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Definition: Let $R, S$ be two rings. A classical morphism $\phi : R \to S$ is a function from elements of $R$ to elements of $S$ which restricts to a homomorphism (of rings, in the usual sense) on commutative subrings of $R$.

This definition is motivated by quantum mechanics; roughly speaking $\phi$ preserves what a classical observer can observe about the noncommutative spaces $\text{Spec } R$ and $\text{Spec } S$. See the discussion at the nLab page on the Bohr topos. Actually it should be more like this:

Definition: Let $R, S$ be two $^\ast$-rings. A classical morphism $\phi : R \to S$ is a function from normal elements of $R$ (elements such that $r^{\ast} r = r r^{\ast}$) to normal elements of $S$ which restricts to a $^{\ast}$-homomorphism on commutative $^{\ast}$-subrings of $R$.

This definition allows, among other things, an elegant statement of the Kochen-Specker theorem, which can be restated as the claim that if $H$ is a Hilbert space of dimension at least $3$, then the algebra $B(H)$ of bounded linear operators $H \to H$ does not admit a classical morphism to $\mathbb{C}$.

Has this definition been studied from a purely ring theory or noncommutative geometry point of view? Have basic properties of the corresponding category been worked out somewhere?

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Is there an example (for general rings, using the first definition) that is not a ring homomorphism? – zyx Jun 20 '12 at 5:24
@zyx: any anti-homomorphism ($\phi(ab) = \phi(b) \phi(a)$) whose image is noncommutative gives an example. I think a counterexample to Kochen-Specker when $\dim H = 2$ should also give an example. But of course it would be good to have more sources of examples! – Qiaochu Yuan Jun 20 '12 at 5:55
Are the rings unital? And do subrings share unities with their superrings? – user23211 Jun 30 '12 at 13:23
@ymar: I would be happy to hear mentions of any unit-related variation on this definition. Probably it is most physically natural to say no and no. – Qiaochu Yuan Jun 30 '12 at 16:07
Then I think I have an example that's neither a homomorphism nor an antihomomorphism. I'm not sure how well it generalizes but for what it's worth: let $R=\{0,a,b,c\}$ be the Klein 4-ring. Let $S=M_{2\times 2}(\mathbb F_2).$ Let $f:R\to S$, $0\mapsto 0,\,a\mapsto\begin{pmatrix}0&1\\0&1\end{pmatrix},\,b\mapsto\begin{pmatrix}0&1\\0&0‌​\end{pmatrix},\,c\mapsto\begin{pmatrix}1&1\\0&0\end{pmatrix}.$ It's not a homomorphism or an antihomomorphism because $a+b=c$ and $f(a)+f(b)\neq f(c).$ – user23211 Jun 30 '12 at 16:50
up vote 6 down vote accepted

In a recent paper, I adapted some of the terminology of Kochen and Specker's original paper to a more ring-theoretic context. I would refer to your first type of morphism as a morphism of partial $\mathbb{Z}$-algebras from $R$ to $S$, or even better as a morphism of partial rings. The point is that every ring has the underlying structure of a partial ring (i.e., there is are categories of rings and of partial rings, and a forgetful functor from rings to partial rings). The functions you are thinking about are the morphisms of the second category.

Regarding your second definition, it may be helpful for you to look at this paper of Benno van den Berg and Chris Heunen. They define the category of partial C*-algebras. Given a C*-algebra $A$, one may only consider $A$ itself as a partial C*-algebra if $A$ is commutative. But the subset $N(A)$ of normal elements of $A$ is always a partial C*-algebra, even if $A$ is not commutative. In fact, "taking the normal part" forms a functor from the category of C*-algebras to the category of partial C*-algebras.

So a clean way to fit all of these ideas together would be to define the category of "partial *-rings" in the obvious way, so that the set $N(R)$ of normal elements of any *-ring $R$ is a partial *-ring and so that the functions you describe above are exactly the morphisms of partial *-rings from $N(R)$ to $N(S)$.

Heunen and van den Berg have worked out some interesting properties of the category of partial C*-algebras. If you need particular facts of theirs for partial *-rings, I would imagine that many of their results should translate easily.

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Thanks! This is exactly what I was looking for. – Qiaochu Yuan Aug 13 '12 at 18:39

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