Proposition $3.45$. from Gracia-Bondia, Varilly and Figueroa's book named "Elements of noncommutative geometry" states that given a $C^*$-algebra $\mathcal{A}$, a Lie group $G$ and a strongly continuous action $\alpha:G\rightarrow\mathcal{A}$, we can construct a Frechet pre $C^*$-algebra of $\mathcal{A}$ as follows:
Let $A^\infty$ be the elements $a\in\mathcal{A}$ such that the map $t\mapsto\alpha_t(a)$ is smooth.
Consider the Lie algebra $\mathfrak{g}$ of $G$ and its universal enveloping algebra $\mathcal{U}(\mathfrak{g})$. Then any element $D\in\mathcal{U}(\mathfrak{g})$ act on $\mathcal{A}^{\infty}$ in the obvious way, and so we can define the seminorms $$a\mapsto||D(t\mapsto\alpha_t(a))|_{t=e}||,$$ which induce a Frechet-space topology on $\mathcal{A}^{\infty}$.
It follows from other properties that $\mathcal{A}^{\infty}$ is a pre $C^*$-algebra. My questions are the next:
What exactly is a smooth element respect to the map described on the first step?
What is the obvious way in that an element of $\mathcal{U}(\mathfrak{g})$ acts on $\mathcal{A}^{\infty}$?
I need this proposition in a very specific case, which is when the Lie group is just $\mathbb{S}^1\oplus\mathbb{S}^1$. I haven't had a formal course on Lie groups, but because of the need of this result, I've been teaching myself the correspondance from Lie groups of matrices and Lie algebras (which seems to be enough to my case) and the construction of the universal enveloping algebra, however those constructions don't make entirely clear how can I lift an action from the group $G$ to $\mathcal{U}(\mathfrak{g})$. My guess is that there is a differential structure that I can't understand by the things that I've studied.
In the future I'd like to teach myself more of Lie groups, but for the moment (and because I'm in a rush) if someone could explain me or give me some references of at least the case $G=\mathbb{S}^1\oplus\mathbb{S}^1$, it would be great.
I am not expert on Lie algebras, however I recently encountered the article On a class of smooth Frechet subalgebras of C∗-algebras that gives a little explanation on that, but gives a brief desription of how the seminorms would look like.
Disclaimer: I am new to Lie algebras also, but this would be my best guess:
I think the smoothness is taken with respect to actions of one parameter subgroups of $G$ i.e. you must have an homomorphism $\gamma : \mathbb{R} \to G$. Those homomorphisms always exists due to the exponetial map on Lie groups. Thus, I believe that the author is requesting that the map has infinite derivatives, with the derivative taken as a derivative of a function of on real variable i.e. the usual limit. Then, the derivatives of all those maps evaluated at the identity of the group $G$ i.e. $\gamma(0)=1_G$ would have a structure of a vector space (tangent space at the identity), and also, will have an algebra structure given by a Lie bracket $[\dot,\dot]: T_{1_G} \times T_{1_G} \to T_{1_G}$ that tells us how those derivations commute. Moreover, that will be a a vector space with a finite base, that we will call $\{\delta_1, ..., \delta_n\}$.
An element of the universal enveloping algebra would look something like $\sum_{n} \delta_{i_{n_1}} \delta_{i_{n_2}} \cdots \delta_{i_{n_k}} $ were this is a finite sum, and $\delta_{i_{n_1}} \delta_{i_{n_2}} \cdots \delta_{i_{n_k}}$ acts on $\gamma: t \to \alpha_t(a)$ by applying the the derivations of the path. Sorry it this part sounds confusing, I am still trying to get my head around that, I'll improve my answer as soon as I get a better understanding of this.
Nevertheless, I recommend you to give a look at example 1.6 on On a class of smooth Frechet subalgebras of C∗-algebras, which describes how the seminorms look like.