Tag Info

Hot answers tagged

3

The fact that such a $ * $-homomorphism exists is the statement of the Gelfand-Naimark-Segal (GNS) Construction.


2

Without loss of generality $A$ acts non-degenerately on $H$, i.e. $$\cap_{X\in A}\ker X=\{0\},$$ otherwise consider the restriction of $A$ onto the orthogonal complement of $\cap_{X\in A}\ker X$. Let $\tilde A$ be the unitization of $A$. Then $\tilde A$ is naturally isomorphic to $A+\mathbb C I_H\subseteq B(H)$. The notion of spectrum of an element $X\in A$ ...


2

Yes, it follows from the following general theorem (see Dixmier "$C^*$-algebras and representations", 2.10.2): Every irreducible representation $\rho$ of a $C^*$-subalgebra $C$ of a $C^*$-algebra $A$ can be continued to an irreducible representation $\pi$ of $A$ in a possibly larger Hilbert space. You can also apply the Lemma 2.10.1 from Dixmier directly. ...


1

I now see the following link: Let $f\in L^1(\mathbb{R})$ and $ x\in\mathcal{H}$ $$ U_f(x) = \int_{\mathbb{R}} f(t)\ e^{-it H}\cdot x\ dt = \int_{\mathbb{R}} \left(\int_{\mathbb{R}} f(t)\ e^{-it p}\ dt\ dP(p) \right) \cdot x $$ $$= \left(\int_{\mathbb{R}} \hat{f}(p)\ dP(p) \right) \cdot x $$ (Convention of - sign in the exponential in Bratteli, Robinson, ...


1

If $x\geq0$, let $f$ be a state on $B(H_2)$; then the functional $f\circ\phi$ is unital and contractive. It is well-known that a unital contractive functional is positive (I know of proofs in Paulsen and Davidson's books). So $f(\phi(x))\geq0$ for all states on $B(H_2)$, which means that $\phi(x)\geq0$. Finally, this can be done for each amplification of ...


1

Because the unit ball of the C$^*$-algebra generated by $A$ sits in between the unit ball of $A$ and the unit ball of the weak closure of $A$; so if the former is dense in the latter, so is the unit ball of C$^*$(A)$. Because $A$ is non-degenerate, the identity is in it weak closure. So if the unit ball of $A$, together with the identity, is dense in the ...


1

I try to answer the question by myself: In order to prove the "$\geq$", from the definition of universal group C*-algebra $C^{*}(F_{n})$, we need only to find a special *-representation $\pi:\mathbb{C}(F_{n})\rightarrow B(H)$, such that $||\pi(\sum_{k=0}^{n-1}\alpha_{k} U_{k})||=\sum_{k=0}^{n-1}|\alpha_{k}|$. Well, It is natural to consider the ...


1

There is an error in your edited material. In general, $ (x x^{*})^{k} \neq x^{k} (x^{*})^{k} $ unless we know that $ x $ commutes with $ x^{*} $. The justification that you seek is simple. When $ k = 0 $, you get $ x = x $, which is true. When $ k = 1 $, you get $ x (x^{*} x) = (x x^{*}) x $, which is also true as $ A $ is an associative algebra. When $ ...


1

In what follows, we assume that $ G $ is a locally compact Hausdorff group that does not have to be abelian. You can certainly define an involution $ ^{*} $ on $ {L^{1}}(G) $ by $$ \forall f \in {L^{1}}(G), ~ \forall x \in G: \quad {f^{*}}(x) \stackrel{\text{df}}{=} \overline{f(x^{-1})} \cdot \Delta(x^{-1}), $$ where $ \Delta $ denotes the modular ...



Only top voted, non community-wiki answers of a minimum length are eligible