# On existence of square root of positive elements of a unital $C^*$-algebra

Given a unital $\mathcal{C}^*$-algebra $A$ and a positive element $a \in A$, I am trying to prove the existence of a square root $a^{\frac{1}{2}}$ i.e. a positive element $b \in A$ such that $b^2 = a$.

My strategy goes like this :

• Restrict to a maximal commutative closed subalgebra $B$ containing the unit, $a$ and $a^*$ to allow using the Gelfand-Naimark theorem (i.e. $B \cong \mathcal{C}(\Phi_B)$ where $\Phi_B$ stands for the set of characters on $B$);

• Show that $\widehat{a} \restriction \Phi_B$ (that is the continuous function that evaluates to $a$ all characters on $B$) has its image in $[0, \infty)$;

• Define $\widehat{b}$ to be the pointwise square root of $\widehat{a}$;

• Show that $\widehat{b}$ is hermitian and positive (easy given previous work);

At some point I got the following :

$$\widehat{a}[\Phi_B] ~=~ \sigma_{\mathcal{C}(\Phi_B)}(\widehat{a}) ~=~ \sigma_B(a)~=~ \sigma_A(a) ~\subseteq~ [0, \infty).$$

The first equality is because the point-spectrum of a continuous function on a compact is it's image. The second equality is due to Gelfand-Naimark theorem. The third equality is true for all maximal commutative subalgebra and the inequality is due to the fact that $a$ is a positive element.

My concern is the following : I'm using the fact that $B$ is maximal commutative subalgebra and I'm also using the fact that $B$ is a closed subalgebra (in Gelfand-Naimark). But are maximal commutative subalgebras always closed ? Is my hypothesis too strong to hold ?

A maximal commutative subalgebra containing the unit, $a$ and $a^*$ surely exists. It's closure surely exists, but is the commuativity preserved ? Because if not, we might get $\sigma_B(a) \supseteq \sigma_A(a)$ in which case the whole proof is no longer valid.

• I guess the maximalily is not necessary if we can prove that the closure of the subalgebra spanned by the unit, $a$ and $a^*$ (call it $C$) preserve the point-spectrum of $a$ (i.e $\sigma_C(a) = \sigma_A(a)$) but this is not obvious to me.
– M.G
Commented May 31, 2015 at 9:14
• Dixmier, $C^*$-algebras, statement 1.3.10, for a ref (even without commutativity the spectrum does not depend on the choice of $C^*$ subalgebra). Commented May 31, 2015 at 10:21
• Proposition (1.23) of Folland's "Abstract harmonic analysis" shows that $\sigma_A (x) =\sigma_B (x)$ if $A$ is any unital $C^\ast$ algebra and $B\subset A$ is a $C^\ast$ subalgebra containing $x$ and the unit of $A$. If you are interested, I can post the proof. Commented May 31, 2015 at 10:23
• Thank you for the references. If you can copy this in an answer I could class this question as "answered"
– M.G
Commented May 31, 2015 at 12:41
• I mean "if you can copy your comment", not the proof in the reference.
– M.G
Commented May 31, 2015 at 12:42

If $B\subset A$ are C$^*$-algebras with the same unit and $b\in B$, then $\sigma_B(b)=\sigma_A(b)$.

Here is the argument as outlined in Murphy's book (very similar argument in Conway's Functional Analysis).

Assume first that $b$ is selfadjoint. Then one can use the fact that in a Banach algebra, if the spectrum has no holes then it does not depend on the algebra. It applies here because, since $b=b^*$, $\sigma(b)\subset\mathbb R$.

For the general case, we want to show that if $b-\lambda 1$ is not invertible in $B$ then it is not invertible in $A$. Or, what is the same, to show that if some element $b$ is invertible in $A$, then it is invertible in $B$. So there exists $a\in A$ with $ab=ba=1$. Taking adjoints, $a^*b^*=b^*a^*=1$. Then $$aa^*b^*b=ab=1,\ \ b^*baa^*=b^*a^*=1$$ making $b^*b$ invertible in $A$. By the first part of the proof, the spectrum of $b^*b$ is the same in $B$ as it is in $A$, so $b^*b$ is invertible in $B$; that is, there exists $c\in B$ with $cb^*b=b^*bc=1$. Now we look, in $A$, at the equations $$(cb^*)b=1=ab$$ and we conclude that $a=cb^*\in B$. So $b$ is invertible in $B$.

• why doesn't have hole Commented Nov 9, 2018 at 1:03
• Because it's a subset of $\mathbb R$. Commented Nov 9, 2018 at 2:14
• but for $(1,2)\cup (5,6)$ have hole. is not conex Commented Nov 9, 2018 at 2:54
• That's not a hole. The complement of your set is connected. Commented Nov 9, 2018 at 3:04
• Yes. Doing operator theory/Banach algebras over the reals is possible but it only adds missery. Commented Nov 9, 2018 at 4:12