If there is already enough room to add all projections, does passing to matrices change anything?

Question

Throughout, $A$ denotes a $*$-algebra. We always assume $A$ is representable in the sense that $A$ can be embedded into $B(H)$ for some Hilbert space $H$. The particular embedding is not important, and it is not assumed that the range of this embedding is closed, i.e. $A$ does need not be a $C^*$-algebra.

Let $P(A)$ be the set of projections (self-adjoint idempotents) in $A$. Let $\sim$ denote the Murray-von Neumann equivalence relation on $P(A)$. That is, $e \sim e'$ when there exists $w \in A$ such that $w^*w = e$ and $ww^*=e'$.

Projections $e_1,e_2 \in A$ are called orthogonal if $e_1 e_2=0$. In this case, $e_2 e_1 = 0$ as well and $e_1 + e_2$ is another projection. That is, $P(A)$ is closed under orthogonal sums. The following fact states that addition of Murray-von Neumann equivalence classes is well-defined, whenever orthogonal representations are to be found.

Fact 1: Suppose that $e_1, e_2 \in P(A)$ are two orthogonal projections, and $e_1',e_2'$ are two more orthogonal projections. If $e_1 \sim e_1'$ and $e_2 \sim e_2'$, then $e_1 + e_2 \sim e_1' + e_2'$.

Proof: Indeed, suppose $w_1,w_2 \in A$ have $w_1^*w_1 = e_1$, $w_1w_1^* = e_1'$ and $w_2^*w_2 = e_2, w_2w_2^* = e_2'$. One has $w_1^*w_2= (e_1'w_1)^* e_2'w_2 = w_1^* e_1' e_2' w_2 = 0$ and, similarly, $w_2^*w_1 = 0$. Using the latter two equalities, one sees $(w_1 + w_2)^*(w_1+w_2) = e_1 + e_2$. Similarly, $(w_1+w_2)(w_1+w_2)^* = e_1' + e_2'$ so that $(e_1+e_2) \sim (e_1' +e_2')$.

As a simple corollary, we see that, if $A$ is spacious in the sense that, for each pair of projections $e_1,e_2 \in P(A)$, there exist $e_1',e_2' \in P(A)$ with $e_1 \sim e_1'$, $e_2 \sim e_2'$ such that $e_1'$ and $e_2'$ are orthogonal, then $P(A) / \sim$ is a commutative monoid with respect to orthogonal direct sum.

Even if $A$ is not already spacious, then $M_\infty(A) = \bigcup_{n=1}^\infty M_n(A)$, which also acts faithfully on a Hilbert space, namely $\bigoplus_{n=1}^\infty H$, is spacious. This is follows easily from the fact that, if $e \in P(M_n(A))$, then $w = \left( \begin{smallmatrix} 0 & 0 \\ e & 0 \\ \end{smallmatrix} \right) \in M_{2n}(A)$ has $w^*w = \left( \begin{smallmatrix} e & 0 \\ 0 & 0 \\ \end{smallmatrix} \right)$, $ww^* = \left( \begin{smallmatrix} 0 & 0 \\ 0 & e \\ \end{smallmatrix} \right)$. We remark that:

• $A \mapsto M_\infty(A)$ is a functor from representable $*$-algebras and $*$-homomorphisms, to the subcategory of representable $*$-algebras that are spacious.

• $A \mapsto P(A) / \sim$ is a functor from spacious, representable $*$-algebras to commutative monoids.

My question is the following:

Question: If $A$ is already spacious so that $P(A)/\sim$ is monoid without passing to matrices, does the corner inclusion $A \to M_\infty(A)$ induce a monoid isomorphism $P(A) / \sim \to P(M_\infty(A)) / \sim$? In other words:

1. Is every $e \in P(M_\infty(A))$ Murray-von Neumann equivalent in $M_\infty(A)$ to an $e' \in A \subset M_\infty(A)$?
2. If $e,e ' \in A$ are Murray-von Neumann equivalent in $M_\infty(A)$, are they already in $A$?

Answer 1

I had forgotten about this question until Jonas Meyer put a bounty on it. In hindsight, I think I now know how to solve parts of it... I'll write down what I've got and maybe someone can give a complete argument building on it.

I'll address the case $1 \in A$ first since it is somewhat simpler. The argument does not seem to be overly difficult, but it is a bit technical and I don't think I have written it up very well. If somebody would like to edit this post into a better state, please feel free!

The nonunital case, I do not have a good grasp on. I think it may be impossible to complete without some extra assumptions, e.g. that $A$ is a C*-algebra, or at least is closed under some flavor of functional calculus.

The unital case

A couple basic things to realize:

1. If $w \in A$ is an isometry (this means $w^*w=1$, which implies $e=ww^*$ is a projection), then $x \mapsto wxw^*$ is $*$-isomorphism of $A$ onto the $*$-subalgebra $eAe = \{ exe : x \in A\} \subset A$ called a "corner". Any projection $e' \in A$ is M-vN equivalent (in $A$) to its image $we'w^*$ in the corner. A partial isometry implementing the equivalence is $we'$.
2. If two isometries $w_1$ and $w_2$ are orthogonal in sense that the range projections $e_1=w_1w_1^*$ and $e_2 = w_2w_2^*$ are orthogonal, then the corners $e_1Ae_1$ and $e_2Ae_2$ multiply to zero so that $e_1Ae_1 + e_2Ae_2$ is a copy of $A \oplus A$ embedded in $A$.

We want to ensure we can find many pairwise-orthogonal isometries (i.e. isometries whose range projections are pairwise-orthogonal). Since $[1]+[1]$ needs to be well-defined, there exist isometries $w_0,w_1 \in A$ such that $e_0 = w_0w_0^*$ and $e_1 = w_1w_1^*$ are orthogonal. Now, for any binary sequence $b = b_1\cdots b_\ell$ of length $\ell$ let's define $w_b = w_{b_1} \cdots w_{b_\ell}$. It's easy to see the $w_b$'s are isometries. One may show they are also pairwise orthogonal, as $b$ ranges over all binary sequences of length $\ell$.

[Note: lurking behind the above argument is a sort of subdivision procedure. We know $A$ contains a copy of $A \oplus A$. But then each of the factors also contains a copy of $A \oplus A$, and so on.]

Now let us consider $M_n(A)$ for some fixed $n$. By the above, we can find pairwise orthogonal isometries $w_1,\ldots,w_n \in A$. The orthogonality implies that $w_i^*w_j = 0$ if $i \neq j$ and $1$ if $i=j$. Thus $$W = \begin{pmatrix} w_1 & w_2 & \cdots & w_n \\ 0 & 0 & \cdots & 0 \\ \vdots & \vdots & \ddots &\vdots \\ 0 & 0 & \cdots & 0 \\ \end{pmatrix} \in M_n(A)$$ is an isometry. Moreover, the range projection of $W$ lies in the upper left corner copy of $A$ in $M_n(A)$. Thus, conjugating by $W$ there is a $*$-isomorphism from $M_n(A)$ onto a sub algebra of the upper left corner copy of $A$ in $M_n(A)$.

Now we can answer the question. Any projection $e \in M_n(A)$ is M-vN equivalent to $WeW^*$ in the upper left corner copy of $A$ in $M_n(A)$. If $e,e'$ in the upper left corner copy of $A$ in $M_n(A)$ are M-vN equivalent via a partial isometry $w \in M_n(A)$, then $WeW^* = w_1 e w_1^*$ and $W e'W^* = w_1 e' w_1^*$ are M-vN equivalent in $A$ via the partial isometry $WwW^* \in A$. Since $e \sim w_1ew_1^*$ in $A$ via $w_1 e$ and $e' \sim w_1 e' w_1^*$ in $A$ via $w_1 e'$, we have $e \sim e'$ in $A$ already.

The Nonunital case

I do not know how to deal with this case. When I attempted it, it seemed to me that the main difficulty was to prove that any projection $$ E = \begin{pmatrix} x_{11} & \cdots & x_{1n} \\ \vdots & \ddots & \vdots \\ x_{n1} & \cdots & x_{nn} \\ \end{pmatrix} \in M_n(A), $$ is dominated by a projection of the form $$ \begin{pmatrix} e & 0 & \cdots & 0 \\ 0 & e & \cdots & 0 \\ \vdots & \vdots & \ddots &\vdots \\ 0 & 0 & \cdots & e \\ \end{pmatrix} $$ where $e$ is a projection in $A$. I think suffices to dominate the restriction of $E$ to the diagonal $$ \begin{pmatrix} x_{11} & 0 & \cdots & 0 \\ 0 & x_{22} & \cdots & 0 \\ \vdots & \vdots & \ddots &\vdots \\ 0 & 0 & \cdots & x_{nn} \\ \end{pmatrix} $$ (something do do with the diagonal map being a "faithful conditional expectation") and the latter should be a positive element with a spectral gap at zero, so that a projection of the form $$ \begin{pmatrix} e_{11} & 0 & \cdots & 0 \\ 0 & e_{22} & \cdots & 0 \\ \vdots & \vdots & \ddots &\vdots \\ 0 & 0 & \cdots & e_{nn} \\ \end{pmatrix} $$ can be found using functional calculus. Then one wants to find a projection dominating $e_1+\ldots+e_n$ which, I think, can also be done using functional calculus.

Answer 2

The answer to question 2 is negative. It is possible that there are projections in $M_\infty(A)$ which are not equivalent to any projection in $A$, even when $A$ is a $C^*$-algebra (so, representable) and spacious.

To do this, I will construct a (non-unitial) $C^*$-algebra which has no non-zero projections, so is trivially spacious. However, $M_2(A)$ will contain non-zero projections, which cannot be equivalent to a projection in $A$.

Let us start by defining $B$ as the universal $C^*$-algebra generated by a pair of unitary elements $u,v$. This is the group $C^*$-algebra over the free group with two generators, and has the following properties. Given any other $C^*$-algebra $B^\prime$ containing unitaries $u^\prime,v^\prime$, then there is a unique *-homomorphism $\theta\colon B\to B^\prime$ satisfying $\theta(u)=u^\prime$ and $\theta(v)=v^\prime$. Furthermore, it can be shown that $B$ has no projections other than $0$ and $1$ (see Theorem 1 of Choi, Man Duen., The full $C^\ast$-algebra of the free group on two generators, Pacific J. Math. Volume 87, Number 1 (1980), 41-48.).

Let us now define the following elements of $B$, \begin{align} &x=\frac12+\frac14\left(v+v^*\right),\\ &y=\frac{i}{4}\left(v-v^*\right),\\ &w=uy,\\ &z=1-uxu^*, \end{align} and let $A$ be the $C^*$-subalgebra generated by $x,w,z$. Note that $x,y,z$ are self-adjoint. Simple computations from the definitions show that \begin{align} &w^*w=y^2=x(1-x)=x-x^2,\\ &ux+zu=u,\\ &ux=(1-z)u,\\ &u(1-x)=zu,\\ &ww^*=uy^2u^*=ux(1-x)u^*=(1-z)uu^*z=z-z^2. \end{align} Right-multiplying the second of these by $y$ gives $wx+zw=w$. It follows that, $$ \left(\begin{array}{cc}x&w^*\\ w&z\end{array}\right)^2=\left(\begin{array}{cc}x&w^*\\ w&z\end{array}\right), $$ so $M_2(A)$ contains a non-zero projection. It just remains to show that $A$ contains no non-zero projections. As it is a sub-*-algebra of $B$, and we know that $B$ does not have any projections besides $0$ and $1$, it is enough to show that $A$ does not contain the identity of $B$. Using the universal property, there is a unique *-homomorphism $\theta\colon B\to M_2(\mathbb{C})$ with, $$ \theta(u)=\left(\begin{array}{cc}0&1\\ 1&0\end{array}\right),\ \theta(v)=\left(\begin{array}{cc}1&0\\ 0&-1\end{array}\right). $$ It can be checked that $\theta(w)=\theta(y)=0$ and, $$ \theta(x)=\left(\begin{array}{cc}1&0\\ 0&0\end{array}\right). $$ It follows that $\theta$ maps all of $A$ into $\{(a,0;0,0)\colon a\in\mathbb{C}\}$ and, in particular, does not map any element of $A$ to the identity in $M_2(\mathbb{C})$. So, $A$ cannot contain the identity of $A$.

Answer 3

Inspired by George Lowther's solution, here is another more elementary example of a (nonunital) C*-algebra $A$ such that $A$ has no nonzero projections, but $M_2(A)$ does have nonzero projections.

• Let $D$ be the sub algebra of $M_2(\mathbb{C})$ consisting of scalar multiples of the identity.
• Let $C$ be the sub algebra of $M_2(\mathbb{C})$ consisting of matrices of the form $\begin{bmatrix} * & 0 \\ 0 & 0 \end{bmatrix}$.
• Let $A$ be the sub algebra of $C([0,1],M_2(\mathbb{C}))$ consisting of all paths $f$ for which $f(0)\in D$ and $f(1)\in C$.

I claim there are no nonzero projections in $A$. To see this, note that a projection in $C([0,1],M_2(\mathbb{C}))$ is just a path of projections in $M_2(\mathbb{C}$), and the rank (which equals the trace) must stay constant along such a path. Since the only nonzero projection in $D$ has rank 2, while the only nonzero projection in $C$ has rank 1, there cannot be a nonzero projection in $A$.

On the other hand, $M_2(A)$ has nonzero projections. Take, for example: $$P(t) = U(t) \begin{bmatrix} 1 & 0 & 0 & 0\\ 0 & 1 & 0 & 0\\ 0& 0 & 0 & 0 \\ 0& 0 & 0 & 0 \\ \end{bmatrix} U(t)^{-1}$$ where $U : [0,1] \to M_4(\mathbb{C})$ is a path of unitaries with \begin{align*} U(0)=I_4 && U(1)= \begin{bmatrix} 1 & 0 & 0 & 0\\ 0 & 0 & 1 & 0\\ 0& 1 & 0 & 0 \\ 0& 0 & 0 & 1 \\ \end{bmatrix} \end{align*}