TITLE: Does antisymmetricity of $\preceq$ fail in general C*-algebras?
QUESTION [5 upvotes]: 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 $p\sim q$.

REPLY [3 votes]: 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 $n\geq 3$.