# Apparently a catastrophe: unital injective *-Hom implies factorization of Algebra into tensor product

In a recent lecture we saw the following result, which was hailed as somewhat of a catastrophe:

We denote by $$\operatorname{M}_m$$ be the set of $$m\times m$$ matrices and by $$(\hat A,\hat\varphi)$$ a non-commutative probability space (for simplicity's sake throughout the lecture we assumed $$\hat A\simeq\operatorname{M}_n\otimes\operatorname{L}^\infty(\Omega,\Sigma,\mu)$$, but it sounded like that was just a simplifying assumption due to the fact that some people in the lecture had no previous knowledge on von-Neumann algebras or even C$$^\ast$$-Algebras).

Let $$i_0\colon \operatorname{M}_m\to \hat A$$ be a $$\operatorname{M}$$-valued random variable, i.e., a unital, injective $$\ast$$-homomorphism such that the conditional expectation of $$(\hat A,\hat\varphi)$$ onto $$i_0(\operatorname{M}_m)$$ exists.

• In the above situation there exists a subalgebra $$\hat C\subseteq\hat A$$ such that $$\hat A\simeq \operatorname{M}_m\otimes \hat C$$ with $$i_0(x)=x\otimes \mathbb{1}_{\hat C}$$ for all $$x\in\operatorname{M}_m$$.
• Let $$\hat A =\operatorname{M}_m\otimes \hat C$$ and $$P\colon (\hat A,\hat\varphi)\to \operatorname{M}_m\otimes\mathbb{1}_{\hat C}$$ the conditional expectation. Then there exist states $$\varphi$$ on $$\operatorname{M}_m$$ and $$\hat\psi$$ on $$\hat C$$ such that $$\hat\varphi=\varphi\otimes\hat\psi$$ and $$P(x\otimes y)=\hat\psi(y)\cdot (x\otimes\mathbb{1}_{\hat C})$$.

Apparently this result seemed like a big catastrophe when it was first discovered because of reasons I did not quite grasp. In the lecture the consequence that $$\hat A$$ could be factorized in different ways was mentioned and, apparently even more severe, $$\hat\varphi=\varphi\otimes\hat\psi$$. The lecturer explained something along the lines of this resulting in some people completely discarding the idea of stationary quantum markov-processes. From what I gathered, it means that the naive way of constructing such a process fails, but I am not sure what exactly was meant by that.

I'd be very happy if anyone could shed some light into this, it has bugged me all week and I just could not put my finger on where exactly the problem lies. It certainly feels like something inherently fishy is going on, but I guess that this feeling could as well just be based on the impression that the lecturer gave us.

Apologies in advance if my notation/nomenclature is not proper, the lecture was given in german and I am not well-versed with the "correct" terminology in the broader german community let alone the international one.

Just in case anybody is ever interested in the solution to this:

The naive construction of a stationary Markov-process in the commutative case was to have a finite measure space $$(\Omega,\mu)$$ with a initial distribution $$\mu$$ and a transition matrix $$T$$. This gives rise to a measure $$\hat{\mu}$$ on the space of paths of length $$k$$ given by $$(\omega_1,\dots,\omega_k)\in\Omega^k$$ via $$\hat\mu (\omega_1,\dots,\omega_k)=\mu(\omega_1)\cdot t_{\omega_1 \omega_2}\cdot\dots\cdot t_{\omega_{k-1}\omega_k}$$, where $$t_{\omega_j\omega_k}$$ is the transition probability from state $$\omega_j$$ to $$\omega_k$$ given by the entry $$t_{jk}$$ of $$T$$.

By the theorem of Kolmogorov-Caratheodory there is a unique measure on $$\Omega^{\mathbb{N}}$$ that extends this measure, i.e., it coincides with the so constructed $$\hat\mu$$ if restricted to cylinder sets.

Then the coordinate projection $$Y_k\colon \Omega^{\mathbb{N}}\to\Omega$$ given by $$(\omega_1,\omega_2,\dots)\mapsto \omega_k$$ form a stationary markov process. It is worth mentioning that the whole structure of the process is somehow encoded in the measure, as the maps themselves are as simple and independent as they could possibly be.

Now, this is where the so-called No-Go Theorem comes in. If we simply try to mimick this basic construction in the non-commutative case, then the corresponding non-commutative counterpart would be a family of random variables $$i_k\colon A\to \hat A$$, where $$\hat A = A\otimes A\dots\otimes A$$ and the map is given by $$x\mapsto 1\otimes 1\otimes\cdots x\otimes 1\otimes\dots\otimes 1$$, where the $$x$$ stands in the $$k$$-th position.

Now, if we are in the case that $$A$$ is indeed a $$\operatorname{M}_n$$, the second part of the No-Go Theorem applies and because of our explicit knowledge of how the random variables map $$A$$ into $$\hat A$$, we indeed see that the random variables must be independent.

My mistake was overlooking the difference between the first and second point. In the second point we actually have equality and not just isomorphy, which means that we know that the Algebra \emph{is} of product type, which means that our knowledge of how the second random variable then embeds into this product allows us to factorise $$\hat \varphi=\varphi_1\otimes\varphi_2\otimes\hat\psi$$ etc. (I know that this point could probably use further elaboration, but given the limited interest the question itself received, I don't see a reason to write more about this than I already have for sake of completeness).