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Thread: constructing a C*-algebra

  1. #1
    Member Mauritzvdworm's Avatar
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    constructing a C*-algebra

    Is it possible to construct a C*-algebra $\displaystyle \mathcal{A}$ such that there exists $\displaystyle x\in\mathcal{A}$ which cannot be decomposed into $\displaystyle x=x_1x_2$ with $\displaystyle x_1\geq0$ and $\displaystyle x_2$ a partial isometry?

    It does not seem so, since any C*-algebra can be seen as a C*-subalgebra of some $\displaystyle B(H)$ and any operator $\displaystyle T\in B(H)$ can be decomposed using the polar decomposition, or am I missing something?
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  2. #2
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    Opalg's Avatar
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    Quote Originally Posted by Mauritzvdworm View Post
    Is it possible to construct a C*-algebra $\displaystyle \mathcal{A}$ such that there exists $\displaystyle x\in\mathcal{A}$ which cannot be decomposed into $\displaystyle x=x_1x_2$ with $\displaystyle x_1\geq0$ and $\displaystyle x_2$ a partial isometry?

    It does not seem so, since any C*-algebra can be seen as a C*-subalgebra of some $\displaystyle B(H)$ and any operator $\displaystyle T\in B(H)$ can be decomposed using the polar decomposition, or am I missing something?
    The decomposition certainly works in B(H), giving a positive operator and a partial isometry in B(H). The positive operator will be in $\displaystyle \mathcal{A}$, but the partial isometry does not necessarily belong to $\displaystyle \mathcal{A}$.

    For example, if $\displaystyle \mathcal{A}$ is the commutative C*-algebra of continuous functions on [1,1] and x is the function x(t) = t, then the positive part $\displaystyle x_1$ will be the function $\displaystyle x_1(t) = |t|$. But the partial isometry will be the function $\displaystyle x_2$, where $\displaystyle x_2(t)$ is 1 when x<0 and +1 when x>0. Obviously $\displaystyle x_2$ is not continuous and so cannot belong to $\displaystyle \mathcal{A}$.
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  3. #3
    Member Mauritzvdworm's Avatar
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    That is what I thought, interestingly in the case of von Neumann algebras the partial isometry is part of the von Neumann algebra. Thank you.
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