constructing a C*-algebra

Aug 2009
122
19
Pretoria
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?
 

Opalg

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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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Aug 2009
122
19
Pretoria
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.