2
$\begingroup$

Let $\theta$ be a real number. We define $A_{\theta}$, the algebra of continuous functions on a noncommutative $2$-torus, to be the universal $C^*$-algebra generated by two generators $U$ and $V$ which satisfy $$ UV=e^{2\pi i\theta}VU, UU*=U^*U=VV^*=V^*V=1. $$

In literature people define smooth functions on a noncommutative torus to be a dense subalgebra $A^{\infty}_{\theta}$of $A_{\theta}$ as $$ A^{\infty}_{\theta}:=\{\sum_{r,s\in \mathbb{Z}}a_{rs}U^rV^s|\{a_{rs}\}\in \mathcal{S}(\mathbb{Z^2})\} $$ where $\mathcal{S}(\mathbb{Z^2})$ denotes the space of rapid decreasing functions on $\mathbb{Z}^2$.

I'm not sure if the definition of $A^{\infty}_{\theta}$ is a convenient choice. In the commutative case, i.e. when $\theta=0$. $A^{\infty}_{\theta}$ coincides with the algebra of smooth functions on $T^2$ by Fourier transformation. In the noncommutative case, we can still do lots of differential operations on $A^{\infty}_{\theta}$. For example we have basic derivations on $A^{\infty}_{\theta}$. This means that elements in $A^{\infty}_{\theta}$ must be smooth but I wonder if there exist "smooth functions" outside $A^{\infty}_{\theta}$.

My question is: Is $A^{\infty}_{\theta}$ also the maximal noncommutative smooth functions on a noncommutative $2$-torus, i.e. the maximal subalgebra in $A_{\theta}$ such that arbitrary composition of basic derivations could be defined?

$\endgroup$

1 Answer 1

2
$\begingroup$

The two basic derivations are infinitesimal generators of an action $\phi$ of $\mathbb{T}^2$ on $A_\theta$, and the Fourier coefficients of $A \in A_\theta$ are recovered by $a_{rs}U^rV^s = \int_{\mathbb{T}^2} \phi_{xy}(A) e^{rx + sy} dxdy$. This shows that the $a_{rs}$ are bounded by $\|A\|$. The two generators act on the Fourier coefficients by multiplication by $r$ and $s$, so if $A$ is in the domain of compositions of basic derivations then the values $r^k s^l a_{rs}$ must be uniformly bounded, for each $k$ and $l$. I may be missing some factors of $2\pi i$, etc., but that's the idea. This is covered in Section 5.5 of my book Mathematical Quantization.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .