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Let $k$ be a field of characteristic zero, and let $A_1(k)$ be the first Weyl algebra, namely, the associative non-commutative $k$-algebra generated by $x$ and $y$ subject to the relation $yx-xy=1$.

Suppose that the following map $f$ is a $k$-algebra endomorphism of $A_1$: $(x,y) \mapsto (f(x):=p,f(y):=q)$, where $p=uy$ and $q=x+vy$, $u,v \in A_1(k)$, $y$ does not divide $u$, namely, $u \notin A_1y$.

As a $k$-algebra endomorphism of $A_1(k)$, we have $[q,p]=1$; indeed, just apply $f$ to $yx-xy=1$.

Is it true that $f$ is actually an automorphism of $A_1(k)$? In particular, is it true that necessarily $p=- y$?

My partial answer: $1=[q,p]=[x+vy,uy]=[x,uy]+[vy,uy]=-[uy,x]+[vy, uy]$ $=-(u[y,x]+[u,x]y)+[vy,uy]=-(u+[u,x]y)+[vy,uy]=-u-[u,x]y+[vy,uy]$

Denote: $E:=[vy,uy]$. Then, $E=[vy,uy]=v[y,uy]+[v,uy]y=-v[uy,y]-[uy,v]y=$ $-v(u[y,y]+[u,y]y)-[uy,v]y=-v[u,y]y-[uy,v]y=$ $-v[u,y]y-(u[y,v]y+[u,v]y^2)=$ $-v[u,y]y-u[y,v]y-[u,v]y^2$.

So we have $E=-v[u,y]y-u[y,v]y-[u,v]y^2$. Then, $1=-u-[u,x]y-v[u,y]y-u[y,v]y-[u,v]y^2$.

Write $u=a_ny^n+\cdots+a_1y+a_0$, $a_j \in k[x]$, $a_0 \neq 0$ (since we have assumed that $y$ does not divide $u$). We see that $a_0=-1$.

Now, the highest $(0,1)$-term of $-u-[u,x]y-v[u,y]y-u[y,v]y-[u,v]y^2$ should be zero, and by considerations of $(0,1)$-degrees, it equals the $(0,1)$-highest term of $-[u,v]y^2$.

Motivation: Please see this question, in order to understand the motivation for my above question.

Any hints and comments are welcome! (I have also asked the above question in MSE, but have not received any comments yet).

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  • $\begingroup$ If $B=0$ and $A=x$ then $y$ does not divide $A$, but relations are not preserved $\endgroup$
    – Marco Farinati
    Commented Jul 2, 2019 at 21:24
  • $\begingroup$ @MarcoFarinati, thank you for your comment (but $[q,p] \neq 1$, as you mentioned). $\endgroup$
    – user237522
    Commented Jul 2, 2019 at 21:28
  • $\begingroup$ Maybe I didn't understand your question properly. Do you begin with an endomorphism $f$ or you begin with $A$ and $B$ and try to define $f$? You probable know that $f$ Endo implies $f$ auto is an old conjecture $\endgroup$
    – Marco Farinati
    Commented Jul 2, 2019 at 22:03
  • $\begingroup$ @MarcoFarinati, thanks. I begin with a specific $f: (x,y) \mapsto (Ay,x+By)$. Yes, I am familiar with the Dixmier Conjecture. My above question is about a special case. $\endgroup$
    – user237522
    Commented Jul 3, 2019 at 9:41
  • $\begingroup$ did you try the Z-grading $|x|=1$ and $|y|=-1$, introducing $h=yx$, and writing $A$ and $B$ in terms of polynomials on $h$ and powers of $x$ + polynomial on $h$ times powers of $y$? Bavula had a lot of success using this grading $\endgroup$
    – Marco Farinati
    Commented Jul 3, 2019 at 12:41

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