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Let $T: E \rightarrow E$ be an endomorphism of a finite-dimensional vector space, and let $S$ be a circle in the complex plane that does not intersect any eigenvalues of $T$. Now let $Q = \frac{1}{2\pi i} \int_S (z-T)^{-1} \, dz$.

Why is $Q$ a projection operator?

The motivation behind this question is that the above situation occurs in a proof of Bott's periodicity theorem, but it's not clear to me that $Q$ is a projection...

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(This has nothing to do with k-theory, by the way.) –  Mariano Suárez-Alvarez Dec 23 '10 at 4:22
    
(You're right, but I figured that if anyone else read Atiyah's proof in his book, then they may have come across this issue as well. I'll remove the tag.) (Parenthetical conversations feel like whispering.) –  Dylan Wilson Dec 23 '10 at 4:27
    
I would definitively use the tag (functional-analysis) –  AD. Dec 23 '10 at 6:11
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2 Answers

Show that the integral depends continuously on $T$, and show that $Q^2=Q$ when $T$ is diagonalizable, by finding how $Q$ changes change you change $T$ by a similar matrix, and then reducing to the one dimensional case. Then use the fact that diagonalizable matrices are dense in the space of all matrices, and that $Q^2$ and $Q$ are continuous functions of $Q$.

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huh... so it seems like this is fairly nontrivial? I thought I was just being dumb since Atiyah just defines Q and then says, "which is a projection operator that commutes with T." Thanks! –  Dylan Wilson Dec 23 '10 at 4:00
    
Also I think you mean T at the very end –  Dylan Wilson Dec 23 '10 at 4:02
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"Trivial" is a relative notion! (I mean $Q$, but I also have in mind the fact that the composition of continuous functions is continuous :) ) –  Mariano Suárez-Alvarez Dec 23 '10 at 4:02
    
@Dylan: by the way, can you see what is the subspace $Q$ projects on to? –  Mariano Suárez-Alvarez Dec 23 '10 at 4:13
    
Ah, for that I think Jonas's answer gives it away... I think that its image should contain all the eigenvectors corresponding to eigenvalues of T inside S? –  Dylan Wilson Dec 23 '10 at 4:18
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If $A$ is any Banach algebra (such as the algebra of endomorphisms of a finite dimensional complex vector space), then for each subset $\Omega$ of the complex plane and each element $T$ of $A$ whose spectrum is contained in $\Omega$, holomorphic functional calculus yields a homomorphism $f\mapsto f(T)$ from the algebra of functions holomorphic in an open set containing $\Omega$ (identified if they agree on some neighborhood of $\Omega$) into $A$. Since the function $f:(\mathbb{C}\setminus S)\to\mathbb{C}$ defined by $f(w)=\frac{1}{2\pi i}\int_S(z-w)^{-1}dz$ takes on only the values $0$ and $1$ (it gives the winding number of $S$ about $w$), $f$ is idempotent (i.e., $f(w)^2=f(w)$ for all $w\in\mathbb{C}\setminus S$), and thus $f(T)$ is an idempotent element of $A$ for each $T$ whose spectrum is disjoint from $S$.

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ooooh... I like both of these answers equally. I don't know which one to accept... –  Dylan Wilson Dec 23 '10 at 4:08
    
And am I to understand "0" and "1" as the 0 map E --> E and the identity on E? Also, I assume that the verification that the fact that this winding number acts like the more familiar winding number is relatively straightforward... –  Dylan Wilson Dec 23 '10 at 4:10
    
@Dylan: Sorry for the lack of clarity. By $f(w)$ I meant that $w$ is in $\mathbb{C}\setminus S$, and $f:\mathbb{C}\setminus S\to \mathbb{C}$ is the ordinary winding number. So $f$ is a plain old function of a complex variable, where the operations are pointwise. Its values are the complex numbers $0$ and $1$, hence it equals its (pointwise) square. The homomorphism property of holomorphic functional calculus then guarantees that $f(T)^2=f(T)$. I have edited to clarify. –  Jonas Meyer Dec 23 '10 at 4:16
    
@Dylan: But there is certainly work involved in verifying the relevant properties of holomorphic functional calculus for Banach algebras, and this approach to your particular problem is probably overkill. –  Jonas Meyer Dec 23 '10 at 4:21
    
Ah! Now I understand... very clever, and very helpful intuitively. Thank you! –  Dylan Wilson Dec 23 '10 at 4:23
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