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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
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2 Answers

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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Have a look at (this). It might be helpful.

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