An element $a\in\mathfrak{A}$ (unital C*-algebra) is a partial isometry if $a^*\cdot a $ is projection.

Can one recover the equivalent caracterizations of a partial isometry in $\mathcal{B}(\mathcal{H})$ but for an abstract C*-algebra? namely:

$a\ \text{partial isometry} \Leftrightarrow a^*\ \text{p. iso} \Leftrightarrow a^*a\ \text{projection} \Leftrightarrow aa^*\ \text{projection}\Leftrightarrow aa^*a=a \Leftrightarrow a^*aa^*=a^*$

The only thing I can see is that the last two ones (are of course equivalent and) imply the others

  • Just embed $\mathfrak{A}$ into some $\mathcal{B}(\mathcal{H})$ – Norbert Dec 20 '13 at 21:04
  • I guess one can proove it by saying that it holds in a faithful representation, but... – Noix07 Dec 20 '13 at 21:17
up vote 11 down vote accepted

Yes, this can be done without making the $C^*$-algebra concrete in $B(H)$. All we need is that $z=0$ iff $z^*z=0$, which follows from the $C^*$-identity $\|z^*z\|=\|z\|^2$.

Lemma Let $a\in A$. The following assertions are equivalent.

1 - $p=a^*a$ is idempotent, i.e. $p^2=p$.

2 - $aa^*a=a$

3 - $a^*aa^*=a^*$

4 - $q=aa^*$ is idempotent.

Proof
$1\Rightarrow 2:$ Here is the main trick. Set $z:=aa^*a-a$. Note that $z^*z=p^3-p^2-p^2+p=0$, whence $z=0$.

$2\Rightarrow 3:$ Take the adjoint.

$3\Rightarrow 4:$ Left-multiply $3$ by $a$.

$4\Rightarrow 1:$ Applying $1\Rightarrow 2$ to the element $a^*$ instead of $a$, we get $a^*aa^*=a^*$. Right-multiplication by $a$ yields $1$. $\Box$

By definition, an element $a$ is a partial isometry in an abstract $C^*$-algebra if $a^*a$ is a projection. Of course, the latter is always self-adjoint. So $a$ is a partial isometry iff $a^*a$ is idempotent. It follows in particular that $a$ is a partial isometry iff $a^*$ is a partial isometry.

Remark Two projections $p,q$ as above are called Murray-von Neumann equivalent. The lemma above helps prove that the latter is an equivalence relation (namely transitivity, which is the slightly tricky part).

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