Author: Eiko

Tags: Mathematics, connections, differential equations

Time: 2024-09-14 20:34:38 - 2025-01-22 13:07:52 (UTC)

Reviewing Connection

Let $\pi :X\to S$ be an $S$-scheme and $\mathcal{E}$ a quasi-coherent ${\mathcal{O}}_X$-module. An $S$-connection on $\mathcal{E}$ is a ${\pi }^{-1}{\mathcal{O}}_S$-linear map

$$\mathrm{\nabla }:\mathcal{E}\to {\mathrm{\Omega }}_{X/S}^1{\otimes }_{{\mathcal{O}}_S}\mathcal{E}$$

Such that for $f\in {\mathcal{O}}_X,e\in \mathcal{E}$, we have

$$\mathrm{\nabla }(fe)=df\cdot e+f\cdot \mathrm{\nabla }e$$

Choosing frame sections to (locally) trivialize connection

For locally free sheaves, locally one can choose an isomorphism $\phi :{\mathcal{O}}^n\to \mathcal{E}$ to trivialize it. This is the same as choosing $n$ sections $e_1,\dots ,e_n$ locally so that they form a basis $\mathcal{E}=⨁{\mathcal{O}}_X{e}_i$. It allows us to associate any section of $\mathcal{E}$ into $n$ coordinate functions $(f_i)$, and allows us to operate the connection on coordinate functions,

We obtain the coordinate representation of the connection

$${\mathrm{\nabla }}^{\prime }:{\mathcal{O}}^n\to {\mathrm{\Omega }}_{X/S}^1{\otimes }_{{\mathcal{O}}_S}{\mathcal{O}}^n=({\mathrm{\Omega }}_{X/S}^1{)}^n$$

which is explicitly

$$\begin{array}{rl}{\mathrm{\nabla }}^{\prime }(f^1,\dots ,f^n{)}^t& =\mathrm{\nabla }(f^i{e}_i)\\ & =d{f}^i\otimes e_i+f^i\mathrm{\nabla }{e}_i\\ & =(d{f}^i+f^j{\omega }_j^i)\otimes e_i\\ & =d(f^1,\dots ,f^n{)}^t+\mathrm{\Lambda }\cdot (f^1,\dots ,f^n{)}^t,\end{array}$$

where $\mathrm{\Lambda }=({\omega }_i^j)$ is the matrix of 1-forms of the $i$-th component of $\mathrm{\nabla }{e}_j$.

Matrix Decomposition

So in general we have a decomposition for any locally free connection $\mathrm{\nabla }$,

$${\mathrm{\nabla }}^{\prime }=d^{\oplus n}+\mathrm{\Lambda }\in \mathrm{Hom}({\mathcal{O}}_X^n,({\mathrm{\Omega }}_{X/S}^1{)}^n).$$

Flat sections in coordinate

A flat section $\mathrm{\nabla }s=0$ when viewed in coordinate says

$$\mathrm{\nabla }(s^i{e}_i)=(d{s}^i+s^j{\omega }_j^i)\otimes e_i=0\iff ds=-\mathrm{\Lambda }s.$$

where $s=(s^1,\dots ,s^n{)}^t$ is the coordinate representation of $s$ and $ds+\mathrm{\Lambda }s=0$ is a system of linear differential equations.

Example

Consider the family of elliptic curves $\pi :E\to S=\mathrm{Spec}(k[t])$ given by the Weierstrass equation

$$y^2=x^3+tx+1.$$

Choose $\omega =\frac{dx}{y},x\omega$ as a basis section of ${\mathcal{H}}_{dR}^1(E/S)$ which is a sheaf on $S$, then there is a $k$-linear Gauss-Manin connection defined with ${\mathcal{H}}_{dR}^1(E/S)$ on the space $S\to k$ by

$$\mathrm{\nabla }:{\mathcal{H}}_{dR}^1(E/S)\to {\mathrm{\Omega }}_{S/k}^1{\otimes }_k{\mathcal{H}}_{dR}^1(E/S)$$

given by (using $2y\mathrm{d}y=3x^2\mathrm{d}x+t\mathrm{d}x+x\mathrm{d}t$)

$$\begin{array}{rl}\mathrm{\nabla }\omega & =d\frac{dx}{y}=-\frac{1}{y^2}dy\wedge dx\\ & =\frac{x}{2y^3}dx\wedge dt\\ & =(...\text{ performing reduction algorithms})\\ & =(\frac{2t^2}{27+4t^3}\omega +\frac{-3t}{27+4t^3}x\omega +dg)\wedge dt.\end{array}$$

Combining with another similar computation on $\mathrm{\nabla }(x\omega )$, we obtain the connection matrix

$$\mathrm{\Lambda }=\frac{1}{\mathrm{\Delta }}\left(\begin{array}{cc}-2t^2& 3t\\ 9& 2t^2\end{array}\right)\mathrm{d}t$$

where $\mathrm{\Delta }=-4t^3-27$.

Cyclic Vector Theorem

In the case $S=\mathrm{Spec}(k[t])$ there is essentially only one tangent direction, so we can divide everything in ${\mathrm{\Omega }}_{S/k}^1$ by $dt$ to obtain a derivation ${\partial }_t:k[t]\to k[t]$.

In this special case a connection can be simplified into a differential operator on the $k[t]$-module $\mathcal{E}={\mathcal{H}}_{dR}^1(E/S)$, making it a differential module $(\mathcal{E},D)$ over the differential ring $(k[t],{\partial }_t)$, or equivalently a $D_S=k[t,{\partial }_t]$-module.

The fact that its fibres / sections are finite dimensional vector space means, it has a cyclic vector by the cyclic vector theorem (over the differential ring $k(t)$), which will enable us to get a single differential equation describing all flat sections.

And as we’ve seen above, we will need to invert some functions (for example $\mathrm{\Delta }$) and restrict to an open subset of $S$ for these things to work.

Abstraction

Let’s consider now a two dimensional differential module $M$ generated by $e_1,e_2$ over the differential ring $k(t)$, or equivalently a two dimensional $k(t)[{\partial }_t]$-module with connection matrix

$$\mathrm{\Lambda }={((D{e}_j{)}_i)}_{1\le i,j\le 2}=\frac{1}{\mathrm{\Delta }}\left(\begin{array}{cc}-2t^2& 3t\\ 9& 2t^2\end{array}\right).$$

Our goal will be trying to find a cyclic vector. By the dense generation result, most vectors should work, for example let’s try $e_1$, for it to be cyclic it suffices that $e_1$ and $D{e}_1$ are linearly independent over $k(t)$. We have

$$D{e}_1=-\frac{2t^2}{\mathrm{\Delta }}{e}_1+\frac{9}{\mathrm{\Delta }}{e}_2$$

and this is clearly true, so $e_1$ is a cyclic vector. Now differentiate again we have

$$\begin{array}{rl}{D}^2(e_1)& ={\partial }_t(-\frac{2t^2}{\mathrm{\Delta }})e_1+(-\frac{2t^2}{\mathrm{\Delta }})D(e_1)+{\partial }_t\left(\frac{9}{\mathrm{\Delta }}\right)e_2+\left(\frac{9}{\mathrm{\Delta }}\right)D(e_2)\end{array}$$

Simple Example

We consider the following simple example as an instructive case

$$\mathrm{\Lambda }=\left(\begin{array}{cc}1& 0\\ t& 1\end{array}\right)$$

i.e. we have

$$D{e}_1=e_1+t{e}_2,D{e}_2=e_2.$$

Clearly $e_1$ is a cyclic vector here, by computing $D^2{e}_1$ we have

$$D^2{e}_1=e_1+(1+2t)e_2.$$

This gives us a linear dependence relation

$$(1+t)e_1-(1+2t)D{e}_1+t{D}^2{e}_1=0.$$

Now the generation map gives an coimage-image isomorphism

$$k(t)[D]=D_S\to M,Q\mapsto Q{e}_1$$

$$\Rightarrow M\cong D_S/D_{S}P,P=(1+t)-(1+2t)D+t{D}^2.$$

But there is one thing that is unsatisfying, the quotient we obtained is of the form $D_S/D_{S}P$, if we write it in terms of its canonical basis $e_1,\dots ,D^{n-1}{e}_1$, we have a left cyclic matrix

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

where $a_i$ are the coefficients of (the monic version of) $P$. Under this basis the equations for flat sections $(s_0,\dots ,s_{n-1})$ are

$$s_0^{\prime }=a_0{s}_{n-1},s_i^{\prime }=s_{i-1}+a_i{s}_{n-1},i=1,\dots ,n-1,$$

which is not very clear. We would like to have a ‘right cyclic’ matrix that looks like the transpose of ${\mathrm{\Lambda }}^{\prime }$,

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

for which the equations for flat sections $(s_0,\dots ,s_{n-1})$ are simply

$$s_{n-1}^{\prime }+a_{n-1}{s}_{n-1}+\cdots +a_0{s}_0=0,s_i^{\prime }=s_{i+1},i=0,\dots ,n-2.$$

Cyclic Vector Theorem-Like Method

I’m still thinking about the formal meaning of the following intuitive method, it should be related to something about the cyclic vector theorem / D-module theory.

Consider a system which comes from a flatness equation $\mathrm{\nabla }(f_i{e}_i)=(d{f}_i+f_j{\omega }_i^j)e_i=0$, if we are in dimension $1$ we can divide by a local parameter, denote ${\mathrm{\Lambda }}_1=-{\mathrm{\Lambda }}_{\mathrm{\nabla }}/\mathrm{d}x$ and ${\mathrm{\Lambda }}_0=I_n$, so

$$f^{\prime }={\mathrm{\Lambda }}_{1}f$$

keep differentiation we have

$$f^{(2)}={\mathrm{\Lambda }}_1^{\prime }f+{\mathrm{\Lambda }}_1{f}^{\prime }={\mathrm{\Lambda }}_{2}f,{\mathrm{\Lambda }}_2:={\mathrm{\Lambda }}_1^{\prime }+{\mathrm{\Lambda }}_1^2$$ $$f^{(n+1)}=({\mathrm{\Lambda }}_n^{\prime }+{\mathrm{\Lambda }}_n{\mathrm{\Lambda }}_1)f,{\mathrm{\Lambda }}_{n+1}={\mathrm{\Lambda }}_n^{\prime }+{\mathrm{\Lambda }}_n{\mathrm{\Lambda }}_1.$$

If we are only interested in $f_1$, we can extract all the first lines and pack them into another matrix

$$\left(\begin{array}{c}{f}_1^{(n)}\\ f_1^{(n-1)}\\ ⋮\\ f_1\end{array}\right)=\left(\begin{array}{c}({\mathrm{\Lambda }}_n{)}_1\\ ({\mathrm{\Lambda }}_{n-1}{)}_1\\ ⋮\\ ({\mathrm{\Lambda }}_0{)}_1\end{array}\right)f$$

From here we can deduce a linear dependency relation between $f_1^{(n)},\dots ,f_1$ in terms of function entries of (the first rows of) ${\mathrm{\Lambda }}_n,\dots ,{\mathrm{\Lambda }}_0$ by using adjugate matrix, i.e.

$$a_n{f}^{(n)}+\cdots +a_{0}f=0$$

where

$$a_i=(-1{)}^{i}det\left(\begin{array}{c}({\mathrm{\Lambda }}_n{)}_1\\ ⋮\\ ({\mathrm{\Lambda }}_{\hat{i}}{)}_1\\ ⋮\\ ({\mathrm{\Lambda }}_0{)}_1\end{array}\right).$$

Formal Meaning?

I think it has to do with some kind of dual module.