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Stream: theory: mathematics

Topic: Bases and isometry copy-maps

Sam Staton (May 12 2025 at 11:48):

Hi Suppose I have an associative and commutative comultiplication $X\to X\otimes X$ that is an isometry in $\mathbf{FdHilb}$ (finite dimensional Hilbert spaces). Does this necessarily correspond to an orthonormal basis for $X$?

It is well known that a special dagger commutative Frobenius algebra on a finite dimensional Hilbert space amounts to an orthonormal basis, from the new description of orthonormal bases by @Bob Coecke @dusko and @Jamie Vicary. 

Such an algebra in particular has a comultiplication $\Delta:X\to X\otimes X$ that is an isometry, as well as associative and commutative. The basis comprises the copyable elements for $\Delta$, i.e. $\{x~|~\Delta(x)=(x\otimes x) \& |x|=1\}$. 

So my question is: is the dagger Frobenius requirement (and unit) known to be necessary? Or will an associative and commutative comultiplication that is an isometry suffice?

I can see where the dagger Frobenius structure is used in the neat argument by Bob, Dusko and Jamie, which goes via C star algebras. However, I think I can show that in $\mathbb{C}^2$ the extra Frobenius assumption is not necessary. Before I go to higher dimensions, since my linear algebra is slow, I thought to check whether it is well known, or any suggestions where to look?

John Baez (May 12 2025 at 12:28):

Sam Staton said:

Suppose I have an associative and commutative comultiplication $X\to X\otimes X$ that is an isometry in $\mathbf{FdHilb}$ (finite dimensional Hilbert spaces). Does this necessarily correspond to an orthonormal basis for $X$?

I don't know! It would indicate a surprising suboptimality in existing results, but I haven't heard of anything about this, or thought about it.

If any such comultiplication does correspond to an orthonormal basis for $X$, I think you should be able to systematically extend the comultiplication to a full-fledged dagger Frobenius structure on $X$. Have you attempted that?

It seems worth trying, since if your attempt fails, the failure may lead you a counterexample.

If I were making this attempt, I would start by defining a multiplication

$m = \Delta^\ast: X \otimes X \to X$.

Then there's a standard trick for defining a counit

$\eta : X \to \mathbb{C}$

namely

$\eta(x) = \text{tr}(L_x)$

where $L_x: X \to X$ is left multiplication by $x$, and $\text{tr}$ is the usual trace of a linear map from a finite-dimensional vector space to itself. You may need to rescale $\eta$ by a cleverly chosen constant, e.g. something involving the dimension of $X$.

Then you can define a unit $\epsilon : \mathbb{C} \to X$ by

$\epsilon = \eta^\ast$

So the question then becomes whether you can derive all the special dagger commutative Frobenius axioms from the associativity and commutativity and isometricity (is that a word? :thinking: ) of $\Delta$.

Tobias Fritz (May 12 2025 at 15:40):

Dually speaking, you're asking whether there is a finite-dimensional Hilbert space can be equipped with an associative and commutative multiplication that is a coisometry, but is not isomorphic to the usual $\mathbb{C}^n$, or equivalently not separable.

This reminds me of the concept of subproduct system, which is a generalization of that graded by a semigroup:
image.png
I'm not sure what kinds of examples they consider as I don't really know anything about this, but perhaps it's worth checking out and seeing if any of them are commutative and would yield the sort of example you're looking for.

Tobias Fritz (May 13 2025 at 07:59):

I've done a bit more digging and found that recent work by Laurent Poinsot is highly relevant to the question:

• Hilbertian (function) algebras
• Hilbertian Frobenius algebras

Perhaps you've seen these already?

Tobias Fritz (May 13 2025 at 08:01):

As far as I understand, in his terminology the question is whether every finite-dimensional commutative special Hilbertian algebra is semisimple, or equivalently unital Frobenius (using Corollay 35 from the second paper). Note that his "algebras" are not generally assumed unital, which matches your question.

Tobias Fritz (May 13 2025 at 08:04):

Theorem 22 of the first paper is especially relevant: it shows that the "group-like" elements $x$ with $\Delta(x) = x \otimes x$ form an orthonormal basis of the orthogonal complement of the Jacobson radical of the algebra. So the question is whether the Jacobson radical must vanish, or equivalently whether the algebra must be semisimple.

Sam Staton (May 13 2025 at 19:26):

Thank you both! @John Baez , from abstract stuff that I know, so far I could easily show from specialness (isometry) that the image of $\Delta \circ \mu$ is contained in $(X\otimes \mu)\circ (\Delta\otimes X)$, but not the other way round, so sort-of half the Frobenius law.

Sam Staton (May 13 2025 at 19:29):

@Tobias Fritz thank you very much, these references look very relevant and I'm reading them!
[My argument for two dimensions was indeed that there can be no nilpotents if it's commutative and isometric, but I showed this by splitting on cases, a low level argument, and it's not yet clear whether it will generalize to other dimensions.]

Sam Staton (May 13 2025 at 19:40):

Perhaps you are interested in the backstory. With @Paolo Perrone and Razin Shaikh, we are looking at how Markov cats can arise from SMCs.

We first take the affine reflection of an SMC (e.g. isometries), so that it is semicartesian (e.g. CPTP). Then any comultiplication $\Delta$ can be made into an idempotent $(X\otimes \mathit{discard})\circ \Delta$, and if we split the idempotents coming from commutative associative comultiplications, then we will get a Markov category. 

Starting with isometries, ie pure quantum theory, the affine reflection is CPTP, and splitting those idempotents leads us to $\mathbf{FinStoch}$ if they all come from bases. 

I wanted to check, to derive probability from quantum theory in this way do we need to do something special about the bases? If so, is there a different Markov category that could be extracted from quantum theory instead?

John Baez (May 13 2025 at 19:56):

Sam Staton said:

Thank you both! John Baez , from abstract stuff that I know, so far I could easily show from specialness (isometry) that the image of $\Delta \circ \mu$ is contained in $(X\otimes \mu)\circ (\Delta\otimes X)$, but not the other way round, so sort-of half the Frobenius law.

This "semi-Frobenius law" rings vague bells in my mind. It dimly reminds me of some things I've heard from @Todd Trimble about cartesian bicategories and Frobenius monoids in poset-enriched categories. I could be imagining it.

Jamie Vicary (May 13 2025 at 20:00):

OK I think I have a counterexample. [OK I retract this]

Todd Trimble (May 14 2025 at 00:58):

John Baez said:

Sam Staton said:

Thank you both! John Baez , from abstract stuff that I know, so far I could easily show from specialness (isometry) that the image of $\Delta \circ \mu$ is contained in $(X\otimes \mu)\circ (\Delta\otimes X)$, but not the other way round, so sort-of half the Frobenius law.

This "semi-Frobenius law" rings vague bells in my mind. It dimly reminds me of some things I've heard from Todd Trimble about cartesian bicategories and Frobenius monoids in poset-enriched categories. I could be imagining it.

Just to reply to John, and not adding particularly to the current discussion: in bicategories like $\mathsf{Rel}$ or $\mathsf{Span}$, those relations or spans $X \nrightarrow Y$ that come from functions $X \to Y$ have right adjoints, and so diagonal maps $\Delta: X \to X \times X$ (which I will rewrite as $\Delta: X \to X \otimes X$ since the setwise product $\times$ is not a cartesian product in those bicategories), which are the comultiplications for cocommutative comonoid structures on such $X$, will have right adjoints $\mu: X \otimes X \to X$. Those right adjoints are multiplications for commutative monoid structures on $X$, mated to the cocommutative comonoid structures we started with.

By playing with adjoints like $\Delta \dashv \mu$, coassociativity isomorphisms like

$(1 \otimes \Delta) \circ \Delta \overset{\sim}{\to} (\Delta \otimes 1)\Delta$

induce morphisms

$\Delta \mu \to (1 \otimes \mu)(\Delta \otimes 1)$

which looks like "one half" of Frobenius reciprocity. You get the other half if these morphisms are invertible. That turns out to be the case in $\mathsf{Rel}$ and $\mathsf{Span}$, but not generally for other cases of cartesian bicategories, like the bicategory consisting of categories and profunctors.

Part of the groupoidification story is that these merely "semi-Frobenius" morphisms, which are always present in cartesian bicategories, turn out to be invertible if you work not with categories and profunctors, but instead with groupoids and profunctors between them.

Tobias Fritz (May 14 2025 at 08:55):

So here's a promising-looking approach towards proving that for every finite-dimensional Hilbert space $\mathcal{H}$, every commutative, associative and isometric $\Delta : \mathcal{H} \to \mathcal{H} \otimes \mathcal{H}$ is of the form $\Delta(e_i) = e_i \otimes e_i$ for a suitable orthonormal basis $\{e_i\}$.

• Construct a quantum channel $\Phi : \mathcal{B}(\mathcal{H}) \to \mathcal{B}(\mathcal{H})$ by mapping a density matrix $\rho$ to the partial trace of $\Delta \rho \Delta^*$. The idea is that $\Phi$ first uses $\Delta$ to "copy" the state and then forgets one of the copies.
• Using associativity and isometricity of $\Delta$, show that $\Phi$ is idempotent. (I think I can see how to do this but would need to check the details.)

In the case $\Delta(e_i) = e_i \otimes e_i$, the channel $\Phi$ is precisely the "dephasing" channel which erases all off-diagonal elements from a density matrix. In other words, it conducts a measurement in the given basis and forgets the result. So the fixed points of $\Phi$ are precisely those density matrices that are diagonal in the given basis. Hence for general $\Delta$, it's natural to look at the fixed points of $\Phi$. The pure states in this set of fixed points form the basis that we're looking for.

• Based on the idempotency, $\Phi$ should be a conditional expectation onto a sub-C*-algebra of $\mathcal{B}(\mathcal{H})$. Use the commutativity of $\Delta$ to show that this subalgebra is commutative. Define the basis $\{e_i\}$ now as the basis with respect to which this algebra is precisely the diagonal matrices.
• Then $\Delta(e_i) = e_i \otimes e_i$ will be clear.

I guess the main difficulty lies in the penultimate step, where one needs to show that $\Phi$ has sufficiently many fixed points in order to show that the dimension of the algebra coincides with $\mathrm{dim}(\mathcal{H})$, and I don't yet know how to go about this. But overall, the appeal of this approach is that it would give a relatively concrete way to get at the $e_i$.

Sam Staton (May 14 2025 at 20:58):

Thanks! Yes this is the kind of thing we had in mind to arrive at, so it's an interesting idea to try to resolve it there directly. Although I'm still not sure how to tackle the penultimate step, as you said.
In the paper by Bob, Dusko and Jamie, their proof (sec 4) is by constructing a commutative C star-subalgebra of B(H). I suppose it's the same one? But without the Frobenius law, I can only show that their construction gives a sub-algebra, not a star sub-algebra.