Tag Info

Hot answers tagged

2

This answer is concerned with the first question. The $C^\ast$-subalgebra $B$ you are considering is in fact a closed $^\ast$-ideal and hence given by $$ \{f\in C[0,1]\mid f_C=0\} $$ for some closed subset $C\subseteq [0,1]$. Clearly all functions in $B$ vanish at $x$ when $g(x)=0$. Conversely, if $g(x)\neq 0$, then we have already found a function in $B$ ...


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

Without more context I cannot be certain, but I don't think your interpretation is the one your author has in mind. Given an associative algebra (like a Banach algebra), say $\mathcal{A}$, one gets a Lie algebra using the commutator bracket: $[x,y]=xy-yx$ where $x,y \in \mathcal{A}$. Here "$xy$" denotes $\mathcal{A}$'s (associative) multiplication. So ...



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