# On the definition of ‘smooth vectors’ in Rieffel's “Deformation Quantization for Actions of $\mathbb{R}^{d}$”.

On the first page of Chapter 1 of Rieffel's Deformation Quantization for Actions of $\mathbb{R}^{d}$, Rieffel defines a family of seminorms on the space $A^{\infty}$ of smooth vectors of a Fréchet space $A$, for some action $\alpha$ of the Lie group $\mathbb{R}^{d}$ on $A$, as follows. Suppose we already have a family $(\| \cdot \|_{i})_{i \in \mathbb{N}}$ of seminorms on $A$ that determine its topology. Choose a basis $\lbrace X_{1},\ldots,X_{d} \rbrace$ of $\mathbb{R}^{d}$. Then for each $k \in \lbrace 1,\ldots,d \rbrace$, let $\alpha_{X_{k}}$ denote the operator of partial differentiation on $A^{\infty}$ in the direction of $X_{k}$; we thus identify $\mathbb{R}^{d}$ with its Lie algebra in the usual way. For convenience, denote $\alpha_{X_{k}}$ simply by $\partial_{k}$. Next, for any multi-index $\mu = (\mu_{1},\ldots,\mu_{d}) \in \mathbb{N}_{0}^{d}$, let $\partial^{\mu}$ denote the higher-order partial derivative $\partial_{1}^{\mu_{1}} \cdots \partial_{d}^{\mu_{d}}$. Then equip $A^{\infty}$ with the seminorms $$\forall (j,k) \in \mathbb{N} \times \lbrace 1,\ldots,d \rbrace ~~ \& ~~ \forall a \in A^{\infty}: \quad \| a \|_{j,k} \stackrel{\text{def}}{=} \sup_{1 \leq i \leq j} \sum_{|\mu| \leq k} \frac{\| \partial^{\mu} a \|_{i}}{\mu!},$$ where $|\mu| \stackrel{\text{def}}{=} \mu_{1} + \cdots + \mu_{d}$ and $\mu! \stackrel{\text{def}}{=} \mu_{1}! \cdots \mu_{d}!$.

My question is: As we are applying partial derivatives to $a \in A^{\infty}$, are we identifying $a$ with the function $f_{a}: \mathbb{R}^{d} \rightarrow A$ defined by ${f_{a}}(\mathbf{x}) \stackrel{\text{def}}{=} \alpha(\mathbf{x},a)$?

Thank you very much in advance!

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By definition, $a\in A^\infty$ if $f_a$ a differentiable function from $\mathbb{R}^n$ into $A$. So, the answer is "yes" regarding the definition of $A^\infty$. –  Vahid Shirbisheh Dec 22 '12 at 20:45

I am recording some observations that I made while trying to understand Professor Rieffel’s definition of $\partial_{k}$, which I managed to do in the end. :)

Let

• $G$ be a finite-dimensional Lie group,

• ${\frak{g}}$ the Lie algebra of $G$,

• $A$ a Fréchet space (over $\mathbb{C}$) and

• $\alpha$ a strongly continuous group action of $G$ on $A$.

Define $$A^{\infty} \stackrel{\text{def}}{=} \lbrace a \in A ~|~ \pi(\bullet,a): G \to A \text{ is a smooth function} \rbrace,$$ which we call the space of smooth vectors for the action $\alpha$ of $G$ on $A$. It is a linear subspace of $A$, and by considering the Gårding space for $\alpha$, it can be shown to be dense in $A$.

For each $v \in {\frak{g}}$, define a linear subspace ${\frak{D}}(v)$ of $A$ by $${\frak{D}}(v) \stackrel{\text{def}}{=} \left\lbrace a \in A ~ \Bigg| ~ \lim_{t \to 0} \frac{\alpha(\exp(tv),a) - a}{t} \text{ exists} \right\rbrace.$$ For each $v \in {\frak{g}}$, we have $A^{\infty} \subseteq {\frak{D}}(v)$. This then allows us to define a mapping $\pi: {\frak{g}} \to \mathcal{L}(A^{\infty},A)$ by \begin{align} \forall v \in {\frak{g}}, ~ \forall a \in A^{\infty}: \quad [\pi(v)](a) & = \frac{d}{dt} \Bigg|_{t=0} \alpha(\exp(tv),a) \\ & = \lim_{t \to 0} \frac{\alpha(\exp(tv),a) - a}{t}. \end{align} In fact, we have $\pi: {\frak{g}} \to \mathcal{L}(A^{\infty},A^{\infty}) = \text{End}(A^{\infty})$, and with a little more work, one can show that this is a Lie-algebra homomorphism, i.e., $$\forall v,w \in {\frak{g}}: \quad \pi([v,w]_{\frak{g}}) = [\pi(v),\pi(w)]_{\text{End}(A^{\infty})}.$$

In the case where $G = \mathbb{R}^{d} = {\frak{g}}$, we have (after fixing an ordered basis $(X_{1},\ldots,X_{d})$ of $\mathbb{R}^{d}$) \begin{align} \forall k \in \lbrace 1,\ldots,d \rbrace, ~ \forall a \in A^{\infty}: \quad [\pi(X_{k})](a) & = \frac{d}{dt} \Bigg|_{t=0} \alpha(\exp(t X_{k}),a) \\ & = \lim_{t \to 0} \frac{\alpha(\exp(t X_{k}),a) - \alpha(\exp(0_{\mathbb{R}^{d}}),a)}{t} \\ & = \lim_{t \to 0} \frac{\alpha(t X_{k},a) - \alpha(0_{\mathbb{R}^{d}},a)}{t} \\ & = \lim_{t \to 0} \frac{{f_{a}}(t X_{k}) - {f_{a}}(0_{\mathbb{R}^{d}})}{t} \\ & = {D_{X_{k}} f_{a}}(0_{\mathbb{R}^{d}}). \end{align} Then by $\partial_{k}: A^{\infty} \to A^{\infty}$, Professor Rieffel simply means $\pi(X_{k})$.

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