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Thread: Constructing a projection

  1. #1
    Member Mauritzvdworm's Avatar
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    Constructing a projection

    Let $\displaystyle B$ be a C*-algebra and let $\displaystyle p$ be a rank-one projection in $\displaystyle K$, the space of compact operators on some Hilbert space $\displaystyle H$.

    I would like to construct a projection $\displaystyle \mathcal{P}$ in the multiplier algebra $\displaystyle M(B\otimes K)$ such that $\displaystyle \mathcal{P}(B\otimes K)\mathcal{P}=B\otimes p$

    here is my current idea:
    consider the mapping $\displaystyle \eta:B\otimes K\rightarrow B\otimes p$ such that $\displaystyle b\otimes k\mapsto b\otimes p$. Here $\displaystyle b\in B,k\in K$
    It can easily be shown that $\displaystyle \eta^2=\eta$, however I'm having some trouble in showing that $\displaystyle \eta^*=\eta$. Is it possible?
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  2. #2
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    Quote Originally Posted by Mauritzvdworm View Post
    Let $\displaystyle B$ be a C*-algebra and let $\displaystyle p$ be a rank-one projection in $\displaystyle K$, the space of compact operators on some Hilbert space $\displaystyle H$.

    I would like to construct a projection $\displaystyle \mathcal{P}$ in the multiplier algebra $\displaystyle M(B\otimes K)$ such that $\displaystyle \mathcal{P}(B\otimes K)\mathcal{P}=B\otimes p$

    here is my current idea:
    consider the mapping $\displaystyle \eta:B\otimes K\rightarrow B\otimes p$ such that $\displaystyle b\otimes k\mapsto b\otimes p$. Here $\displaystyle b\in B,k\in K$
    It can easily be shown that $\displaystyle \eta^2=\eta$, however I'm having some trouble in showing that $\displaystyle \eta^*=\eta$. Is it possible?
    The trouble with the mapping $\displaystyle \eta$ is that it is not well defined (for example, $\displaystyle (2b)\otimes k = b\otimes(2k)$, but $\displaystyle \eta$ takes the first of these to $\displaystyle 2b\otimes p$ and the second to $\displaystyle b\otimes p$). Try the map $\displaystyle b\otimes k\mapsto b\otimes pkp$.
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  3. #3
    Member Mauritzvdworm's Avatar
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    Now I see that my original idea was wrong. For the mapping
    $\displaystyle \eta:b\otimes k\mapsto b\otimes pkp$
    it is easy to show that $\displaystyle \eta=\eta^2$, but once again I am missing something in determining $\displaystyle \eta^*$?

    The way I am thinking about this is to have some kind of inner product in order to determine$\displaystyle \eta$.

    Another thing is if this $\displaystyle \eta$ is in fact the projection I am looking for then
    $\displaystyle \mathcal{P}(b\otimes k)\mathcal{P}=(b\otimes pkp)\mathcal{P}$
    now if we let $\displaystyle b\otimes pkp:=a$ then
    $\displaystyle a\mathcal{P}=\left(\mathcal{P}a^*\right)^*=\left(\ mathcal{P}\left(b^*\otimes pk^*p\right)\right)^*=\left(b^*\otimes pk^*p\right)^*=b\otimes pkp$

    but is $\displaystyle pkp=p$?
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  4. #4
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    Looking at the problem again, I don't see why you need the map $\displaystyle \eta$ at all. Simply define $\displaystyle \mathcal{P} = 1\otimes p$, where 1 is the identity element in $\displaystyle M(B)$. The point is that if $\displaystyle p$ is a rank-one projection then $\displaystyle pkp$ is always a scalar multiple of $\displaystyle p$. In fact, if $\displaystyle \xi$ is a unit vector in the range of $\displaystyle p$ then $\displaystyle pkp = \varphi(k)p$, where $\displaystyle \varphi$ is the state on $\displaystyle K$ defined by $\displaystyle \varphi(k) = \langle k\xi,\xi\rangle.$
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  5. #5
    Member Mauritzvdworm's Avatar
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    Aaaaa, I see! Thanks a lot!
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