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This question may sound a bit odd, but basically I want to ask if the following algebras have any meaning in some area of math: $$ A=\mathbb{C} \langle X,Y,Z \rangle /(XY+i YX,YZ+i ZY, XZ+ZX)\\ B= A / (Y^2Z-i(4X^3-XZ^2)).$$ I got this algebra from considering the elliptic curve $E \leftrightarrow y^2 = 4x^3-x$, which has a non-trivial automorphism (a morphism as well as a homomorphism) of order 4 defined by $\phi: (x,y) \mapsto (-x,i y)$. Let $O(E) = \mathbb{C}[X,Y,Z]/(Y^2Z - 4X^3+XZ^2)$ be the homogenous coordinate ring of $E$ ($\mathbb{Z}$-graded) and let $$\mathbb{C}(E) = \left\{ \frac{f}{g} \, \middle| \, f,g \in O(E)_i \text{ for some } i \in \mathbb{N}, g \neq 0 \right\}.$$ Next, I took the twisted coordinate ring $\mathbb{C}(E)[t,\phi]$, given by $t f = \phi(f)t \forall f \in \mathbb{C}(E)$. Since this is not a commutative algebra, I was looking for the kernel of the function $$ \psi: \mathbb{C} \langle X,Y,Z \rangle \rightarrow \mathbb{C}(E)[t,\phi],\\\psi(X) = xt,\psi(Y)=yt,\psi(Z) = 1t,$$ where $x = \frac{X}{Z}, y= \frac{Y}{Z}$ and this is where I found my relations for $A$ and $B$ and I was wondering if these algebras pop up somewhere.

Any help would be appreciated.

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Please use \langle and \rangle instead of inequality signs. You will love the improved spacing. – Rasmus Sep 3 '12 at 10:40
$A$ reminds me of the quantum plane – Julian Kuelshammer Sep 3 '12 at 11:12
Based on what I've heard, "twisted" stuff is used in quantum theory. That's vague because I don't understand it myself. I got the idea while reading Goodearl and Warfield's blurb on quantum algebras in their noncommutative noetherian rings book. – rschwieb Sep 3 '12 at 12:46
up vote 1 down vote accepted

This is probably more of a comment than an answer, but I don't have enough points to comment yet.

Algebra $A$ is a quantum 3-space. It's an example of an Artin-Schelter regular algebra of dimension 3. Moreover, it is a Zhang twist of the polynomial ring $R=\mathbb{C}[x,y,z]$. Hence, the categories of graded rights modules $Gr-A$ and $Gr-R$ are equivalent.

The algebra $B$ does not look familiar to me, but you should probably take a look at the paper "Some algebras associated to automorphisms of elliptic curves" by Artin, Tate, and Van den Bergh. My guess is that the relations defining $B$ condense into two cubic relations and $B$ is again an AS-regular algebra.

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