Isomorphisms of quantum planes - MathOverflow

Isomorphisms of quantum planes

2012-08-28T14:32:27Z

Let $k$ be a field and $q\in k^{*}$. The quantum plane $k_{q}[x,y]$ is the algebra $k\langle x,y\rangle/\langle xy=qyx \rangle$ (i.e. the quotient of the free non-commutative $k$-algebra on two variables $x$ and $y$ modulo the ideal given).

Question: For $q,r\in k^{*}$ and $q\neq r$, when is $k_{q}[x,y]$ isomorphic (as an algebra) to $k_{r}[x',y']$?

I fully expect this is known but after (what I think is) fairly comprehensive literature searching, including a large proportion of the best-known quantum groups texts, I have been unable to find an answer. A reference would be appreciated just as much as a proof.

Some comments:

I know the (algebra) automorphism group: by work of Alev-Chamarie this is $(k^{*})^2$ unless $q=-1$ (when it is a semi-direct product of the torus with the group of order two generated by the map that interchanges the two variables). Hence I don't need to worry about $q=r$.
I want algebra isomorphisms but information on Hopf algebra maps would be nice too (NB. the Hopf automorphisms for the usual Hopf structure are also those just described)
if $q$ has finite order $N$ in $k^{*}$ and $r$ is of infinite order then the corresponding quantum planes are not isomorphic, as in the first case the centre is non-trivial (generated by $x^{N}$ and $y^{N}$) but in the second the centre is just $k$
if $q$ has order $M$ and $r$ has order $N\neq M$, then the quotients by the centres are both finite-dimensional but of different dimension, hence the quantum planes are not isomorphic
I would be happy to know the answer just for $k=\mathbb{C}$

Answer by eithil for Isomorphisms of quantum planes

2012-08-28T17:16:57Z

This answer feels so glib I'm quite worried it's wrong, but anyway:

Write $D_q$ for the full ring of fractions of $k_q[x,y]$. By Alev-Dumas, "Sur le corps des fractions de certaines algebres quantiques", Corollary 3.11c, we know that for $q$, $r$ non-roots of unity, $D_q \cong D_r$ iff $r = q^{\pm1}$.

It's clear that $k_q[x,y] \cong k_r[x',y'] \Rightarrow D_q \cong D_r$, so we have \[k_q[x,y] \cong k_r[x',y'] \Rightarrow r = q^{\pm1},\] and the converse should also be clear.

Answer by B. Bischof for Isomorphisms of quantum planes

2012-08-31T19:21:12Z

While the above is a good way to see it, I kinda like to use the natural representation of $U_q(sl_2)$ on the quantum planes.

Recall that $U_q(sl_2)\simeq_{\mathfrak{Hopf}}U_p(sl_2)$ iff $q=\pm p^{\pm1}$. Now recall there are faithful Hopf representations $\rho_q:U_q(sl_2)\hookrightarrow End_{\mathbb{C}}(\mathbb{C}_q[x,y])$ and $\rho_p:U_p(sl_2)\hookrightarrow End_{\mathbb{C}}(\mathbb{C}_p[x,y])$ given by $E(1)=0$, $F(1)=0$, $K(1)=1$, $ E(x)=0$, $E(y)=x$, $F(x)=y$, $F(y)=0$, $K(x)=qx$, $K(y)=q^{-1}y$. So the quantum groups are Hopf subalgebras endomorphism rings of the quantum planes. Now if there exists an isomorphism of these quantum planes it would carry one quantum group to the other.

My only concern is that this uses the Hopf structure on the quantum plane.

EDIT: As Mariano points out, I was a bit sloppy and need to think a bit more. I will try to revise this in the next couple days. Sorry for the stupidity.

Answer by linearfish for Isomorphisms of quantum planes

2012-12-27T20:49:42Z

You may want to have a look at this paper: Isomorphisms of some quantum spaces. The techniques rely on the graded structure of the quantum planes and, more generally, quantum affine spaces.