somewhere near the beginning.

Cauchy’s theorem via Homotopy

Filed under: Mathematics — Alex @ 12:25 pm 8/8/2007

I just finished chapter 2 in Fulton’s algebraic topology book, which covers winding numbers, homotopy, and vector fields. Along the way, I saw a theorem that reminded me an awful lot of the Cauchy-Goursat theorem in complex analysis: namely that if a function is sufficiently special, integrating it along two different contours which can be smoothly deformed into each other gives the same result. Specifically:

If \gamma and \delta are smoothly homotopic closed paths in an open set U, and \omega is a closed 1-form in U, then

\displaystyle \int_\gamma \omega = \int_\delta \omega.

Strictly speaking, the Cauchy-Goursat theorem says that the integral of an analytic function in a simple domain around a closed curve is zero:

Let f be holomorphic in a simply connected domain D and let \gamma be a piece-wise continuous closed path in D, then

\displaystyle \cint_\gamma f(z)\;dz = 0.

but some simple contour manipulations show that an easy equivalent statement is: if \gamma and \delta are piece-wise continuous closed paths in D, then \int_\gamma f(z)\; dz = \int_\delta f(z)\; dz.

The connection between the two theorems was so striking that I did a little investigation, and found that homotopy theory provides a very concise and palatable proof of a (weaker?) version of the Cauchy-Goursat theorem:

Assume f is analytic in an open set U and C_1, C_2 are smoothly homotopic in U with homotopy H(t,s). Then \int_{C_1} f(z) \; dz = \int_{C_2} f(z) \; dz .

To prove this, let \gamma_s(t) = H(t,s) represent the curve intermediate between C_1,C_2 determined by the blending parameter s. We want to show that for all s, I(s) = \int_{\gamma_s} f(z)\; dz is a constant. Note that I(s) = \int_0^1 f(H(t,s)) \frac{\partial H(t,s)}{\partial t} \; dt . Since we’re assuming smoothness of H, Liebniz’s rule gives

\displaystyle \frac{\partial I}{\partial s} = \int_0^1 \frac{\partial f}{\partial z} \circ \frac{\partial H}{\partial s}\frac{\partial H}{\partial t} + (f\circ H)\frac{\partial^2 H}{\partial t \partial s} \; dt
\displaystyle = \int_0^1 \frac{\partial}{\partial t} \left( (f \circ H) \frac{\partial H}{\partial s} \right) \; dt = \left. (f \circ H) \frac{\partial H}{\partial s} \right|_0^1 = 0

since H(1,s) = H(0,s). QED.

A powerful advantage of this version of the Cauchy-Goursat theorem is that the domain doesn’t need to be simply connected for the theorem to apply. It’s also clearer than the standard proofs I’ve seen.

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