# Minimal projections in a C* algebra

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."

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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

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 $\mathbb{C}e$.
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