In reading through Khalkhali's Noncommutative Geometry text, I came across something I don't understand. Let $\mathfrak{G}$ be a discrete groupoid, and for each $x\in Obj(\frak{G})$, define the *-representation $\pi_x:\mathbb{C}\frak{G}\to\mathscr{L}(\mathscr{l}^2(\frak{G}_{x}))$ via $(\pi_x\gamma )(\gamma ')=\gamma\circ\gamma'$ if the composition is defined and 0 else; here $\frak{G}_x$ is $s^{-1}(x)$ for $s$ the source map (i.e. sends a morphism to its source). We also let $\frak{G}$ be the set of morphisms on its objects. My question is: $\gamma\circ\gamma$ needn't be in $\mathscr{l}^2(\frak{G}_x)$, need it? So how is this thing properly defined?
Representations of groupoid algebras
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functional-analysis
representation-theory
operator-algebras
noncommutative-geometry
groupoids
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0and what is the canonical norm anyway? – 2017-02-01
1 Answers
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If I am understanding your notation correctly, then $l^2(\frak{G_x})$ is just the space of square summable sequence on $\frak{G_x}$ (ie. $l^2(\frak{G_x})=\{\sum_{i}a_i\gamma_i : a_i\in\mathbb{C}, \sum_i|a_i|^2<\infty, \gamma_i\in\frak{G_x}\}$. In this case the norm of such an element would just be $\sqrt{\sum_i|a_i|^2}$ given by the inner product. Since in your case $\gamma \circ\gamma'\in\frak{G_x}$ it would have norm 1.
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0thank you. my concern is that im($\pi_x$) is not in its claimed space – 2017-02-01
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0since the source of the composition needn't be the source of the rightmost morphism – 2017-02-01
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0It should be. If we define the notion appropriately then $s(\gamma')=s(\gamma\circ\gamma')$. Here we read $\gamma\circ\gamma'$ as $\gamma'$ first $\textit{then}$ $\gamma$. – 2017-02-02