Does antisymmetricity of $\preceq$ fail in general C*-algebras?

Question

For projections $p,q$, write $p\preceq q$ if there is $u$ such that $u^*u=p$ and $uu^*\leq q$.

Write $p\sim q$ if $u^*u=p$ and $uu^*=q$.

It is a theorem that $\preceq $ is a partial order in a Von Neumann algebra. But does antisymmetricity hold in general C*-algebra?

And antisymmetricity means, $p\preceq q$ and $q\preceq p$ implies $p\sim q$.

Answer 1

Let $n\geq 3$ and consider the Cuntz algebra $\mathcal {O}_n$, which is generated by isometries $\{S_1,\ldots , S_n\}$, satisfying $\sum_{i=1}^n S_iS_i^*=1$. Working within $M_2( \mathcal {O}_n)$ we have $$ \pmatrix{S_1^* & 0 \cr S_2^* & 0}\pmatrix{S_1 & S_2 \cr 0 & 0} = \pmatrix{S_1^*S_1 &S_1^*S_2 \cr S_2^*S_1 & S_2^*S_2} = \pmatrix{1 & 0 \cr 0 & 1}, $$ while $$ \pmatrix{S_1 & S2 \cr 0 & 0} \pmatrix{S_1^* & 0 \cr S_2^* & 0} = \pmatrix{S_1S_1^* + S_2S_2^* & 0 \cr 0 & 0} \leq \pmatrix{1 & 0 \cr 0 & 0}. $$ So $$ \pmatrix{1 & 0 \cr 0 & 1} \preceq \pmatrix{1 & 0 \cr 0 & 0} $$ and it is clear that $$ \pmatrix{1 & 0 \cr 0 & 0}\preceq \pmatrix{1 & 0 \cr 0 & 1}. $$

However these are not equivalent projections because their class in $K_0(\mathcal {O}_n) = \mathbb {Z}_{n-1}$ differ. Precisely, the $K_0$ class of $\pmatrix{1 & 0 \cr 0 & 0}$ is 1, while the $K_0$ class of $\pmatrix{1 & 0 \cr 0 & 1}$ is 2, and $1\neq 2$ in $\mathbb {Z}_{n-1}$ for $n\geq 3$.

Answer 2

I think you ought to build on the Toeplitz extension $$ 0 \to \mathbb{K} \to \mathcal{T} \to C(S^1) \to 0.$$ Let $v$ be the generating isometry and let $e$ be the rank-1 projection such that $vv^* = 1 - e$. I'm pretty sure that antisymmetry does hold in $\mathcal{T}$ itself, but the fact that $v$ renders $1$ and $1-e$ equivalent should be expoitable.

Here's an idea without all details. If your C*-algebra contains $(1,1-e)$ and also $(v,v)$, then you clearly have $(1,1-e) \preceq (1,1)$, but you also have $(1,1) \preceq (1,1-e)$ because $(v,v)$ witnesses an equivalence between $(1,1)$ and $(1-e,1-e)$.

I think that you aren't ever going to to make $(1,1)$ equivalent to $(1,1-e)$ using $(v,v)$ though, basically because the projections you make with $(v,v)$ should have even dimension/codimension.

To actually make this program work, you might want to work in something other than $C^*((v,v),(1,1-e))$ for convenience, though. I think that in either a degree-2 extension $$0 \to \mathbb{K}(H^2) \to A \to C(S^1) \to 0$$ (so, twice the Toeplitz extension in the relevant $\mathrm{Ext}$-group) or in a pullback $$\begin{matrix} B & \to & \mathcal{T} \\ \downarrow & & \downarrow \\ \mathcal{T} & \to & C(S^1) \\ \end{matrix}$$ (so a kind of doubled Toeplitz algebra), you would have good access to tools for proving the nonequivalence of $(1,1)$ and $(1,1-e)$.