# constructing a C*-algebra

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• May 26th 2010, 11:41 PM
Mauritzvdworm
constructing a C*-algebra
Is it possible to construct a C*-algebra $\mathcal{A}$ such that there exists $x\in\mathcal{A}$ which cannot be decomposed into $x=x_1x_2$ with $x_1\geq0$ and $x_2$ a partial isometry?

It does not seem so, since any C*-algebra can be seen as a C*-subalgebra of some $B(H)$ and any operator $T\in B(H)$ can be decomposed using the polar decomposition, or am I missing something?
• May 27th 2010, 01:03 AM
Opalg
Quote:

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

It does not seem so, since any C*-algebra can be seen as a C*-subalgebra of some $B(H)$ and any operator $T\in B(H)$ 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 $\mathcal{A}$, but the partial isometry does not necessarily belong to $\mathcal{A}$.

For example, if $\mathcal{A}$ is the commutative C*-algebra of continuous functions on [–1,1] and x is the function x(t) = t, then the positive part $x_1$ will be the function $x_1(t) = |t|$. But the partial isometry will be the function $x_2$, where $x_2(t)$ is –1 when x<0 and +1 when x>0. Obviously $x_2$ is not continuous and so cannot belong to $\mathcal{A}$.
• May 27th 2010, 01:07 AM
Mauritzvdworm
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.