Author: Eiko

Tags: D-modules, cyclic modules, differential modules, dual modules

Time: 2024-09-22 20:20:56 - 2024-09-23 00:20:56 (UTC)

Motivation

When dealing with connections and cyclic $D$-modules, there are two possible ways you can write the connection as cyclic matrices,

$${\mathrm{\Lambda }}^{\prime }=\left(\begin{array}{cccc}& & & -a_0\\ 1& & & -a_1\\ & \ddots & & ⋮\\ & & 1& -a_{n-1}\end{array}\right)\text{and}{\mathrm{\Lambda }}^{″}=\left(\begin{array}{cccc}& 1& & \\ & & \ddots & \\ & & & 1\\ -a_0& -a_1& \dots & -a_{n-1}\end{array}\right).$$

We will call the left hand side a left cyclic matrix and right hand side a right cyclic matrix.

• For a cyclic $D$-module of the form $M=D/DP$, where $P={\partial }^n+a_{n-1}{\partial }^{n-1}+\cdots +a_0$, if we take the basis $e_0=1,e_1=\partial ,\dots ,e_{n-1}={\partial }^{n-1}$, then the matrix of left multiplication $L_{\partial }:M\to M$ on $M$ with respect to this basis is ${\mathrm{\Lambda }}^{\prime }$.

• The question is, what module structure does ${\mathrm{\Lambda }}^{″}$ correspond to?

Dual Differential Module

For a finite $K$-dimensional differential module $M$ generated by $e_1,\dots ,e_n$, we can form the dual module $M^{\ast }={\mathrm{Hom}}_K(M,K)$ consisting of $K$-linear functions on $M$. The module has a dual basis given by $e_1^{\ast },\dots ,e_n^{\ast }$, where $e_i^{\ast }(e_j)={\delta }_{ij}$.

In general, for any two left $D$-modules $M,N$, the module ${\mathrm{Hom}}_O(M,N)$ has a natural left $D$-module structure given by, for any derivation $\theta \in D$ and $f\in {\mathrm{Hom}}_O(M,N)$,

$$(\theta f)(m):=\theta (f(m))-f(\theta m).$$

This means $M^{\ast }={\mathrm{Hom}}_K(M,K)$ has left $D$-module structure since $K$ and $M$ are left $D$-modules.

In terms of matrices

Let’s consider that $M$ has a connection matrix $\mathrm{\Lambda }$, i.e. $D{e}_j=\sum _i{\mathrm{\Lambda }}_{ij}{e}_i$ which is the matrix of certain chosen derivation $\partial$ under the given basis.

Then the matrix element ${\mathrm{\Lambda }}_{ij}^{\prime }$ of the connection on $M^{\ast }$ is given by

$$\begin{array}{rl}(D{e}_j^{\ast })(e_i)& =D(e_j^{\ast }{e}_i)-e_j^{\ast }(D{e}_i)\\ & =D({\delta }_{ij})-e_j^{\ast }({\mathrm{\Lambda }}_{ji}{e}_j)\\ & =-{\mathrm{\Lambda }}_{ji}\end{array}$$

So actually the matrix of a dual connection is the negated transpose ${\mathrm{\Lambda }}^{\prime }=-{\mathrm{\Lambda }}^T$.