14
$\begingroup$

We consider potentially non-linear differential equations on the formal one dimensional disc $\Delta$. Such equations are given by expressions $$P(z,f,f',f'',...)=0,$$ where $P$ is an element of the commutative algebra $$\mathcal{D}:=\mathbb{C}[[z]][x_{0},x_{1},...,x_{n},...]$$ of polynomials in infinitely many variables with coefficients in $\mathbb{C}[[z]]$. Note that $\mathcal{D}$ is endowed with a derivation $\delta$ which acts on $z$ as $\partial_{z}$ and satisfies $\delta(x_{i})=x_{i+1}$. The data of the pair $(\mathcal{D},\delta)$ will be called a differential algebra. The element $P\in \mathcal{D}$ generates a differential ideal $\langle P \rangle_{\delta}$ We call a map of differential algebras $$S: \mathcal{D}/\langle P\rangle_{\delta}\rightarrow\mathbb{C}[[z]]$$ a solution of $P$. Note that $\mathbb{C}[[z]]$ is considered as a differential algebra with differential $\partial_{z}$. We remark that $S$ is uniquely determined by the value $S(x_{0})=f$ and that indeed we have $P(z,f,f',f'',...)=0$.

The space of solutions is defined as the pre-sheaf of spaces, $\mathbb{S}(P)$, which takes a derived algebra $B$ to the space of maps of differential algebras, $$\mathcal{D}/\langle P \rangle_{\delta}\otimes B\rightarrow\mathbb{C}[[z]].$$

Remark: In down to earth terms $\mathbb{S}(P)$ is the space cut out (in the derived sense) by the various equations on coefficients implied by the differential equation.

It is not hard to show that in simple cases $\mathbb{S}(P)$ is a classical space, ie there is no derived structure. For example this is the case if $P$ is given by a linear ODE. Less obviously there are also equations for which $\mathbb{S}(P)$ does have some non-trivial derived structure. A nice example is given by taking $P$ to be the minimal algebraic differential equation satisfied by the function $\exp(\frac{1}{\sqrt{z}})$.

Question: Do these higher invariants measure something of interest, ideally something that can be formulated in terms of the classical geometry of the differential equation?

$\endgroup$

0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.