## November 27, 2010

### State-Observable Duality (Part 2)

#### Posted by John Baez

This is the second part of a little story about the foundations of quantum mechanics.

In the first part, I introduced the heroes of our drama: the real numbers $\mathbb{R}$, the complex numbers $\mathbb{C}$, or the quaternions $\mathbb{H}$. I also mentioned their crazy uncle, who mainly stays locked up in the attic making strange noises: the octonions, $\mathbb{O}$.

When our three heroes were sent down from platonic heaven to tell the world about the algebraic structure of quantum mechanics, they took on human avatars and wrote this paper:

• Pascual Jordan, John von Neumann and Eugene Wigner, On an algebraic generalization of the quantum mechanical formalism, Ann. Math. 35 (1934), 29–64.

Then, in the final episode, we’ll meet the Koecher–Vinberg classification of convex homogeneous self-dual cones, and see how it’s really all about state-observable duality.

In 1932, Pascual Jordan tried to isolate some axioms that an ‘algebra of observables’ should satisfy. The unadorned phrase ‘algebra’ usually signals an associative algebra, but this not the kind of algebra Jordan was led to. In both classical and quantum mechanics, observables are closed under addition and multiplication by real scalars. In classical mechanics we can also multiply observables, but in quantum mechanics this becomes problematic. After all, given two bounded self-adjoint operators on a complex Hilbert space, their product is self-adjoint if and only if they commute!

However, in quantum mechanics one can still raise an observable to a power and obtain another observable. From squaring and taking real linear combinations, one can construct a commutative product: $a \circ b = \frac{1}{2}((a+b)^2 - a^2 - b^2) = \frac{1}{2}(a b + b a) .$ This product is not associative, but it is power-associative: any way of parenthesizing a product of copies of the same observable $a$ gives the same result. This led Jordan to define what is now called a formally real Jordan algebra: a real vector space with a bilinear, commutative and power-associative product satisfying $a_1^2 + \cdots + a_n^2 = 0 \quad \implies \quad a_1 = \cdots = a_n = 0$ for all $n$. The last condition gives $A$ a partial ordering: if we write $a \le b$ when the element $b - a$ is a sum of squares, it says $a \le b \; and \; b \le a \; \quad \implies \quad a = b .$ So, in a formally real Jordan algebra it makes sense to speak of one observable being ‘greater’ than another.

In 1934, Jordan published a paper with von Neumann and Wigner classifying finite-dimensional formally real Jordan algebras. They began by proving that any such algebra is a direct sum of ‘simple’ ones. A formally real Jordan algebra is simple when its only ideals are $\{0\}$ and $A$ itself, where an ideal is a vector subspace $B \subseteq A$ such that $b \in B$ implies $a \circ b \in B$ for all $a \in A$.

And then, they proved:

Theorem: Every simple finite-dimensional formally real Jordan algebra is isomorphic to one on this list:

• The algebras $\mathrm{h}_n(\mathbb{R})$ of $n \times n$ self-adjoint real matrices with the product $a \circ b = \frac{1}{2}(a b + b a)$.
• The algebras $\mathrm{h}_n(\mathbb{C})$ of $n \times n$ self-adjoint complex matrices with the product $a \circ b = \frac{1}{2}( a b + b a)$.
• The algebras $\mathrm{h}_n(\mathbb{H})$ of $n \times n$ self-adjoint quaternionic matrices with the product $a \circ b = \frac{1}{2}(a b + b a)$.
• The algebras $\mathrm{h}_n(\mathbb{O})$ of $n \times n$ self-adjoint octonionic matrices with the product $a \circ b = \frac{1}{2}(a b + b a)$but only for $n \le 3$!
• The spin factors, $\mathbb{R}^n \oplus \mathbb{R}$ with the product $(x,t) \circ (x', t') = (t x' + t' x, x \cdot x' + t t').$

Here we say a square matrix $T$ is self-adjoint if $T_{j i} = (T_{i j})^*$. Remember, last time we set up a theory of Hilbert spaces over $\mathbb{K} = \mathbb{R},\mathbb{C},$ or $\mathbb{H}$. In these cases, we can identify a self-adjoint $n \times n$ matrix with an operator $T : \mathbb{K}^n \to \mathbb{K}^n$ that is self-adjoint in the sense that $\langle T v, w \rangle = \langle v, T w \rangle$ for all $v,w \in \mathbb{K}^n$. In the octonionic case we do not know what Hilbert spaces and operators are, but we can still work with matrices. Curiously, in this case we cannot go beyond $3 \times 3$ self-adjoint matrices and still get a Jordan algebra. The $1 \times 1$ self-adjoint octonionic matrices are just the real numbers, and the $2 \times 2$ ones form a Jordan algebra that is isomorphic to a spin factor. The $3 \times 3$ self-adjoint octonionic matrices are the really interesting case: these form a 27-dimensional formally real Jordan algebra called the exceptional Jordan algebra.

What does all this mean for physics? The spin factors have an intriguing relation to special relativity, since $\mathbb{R}^n \oplus \mathbb{R}$ can be identified with $(n+1)$-dimensional Minkowski spacetime, and its cone of positive elements is then revealed to be none other than the future lightcone. Furthermore, we have some interesting coincidences:

• The Jordan algebra $\mathrm{h}_2(\mathbb{R})$ is isomorphic to the spin factor $\mathbb{R}^2 \oplus \mathbb{R}$.
• The Jordan algebra $\mathrm{h}_2(\mathbb{C})$ is isomorphic to the spin factor $\mathbb{R}^3 \oplus \mathbb{R}$.
• The Jordan algebra $\mathrm{h}_2(\mathbb{H})$ is isomorphic to the spin factor $\mathbb{R}^5 \oplus \mathbb{R}$.
• The Jordan algebra $\mathrm{h}_2(\mathbb{O})$ is isomorphic to the spin factor $\mathbb{R}^9 \oplus \mathbb{R}$.

This sets up a relation between the real numbers, complex numbers, quaternions and octonions and the Minkowski spacetimes of dimensions 3,4,6 and 10. These are precisely the dimensions where a classical superstring Lagrangian can be written down! Far from being a coincidence, this is the tip of a huge and still not fully fathomed iceberg, which John Huerta is digging into.

The exceptional Jordan algebra remains mysterious. Practically ever since it was discovered, physicists have looked for some application of this entity. For example, when it was first found that quarks come in three colors, Okubo and others hoped that $3 \times 3$ self-adjoint octonionic matrices might serve as observables for these exotic degrees of freedom. Alas, nothing much came of this. More recently, people have discovered some relationships between 10-dimensional string theory and the exceptional Jordan algebra, arising from the fact that the $2 \times 2$ self-adjoint octonionic matrices can be identified with 10-dimensional Minkowski spacetime. This lets us think of the exceptional Jordan algebra as built from scalars, spinors and vectors in 10d spacetime. But the full significance of this remains mysterious, at least to me.

In 1983, Zelmanov generalized the Jordan–von Neumann–Wigner classification to the infinite-dimensional case, working with Jordan algebras that need not be formally real. In any formally real Jordan algebra, the following peculiar law holds: $(a^2 \circ b) \circ a = a^2 \circ (b \circ a).$ Any vector space with a commutative bilinear product obeying this law is called a Jordan algebra. Zelmanov classified the simple Jordan algebras and proved they are all of three kinds: a kind generalizing the Jordan algebras of self-adjoint matrices, a kind generalizing the spin factors, and a kind generalizing the exceptional Jordan algebra.

This theorem is part of a massive development of Jordan algebra theory carried out by Zelmanov in the 1970’s and 1980’s. For a good introduction to this, try:

• Kevin McCrimmon, A Taste of Jordan Algebras, Springer, Berlin, 2004, 562 pages.

If you look at the size of this book, you may wonder why it’s called “a taste” — it looks like quite a large helping! But in fact it doesn’t contain a proof of Zelmanov’s classification theorem, which is apparently very deep; instead, after a very readable account of the basics, it goes through the proof of a preliminary result, in which Zelmanov showed that no infinite-dimensional simple Jordan algebras are exceptional: that is, they all live inside associative algebras, with a product of the form $a \circ b = \frac{1}{2}(a b + b a)$. There are, however, two 27-dimensional exceptional Jordan algebras. There’s the one described above, and its sister, which is defined the same way, but with the so-called split octonions taking the place of the octonions.

For a very enjoyable tour of Jordan algebras before Zelmanov did his thing, try:

• Kevin McCrimmon, Jordan algebras and their applications, Bull. Amer. Math. Soc. 84 (1978), 612–627. Freely available online from the AMS and Project Euclid.

As you’ll see, there are important connections between Jordan algebras and geometry that I haven’t mentioned here. This fun essay has been refashioned and expanded into a long introductory section in McCrimmon’s book.

Alas, Zelmanov’s classification theorem does not highlight the special role of the reals, complex numbers and quaternions. Indeed, when we drop the ‘formally real’ condition, a host of additional finite-dimensional simple Jordan algebras appear, beside those in Jordan, von Neumann and Wigner’s classification theorem. These seem to have little to do with the foundations of quantum theory, although the exceptional Jordan algebra built from split octonions does make an appearance in string theory.

So, do $\mathbb{R}$, $\mathbb{C}$ and $\mathbb{H}$ play some privileged role in the study of quantum systems with infinitely many degrees of freedom, like quantum fields — or not? The best result along these lines is the amazing theorem of Maria Pia Solèr:

• S. 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.

Starting with simple hypotheses that don’t even mention topology or the real numbers, she is led to three choices: infinite-dimensional Hilbert spaces over $\mathbb{R}$, $\mathbb{C}$ or $\mathbb{H}$. Zounds!

Since David Corfield has expressed interest in this theorem, I may try to describe it in a future post. But my goal for next time is to tell you how the result of Jordan, von Neumann and Wigner reappears as a classification of certain cones: cones that can be used to describe nonnegative observables, but also ‘mixed states’. And here is where we’ll meet state-observable duality.

Posted at November 27, 2010 1:45 AM UTC

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### Re: State-Observable Duality (Part 2)

“carried about Zelmanov” should be “carried out by Zelmanov”?

Posted by: Stuart on November 27, 2010 7:43 AM | Permalink | Reply to this

### Re: State-Observable Duality (Part 2)

Thanks! Fixed!

Posted by: John Baez on November 27, 2010 7:54 AM | Permalink | Reply to this

### Re: State-Observable Duality (Part 2)

“… human avatars” - oh the poetry of maths…
Thanks for this rapture!

Posted by: Florifulgurator (Martin Gisser) on November 27, 2010 2:58 PM | Permalink | Reply to this

### Re: State-Observable Duality (Part 2)

John wrote:

In both classical and quantum mechanics, observables are closed under addition and multiplication by real scalars. In classical mechanics we can also multiply observables, but in quantum mechanics this becomes problematic. After all, given two bounded self-adjoint operators on a complex Hilbert space, their product is self-adjoint if and only if they commute!

I’ve always been a bit confused by this way of setting up the foundations. I understand operationally how one can multiply an observable by a scalar $c$: when you make the observation $x$, just write $c x$ in your notebook.

And I understand why multiplying observables is problematic if they don’t commute: in order to write down the product of two observables, I need to measure both of them, but if they don’t commute this depends on the order. But why doesn’t this still allow me to define a non-commutative product? And what does this have to do with composing operators?

What I’m even more confused about is why we can add observables: to do so operationally, we still need to measure both of them, which in general depends on the order. So why do people emphasize that we need to adjust our expections about the multiplicative structure, but throw in a commutative addition without any concern?

(Of course, if you assume that your observables are self-adjoint operators on a Hilbert space, you see that you can add them but not multiply them, but I’m interested in deducing that observables have this structure from basic principles motivated by physical intuition.)

Posted by: Dan Christensen on November 27, 2010 5:15 PM | Permalink | Reply to this

### Re: State-Observable Duality (Part 2)

I know nothing about the physics, but if you consider things from the perspective of quantum probability theory then the difference makes sense.

Addition of probabilities corresponds to the undistinguished possibility of either outcome being the case, i.e. the union of the events behind the probability measures: $\Pr(x) + \Pr(y) = \Pr(x \cup y)$ for the appropriate definition of $+$ on the probability measure (which is not $+$ on the reals unless $x$ and $y$ are disjoint). Thus, you don’t need to measure the likelihood of both events, you need to measure the likelihood of either event.

Whereas the product of probabilities corresponds to the conjoined possibility of both outcomes being (simultaneously) the case. In Bayesian probability this means we can use the chain rule: $\Pr(x \cap y) = \Pr(x) * \Pr(y\mid x) = \Pr(y) * \Pr(x\mid y)$ where $*$ is just the product of reals. However, as you can see, there isn’t quite a homomorphism between intersections and products the way there is between unions and sums. The complication is more about “pushing the product in” than it is about “pulling the intersection out”. Under the assumptions of Bayesian probability theory, the joint probability comes out to always be equal to both products of conditional probabilities. However, in quantum probability theory this is only true when $x$ and $y$ are orthogonal, which is precisely when order effects do not occur. (And note that this is different than when $x$ and $y$ are independent, i.e. when $\Pr(x \cap y) = \Pr(x) * \Pr(y)$.)

Posted by: wren ng thornton on November 28, 2010 2:00 AM | Permalink | Reply to this

### Re: State-Observable Duality (Part 2)

Hi! Great questions, Dan.

Your ‘confusions’ are in fact shared by everyone who thinks hard about this stuff. Most physicists are so used to adding and multiplying operators that they do it with scarcely a thought about what it means operationally if we think of these operators as ‘observables’. Indeed it’s tricky to say what it means. So people who are serious about deriving something like quantum theory — or some more general class of theories — from intuitively plausible axioms typically avoid starting from the assumption that observables form some sort of ‘algebra’.

They instead tend to start from either a logical approach, where the primitive concepts are ‘propositions’, ‘and’, ‘or’ and ‘not’:

• C. Piron, Foundations of Quantum Physics, W. A. Benjamin, New York, 1976.

or a probabilistic approach, where the primitive concepts are ‘observable’ and ‘state’, and measuring an observable in a state gives a probability distribution of outcomes:

• G. W. Mackey, The Mathematical Foundations of Quantum Mechanics, W. A. Benjamin, New York, 1963.

Many variations of both approaches have been studied, and their relation has also been deeply explored; these books are just two classics written by famous advocates of the logical approach and the probabilistic approach. For a broad overview I recommend:

• J. M. Jauch, Foundations of Quantum Mechanics, Addison-Wesley, Reading, MA, 1968.

Anyway, here’s an approach I’ll sort of implicitly be hinting at next time.

Suppose we take states as primitive. We can then operationally define convex linear combinations of states as follows. Say we’re given $0 \le p \le 1$ and two states, $S$ and $S'$. Then to prepare the system in state $p S + (1-p) S'$ we flip a coin that has probability $p$ of landing heads-up. If it lands heads up we prepare the system in state $S$; if it lands tails up we measure $S'$.

Note that this sort of convex linear combination is a mixture of states — it has nothing, or almost nothing, to do with the superposition of pure states we see in quantum mechanics!

If we assume some axioms, we can take the set of states and extend it by taking all formal linear combinations to obtain a vector space $V$. We can think of observables as elements of the dual space of $V$: to any state and any observable, we get a number that’s the expected value of that observable in that state.

Indeed, next time I’ll just act as if every element of the dual space of $V$ is an observable. That sort of fails to confront the problem of what we mean, operationally, by adding observables. But the interesting thing it that we don’t need to assume a product of observables. Instead we focus on the fact that our space $V$ has a convex cone $C$ in it, consisting of the nonnegative linear combinations of states. The intersection of this cone with some hyperplane will be our original set of states.

In my next post, which I’ve already written, I’ll assume everyone in the universe is familiar with these ideas, merely because I am. This is typical of bad exposition. To confuse you even more, I’ll use $V$ to stand for the vector space of observables and $V^*$ to stand for the vector space containing the states! And I’ll focus on explaining something I just learned: how, with a few extra axioms, we can conclude that our system is described by a version of quantum theory built using these Jordan algebras:

• The algebras $\mathrm{h}_n(\mathbb{R})$ of $n \times n$ self-adjoint real matrices with product $a \circ b = \frac{1}{2}(a b + b a)$.
• The algebras $\mathrm{h}_n(\mathbb{C})$ of $n \times n$ self-adjoint complex matrices with product $a \circ b = \frac{1}{2}( a b + b a)$.
• The algebras $\mathrm{h}_n(\mathbb{H})$ of $n \times n$ self-adjoint quaternionic matrices with product $a \circ b = \frac{1}{2}(a b + b a)$.
• The algebras $\mathrm{h}_n(\mathbb{O})$ of $n \times n$ self-adjoint octonionic matrices with product $a \circ b = \frac{1}{2}(a b + b a)$, but only for $n \le 3$.
• The algebras $\mathbb{R}^n \oplus \mathbb{R}$ with product $(x,t) \circ (x', t') = (t x' + t' x, x \cdot x' + t t').$

This is actually a famous old theorem due to Koecher and Vinberg.

(Actually the reason why I assumed everyone in the world knows the convex cone approach to quantum theory is that my post is based on a paper I’m submitting to a conference proceedings on quantum physics and logic, and lots of people in that world are familiar with the convex cone approach. There’s a lot to say about it, and I don’t want to say any of that stuff.)

Posted by: John Baez on November 28, 2010 3:12 AM | Permalink | Reply to this

### Re: State-Observable Duality (Part 2)

The same ending to the story, but a different path, is in this nice survey by Isar Stubbe and Bart Van Steirteghem.

Posted by: bob on November 27, 2010 5:21 PM | Permalink | Reply to this

### Re: State-Observable Duality (Part 2)

Thanks!

Posted by: John Baez on November 28, 2010 4:08 AM | Permalink | Reply to this

### Re: State-Observable Duality (Part 2)

John wrote:

The best result along these lines is the amazing theorem of Maria Pia Solèr…

Now that’s a surprising theorem!

The assumptions are that we have

• a field $k$ with involution, a “*-field”,

• a $k-$ vector space $E$ with a Hermitian form,

• for every closed subspace $M$ we have $M + M^{\perp} = E$, that is $E$ is orthomodular and

• there is an infinite orthonormal sequence.

Then $k$ is either $\mathbb{R}, \mathbb{C}$ or $\mathbb{Q}$.

The author writes:

Contrary to what our early graduate education would lead us to believe, $*$-fields exist in incredible abundance and variety; the three classical number fields analysis, described above, are extraordinarily special within the class of general $*-$fields.

Is there a list of examples of $*-$fields I could look up?

Later the author writes about the theorem:

Also, no amount of computer simulation would suggest this result.

Is this an in-joke I don’t get?

Posted by: Tim van Beek on November 28, 2010 2:36 PM | Permalink | Reply to this

### Re: State-Observable Duality (Part 2)

Then k is either $\mathbb{R}$, $\mathbb{C}$ or $\mathbb{Q}$.

I missed this before; it really is unbelievable! But this $\mathbb{Q}$ stands for “quaternions”, not “quotients (of integers)”. Normally I’d use $\mathbb{H}$.

I’d not thought of $*$-fields before, but if there’s no easily available list out there, maybe we could start one here. First example that comes to mind is a quadratic extension over a prime field, like $\mathbb{Q}[\sqrt{2}]$ (rationals this time), or the field with nine elements. Heck, a quadratic extension of any field, although fields with just one nontrivial involution might be more interesting.

I guess I’ll have to read the paper, but I’m confused by the condition

• for every closed subspace $M$ we have $M + M^\perp = E$, that is E is orthomodular

Posted by: Todd Trimble on November 28, 2010 4:20 PM | Permalink | Reply to this

### Re: State-Observable Duality (Part 2)

Answering my own question, “closed” means

$M^{\perp\perp} = M$

Posted by: Todd Trimble on November 28, 2010 4:24 PM | Permalink | Reply to this

### Re: State-Observable Duality (Part 2)

For a subspace $M$ of a Hilbert space over $\mathbb{R}, \mathbb{C}$ or $\mathbb{H}$, $M$ is closed in the usual topological sense iff $M^{\perp \perp} = M$. So this is a clever way of introducing the flavor of topology without actually introducing a topology.

This is where the rabbit gets slipped into the hat.

Posted by: John Baez on November 29, 2010 2:55 AM | Permalink | Reply to this

### Re: State-Observable Duality (Part 2)

Orthomodularity is the classical (works from v. Neumann etc) condition of the logic of quantum mechanics.

Posted by: Alejandro Rivero on November 28, 2010 7:23 PM | Permalink | Reply to this

### Re: State-Observable Duality (Part 2)

My understanding is that we need orthomodularity to be able to say what an orthogonal projection is. The orthogonal projections correspond to answers to yes-no-questions one can ask about a quantum system in a certain quantum state, so orthomodularity means we are able to ask yes-no-questions.

And every quantum state is completely specified by the answers to all yes-no-questions by Gleason’s theorem (in Hilbert spaces where it is valid, that is).

The paper of S.S. Holland also discusses this axiom of quantum mechanics:

Axiom VII: The partially ordered set of all questions in quantum mechanics is isomorphic to the partially ordered set of all closed subspaces of a separable, infinite dimensional Hilbert space.

…and how one could use the Solèr theorem to derive this axiom as a theorem from more basic axioms.

Posted by: Tim van Beek on November 29, 2010 12:31 PM | Permalink | Reply to this

### Re: State-Observable Duality (Part 2)

Tim wrote:

My understanding is that we need orthomodularity to be able to say what an orthogonal projection is.

You can define an orthogonal projection on any vector space $E$ with a Hermitian form: it’s an operator $p : E \to E$ with $p^2 = p$ and $p^\dagger = p$. You don’t the vector space to be orthomodular.

My understanding is that Soler needs orthomodularity in her sense ($M + M^\perp = E$) to make the lattice of closed subspaces of $E$ be orthomodular in the usual sense of quantum logic.

And, of course, to prove her theorem!

Posted by: John Baez on November 29, 2010 12:53 PM | Permalink | Reply to this

### Re: State-Observable Duality (Part 2)

John wrote:

My understanding is that Soler needs orthomodularity in her sense ($M+M^{\perp} =E$) to make the lattice of closed subspaces of E be orthomodular in the usual sense of quantum logic. And, of course, to prove her theorem!

I meant “we need orthomodularity” in the following sense: Let’s say we would like to play “quantum mechanics in infinite dimensions”, what would the space look like that we need? In this sense, what would we need “orthomodularity” for? Can’t we play “quantum mechanics” without it? (It seems to me to be very much harder to question the other assumptions of the Solèr theorem).

If we don’t assume “orthomodularity”, then there will be an orthogonal projection p - which is an observable! - that does not correspond to a yes-no-question, because there is a state that is neither in the range nor in the orthogonal complement of the range of p. Seems hard to make sense of that, at least to me :-) Or am I missing something?

Posted by: Tim van Beek on November 29, 2010 3:56 PM | Permalink | Reply to this

### Re: State-Observable Duality (Part 2)

I wrote:

Then k is either $\mathbb{R}, \mathbb{C}$ or $\mathbb{Q}$.

That should be $\mathbb{H}$ instead of $\mathbb{Q}$, of course, I must have been thinking “Quaternions” when I typed this.

Todd wrote:

closed means $M^{\perp \perp} = M$

With this definition there really is no topological concept involved in the formulation of the assumptions of the Solèr theorem, all is algebra. In particular there is no assumption that the field $k$ cannot be finite, for example. It would be interesting to learn how

a) one proves that $k$ cannot be finite and

b) that $k$ cannot be $\mathbb{Q}$ (this time this is supposed to be the rationals, and not a typo :-)

Posted by: Tim van Beek on November 28, 2010 7:48 PM | Permalink | Reply to this

### Re: State-Observable Duality (Part 2)

Jacques Faraut, Adam Korányi Analysis on symmetric cones

is a nice book about the classification and lots of other useful math related to symmetric cones and Jordan algebras.

Posted by: Arnold Neumaier on November 29, 2010 5:26 PM | Permalink | Reply to this

### Re: State-Observable Duality (Part 2)

Thanks! For some reason that book is only available from the National University of Singapore library if one orders it, so I haven’t read it. I’ll order it now.

I’m finding this book pretty interesting:

• Wolfgang Bertram, The Geometry of Jordan and Lie Structures.

And then there are the classics by Max Koecher and Ottmar Loos. I wish I really understood better the relations between Lie triple systems, and Jordan triple systems, and Jordan pairs, and quandles, and symmetric spaces, and other such things. I really want to see a big diagram with functors going between all these categories, and then nice explanations of all these functors. I may have to draw this diagram myself.

Posted by: John Baez on December 1, 2010 2:49 AM | Permalink | Reply to this

### Re: State-Observable Duality (Part 2)

There is a statement by John Harding and Andreas Döring relating Jordan algebras and their topos-theoretic description of quantum algebras, which might be noteworthy:

for $A$ and $B$ von Neumann algebras, write $A_J$, $B_J$ for the corresponding Jordan algebras (symmetrized product operation) and $ComSub(A)$, $ComSub(B)$ for their posets of commutative vN subalgebras.

They show that

• Jordan algebra isomorphisms $A_J \to B_J$

correspond to

• poset isomorphisms $ComSub(A) \to ComSub(B)$.

More on this is at semilattice of commutative subalgebras.

Posted by: Urs Schreiber on December 14, 2010 10:08 PM | Permalink | Reply to this

### Re: State-Observable Duality (Part 2)

Interesting! Thanks, Urs!

I may not get around to reading their work and finding out the answer to this question very soon, so I’ll ask it out loud here:

Why is passing to Jordan algebras necessary here? What’s an example where we have a poset isomorphism $ComSub(A) \to ComSub(B)$ that does not come from a unique a von Neumann algebra isomorphism $A to B$?

Or equivalently, given the Döring–Harding result: what’s an example of a Jordan algebra isomorphism $A_J \to B_J$ that extends to a von Neumann algebra isomorphism $A \to B$ in either more than one way, or not at all?

Posted by: John Baez on December 15, 2010 1:51 AM | Permalink | Reply to this

### Re: State-Observable Duality (Part 2)

what’s an example of a Jordan algebra isomorphism $A_J \to B_J$ that extends to a von Neumann algebra isomorphism $A \to B$ in either more than one way, or not at all?

The following is the example mentioned by Harding and Döring:

In the article A factor not anti-isomorphic to itself Alain Connes describes a von Neumann algebra $A$ that is not isomorphic to $A^{op}$.

But for any algebra we have $A_J \simeq (A^{op})_J$.

Posted by: Urs Schreiber on December 15, 2010 10:12 AM | Permalink | Reply to this

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