# Commutativity criteria for $C^*$-algebras

If for any $x, y\in A$ a $C^*$-algebra, $0\leq x\leq y\implies x^2\leq y^2$, then it is true that A is commutative ?

It is easy to show that the implication $0\leq x\leq y\implies x^2\leq y^2$ is not true in general, for example in $\mathcal{M}_2(\mathbb{R})$ take for $x$ \begin{bmatrix}1&1\\1&1\end{bmatrix} and for $y$ \begin{bmatrix}2&1\\1&1\end{bmatrix} Moreover if $A$ is commutative then $y^2-x^2=(y-x)(y+x)$ is a product of two positive elements that commute with each other, thus $y^2-x^2\geq 0$. Indeed, if $a, b\in A^+$ and $ab=ba$ then $(ab)^*=ab$ and $\sigma (ab)=\sigma (a^{1/2}ba^{1/2})\subset [0, \infty [$.

• The result ("Ogasawara's theorem") seems to be proved in "Quantum Measure Theory, Volume 36" p.236. Unfortunately I can't read the whole proof on Google books. – Patissot Jun 22 '15 at 12:24
• I think you are giving a pure algebraic proof for the commutative case. But it is obvious since a commutative algebra is C(X) or $C_{0}(X)$. – Ali Taghavi Mar 11 '16 at 4:41

Indeed this is known as Ogasawara's theorem, as you pointed out. I like to think of this in terms of representation theory, though this might be total overkill. Basically, I can prove the result in three steps:

1. Prove that $$B(H)$$ contains self-adjoint operators $$x,y$$ satisfying $$0 \leq x \leq y$$ and $$x^2 \not\leq y^2$$ whenever $$\dim(H) > 1$$. You already took care of that by providing an example.
2. Prove that any strongly dense $$C^*$$-subalgebra $$A\subseteq B(H)$$ with $$\dim(H) > 1$$ contains self-adjoint $$x,y$$ satisfying $$0 \leq x \leq y$$ and $$x^2 \not\leq y^2$$. This is the hardest part. I will get back to this.
3. Use representation theory: an arbitrary $$C^*$$-algebra $$A$$ is commutative if and only if every irreducible representation is one-dimensional. (Furthermore, recall that a non-zero representation $$\varphi : A \to B(H)$$ is irreducible if and only if $$\varphi[A]$$ is strongly dense in $$B(H)$$.) Thus, if $$A$$ is not commutative, then there exists an irreducible representation $$\varphi : A \to B(H)$$ with $$\dim(H) > 1$$. Now by part 2 there are $$x,y\in A$$ such that $$\varphi(x)$$ and $$\varphi(y)$$ are self-adjoint and satisfy $$0 \leq \varphi(x) \leq \varphi(y)$$ as well as $$\varphi(x)^2 \not\leq \varphi(y)^2$$. Now it is not so hard to see that we may choose $$x$$ and $$y$$ self-adjoint and satisfying $$0 \leq x \leq y$$, though this is not completely trivial either. I'll leave this as an exercise. (Hint: first prove that $$\varphi(a^-) = \varphi(a)^-$$ holds for all self-adjoint $$a\in A$$. Then for self-adjoint $$a,b\in A$$ we have $$\varphi(a) \leq \varphi(b)$$ if and only if $$(b - a)^- \in \ker(\varphi)$$ holds.)

I find that most results relating some order-theoretic property of a $$C^*$$-algebra with its commutativity can be proven this way. (For instance, a similar approach can be used to prove Sherman's theorem: the set $$A^{\text{sa}}$$ of self-adjoint elements of a $$C^*$$-algebra $$A$$ is a lattice if and only if $$A$$ is commutative.) It may not be the easiest proof of Ogasawara's theorem, but at least it doesn't really require any deep insight or clever trick, as it relies on the familiar theorems.

Now let's get into step 2 of the solution. There might well be an easier way to do this (perhaps even constructively), but here's what I've come up with. We shall use Kaplansky's density theorem as well as the following theorem used in the proof of said theorem:

Theorem 1. If $$f : \mathbb{R} \to \mathbb{R}$$ is a bounded continuous function and $$\{a_\lambda\}_{\lambda\in\Lambda}$$ is a net of self-adjoint operators converging strongly to $$a$$, then the net $$\{f(a_\lambda)\}_{\lambda\in\Lambda}$$ converges strongly to $$f(a)$$.

This has the following very useful consequence:

Corollary 2. If $$f : \mathbb{R} \to \mathbb{R}$$ is a (not necessarily bounded) continuous function and $$\{a_\lambda\}_{\lambda\in\Lambda}$$ is a norm-bounded net of self-adjoint operators converging strongly to $$a$$, then the net $$\{f(a_\lambda)\}_{\lambda\in\Lambda}$$ converges strongly to $$f(a)$$.

Proof. There is some $$M \in \mathbb{R}_{>0}$$ such that $$\sigma(a_\lambda) \subseteq [-M,M]$$ holds for all $$\lambda\in\Lambda$$, as well as $$\sigma(a) \subseteq [-M,M]$$. Thus, we may cut $$f$$ off at $$x = -M$$ and $$x = M$$ so that $$f$$ becomes bounded. More precisely, we may define $$g : \mathbb{R} \to \mathbb{R}$$ by $$g(x) = \begin{cases} f(-M), & \text{if x < -M};\\ f(x), & \text{if -M \leq x \leq M};\\ f(M), & \text{if x > M}.\end{cases}$$ Now $$f$$ and $$g$$ coincide on $$\sigma(a_\lambda)$$ for all $$\lambda\in\Lambda$$, so we have $$f(a_\lambda) = g(a_\lambda)$$.$$\quad\Box$$

The version of Kaplansky's density theorem that we need is the following:

Theorem 3. Let $$A \subseteq B(H)$$ be a $$C^*$$-subalgebra with strong closure $$B$$. Then the unit ball of $$A^+$$ is stongly dense in the unit ball of $$B^+$$.

Proof (assuming some other form of Kaplansky's density theorem). Should this result not be included in the version of Kaplansky's density theorem that you're used to, let $$b\in B^+$$ be given with $$||b|| \leq 1$$. By some other form of Kaplansky's density theorem, there exists a net $$\{a_\lambda\}_{\lambda\in\Lambda}$$ of self-adjoint elements of $$A$$ converging strongly to $$b$$. Consider the function $$f : \mathbb{R} \to \mathbb{R}$$ given by $$f(x) = \begin{cases} 0, & \text{if x < 0}; \\ x, & \text{if 0\leq x \leq 1} \\ 1, & \text{if x > 1}. \end{cases}$$ By theorem 1 above, the net $$\{f(a_\lambda)\}_{\lambda\in\Lambda}$$ converges strongly to $$f(b)$$. But we have $$f(b) = b$$, and $$f(a_\lambda)$$ is an element of the unit ball of $$A^+$$ for all $$\lambda\in\Lambda$$.$$\quad\Box$$

Finally, we need the following easy lemmas, which I will not prove.

Lemma 4. The positive cone of $$B(H)$$ is strongly closed.

Definition 5. Let $$A$$ be any $$C^*$$-algebra. For any two $$a,b\in A^{\text{sa}}$$ we define $$a \curlyvee b \in A^{\text{sa}}$$ by $$a\curlyvee b = \tfrac{1}{2}(a + b + |a - b|).$$

Lemma 6. For any $$a,b\in A^{\text{sa}}$$, the element $$a \curlyvee b$$ is an upper bound for $$a$$ and $$b$$. Furthermore, if $$a$$ and $$b$$ are comparable, then $$a\curlyvee b$$ is simply the largest of the two.

Now we are ready to prove the result. Suppose that $$A \subseteq B(H)$$ is a strongly dense $$C^*$$-subalgebra of $$B(H)$$, where $$\dim(H) > 1$$ holds. By part 1 of the general solution, there are self-adjoint $$x,y\in B(H)$$ satisfying $$0 \leq x \leq y$$ and $$x^2 \not\leq y^2$$. By theorem 3, there exist norm-bounded nets $$\{a_\lambda\}_{\lambda\in\Lambda}$$ and $$\{b_\mu\}_{\mu\in M}$$ consisting of positive elements in $$A$$ that converge strongly to $$x$$ and $$y$$, respectively. Without loss of generality, we may assume that these nets are indexed by the same directed set $$D$$ (take, if necessary, the direct product of $$\Lambda$$ and $$M$$). The net $$\{a_\lambda - b_\lambda\}_{\lambda\in D}$$ is norm-bounded as well, so by corollary 2 we have $$\lim_{\lambda\in D} a_\lambda \curlyvee b_\lambda = \lim_{\lambda\in D} \tfrac{1}{2}(a_\lambda + b_\lambda + |a_\lambda - b_\lambda|) = \tfrac{1}{2}(x + y + |x - y|) = x\curlyvee y = y.$$ For convenience, write $$c_\lambda = a_\lambda \curlyvee b_\lambda$$. The point of this is that we have $$0 \leq a_\lambda \leq c_\lambda$$ for all $$\lambda\in D$$. This way we can approximate $$x$$ and $$y$$ using two norm-bounded nets $$\{a_\lambda\}_{\lambda\in D}$$ and $$\{c_\lambda\}_{\lambda\in D}$$ such that $$0 \leq a_\lambda \leq c_\lambda$$ holds for all $$\lambda\in\Lambda$$. Using corollary 2 once more, we find $$\lim_{\lambda\in D} c_\lambda^2 - a_\lambda^2 = y^2 - x^2 \notin B(H)^+.$$ Since $$B(H)^+$$ is strongly closed, there must be at least one $$\lambda\in D$$ such that $$c_\lambda^2 - a_\lambda^2 \not\geq 0$$ holds, which proves the claim.

• Thank you for your answer. I got the general idea, however I can't prove some claims. For example, how do you show that if every irreductible representations are one-dimensional then $A$ is commutative (I have found a proof for the converse) ? – Patissot Jul 4 '15 at 9:01
• This is done by proving that the direct sum of all irreducible representations is faithful. This way we obtain a faithful diagonal representation of $A$, so it follows that $A$ is commutative. – Josse van Dobben de Bruyn Jul 4 '15 at 14:19
• Thank you. I only knew the case of a finite dimensional $C*$-algebra, so I was not sure that I can use this generalized result. I guess this is a straightforward application of Zorn's lemma. It remains me to prove the last step of the proof : $\varphi(a^-) = \varphi(a)^-$ for all $s.a.$ operators. – Patissot Jul 5 '15 at 7:48
• I'm not sure you need Zorn's lemma. For arbitrary $a\in A$ there exists an irreducible representation $\varphi : A \to B(H)$ such that $||\varphi(a)|| = ||a||$ holds. Thus, if we have $ab - ba \neq 0$, then there is an irreducible representation $\varphi : A \to B(H)$ such that $||\varphi(ab - ba)|| = ||ab - ba|| > 0$ holds. Therefore $\varphi(a)$ and $\varphi(b)$ do not commute, so it follows that $\dim(H) > 1$ must hold. Where exactly did you use Zorn's lemma? I don't think I see the connection... (So this may be a gap in my knowledge.) – Josse van Dobben de Bruyn Jul 5 '15 at 16:30