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Background

The Quantum Torus

Let $q$ be an arbitrary complex number, and define (the algebra of) the quantum torus to be $$T_q:=\mathbb{C}\langle x^{\pm 1},y^{\pm 1}\rangle/xy-qyx$$ For $q=1$, this is the commutative ring of functions on the torus $\mathbb{C}^\times\times \mathbb{C}^\times$; hence, for general $q$, this is regarded as a quantization of the torus.

Hall Algebras

Consider a small abelian category $A$, with the property that $Hom_A(M,N)$ and $Ext^i_A(M,N)$ are always finite sets for any $M,N\in A$ and $i\in \mathbb{Z}$. Let $\overline{A}$ denote the set of isomorphism classes in $A$, and let $$H(A)=\oplus_{[M]\in \overline{A}}\mathbb{C}[M]$$ denote the complex vector space spanned by $\overline{A}$. Endow $H(A)$ with a multiplication by the formula $$ [M]\cdot [N]=\sqrt{\langle [M],[N]\rangle)}\sum_{[R]\in \overline{A}}\frac{a_{MN}^R}{|Aut(M)||Aut(N)|}[R]$$ where $a_{MN}^R$ is the number of short exact sequences $$0\rightarrow N\rightarrow R\rightarrow M\rightarrow 0$$ and $$\langle [M],[N]\rangle = \sum (-1)^i |Ext^i_A(M,N)|$$ is the Euler form. This multiplication makes $H(A)$ into an associate algebra called the Hall algebra of $A$; the proof can be found e.g. here.

Finite Fields and Quantization

The categories $A$ appearing in the construction of a Hall algebra are usually linear over some finite field $\mathbb{F}_q$. Often, it is possible to simultaneously define a category $A_q$ for each finite field $\mathbb{F}_q$; usually by considering modules on the $\mathbb{F}_q$-points of some scheme over $\mathbb{Z}$. The corresponding Hall algebras $H(A_q)$ will then usually be closely related, and can often be defined by relations that are functions in $q$.

The Question

I know that there are cases where an algebra is deformed by a parameter $q$, and then the resulting family of algebras `magically' coincides with a family of Hall algebras $H(A_q)$ in the special cases when $q$ is a prime power. I think this happens in the case of the Hecke algebra (discussed here), and the case of quantum universal enveloping algebras (discussed here). I somewhat understand that this is a symptom of a related convolution algebra on the scheme used to define $A_q$.

Is there a family of categories $A_q$ such that the corresponding Hall algebras $H(A_q)$ are isomorphic to the Quantum Torus $T_q$ for all $q$ a prime power? If so, is there a convolution algebra realization of the Quantum Torus?

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2 Answers 2

As far as I understand, the Hall algebra of a category (say, with finite length of objects) is graded by the Grothendieck monoid of this category, spanned by simple objects over $\Bbb Z_+$, and it must have the ground field in degree $0$. The quantum torus algebra does not seem to have such a grading (it has a $\Bbb Z^2$-grading, not a $\Bbb Z_+^m$-grading). Maybe one should ask this question for the q-Weyl algebra $xy=qyx$ (not allowing negative powers of x and y)? Note that this algebra appears as a subalgebra of a Hall algebra (the Hall algebra for the quiver $A_2$ is $U_q(n_+)$, where $n_+$ is the nilpotent subalgebra of $sl(3)$; the q-Weyl algebra is generated by $e_{12}$ and $e_{13}$ inside this algebra).

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[EDIT: ignore this post. I wrote it last night, and this morning I see that it's nonsense.]

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How about element $1$? It is not a commutator. –  Pavel Etingof Feb 9 '10 at 13:23
    
Fair enough. The product of things that are trivial in the Grothendieck group are still trivial in the Grothendieck group, so being a product of commutators is enough to show something goes to 0 in the Grothendieck group. –  Ben Webster Feb 9 '10 at 13:42
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I must admit I also did not understand the argument with commutators. For example, I don't see any connections between extensions of $X$ by $Y$ and extensions of $Y$ by $X$. Also I don't understand what your map to the Grothendieck groups is. Do you rather mean that the Hall algebra is graded by the Grothendieck group (and even by the Grothendieck semigroup?) –  Pavel Etingof Feb 9 '10 at 13:54
    
You're right, that didn't make any sense. It sounded right last night, but now that I look, it's complete nonsense. –  Ben Webster Feb 9 '10 at 14:47
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