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I am starting with $C^*$ algebras. There are some notations that I don't understand. Please help me.

  1. What does the identity representation of $C^*$ algebras mean?
  2. Let $A$ be $C^*$ algebra generated by the irreducible operator A and the identity. may be representation is also irreducible?
  3. For the operator $T$, then $T^{(n)}$ means that?
  4. For the Hilbert space, the notation $H^{(n)}$ means that? Sorry if it is too easy for readers. I'm new in this area.
Martin Argerami
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user120127
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1 Answers1

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  1. If your C$^*$-algebra is already represented inside some $B(H)$, then the identity map (i.e. "doing nothing") is the identity representation.

  2. If a represented C$^*$-algebra contains an irreducible operator, then it is of course irreducible.

  3. $H^{(n)}$ is the direct sum of $n$ copies of $H$.

  4. $T^{(n)}$ is the operator in $B(H^{(n)})$ that acts by $T^{(n)}(h_1,\ldots,h_n)=(Th_1,\ldots,Th_n)$.

Martin Argerami
  • 205,756
  • If $T_i$, $i \in N$ is irreducible then may be the direct sum $\sum_{i\in N}T_i$ also is irreducible? – user120127 Jan 11 '14 at 01:34
  • No. That fails even in dimension one: $1,2,3$ are each irreducible on $\mathbb C$, but $\begin{bmatrix}1&0&0\0&2&0\0&0&3\end{bmatrix}$ is not irreducible in $M_3 (\mathbb C) $. – Martin Argerami Jan 11 '14 at 04:18
  • Thank you. I have another question. If the direct sum $H=H_1 + H_2$ such that $T$ is completely reducible then may be $T|_{H_1}$ is also completely reducible. – user120127 Jan 11 '14 at 16:43
  • Yes, of course. Just form the definition of completely reducible, the operator is completely reducible when restricted to any subspace. – Martin Argerami Jan 11 '14 at 16:59
  • It looks the same of irreducible operators? – user120127 Jan 11 '14 at 17:09
  • Yes, because if an operator is reducible, then so it any amplification of it. – Martin Argerami Jan 11 '14 at 17:12
  • If $T$ is irreducible on $H$ then $T^{(n)}$ is also irreducible on $H^{(n)}$, isn't it? – user120127 Jan 12 '14 at 13:52
  • Not at all. Same as before, $1$ is irreducible on $\mathbb C$ but $\begin{bmatrix} 1&0\0&1\end{bmatrix}$ is not irreducible on $\mathbb C^2$. – Martin Argerami Jan 12 '14 at 14:57
  • I am reading a theorem on the structure of opreators, which says that if $T$ is a bounded linear operator then we have the direct sum $T=T_0+T_1^{(n_1)}+...+T_i^{(n_i)}+...$. I guess $T_0$ is completely reducible, $T_i^{(n_i)}$ is irreducible. Is it correct? – user120127 Jan 12 '14 at 15:12
  • If you don't say what $T_0,T_1,\ldots$ are it is impossible to say anything. – Martin Argerami Jan 12 '14 at 16:29
  • $T_j \in L(H_j)$. – user120127 Jan 13 '14 at 03:03
  • That is still meaningless. The Theorem you are quoting is likely a multiplicity theorem, but just writing a decomposition without saying what conditions the components are supposed to satisfy is meaningless. – Martin Argerami Jan 13 '14 at 04:09
  • If there exists an unitary operator $U$ such that $UT^{(m)}=T^{(n)}U$ for irreducible operator $T$, how to prove $m=n$. – user120127 Jan 13 '14 at 08:35
  • @Martin Argerami, is the identity representation be non-degenerate? – math112358 Feb 22 '20 at 03:08
  • Not necessarily. If you have for example $A=A_0\oplus 0\subset B(H)\oplus B(H)$, then the identity representation is degenerate (since $A(H\oplus H)=H\oplus 0$). – Martin Argerami Feb 22 '20 at 04:18