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Thread: isomorphism

  1. #1
    Member Mauritzvdworm's Avatar
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    isomorphism

    Let $\displaystyle K=K(H)$ with $\displaystyle H$ a infinite dimensional, separable Hilbert space. Let $\displaystyle (e_n)^{\infty}_{n=1}$ be an orthonormal basis for $\displaystyle H$. Let $\displaystyle e_{ij}$ be an operator in $\displaystyle B(H)$ defined by $\displaystyle e_{ij}(x)=\langle x,e_j\rangle e_i$
    set $\displaystyle p_n=\sum^{n}_{j=1}e_{jj}$ now show that the following map is a *-isomorphism (bijective *-homomorphism)

    $\displaystyle \psi_{n}:M_n(A)\rightarrow p_nKp_n\otimes A, (a_{r,s})\mapsto\sum^{n}_{i,j=1}e_{ij}\otimes a_{r,s}$
    Last edited by mr fantastic; Jul 19th 2010 at 03:18 AM.
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  2. #2
    Member Mauritzvdworm's Avatar
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    I managed to show that $\displaystyle \psi_n$ is linear, multiplicative and a bijection (though it took some work)

    The only missing part is to show that $\displaystyle \psi(a^*)=\psi(a)^*$
    What does $\displaystyle e^*_{ij}$ look like?
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  3. #3
    Member Mauritzvdworm's Avatar
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    Let $\displaystyle y\in H$ then consider
    $\displaystyle \langle e^{*}_{ij}y,x \rangle=\langle y,e_{ij}x \rangle=\langle y,\langle e_j,x \rangle e_i \rangle=\langle e_j,x \rangle \langle y,e_i \rangle=\langle \langle e_i,y \rangle e_j,x \rangle$

    so $\displaystyle e^{*}_{ij}=e_{ji}$

    Then everything works out
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