Satisfying Solution

In my previous post, I posed the problem:

Problem Let f(x,y) and g(x,y) be polynomials in two variables such that \lim_{t \to \infty} f(t,t^{-1}) and \lim_{t \to \infty} g(t,t^{-1}) exist. Show that the determinant

(\partial f/\partial x) (\partial g/\partial y) - ( \partial f/\partial y) (\partial g/\partial x ) is not a nonzero constant.

Dennis gave a nice, if brief solution, and Pace fleshed it out a bit in an e-mail to me. I now understand where the mysterious “trace operator” in the solution comes from, and that’s what I want to explain here.

First, Dennis’ solution. For any polynomial h(x,y) = \sum h_{pq} x^p y^q, define T(h) = \sum h_{pp}/(p+1). We now observe the mysterious identity

T( f_x g_y - g_x f_y) = \sum_{m = -\infty}^{\infty} m \left( \sum f_{(i+m)i} \right) \left( \sum g_{j(j+m)} \right).

Now, if f(t, t^{-1}) contains no positive powers of t, then $\sum f_{(i+m)i$ is zero for positive m. Similarly, the hypothesis on g forces the third factor to be zero for negative m. And, when m=0, the m in front kills the summand. So the conditions on f and g force T(f_x g_y - f_x g_y) to be zero. In particular, f_x g_y - f_x g_y is not a nonzero constant.

It’s very slick. The only question is, what is T? The explanation, as it turns out, gives us a nice chance to practice our differential forms.

Without futher ado, the answer.

T(h) = (1/\pi) \int_{|z|<1} h(z, \bar{z}) dA.

That is, we are averaging the function h(z, \bar{z}) over the unit disc. I always get confused when \bar{z}‘s show up, so let me just emphasize that, although we are talking about a complex valued function on the complex plane (namely h(z, \bar{z})), this is a perfectly ordinary integral over a disc.

We can check this quickly in polar coordinates:

(1/\pi) \int_{r=0}^1 \int_{\theta=0}^{2 \pi} h_{pq} r^{p+q} e^{(p-q) i \theta} r dr d \theta = (1/\pi) (2 \pi) \sum h_{pp}/(2p+2).

So, why is T(f_x g_y - f_y g_x)=0? Consider (f,g) as a map from one 2-dimensional vector space to another. I want to distinguish the domain and the range, so I’ll write (f,g) : U \to V. I’ll use the coordinates (x,y) on U and (s,t) on V. Consider the two form ds \wedge dt on V.

A reassuring sidenote: s is a smooth (complex valued) function on V, so there is no reason we can’t take its differential as a function on \mathbb{R}^4. Later, we’ll see d \bar{z}‘s showing up, and there is nothing deep about them either. The definitions of differential forms and integrals which you learned in real analysis are still valid in complex analysis. (If you wonder why I make such a point of this, remember that differential operators do mean something different; try explaining the computation of (\partial/\partial z) |z| in terms of small movements along a path!)

In any case, notice that (f,g)^* (ds \wedge dt) = (f_x g_y - f_y g_x) dx \wedge dy. Indeed, this is almost how the Jacobian is defined. So T( f_x g_y - f_y g_x) is the integral of (f,g)^* (ds \wedge dt) over the disc D:=\{ (z, \bar{z}) : |z| \leq 1 \} in U. (Up to the factor of \pi, and the ratio between dz \wedge d \bar{z} and dA.)

But ds \wedge dt is closed, so (f,g)^* (ds \wedge dt) is closed as well. So I can move that integral to any disc with the same boundary. In particular, if \lim_{z \to \infty} (f,g)(z, z^{-1}) is well defined, then I can push the integral onto the disc D':=\{ (z, z^{-1}) : |z| \leq 1 \}. But D' is a holomorphic disc of one complex dimension! So the pull back of ds \wedge dt to D', along a holomorphic map, is a wedge of two proportional 1-forms. In short, it must vanish.

When f and g do not obey the above limiting conditions, then we can’t push the integrating surface off to infinity, as there is a pole near the origin of D'. I leave it to the reader to see why we get the no-longer-mysterious identity above in that setting. Hint: (f,g)^* (ds \wedge dt) isn’t just closed, it’s also exact.

I’ll close by wondering whether there is any good theory of discs like D; discs in \mathbb{C}^2 such that the restriction of dx \wedge dy is the area form. For example, are such discs determined by their boundary? Which curves bound such discs?

4 thoughts on “Satisfying Solution

  1. Wow— the picture I had developed of T was nowhere near as clear as this. (I kept the statement as brief as possible as not to confuse anybody with illusory connections I kept almost seeing in that identity. Now I understand it.) I like it when a story has a happy ending.

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