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.

1$\begingroup$ There must be noncommutative C* algebras with exactly one projection, right? $\endgroup$ – Johannes Hahn Jan 27 '14 at 0:35

$\begingroup$ You mean one nonzero (and nonunital, if $A$ is unital) projection right? Then yes, but that projection might still be central. But if you could find a simple C*algebra with precisely one nonzero nonunital projection then that would answer my question. $\endgroup$ – Tristan Bice Jan 27 '14 at 0:49

$\begingroup$ @Tristan: if there is one then there are at least two, since if $p$ is a projection then so is $1  p$. Also, I think if $A$ is a von Neumann algebra then the answer is yes because in this case the subspace spanned by projections is dense. $\endgroup$ – Qiaochu Yuan Jan 27 '14 at 0:50

2$\begingroup$ @Qiaochu: Yes, if $A$ is unital then the existence of one would imply the existence of two. And if $A$ is real rank zero (as von Neumann algebras are) or even just generated by its projections then the answer is indeed yes. $\endgroup$ – Tristan Bice Jan 27 '14 at 0:55

4$\begingroup$ The question in the title is the negation of the question in the text. (This is a bit confusing when reading Narutaka's answer.) $\endgroup$ – j.p. Jan 27 '14 at 10:43
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, $$q:=\left(\begin{array}{cc} \frac{1+\sqrt{14xx^*}}{2} & x\\ x^*& \frac{1\sqrt{14x^*x}}{2}\end{array}\right)\quad{\rm in}\quad \left(\begin{array}{cc} pAp & pAp^\perp \\ p^\perp Ap & p^\perp A p^\perp\end{array}\right),$$ or more formally, $$q=\frac{p+\sqrt{p4xx^*}}{2}+x+x^*+\frac{p^\perp\sqrt{p^\perp4x^*x}}{2}\in A$$ is a selfadjoint projection which doesn't commute with $p$. Here $p^\perp=1p$ 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 nonunital $\mathrm{C}^*$algebra $p^\perp A p^\perp$, because the function $f(t)=(1(14t)^{1/2})/2$ vanishes at $t=0$.

$\begingroup$ Perfect answer, thank you professor Ozawa! Just one minor point  as $p$ is not central, I think you mean that $p^\perp Ap^\perp$ is a hereditary C*subalgebra, rather than a twosided ideal, but regardless the continuous functional calculus is still perfectly valid as you say. Incidentally, does anyone know any reference for this or is it just common knowledge among experts? $\endgroup$ – Tristan Bice Jan 27 '14 at 12:29

$\begingroup$ @Tristan: Thank you. I corrected the mistake. BTW, I'm not aware of any literature about this, but since the proof is rather standard (particularly to von Neumann algebraists), one can probably say it's a common knowledge. $\endgroup$ – Narutaka OZAWA Jan 27 '14 at 16:18
[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 nonorthogonal 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 latticecomplement of this immage to
its orthocomplement). [Starting from a idempotent $e$, to obtain the
projection $p$ with the same image Kaplansky uses $p=ee^*/(1+ee^*+e^*eee^*)$ (note that
$1+(ee^*)(ee^*)^*=1(ee^*)^2=1+ee^*+e^*eee^*=1+(ee^*)^*(ee^*)$ 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 $f=e+(1e)xe$ (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
$f=p+(1p)xp$ (which is trivial since it brings back to $p$) and also
the projections $p'$ associated to $1f$ (which is nontrivial, since the
image of $1f$ 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 $A=pAp+pA(1p)+(1p)Ap+(1p)A(1p)$ only the two diagonal
terms $pAp$, $(1p)A(1p)$ 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
idempotentgenerated right ideal is projectiongenerated. 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.
S00029939199110557673 (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").

$\begingroup$ Nice, thanks for the detailed answer. I don't quite see how complements are relevent  $p+p^\perp xp$ can be different from $p$ while its range is still not a complement of $p$. And by $eA=fA$ I think you means $Ae=Af$, and even here I don't think you need that formula. But I can put the following argument together from what you've mentioned, which I think might be along the lines are you are expressing: $\endgroup$ – Tristan Bice Feb 14 '14 at 14:38

$\begingroup$ As you say, take a projection $p\in A$ which commutes with all other projections in $A$. For any $x\in A$, consider the idempotent $e=p+p^\perp xp$ and assume there is a projection $q\in A$ with $qA=eA$. Then, for some $a\in A$, we have $q=ea=eea=eq$ and hence, as $pq=qp$, $qp^\perp=(qeq)p^\perp=0$, i.e. $q\leq p$. Likewise, we have $e=qe$ and hence, as $q\leq p$, $p^\perp xp=p^\perp(eqe)=0$. So if all idempotentgenerated right ideals are projectiongenerated in $A$, this shows that $p^\perp Ap=\{0\}$ and hence $p$ is central. $\endgroup$ – Tristan Bice Feb 14 '14 at 14:39

$\begingroup$ Nice, you essentially digested the 7lines proof in Chevalier following Maeda. I guessed right, Kaplansky got it first (1955): rings of operators, ex. 2 pag. 37. My objective was not only to give old published references, but also to give a taste of the interplay between lattices and rings of operators. Unfortunately I sometimes mix the right/left conventions (dualizing kernels and images), but you can correct this. $\endgroup$ – user46855 Feb 15 '14 at 15:02