A $\star$-ring is a ring with an involutive anti-automorphism. The simplest example of a noncommutative $\star$-ring is perhaps $B(l^2)$, the ring of bounded linear functions on the sequence space $l^2$. (This ring is in fact a C$\star$-algebra.)

Is there a characterization of the commutative $\star$-subrings of $B(l^2)$?

  • $\begingroup$ thank you Ricky. There really some trouble with formatting the stars. $\endgroup$ – user2529 Aug 14 '11 at 5:11
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    $\begingroup$ (2-by-2 matrices seems simpler than $B(\ell^2)$.) You want no additional structure beyond being commutative $\star$-subrings? That is, you don't require them to be algebras (closed under scalar multiplication), and you don't require them to be closed in any topology? $\endgroup$ – Jonas Meyer Aug 14 '11 at 5:29
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    $\begingroup$ You can note is that any abelian $*$-sub-ring of $B(\ell^2)$ will be contained in a unital abelian $C^*$-algebra and so will be a $*$-subring of the ring of continuous complex-valued functions on a compact topological space. So your question reduces to finding a characterization of such $*$-rings -- I am not sure if there is an algebraic one. One positive thing that can be said is that the maximal abelian $*$-subalgebras of $B(\ell^2)$ can be completely understood; they are obtained by choosing an isomorphism $\ell^2\cong L^2(X,\mu)$ and considering $L^\infty(X,\mu)\subset B(L^2(X,\mu))$. $\endgroup$ – Dima Shlyakhtenko Aug 14 '11 at 5:29
  • $\begingroup$ In light of Dima Shlyakhtenko's comment, I suggest that the title be changed to reflect the fact that $B(\ell^2)$ is superfluous. $\endgroup$ – Yemon Choi Aug 14 '11 at 6:54
  • $\begingroup$ Downvoting since I increasingly feel that the question should have had more thought put into it (see my previous comment) $\endgroup$ – Yemon Choi Aug 26 '11 at 8:35

A unital $*$-ring $A$ (commutative or not) is a subring of $B(H)$ if and only if for each $a \in A$, there exists a linear functional $\varphi \colon A \to \mathbb R$, such that

1) $\varphi(1)=1$ and $\varphi(b^*b) \geq 0$, for all $b \in A$,

2) For all $b \in R$, there exists a constant $C(b)$ (not depending on $\varphi$), such that for all $c \in R$ $$\varphi(c^*b^*bc) \leq C(b) \cdot \varphi(c^*c),$$

3) $\varphi(a) \neq 0$.

In principle, this a condition on the cone of sums of hermitean squares $\Sigma^2 A$ inside $A$. For example, you clearly need $\Sigma^2 A \cap (- \Sigma^2 A) = \{0\}$. However, in practice it is difficult to check directly whether $A$ satisfies this property or not.

  • $\begingroup$ Are things noticeably simpler in the commutative case? $\endgroup$ – Yemon Choi Aug 26 '11 at 8:35
  • $\begingroup$ In concrete commutative cases, one can sometimes describe the cone of sums of squares. In general, commutativity does not help as far as I can see. $\endgroup$ – Andreas Thom Aug 26 '11 at 17:02

Let $A$ be a (say finitely generated, unital) commutative complex $\star$-subalgebra of $B(H)$. Then, the self-adjoint elements form a real subalgebra $B:=A_h \subset A$, such that $B[i] = A$. Moroever, in $B$, we have $$a_1^2 + \cdots a_n^2 =0 \quad \Rightarrow \quad a_1= \cdots=a_n =0,\quad \forall n.$$

The second condition is sometimes called real-reduced. (It implies in particular, that there are no nilpotent elements, hence the terminology.) I claim that these two conditions are also sufficient.

Theorem: Let $B$ be any finitely generated, unital, real and real-reduced algebra (with trivial involution). Then, $B[i]$ (with the obvious conjugate-linear involution) is a $\star$-subalgebra of $B(H)$.

Sketch of proof: The $\mathbb R$-points of $B$ form a closed real-algebraic variety, contained in $\mathbb R^n$ for some $n$. We may pick a bounded subset $X \subset \mathbb R^n$ and find $B[i] \subset C(X,\mathbb C) \subset B(L^2(X,\mu))$, for some suitable measure on $X$. q.e.d.

Finite generation of $B$ is necessary to give an easy characterization. Indeed, if $\bf R$ is some proper real closed extension of $\mathbb R$, then $B=\bf R$ is real-reduced, but ${\bf R} \not \subset B(H)$ by Mazur's theorem, which says that every subfield of $B(H)$ is either $\mathbb R$ or $\mathbb C$.


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