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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...

Dylan Wilson
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2 Answers2

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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 $S$.

Jonas Meyer
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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 04:08
  • 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 04:10
  • @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 04:16
  • @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 04:21
  • Ah! Now I understand... very clever, and very helpful intuitively. Thank you! – Dylan Wilson Dec 23 '10 at 04:23
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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 $Q$.

Mariano Suárez-Álvarez
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