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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 $k=\mathbb{C}$
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Note on derivations of graded rings and classification of differential polynomial rings by Awami, Van den Bergh, and Oystaeyen, Observation 2.1 and subsequent discussion should give you what you want I think. –  B. Bischof Aug 28 '12 at 16:14
    
As far as I remember algebras are Morita equivalent for modula trasform tau->-1/tau q=exp(2 pi tau), most probably you know this just for completness... I had an impression that algebras are not isomorphic for other "q", it also called quantum torus algebra... –  Alexander Chervov Aug 28 '12 at 16:52
    
Alexander, the «quantum torus» is the localization of what Jan calls the quantum plane at $x$ and $y$. –  Mariano Suárez-Alvarez Aug 28 '12 at 17:41
    
@Mariano oops, yes you are right, quantum torus allows x^{-1}, y^{-1} –  Alexander Chervov Aug 28 '12 at 18:27
    
@B. Bischof: this paper of Awami, van den Bergh and Oystaeyen seems to be difficult to get hold of. Do you know a location where I might get it? (I can probably get it on an inter-library loan but would be grateful for faster access.) –  Jan Grabowski Aug 29 '12 at 11:43
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3 Answers 3

up vote 11 down vote accepted

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.

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This is not glib at all. This is just it (for the non-root-of-unity case) –  Mariano Suárez-Alvarez Aug 28 '12 at 17:30
    
@Mariano do you expect it is not true for roots of unity ? –  Alexander Chervov Aug 28 '12 at 18:29
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The same is true, but you need a different argument. Alev and Dumas show that the subgroup generated by $q$ in $k^\times$ is an isomorphism invariant. When $q$ is of infinite order, this subgroup is infinite cyclic and allows you to pick its two generators. If $q$ is a root of $1$, the subgroup has many generators and makes your life more complicated. See arxiv.org/abs/1206.4417 –  Mariano Suárez-Alvarez Aug 28 '12 at 18:37
    
Oh, of course! I'd left the root-of-unity case alone since Jan seemed to have pretty much covered it in his post, but I hadn't considered what would happen if $q$ and $r$ were both primitive $n$-th roots of unity, not inverse to each other. They wouldn't be isomorphic as $k$-algebras, but they would be isomorphic as rings, is that right? –  eithil Aug 29 '12 at 10:38
    
Thank you both for this sequence of comments - they are very helpful. A reference dealing with the root of unity case would be very nice, if one exists. If not, would you mind expanding your last sentence a little, @eithil? –  Jan Grabowski Aug 29 '12 at 11:32
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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.

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Thanks, I saw this already. (By one of those coincidences, it appeared not long after I originally asked the question.) But thanks for posting the link for future readers of the question. –  Jan Grabowski Dec 28 '12 at 9:20
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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.

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How do you know that «Now if there exists an isomorphism of these quantum planes it would carry one quantum group to the other»? –  Mariano Suárez-Alvarez Aug 31 '12 at 21:08
    
(Also, your maps do not give you a copy of the quantum group inside the quantum plane but in the endomorphism ring of the quantum plane (as a vector space) I am almost sure that the GKDim of U_q(sl_2) is 3 while that of the quantum plane is 2, and GKdim is monotone wrt inclusion of algebras) –  Mariano Suárez-Alvarez Aug 31 '12 at 21:20
    
Yes I misspoke about the "copy of Uq". I need to think more about your first comment. I was thinking that this would be the same elements making up the image of the other rep, but I don't know this. I'll think a bit more. –  B. Bischof Aug 31 '12 at 22:26
    
A much simpler argument to show that there are no $U_q$'s inside quantum planes is to simply look at units! –  Mariano Suárez-Alvarez Sep 1 '12 at 1:30
    
Thanks for your comments. I will think about this some more. –  B. Bischof Sep 1 '12 at 2:26
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