6
$\begingroup$

In Example 1.1.4 of the book Grobner Deformations of Hypergeometric Differential Equations, it is stated without proof that

$$\partial^2 \in D\cdot \langle x\partial^4, x^3\partial^2 \rangle \tag{$\star$}$$

where $D$ denotes the Weyl algebra over $\mathbb{k}[x]$, and $ D\cdot \langle x\partial^4, x^3\partial^2 \rangle$ denotes the left-ideal generated by the operators $x\partial^4,x^3\partial^2$. Since Example 1.1.4 is at the very beginning of the book, and no proof is given by the authors, I'm presuming that there a simple way to verify this (i.e. without using the machinery of the main text).

My current line of attack:

So far, this is my thinking: since $D$ is a domain, the equation $(\star)$ is equivalent to solving the following linear equation in the non-commutative ring $D$:

$$1=D_1x\partial^2 +D_2 x^3. ~~~~~~~~~~~~~~~~$$

where $D_1,D_2\in D$ are unknowns. I am able to prove (by brute force calculation) that $\text{ord}\,D_2\geq 1$. Obviously, $\text{ord}\, D_2 = \text{ord}\, D_1+2$. The brute-force calculation in the general case quickly spirals out of control.

$\endgroup$
5
$\begingroup$

Following my nose gave the following argument. Writing $I$ be the left ideal generated by $x\partial^2$ and $x^3$ and using $\cdot$ to stress multiplication we get

$$ x^2 \cdot x\partial^2 - \partial^2\cdot x^3 = [ x^3, \partial^2] = -6x^2\partial - 6x\in I$$

So $$\frac{1}{6}x\partial\cdot (-6x^2\partial - 6x) + x^2\cdot x\partial^2 = x[x^2,\partial]\partial - x^2\partial + x[x,\partial] = -3x^2\partial - x\in I$$

Taking a suitable $\mathbb{k}$-linear combinations of these gives $x\in I$ and $x^2\partial\in I$.

Then $$\partial^2\cdot x - 1\cdot x\partial^2 = [\partial^2,x]=2x\partial \in I.$$ Since also $2\partial x\in I$ we conclude $$\partial x- x\partial=[\partial,x]=1\in I$$ as required.

The general strategy at each step is to take two elements in the left ideal with the same principal symbol with respect to the filtration with $F_0=\mathbb{k}$, $F_1=\mathbb{k}\cdot\{1,x,\partial\}$ and $F_n=F_1^n$ and then compute their difference which will live in a lower filtered part. I haven't read the book but I'd imagine that this idea is at the heart of computations throughout.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.