How does one show that the projections in the [CAR algebra][1] do not form a complete lattice?


Background info: my paper with Scholz [2] paper works in the (infinite) CAR algebra and tries to define algorithmic randomness for its states. The analogs of effectively open sets are certain increasing sequences of projections $p\_n\in M\_{2^n}$, and we wondered whether their supremum exists in the algebra.


[1] <https://en.wikipedia.org/wiki/CCR_and_CAR_algebras>


[2] Martin-Löf random quantum states. With Volkher Scholz. Submitted, arXiv: 1709.08422 



Projections in CAR (Canonical Anticommutation Relation) algebra
As expressed in my comment, the finite-dimensional CAR algebras *do* have a complete projection lattice. Here I outline the proof that the CAR algebra with countably many degrees of freedom (equivalently, the CAR algebra of a separable Hilbert space) does not have a complete lattice of projections. The short summary is that $A$ is separable, but if it had a complete projection lattice, it would embed $\ell^\infty$, which is not (norm) separable.


I will take for granted the isomorphism with the infinite tensor product $A \cong M\_2^{\otimes \infty}$. This is described in Takesaki's *Theory of Operator Algebras III*, Exercise XIV.1. This is a separable C$^\*$-algebra -- the set of elements of $A$ that are finite sums of elements of the form $a\_1 \otimes a\_2 \otimes \cdots \otimes a\_n \otimes 1 \otimes \cdots$ where $a\_i$ are matrices with rational entries is a countable dense subset.


Now, $A$ also has a subalgebra isomorphic to $C(2^\omega)$, under the isomorphism $C(2^\omega) \cong (\mathbb{C}^2)^{\otimes \infty}$. We can find a countably infinite family of disjoint projections $(p\_i)\_{i \in \mathbb{N}}$ with supremum 1 in $A$ by taking a countably infinite family of disjoint clopen sets $(C\_i)\_{i \in \mathbb{N}}$ whose join is $1$ in $2^\omega$, and taking $p\_i$ to be the image of $\chi\_{C\_i}$ in $A$ under the aforementioned isomorphism. One example of such a family of clopens we could use is $$
C\_i = \{ f \in 2^\omega \mid f(i) = 1 \text{ and } \forall j < i . f(j) = 0 \},
$$
*i.e.* $C\_i$ is the set of sequences whose first 1 is at index $i$. 


Now suppose for a contradiction that the projection lattice of $A$, which we write as $\newcommand{\Proj}{\mathrm{Proj}}\Proj(A)$, is a complete lattice. It follows from a simple calculation that $\Proj(A)$ is an orthomodular lattice. Define $\newcommand{\N}{\mathbb{N}}\newcommand{\powerset}{\mathcal{P}} g : \powerset(\N) \rightarrow \Proj(A)$ by $g(S) = \bigvee\_{i \in S} p\_i$. By construction this preserves joins, and it also preserves complements (essentially because the join of $(p\_i)\_i$ is $1$), so is a complete homomorphism of orthomodular lattices. Therefore, by extending $g$ to simple functions in $\ell^\infty$ and then by continuity, there is a \*-homomorphism $f : \ell^\infty \rightarrow A$ that restricts to $g$ on projections. 


If there were an $a \in \ell^\infty$ such that $a \neq 0 $ and $f(a) = 0$, then $f(a^\*a) = 0$ as well, so without loss of generality we can take $a$ to be positive. As $a \neq 0$, there exists $x \in \N$ such that $a(x) > 0$. Then $f(a(x)\chi\_{\{x\}}) \leq f(a) = 0$, so $0 = f(\chi\_{\{x\}}) = g(x) = p\_x$, but this contradicts the definition of the family $(p\_i)\_{i \in \N}$. So $f$ is injective, and therefore embeds $\ell^\infty$ as a C$^\*$-subalgebra of $A$. But then we have a contradiction, because $A$ is norm-separable, but $\ell^\infty$ is not.


