C*-algebras 7: the continuous functional calculus of normal elements

[@iman4423]

Thanks very much for the lecture, It was very helpful.
I am sitting thinking about using Continuous function Calculus to prove that the extreme point of the set P1(A) := {x ∈A+ : ‖x‖≤1} is equal to Proj(A). A is C*algebra .
Also I was trying to prove that Proj(A)=P1(A) in finite dim space.
I wanted to use CFC to show that the none-projections are not extreme points, but I could not prove that, Can you help me with this?

[@TheArmchairIntellectual]

Take the approach that Kadison took when he first proved this result in 1950: www.jstor.org/stable/1969534. What you are after is Theorem 4, and instead of proving that none-projections are not extreme points he showed that extreme points are necessarily projections (with a fair bit of handwaving). Basically, if T is an extreme point then you can examine the Gelfand transform of T to realize that this must be a function that is equal to its square (otherwise it will be possible to find a pair of functions T' and T'' that T is a convex combination of). Of course there are some details to iron out, but this is the basic gist of it.

[@iman4423]

@@TheArmchairIntellectual thanks for your reply, I managed to prove the first direction, but for the second, I managed to prove if the projection P is written as a convex combination of X and Y then X=PXP and Y=PYP.
and I have already proved that the unit is extreme point. But I could not complete the prove. i.e, I did not fully understood his last two line in his proof. Can you please elaborate a little bit? thanks very much for your help!

[@iman4423]

also, I was wondering if you can guide me through some resource where I can get the prove of Proj(A)=P1(A) when A=Mn(C). Sorry for all my questions, hopefully you can help me, thanks very much

[@TheArmchairIntellectual]

@@iman4423 If we denote by C our original C*-algebra and by P the projection, then we have that C'=PAP is a sub C*-algebra of A (The notion of B*-algebra in Lemma 2 is a historical artefact, it's really the same as a C*-algebra). Moreover, we see that PX=XP=X for any X in C', meaning that P is the unit of C'. Thus, by Lemma 2 it follows that P is an extreme point of the unit sphere in C', which implies that A=B=P, since both A and B were shown to lie in C'. Thus, we see that P is an extreme point of the unit sphere in C as well.

[@TheArmchairIntellectual]

@@iman4423 I'm not entirely sure that the set of projections is equal to P1(A) when A is a matrix algebra. This because of the fact that ||k*B||=|k|*||B|| for any constant k and any matrix B in A. Hence, if 0