# cauchy integral theorem poincare

It appears to me that there is a tight connection between the Cauchy integral theorem and Poincare's Lemma(in 2 dimensions). While the former requires the function to be holomorphic, i.e. to be differentiable in the real sense and fulfill the Cauchy-Riemann equations, the latter stipulates that the function have a 2-dimensional curl of 0(in vector-analysis formulation, naturally). Additionally, for both to apply, the space has to be contractible, only that this is merely implicitly stated in the Cauchy integral theorem since an everywhere holomorphic function certainly is defined at every point.

Now, i am only a student of physics, but it irks me that this apparently deep connection(especially considering that the CR-equations appear to me to basically be a complex 2-dim-curl equal to 0, or, equivalently, an exterior derivative of the one-form equal to 0) is never commented on in our watered-down mathematics courses.

So here is the question: what is the actual connection between these two marvelous theorems?

• There are a lot of things named after Poincare. What exactly is it that you call Poincare's lemma? – Willie Wong Aug 9 '16 at 9:29
• right, sorry, i could have thought of that. So what I mean is the following: – Leviciviathan Aug 9 '16 at 11:47
• let U be a bounded subset of R^2 that is connected and simply connected. furthermore, let f be a continuously differentiable vector field defined on the closure of U. Then the following are equivalent: 1) f is conservative and 2) the curl of f is 0. I hope this clears it up! – Leviciviathan Aug 9 '16 at 11:55
• and the (maybe existing) connection to the Cauchy Integral theorem is that from a vectorfield being conservative it follows that any integration along a closed curve is 0, which is just very similar to what the Cauchy integral theorem tells us – Leviciviathan Aug 9 '16 at 11:59

Let $f(z) = f(x+iy) = u(x,y) + i v(x,y)$ be your holomorphic function.
Let's think formally about the one form $$f(z)~\mathrm{d}z = (u~\mathrm{d}x - v~\mathrm{d}y) + i (v~\mathrm{d}x + u~\mathrm{d}y)$$
It exterior derivative is $$\mathrm{d} (f ~\mathrm{d}z) = [(-\partial_y u - \partial_x v) + i(\partial_x u - \partial_y v)] ~\mathrm{d}x\wedge\mathrm{d}y$$
In other words, the complex-valued one form $f(z)~\mathrm{d}z$ is closed if and only if the Cauchy-Riemann equations are satisfied.