Friday, February 5, 2010

An exercise in geometric measure theory

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I was given a math paper to read over that includes, in its introduction, the assertion
"the length of an algebraic curve of degree $d$ in the unit disc is less than $4d$". I don't think I'd seen this before, so I've decided to try proving something like it, and got a smaller constant than $4$, but not much smaller. (If you've not seen this problem before, can you guess?!)

The proof involves the following ideas:

* The length $l(s)$ of a line segment $s$ is given by

$\displaystyle l(s) = \frac{1}{2}\int_0^\pi |p_\theta(s)|\,d\theta $

where $p_\theta$ is the orthogonal projection onto a line $L_\theta$ at angle $\theta$ to the ordinate axis (or what you will), and $|X|$ is the length of the set $X$.
This boils down to the identity $\int_a^{\pi+a} |\sin(\theta)|\,d\theta = \int_0^\pi \sin\theta\,d\theta = 2$.

* The length of a measurable curve is approximated by the length of a fine polygonal approximation to the curve.

* An algebraic curve $\gamma$ of degree $d$ (in the plane) meets any line $L$ at most $d$ times, or else $L$ is a component of $\gamma$

The practical upshot of these ideas is that you can measure the length of a decent bounded curve $\gamma$ by
\item for each $\theta$ and for each $x\in L_\theta$, counting the number $n_\theta(x)$ of points on $\gamma$ that project to $x$ orthogonal to $L_\theta$ --- that is, the number of times the line through $x$ and perpendicular to $L_\theta$ intersects the curve $\gamma$.
\item for each $\theta$, calculating the integral $\int_{L_\theta} n_\theta(x)\,dx $.
\item integrating the results of the last calculation:

$\displaystyle l(\gamma) = \frac{1}{2}\int_0^\pi \int_{L_\theta} n_\theta(x)\,dx\,d\theta $

Because of the caveat about $L$ perhaps being a component of $\gamma$, we note that the definition of $n_\theta(x)$ needs a little refinement: strictly speaking, there may be countably many $\theta$ and $x\in L_\theta$ such that $n_\theta(x)$ as described should be $2^{\aleph_0}$, and it's not clear how to define the integral in this case; but these $x$ and $\theta$ correspond to a line segment projected to a point, so that it actually contributes $0$ to the length estimate described by polygonal approximation. This might be a good example to have in mind when carefully discussing what happens when you try to change the order of limit operations...

I should point out that all the integrals are unordered integrals of positive things --- basically, this is still measure theory. Consequently, we are still within the realm of convex analysis, so we can get estimates on $l(\gamma)$ by finding estimates for the numbers at any of steps $1,2,3$. In the present example, we want an upper bound on $l(\gamma)$, and we are given an upper bound $n_\theta(x) \lt d$

This lets us write
$\displaystyle l(\gamma) \lt \frac{1}{2}\int_0^\pi \int_{p_\theta C} d\,dx\,d\theta $

where $C$ is the unit disk (or circle ... ). The length of the projection $p_\theta C$ of the unit disk is simply $2$, which gives the estimate
$\displaystyle l(\gamma) \lt \pi d $

Since $\pi \lt 4$, we're done!

Of course, with an upper bound, it's useful to have examples to compare it to. The unit circle itself is a curve of degree $2$, with length $2\pi = \pi 2$ within the (closed) unit disk; more generally, the curve $(a\cos(n\theta)+b\cos\theta,a\sin(n\theta)+b\sin\theta)$ is algebraic of degree $2n$, with length within the unit disk as close to $2n\pi$ as you like (choosing $a,b$ carefully). I'm having difficulty thinking of a curve of degree $2n+1$ with length in the disk more than $2n\pi+2$.

That's all!



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