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In Alain Connes' $C^*$-algèbres et géométrie différentielle (an English translation is here,), for a $C^*$-algebra $A$, we consider a $C^*$-dynamic system $(A,G,\alpha)$, where $G$ is a Lie group and $\alpha: G\to \text{Aut}(A)$ is a continuous homomorphism. We shall say that $x \in A$ is of $C^\infty$ class iff the map $g \mapsto \alpha_g(x)$ from $G$ to the normed space $A$ is $C^\infty$. The involutive algebra $A^{\infty}= \{ x \in A, x \text{ of class } C^{\infty}\}$ is norm dense in $A$.

For example, let $\theta$ be an irrational number and $A_{\theta}$ be the irrational rotation algebra generated by two unitary element $U_1$ and $U_2$ satisfying $$ U_1U_2=e^{2\pi \theta i}U_2U_1. $$ We can endow $A_{\theta}$ with an action of $\mathbb{T}^2$, where $\mathbb{T}=\{z\in \mathbb{C}|~|z|=1\}$, such that $$ \alpha_{z_1,z_2}U_i=z_iU_i. $$ It is easy to check that the algebra $A_{\theta}^{\infty}$ consists of elements of the form $$ \sum_{p,q\in \mathbb{Z}} a_{p,q}U_1^pU_2^q $$ where $a_{p,q}$ decays rapidly, i.e. for any $n$, $(|p|+|q|)^n |a_{p,q}|$ is bounded.

I still don't quite understand the definition. The subalgebra $A^{\infty}$ seems to be heavily depends on the choice of $G$ and $\alpha$. For example, if I choose $G=\{e\}$ the trivial Lie group, then $A^{\infty}=A$.

I understand that within the same $A$ there may be different $A^{\infty}$'s, but I want to know

  1. Why the definition in the above example is "correct"?
  2. Is there a more systmetic way do define $A^{\infty}$ in a $C^*$-algebra $A$?
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    $\begingroup$ The choice of $A^\infty$ is wholly contextual. If you're working with $C^\ast$-dynamical systems for Lie groups, then this choice is natural; in the special case of strict deformation quantizations à la Rieffel, you use a certain subalgebra of this one. If you're working with regular spectral triples, then there's another choice. These choices are distinct in general but interrelated---in the case of noncommutative tori, they all transparenty coincide. (1/2) $\endgroup$
    – Branimir Ćaćić
    Commented Jul 30 at 18:11
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    $\begingroup$ There's no one notion of noncommutative manifold in NCG à la Connes, but there is a commonality: they all involve a $C^\ast$-algebra $A$ together with additional structure that either induces a canonical choice of "smooth" subalgebra $A^\infty$ or includes (as part of the data) a choice of $A^\infty$. (2/2) $\endgroup$
    – Branimir Ćaćić
    Commented Jul 30 at 18:12
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    $\begingroup$ For what it’s worth, even smooth functions in a commutative $C^\ast$-algebra are subject to choices and an “additional structure” than just the algebra itself. After all, there are smooth manifolds that are homeomorphic (therefore correspond to the same $C^\ast$-algebra) but not diffeomorphic (therefore the identification of $C^\ast$-algebras does not carry smooth functions to smooth functions) to each other. Say, the Milnor spheres. So one really shouldn’t expect the choice of $A^\infty$ to be more than what it is - a choice. $\endgroup$
    – David Gao
    Commented Jul 31 at 0:34
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    $\begingroup$ At least we know what a "commutative" manifold is—in NCG à la Connes, even the would-be notion of noncommutative manifold you use tends to be a contextual choice. $\endgroup$
    – Branimir Ćaćić
    Commented Jul 31 at 11:39

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