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."
Minimal projections in a C* algebra
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functional-analysis
operator-algebras
c-star-algebras
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0Isn't the continuous functional calculus enough here? – 2012-09-25
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0Can you describe how? I wanted to apply step functions. – 2012-09-25
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0Ah, maybe not, I didn't see which functions you were applying. – 2012-09-25
1 Answers
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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 $\mathbb{C}e$.
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2Or 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. – 2012-09-27
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3Good point. It's interesting that according to Takesaki's definition $e$ is **not** a minimal projection in your example (nor mine, for that matter). – 2012-09-27
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1@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. – 2012-09-27