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If $ A $ is a unital C$ ^{*} $-algebra and $ a $ is invertible, then

  1. $ a = u|a| $ for a unique unitary element $ u $ of $ A $.
  2. If $ \| a \| = \| a^{-1} \| = 1 $, what can you say about $ |a| $?

I don’t know how to start!

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You can just say that $ A $ is a C$ ^{*} $-algebra. –  Haskell Curry Mar 3 '13 at 3:15
    
Set $u:=a/|a|$ and show that $u$ is a unitary element of $A$. The rest follows from Haskell Curry's answer. –  Vahid Shirbisheh Mar 3 '13 at 3:30
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@VahidShirbisheh do you mean put $u=a|a|^{-1}$ ?? –  user61965 Mar 3 '13 at 3:41
    
@edab12: Yes, your expression is more correct. –  Vahid Shirbisheh Mar 3 '13 at 4:03
    
@edab12: As $ |a| $ must be invertible, we have $ u = a |a|^{-1} $. Hence, $ u $ is uniquely determined. To show that $ u $ is unitary, simply observe that $$ u^{\ast} u = |a|^{-1} a^{\ast} a |a|^{-1} = |a|^{-1} |a|^{2} |a|^{-1} = \mathbf{1}_{A} $$ and $$ u u^{\ast} = a |a|^{-1} |a|^{-1} a^{\ast} = a |a|^{-2} a^{\ast} = a a^{-1} (a^{*})^{-1} a^{\ast} = \mathbf{1}_{A}. $$ –  Haskell Curry Mar 3 '13 at 7:09

1 Answer 1

up vote 3 down vote accepted

We will make use of the following basic result from the theory of Banach algebras.

Theorem 1 Let $ A $ be a Banach algebra. If $ x \in A $ is invertible, then $ 0 \notin {\sigma_{A}}(x) $ and $$ {\sigma_{A}}(x^{-1}) = \left\{ \frac{1}{\lambda} ~ \Bigg| ~ \lambda \in {\sigma_{A}}(x) \right\}. $$

We also need the following theorem from the theory of C$ ^{*} $-algebras.

Theorem 2 Let $ A $ be a C$ ^{*} $-algebra. Then for any normal element $ x $ of $ A $, we have $$ {r_{A}}(x) = \| x \|_{A}, $$ where $ {r_{A}}(x) $ denotes the spectral radius of $ x $.


Let $ A $ be a C$ ^{*} $-algebra. Let $ a \in A $ be invertible, and suppose that $ \| a \|_{A} = \| a^{-1} \|_{A} = 1 $. Then \begin{align} {\sigma_{A}}(|a|) &= {\sigma_{A}} \left( \sqrt{a^{*} a} \right) \\ &= \sqrt{{\sigma_{A}}(a^{*} a)} \quad (\text{By the Spectral Mapping Theorem applied to $ a^{*} a $.}) \\ &= \{ 1 \}. \quad (\clubsuit) \end{align}

Proof of $ (\clubsuit) $: Firstly, observe that $ {\sigma_{A}}(a^{*} a) \subseteq [0,\infty) $. Next, using the C$ ^{*} $-condition and the fact that involution in a C$ ^{*} $-algebra is an isometry, we have $$ \| a^{*} a \|_{A} = \| (a^{*}) (a^{*})^{*} \|_{A} = \| a^{*} \|_{A}^{2} = \| a \|_{A}^{2} = 1 $$ and \begin{align} \| (a^{*} a)^{-1} \|_{A} &= \| a^{-1} (a^{*})^{-1} \|_{A} \\ &= \| a^{-1} (a^{-1})^{*} \|_{A} \\ &= \| a^{-1} \|_{A}^{2} \\ &= 1. \end{align} By Theorems $ 1 $ and $ 2 $, we therefore obtain $ {\sigma_{A}}(a^{*} a) = \{ 1 \} $.

Let us now work with the Continuous Functional Calculus corresponding to the self-adjoint element $ a^{*} a $. As $ {\sigma_{A}}(a^{*} a) = \{ 1 \} $, applying both the constant function $ 1 $ and the identity function to $ a^{*} a $ yields $$ \mathbf{1}_{A} = a^{*} a. $$ Therefore, $ |a| := \sqrt{a^{*} a} = \mathbf{1}_{A} $ and $ a = u|a| = u $.


Conclusion: If $ a \in A $ is invertible and $ \| a \|_{A} = \| a^{-1} \|_{A} = 1 $, then $ |a| = \mathbf{1}_{A} $ and $ a $ is unitary.


Note: In the argument above, we first proved that $ a^{*} a = \mathbf{1}_{A} $, then used polar decomposition to deduce that $ a $ is unitary and hence $ a a^{*} = \mathbf{1}_{A} $. However, we could have proven directly that $ a $ is unitary by performing the argument with $ a^{*} $ instead of $ a $, as the given conditions already imply that $ a^{*} $ is invertible and that $ \| a^{*} \|_{A} = \| (a^{*})^{-1} \|_{A} = 1 $. Once again, we have to use the fact that involution in a C$ ^{*} $-algebra is an isometry.

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Thank you very much, but does this mean that $|a|$ must be the identity of $A$ ? –  user61965 Mar 3 '13 at 3:28
    
in Gerard Murphy's : $C^{*}$ Algebras and Operator Theory, page 73 , the second condition holds iff $a$ is unitary ! –  user61965 Mar 3 '13 at 3:39
    
@edab12: Yes, $ |a| $ must be the identity. –  Haskell Curry Mar 3 '13 at 7:01
    
Yes, I didn't use the polar decomposition to prove that $a$ is unitary, Thank you for ur help! –  user61965 Mar 3 '13 at 7:28

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