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I am reading a book about C*-algebra. When i study von Neumann algebras in this book, i meet with a problem. In the book,

If $H$ is a Hilbert space, we write $H^{(n)}$ for the orthogonal sum of n copies of $H$. If $a\in M_{n}(B(H))$ (an operator matrice), we define $\phi(a)\in B(H^{(n)})$ by setting $$\phi(a)(x_{1}, x_{2},..., x_{n})=(\Sigma_{j=1}^{n}a_{1j}(x_{j}),...,\Sigma_{j=1}^{n}a_{nj}(x_{j})),$$ for all $(x_{1}, ...,x_{n})\in H^{(n)}$. It is easy to verify that the map $$\phi: M_{n}(B(H)) \rightarrow B(H^{(n)}), a\rightarrow \phi(a),$$ is a *-isomorphism.

My question is how to verify that $\phi$ is a *-isomorphism. I suppose it is not an easy conclusion for me .

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  • $\begingroup$ The first thing is to verify that it is a bijection (linearity should be clear), then check that $\phi(a^\ast) = \phi(a)^\ast$. $\endgroup$ – Daniel Fischer Nov 7 '13 at 15:52
  • $\begingroup$ I know that, but how? $\endgroup$ – Yan kai Nov 7 '13 at 16:02
  • $\begingroup$ What part have you problems with? Bijectivity is most easily proved by giving the inverse. The $^\ast$ property is a direct verification. $\endgroup$ – Daniel Fischer Nov 7 '13 at 16:04
  • $\begingroup$ I do not know what is the specific form of $\phi(a)^{*}$. $\endgroup$ – Yan kai Nov 7 '13 at 16:10
  • $\begingroup$ Well, you have $\langle \phi(a)v,w\rangle = \langle v, \phi(a)^\ast w\rangle$ for all $v,w\in H^{(n)}$. And that's the defining identity, so show $\langle \phi(a)v,w\rangle = \langle v,\phi(a^\ast)w\rangle$ for all $v,w \in H^{(n)}$. $\endgroup$ – Daniel Fischer Nov 7 '13 at 16:17
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I will use the notation $$ (x_1,\ldots,x_n)=\bigoplus_{k=1}^nx_k. $$ So $$ \left\langle\bigoplus_{k=1}^nx_k,\bigoplus_{k=1}^ny_k\right\rangle=\sum_{k=1}^n\langle x_k,y_k\rangle. $$

With the same proof as for complex matrices, one shows that $\phi(a)^*=\phi(a^*)$, where $(a^*)_{kj}=a_{jk}^*$: $$ \langle\phi(a)\bigoplus_{k=1}^nx_k,\bigoplus_{k=1}^ny_k\rangle=\langle\bigoplus_{k=1}^n\sum_{j=1}^na_{kj}x_j,\bigoplus_{k=1}^ny_k\rangle=\sum_{k=1}^n\sum_{j=1}^n\langle a_{kj}x_j,y_k\rangle =\sum_{j=1}^n\sum_{k=1}^n\langle x_j,a_{kj}^*y_k\rangle=\sum_{j=1}^n\langle x_j,\sum_{k=1}^na_{kj}^*y_k\rangle=\langle\bigoplus_{j=1}^nx_j,\bigoplus_{j=1}^n\sum_{k=1}^na_{kj}^*y_k\rangle=\langle\bigoplus_{j=1}^nx_j,\phi(a^*)\bigoplus_{j=1}^ny_j\rangle =\langle\phi(a^*)^*\bigoplus_{k=1}^nx_k,\bigoplus_{k=1}^ny_k\rangle. $$ So $\phi(a^*)^*=\phi(a)$ for all $a$, which implies that $\phi(a^*)=\phi(a)^*$.

We have $$ \phi(a+b)(\bigoplus_{k=1}^nx_k)=\bigoplus_{k=1}^n\sum_{j=1}^n(a+b)_{kj}x_j=\bigoplus_{k=1}^n\sum_{j=1}^n\,a_{kj}x_j+b_{kj}x_j\\=\bigoplus_{k=1}^n\sum_{j=1}^n\,a_{kj}x_j+\bigoplus_{k=1}^n\sum_{j=1}^n\,b_{kj}x_j=\phi(a)(\bigoplus_{k=1}^nx_k)+\phi(b)(\bigoplus_{k=1}^nx_k)=[\phi(a)+\phi(b)]\bigoplus_{k=1}^nx_k, $$ showing that $\phi$ is linear. Also, $$ \phi(ab)(\bigoplus_{k=1}^nx_k)=\bigoplus_{k=1}^n\sum_{j=1}^n(ab)_{kj}x_j=\bigoplus_{k=1}^n\sum_{j=1}^n\sum_{h=1}^na_{kh}b_{hj}x_j=\bigoplus_{k=1}^n\sum_{h=1}^na_{kh}\sum_{j=1}^nb_{hj}x_j =\phi(a)(\bigoplus_{k=1}^n\sum_{j=1}^nb_{hj}x_j)=\phi(a)\phi(b)(\bigoplus_{k=1}^nx_k), $$ and then $\phi(ab)=\phi(a)\phi(b)$.

If $\phi(a)=0$, then $\sum_{j=1}^na_{kj}x_j=0$ for any $k$ and any $x_1,\ldots,x_n$. From here we can see that $a_{kj}=0$ for all $k,j$. That is, $a=0$. So $\phi$ is injective.

Finally, let $T\in B(H^{(n)})$. Let $\pi_j$ be the map $\displaystyle\bigoplus_{k=1}^nx_k\mapsto x_j$. Then define $T_{kj}\in B(H)$ by $$ T_{kj}x=\pi_k T(0,\ldots,0,x,0,\ldots,0), $$ where the $x$ is in the $j^{\rm th}$ place in the $n$-tuple. Then $$ \phi((T_{kj})_{kj})(\bigoplus_{k=1}^nx_k)=\bigoplus_{k=1}^n\sum_{j=1}^nT_{kj}x_j=\sum_{j=1}^n\bigoplus_{k=1}^nT_{kj}x_j=\sum_{j=1}^n\bigoplus_{k=1}^n\pi_kT(0,\ldots,0,x_j,0,\ldots,0)\bigoplus_{k=1}^n\pi_kT(\bigoplus_{k=1}^nx_j)=T(\bigoplus_{k=1}^nx_k). $$ And so $\phi$ is surjective.

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  • $\begingroup$ (+1) though that was a very simple question, you put a lot of efforts $\endgroup$ – Norbert Nov 7 '13 at 18:37
  • $\begingroup$ Thanks. I try to answer with the mentality that answers should be useful to others in the future. $\endgroup$ – Martin Argerami Nov 8 '13 at 1:51
  • $\begingroup$ Thanks for your detailed answer, Martin Argerami. :) $\endgroup$ – Yan kai Nov 8 '13 at 1:58
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I answered a similar question previously, you may (or may not) find this helpful A theorem about operators in Hilbert sapce.

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