Solèr’s Theorem
Posted by John Baez
David Corfield likes theorems that say what’s special about the real numbers — especially theorems where $\mathbb{R}$ emerges unexpectedly at the end, like a rabbit from a magician’s hat. He enjoys guessing how the rabbit got into the hat!
So, his ears perked up when I mentioned my favorite theorem of this type: Solèr’s Theorem. In 1995, Maria Pia Solèr proved a result with powerful implications for the foundations of quantum mechanics. She starts with what sounds like a vast generalization of the concept of infinite-dimensional Hilbert space: a generalization that replaces the complex numbers with an arbitrary ‘division $\ast$-ring’. But then, she shocks us by showing that this ring must be one of these three:
- the real numbers $\mathbb{R}$,
- the complex number $\mathbb{C}$,
- the quaternions $\mathbb{H}$!
These are our old friends, the famous trio: the associative normed division algebras!
Let me tell you what Solèr’s Theorem actually says. I’ll do little more than give the necessary definitions and state the result. I won’t say anything about the proof, and only a bit about the implications for quantum theory. For that, you can read this wonderful paper:
- Samuel S. Holland Jr., Orthomodularity in infinite dimensions; a theorem of M. Solèr, Bull. Amer. Math. Soc. 32 (1995), 205–234. Also available as arXiv:math/9504224.
and then:
- Maria Pia Solèr, Characterization of Hilbert spaces by orthomodular spaces, Comm. Algebra 23 (1995), 219–243.
To state Solèr’s theorem, let’s start by saying what sort of ring her generalized Hilbert spaces will be vector spaces over. We want rings where we can divide, and we want them to have an operation like complex conjugation. A ring is called a division ring if every nonzero element $x$ has an element $x^{-1}$ for which $x x^{-1} = x^{-1} x = 1$. A ring is called a $\ast$-ring if it is equipped with a function $x \mapsto x^*$ for which: $(x + y)^* = x^* + y^*, \qquad (x y)^* = y^* x^* , \qquad (x^*)^* = x.$ A division $\ast$-ring is a $\ast$-ring that is also a division ring.
Of course $\mathbb{R}$, $\mathbb{C}$ or $\mathbb{H}$ are examples of division $\ast$-rings, but there are many more — I’ll list a few later.
Now, suppose $\mathbb{K}$ is a division $\ast$-ring. As before, let’s define a $\mathbb{K}$-vector space to be a right $\mathbb{K}$-module. You can use left modules too — it doesn’t really matter. And as before, let’s say a function $T: V \to V'$ between $\mathbb{K}$-vector spaces is $\mathbb{K}$-linear if $T(v x + w y) = T(v)x + T(w)y$ for all $v,w \in V$ and $x,y \in \mathbb{K}$.
We define a hermitian form on a $\mathbb{K}$-vector space $V$ to be a map $\begin{aligned} V \times V & \to & \mathbb{K} \\ (u,v) &\mapsto& \langle u,v \rangle \end{aligned}$ which is $\mathbb{K}$-linear in the second argument and obeys the equation $\langle v, u \rangle = \langle u, v \rangle^*$ for all $u, v \in V$. We say a hermitian form is nondegenerate if $\langle u, v \rangle = 0 \; for \; all \; u \in V \quad \iff \quad v = 0 .$
Suppose $H$ is a $\mathbb{K}$-vector space with a nondegenerate hermitian form. Then we can define various concepts familiar from the theory of Hilbert spaces. For example:
- We say a sequence $e_i \in H$ is orthonormal if $\langle e_i, e_j \rangle = \delta_{ij}$.
- For any vector subspace $S \subseteq H$, we define the orthogonal complement $S^\perp$ by $S^\perp = \{ v \in H \colon \; \langle u, v \rangle = 0 \; for \; all \; u \in S \} .$
- We say $S$ is closed if $S^{\perp \perp} = S$. (If $H$ is a real, complex, or quaternionic Hilbert space this is true iff $S$ is closed in the usual topological sense, but here we have no topology!)
- And finally, we say $H$ is orthomodular if $S + S^\perp = H$ for every closed subspace $S$. (This is automatically true if $H$ is a real, complex, or quaternionic Hilbert space.)
We are now ready to state Solèr’s theorem!
Theorem: Let $\mathbb{K}$ be a division $\ast$-ring, and let $H$ be a $\mathbb{K}$-vector space equipped with an orthomodular hermitian form for which there exists an infinite orthonormal sequence. Then $\mathbb{K}$ is isomorphic to $\mathbb{R}, \mathbb{C}$ or $\mathbb{H}$, and $H$ is an infinite-dimensional $\mathbb{K}$-Hilbert space.
Nothing in the assumptions mentions the continuum: the hypotheses are purely algebraic. It therefore seems quite magical that $\mathbb{K}$ is forced to be the real numbers, complex numbers or quaternions.
The orthomodularity condition is the key: this is how the rabbit gets into the hat! It has a long history. In quantum logic, unlike classical logic, ‘and’ may fail to distribute over ‘or’. You can see this by pondering subspaces of Hilbert spaces, or more vividly, by pondering a particle on a line. If you think about it a while, you’ll see that the distributive law fails because of the uncertainty principle.
I’m not sure, but I believe it was John von Neumann who proposed the modular law as a weakened substitute. The distributive law says:
$p \; and \;(q \;or \;r) \; = \; (p \; and \; q) \; or \; (p \;and \; r)$
while the modular law says “at least this is true if $(p \;and \; q) = q$”. In other words, the modular law says the distributive law holds when $q$ implies $p$.
You can check that the modular law holds when our propositions are subspaces of a vector space, ‘$and$’ is the intersection of subspaces (usually called $\cap$), and ‘$or$’ is the linear span of subspaces (usually called $+$).
But in quantum theory, our propositions are closed subspaces of a Hilbert space, ‘and’ is the intersection, and ‘or’ is the closure of the linear space. For finite-dimensional Hilbert spaces, this fine print — italicized here — doesn’t make any difference. But for infinite-dimensional Hilbert spaces the fine print matters, and the modular law fails.
Puzzle: find a counterexample.
So, we need a further fallback position, and this is called the orthomodular law. This involves the logical operation “not” as well. The orthomodular law says that the distributive law holds:
$p \; and \; (q \;or \; r) = (p \; and \; q) \;or \; (p \; and \; r)$
if $r$ implies $p$ and also $q = not(r)$. Unraveling this a bit, it says:
$p = (p \; and \;q) \;or \; (p \;and \; not(q))$
When our propositions are closed subspaces of a Hilbert space, “not” is the orthogonal complement. This is also true when our propositions are closed subspaces in a $\mathbb{K}$-vector space $H$ equipped with a hermitian form, where now $\mathbb{K}$ can be any division $\ast$-ring. And I’m pretty sure that in this case, the lattice of closed subspaces is orthomodular if and only if $H$ is orthomodular in Solèr’s sense — namely, for every closed subspace $S \subseteq H$ we have $S + S^\perp = H$.
That would explain her use of the word ‘orthomodular’. But there’s a lot more history behind these ideas. In 1964, Piron had attempted to prove that any orthomodular complex inner product space must be a Hilbert space. His proof had a flaw which Ameniya and Araki fixed a few years later. Their proof also handles the real and quaternionic cases. Eventually these ideas led Solèr to realize that orthomodularity picks out real, complex, and quaternionic Hilbert spaces as special. And her results have implications for other axiomatic approaches to quantum theory: for example, approaches using orthomodular lattices and convex cones. For these, read Holland’s paper.
Some of you were asking about examples of division $\ast$-rings. For starters, you can take any division ring and set $x^* = x$. So, for example: take the field of rational functions of 7 variables, or the field with 7 elements.
But it’s more fun to find examples where $x^* \ne x$. So, for example: take the field with 7 elements, which does not have a square root of $-1$, but then formally adjoin a square root of minus $-1$, and get a field with 49 elements. This new field has an obvious concept of ‘complex conjugation’, so it becomes a division $\ast$-ring. Or take the field of rational functions on some algebraic variety $X$ that has an ‘involution’, a map $f: X \to X$ with $f^2 = 1$, and do the obvious thing.
The examples I’ve given so far are all commutative.
Puzzle: what are some noncommutative division $\ast$-rings?
Of course finite division rings are all commutative, and the only finite-dimensional associative division algebras over the real numbers are our three friends. But I don’t know much about infinite-dimensional noncommutative (but associative) division algebras over the real numbers!
Also, some of you were asking about how the orthomodularity condition rules out finite fields. Let’s look at the simplest example. Remember that $\mathbb{F}_2$, the field with two elements, is what normal mortals call the integers mod two. We can make this field into a division $\ast$-ring with $x^* = x$. Then we can put a hermitian form on the 2-dimensional vector space $H = \mathbb{F}_2^2$ in the obvious way: $\langle u, v \rangle = u_1 v_1 + u_2 v_2$ The vector $(1,1)$ is orthogonal to itself, since $1^2 + 1^2 = 0$. So, if we let $S$ be the subspace of $H$ spanned by this vector, we have $S \subseteq S^\perp$ which seems weird if you’re used to Hilbert spaces. Indeed it’s easy to see $S = S^\perp ,$ so $S$ is closed, but we certainly don’t have $S + S^\perp = H$ So, Solèr’s orthomodularity condition fails!
This example is finite-dimensonal, but it’s easy to see that the same problem shows up in some infinite-dimensional cases too.
It would also be fun to look at some other examples, for example infinite-dimensional vector spaces over the rational numbers…
Puzzle: Pick an infinite-dimensional rational vector space with a hermitian form on it, and show it isn’t orthomodular.
Re: Solèr’s Theorem
A couple of typos:
In the definition of *-ring, “y*x*” should have the second “*” as superscript.
After “And as before, let’s say a function” – “maps” is missing its slash
“And results have” should be “And these results have”?