Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $e$ be a projection in a C* algebra $A$. Is $eAe= \mathbb{C}e$ equivalent to the nonexistence of any projection in between $e$ and $0$? I know it is true if $A$ is a Von Neumann algebra because then you can use the Borel functional calculus. Takesaki states that the definition of minimality of a projection is $eAe= \mathbb{C}e$ "because it means" that there are no projections in between $e$ and $0$. I can't tell if "because it means" means "implies" or "is equivalent to."

share|cite|improve this question
Isn't the continuous functional calculus enough here? – Kevin Carlson Sep 25 '12 at 5:53
Can you describe how? I wanted to apply step functions. – Jeff Sep 25 '12 at 6:04
Ah, maybe not, I didn't see which functions you were applying. – Kevin Carlson Sep 25 '12 at 6:33
up vote 6 down vote accepted

It is easy to see that $eAe=\mathbb{C}e$ implies that there are no projections below $e$.

But the converse is not true. Consider for instance $A=C([0,1]\cup[2,3])$. Then $e=1_{[0,1]}$ is a projection in $A$ that admits no proper subprojection, and $eAe=C[0,1]\subset A$ is not $\mathbb{C}e$.

share|cite|improve this answer
Or even, $A=C[0,1]$, $e=1$. I.e., the point is that there are unital C*-algebras (of dimension greater than $1$) without nonscalar projections. – Jonas Meyer Sep 27 '12 at 0:53
Good point. It's interesting that according to Takesaki's definition $e$ is not a minimal projection in your example (nor mine, for that matter). – Martin Argerami Sep 27 '12 at 1:09
@JonasMeyer Good point. This can't happen when you have a von neumann algebra of dim > 1 because you just take a nonscalar, which is the weak limit of some linear combination of projections by the usual simple function business from measure theory, and since it's nonscalar, one of these projections has to be nonscalar. – Jeff Sep 27 '12 at 3:17

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.