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Math Help - Constructing a projection

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

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

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

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

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

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

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

    Another thing is if this \eta is in fact the projection I am looking for then
    \mathcal{P}(b\otimes k)\mathcal{P}=(b\otimes pkp)\mathcal{P}
    now if we let b\otimes pkp:=a then
    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 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  \eta at all. Simply define \mathcal{P} = 1\otimes p, where 1 is the identity element in M(B). The point is that if  p is a rank-one projection then pkp is always a scalar multiple of  p. In fact, if \xi is a unit vector in the range of  p then pkp = \varphi(k)p, where  \varphi is the state on  K defined by \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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