2
$\begingroup$

I am reading about $C^*$-algebras from Chapter VIII in Conway's A Course in Functional Analysis. I've come across the following proposition which describes the "orthogonal decomposition" for hermitian elements of a $C^*$-algbera:

If $\mathcal{A}$ is a $C^*$-algebra with identity $1$ and $a \in \mathcal{A}$ is hermitian, then there exist unique positive elements $u,v \in \mathcal{A}$ such that $a = u - v$ and $0 = uv = vu$.

Here, an element $b \in \mathcal{A}$ is positive if $b$ is hermitian and the spectrum $\sigma(b) \subseteq [0, \infty)$.

Conway constructs $u$ and $v$ using the continuous functional calculus, and I understand the construction well enough.

However, my difficulty is understanding the uniqueness part of the proof.

Here's is my attempt to understand Conway's uniqueness proof. Suppose we also have positive $u_1, v_1$ that satisfy $a = u_1 - v_1$, $0 = u_1v_1 = v_1u_1$. Then one can use functional calculus to show that $a, u, v, u_1, v_1$ are pairwise commuting elements of $\mathcal{A}$. If we then let $\mathcal{B}$ be the $C^*$-algebra generated by $1, a, u,v, u_1,$ and $v_1,$ then $\mathcal{B}$ is abelian and so we know that the Gelfand transform $\gamma: \mathcal{B} \to C(\Sigma)$, $\mathcal{B} \ni b \mapsto \hat{b}$, is an isometric $\ast$-isomorphism. Here $\Sigma$ is the maximal ideal space of $\mathcal{B}$ (i.e., the set of all non-zero homomorphisms $h: \mathcal{B} \to \mathbb{C}$, equipped with the weak$^*$ subspace topology inherited from $\mathcal{B}^*$).

Conway then asks the reader to use uniqueness in $C(\Sigma)$ to finish the proof, but I am not sure how to do this. One thing I am aware of is that since $u, v, u_1, v_1$ are all positive elements, their Gelfand transforms are nonnegative functions on $\Sigma$ (since $\hat{b}$ takes values in $\sigma(b)$).

Hints or solutions are greatly appreciated.

$\endgroup$
2
$\begingroup$

In the $C^*$-algebra $C(\Sigma)$ every self-adjoint (=real valued) function $f$ has the unique decomposition $f=f_+-f_-,$ where $f_+,f_-$ are positive and $f_+f_-=0.$

Now use the fact that the Gel'fand transform $\hat b$ of $b\in\mathcal B$ is a positive function $\Longleftrightarrow$ $b$ is a positive element of $\mathcal B$ $\Longleftrightarrow$ $b$ is a positive element of $\mathcal A.$

I hope my hint will be enough.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.