of codimension at most $n$, as long as we allow complex solutions (otherwise there may be no solutions at all!). However, this bound need not be attained. Counterexamples tend to be slightly complicated but try Googling "complete intersection." – Justin Campbell Feb 11 '14 at 05:16