Can non-central projections still commute with all other projections? Let $A$ be a C*-algebra and let $\mathcal{P}(A)$ denote the set of projections in $A$.  If $p\in\mathcal{P}(A)$ commutes with everything in $\mathcal{P}(A)$ does it necessarily commute with everything in $A$?  I feel like I should/did know this but I'm currently drawing a blank, so any help would be much appreciated.
 A: The answer is yes, but I don't know where it is written. If $p$ is not central, then $pAp^\perp\neq\{0\}$ and one can take $x\in pAp^\perp$ such that $0<\|x\|<1/2$. Then, $$q:=\left(\begin{array}{cc} \frac{1+\sqrt{1-4xx^*}}{2} & x\\ x^*& \frac{1-\sqrt{1-4x^*x}}{2}\end{array}\right)\quad{\rm in}\quad \left(\begin{array}{cc} pAp & pAp^\perp \\ p^\perp Ap & p^\perp A p^\perp\end{array}\right),$$ or more formally, $$q=\frac{p+\sqrt{p-4xx^*}}{2}+x+x^*+\frac{p^\perp-\sqrt{p^\perp-4x^*x}}{2}\in A$$ is a self-adjoint projection which doesn't commute with $p$.
Here $p^\perp=1-p$ belongs to the multiplier of $A$. Even if $A$ is not unital, $p^\perp A p^\perp$ makes sense as a hereditary subalgebra of $A$ and the functional calculus also makes sense in the non-unital $\mathrm{C}^*$-algebra $p^\perp A p^\perp$, because the function $f(t)=(1-(1-4t)^{1/2})/2$ vanishes at $t=0$.
A: [Below I consider only the unital case. I have to think about the non unital case, but it should be sufficient
to consider the associated $C^*$-algebra with unit $A\oplus K1$, with
$K$ the complex o real $*$-field]
The positive answer, already given by professor Ozawa, is valid in every
$*$-symmetric $*$-ring (each element $1+xx^*$ is invertible). In other
words, a purely algebraic proof (in Kaplansky's style) is available.                                                           
The first basic lemma (in Kaplansky, ring of operators, pag. 34, as
already noted by professor Handelman in another answer on mathoverflow,
"Expression of a non-orthogonal projection in a $C^*$ algebra via an
orthogonal one") is this: if $e$ is an idempotent, then $eA=pA$ for a
suitable (unique) projection $p$. (Geometrically, for any fixed faithful
representation of $A$ as ring of endomorphisms of a module (or any
abelian object), $p$ is the projection with the same image as $e$; the
kernel is changed from an arbitrary lattice-complement of this immage to
its orthocomplement). [Starting from a idempotent $e$, to obtain the
projection $p$ with the same image Kaplansky uses                    $p=ee^*/(1+ee^*+e^*e-e-e^*)$ (note that
$1+(e-e^*)(e-e^*)^*=1-(e-e^*)^2=1+ee^*+e^*e-e-e^*=1+(e-e^*)^*(e-e^*)$ is
invertible).]                                                                                                                  
The second basic and folklore lemma (valid in every ring) is the
description of all and only the idempotents $f$ with the same image as a
given idempotent $e$: $eA=fA$ iff $f=e+(1-e)xe$ (geometrically, one sums
a operator with image contained in the image of $e$ and kernel
containing that image).                                                                                                        
Now take a projection $p$. By hypothesis it must commute with all
projections associated by the first lemma to the idempotents
$f=p+(1-p)xp$ (which is trivial since it brings back to $p$) and also
the projections $p'$ associated to $1-f$ (which is non-trivial, since the
image of $1-f$ i.e. the kernel of $f$ is an arbitrary complement of the
image of $f$ i.e. the image of $p$).                                                                                           
Now this commutation of two projections $p$ and $p'$, with $p'$ having
image a complement of the image of $p$, is only possible when their
product is zero (i.e. orthocomplementary images) since $pp'$ and $p'p$
have images one inside the image of $p$ and the other inside the image
of $p'$. So $p$ must be a projection with only one complement to its
image (the orthocomplement), and dually the orthocomplement of the image
has only one complement. This means that in the Peirce (matrix block)
decomposition $A=pAp+pA(1-p)+(1-p)Ap+(1-p)A(1-p)$ only the two diagonal
terms $pAp$, $(1-p)A(1-p)$ are nonzero, hence $p$ is central (it is the
unit for the corner $eAe$ and it annihilates on both sides the other
corner).                                                                                                                       
So the above proof works in all $*$-rings such that each
idempotent-generated right ideal is projection-generated. This, besides
$*$-symmetric rings (hence real or complex $C^*$-algebras) includes all
Rickart $*$-rings; infact, this is implicit in Berberian's book
Baer$^*$-rings, 8A pag. 39, with complements in Chevalier, proc. ams.
S0002-9939-1991-1055767-3 (prop. 12 pag. 946, already known to Berberian
and to S. Maeda in 1958; I bet that at that time also Kaplansy would
have considered this "folklore").                                                                                            
