# What algebraic structure do self-adjoint operators form?

Consider the $\mathbb{C}$-algebra of all matrices of dimension $n$ over the complex numbers, $Mat_n\mathbb{C}$;

We have here a notion of adjointness which is an involution; and thus of self-adjointness; what algebraic structure do all self-adjoint elements form? It's at least a linear space; but we don't have $(AB)^*=A^*B^*$, so it can't be a sub-algebra.

• It is a closed cone. Google positive cone of a C*-algebra – Norbert May 17 '15 at 15:28

Let $A$ be any $C^*$-algebra and let $A^{\text{sa}}$ denote the set of self-adjoint elements. Note that $A^{\text{sa}}$ is only a real vector space, not a complex one: for $x\in A^{\text{sa}} \setminus \{0\}$ and $\lambda\in\mathbb{C}$ we have $\lambda x \in A^{\text{sa}}$ if and only if $\lambda\in\mathbb{R}$ holds. As you pointed out, $A^{\text{sa}}$ is not a subalgebra, since the product of self-adjoint elements is not in general self-adjoint: $$\begin{pmatrix} -1 & 0 \\ 0 & 1 \end{pmatrix}\begin{pmatrix} 1 & 1 \\ 1 & 1\end{pmatrix} = \begin{pmatrix} -1 & -1 \\ 1 & 1\end{pmatrix}.$$ However, with the Jordan product $x \circ y = \tfrac{1}{2}(xy + yx)$ you enter the realm of Jordan operator algebras. I'm not really familiar with this subject, but you may want to look into it.

Apart from that, the positive semi-definite cone gives rise to some interesting order structure (though one might argue that order structure doesn't classify as algebraic structure). For $x,y\in A^{\text{sa}}$ we write $x \leq y$ whenever $y - x$ is positive semi-definite (that is, we have $\sigma(y - x) \subseteq \mathbb{R}_{\geq 0}$, where $\sigma(y - x)$ denotes the spectrum of $y - x$). This way $A^{\text{sa}}$ becomes a partially ordered real vector space. This ordered vector space turns out to be quite nice: for instance one can prove that $A^{\text{sa}}$ is both Archimedean and directed, and $*$-homomorphisms automatically preserve the order structure (in the sense that $x \leq y$ implies $\varphi(x) \leq \varphi(y)$ whenever $\varphi$ is a $*$-homomorphism between $C^*$-algebra's).

The remaining discussion assumes some familiarity with the theory of $C^*$-algebras. Interestingly, there are several theorems relating the order structure of $A^{\text{sa}}$ to the commutativity of $A$:

• Sherman's theorem: $A^{\text{sa}}$ is a lattice if and only if $A$ is commutative.
• Ogasawara's theorem: we have $0 \leq a \leq b$ implies $a^2 \leq b^2$ if and only if $A$ is commutative.
• Fukamiya–Misonou–Takeda theorem: $A^{\text{sa}}$ has the Riesz decomposition property (for $a,b,z\in A^{\text{sa}}$ with $0\leq z \leq a + b$ there exist $u,v\in A^{\text{sa}}$ with $z = u + v$ and $0 \leq u \leq a$ as well as $0 \leq v \leq b$) if and only if $A$ is commutative.
• We have $a \circ b \geq 0$ for all $a,b\in A^+$ if and only if $A$ is commutative.

For each of these theorems, one of the directions is easy: if $A$ is commutative, then it is $*$-isomorphic to $C_0(\Omega)$ for some locally compact Hausdorff space $\Omega$ by the Gelfand representation. Then all of these properties can easily be proven. Proving the reverse direction is much harder. (An overview of all these commutativity criteria is provided in a paper by David M. Topping, although his proofs are quite difficult. I find it easier to think of these theorems in terms of representation theory, as I did in this proof of Ogasawara's theorem.)

As the $C^*$-algebra $\text{Mat}_n(\mathbb{C})$ is not commutative for $n \geq 2$, it follows that none of these nice order properties hold for the space of self-adjoint elements of $\text{Mat}_n(\mathbb{C})$: it is not a lattice, it does not have the Riesz decomposition property, etcetera.