In Axler's Harmonic Function Theory, he and his coauthors develop the theory of harmonic functions on spheres and discs by considering the restrictions of arbitrary polynomials on the sphere $S^{n-1} = \{x \in \mathbb{R}^n : ||x||^2 = 1 \}$ and taking the Poisson integral to get a harmonic polynomial in the interior ball. One can then take the Kelvin transform to get a harmonic function on the exterior of the sphere. This process yields a canonical projection $\mathscr{P}(\mathbb{R}^n) \to \mathscr{H}(\mathbb{R}^n)$, from the space of polynomials to the space of harmonic functions, factoring through the restriction map to $L^2(S^{n-1})$.

Does this theory generalize to knot complements? Say we have a knot $K \subseteq \mathbb{R}^3$, and we take a small tubular neighborhood $V$ around $K$, whose boundary is topologically a torus $T$. Given a function on the knot complement, one could restrict to $T$ and then solve the Dirichlet problem on the knot complement to get a projection like the one above. However, in the sphere case, there are many nice properties of the harmonic function theory; namely it comes with an efficient algorithm for computation of a harmonic polynomial basis of $L^2(S^{n-1})$ which involves repeatedly differentiating the function $f(x) = |x|^{2-n}$.

Is anyone aware of any theory along this vein? Are there any obstacles to generalizing what happens in the sphere case?


This is more of a comment, but way too long. First, two remarks on the initial part of the question:

  1. The Kelvin transform of a harmonic polynomial is of course harmonic, but it is not a polynomial. For example, the constant $1$ gets transformed into $|x|^{2-n}$.

  2. The extension of the projection $\pi : \mathscr P(\mathbb R^n) \mapsto \mathscr H(\mathbb R^n)$ is not clear to me. Let $\pi = \pi_2 \circ \pi_1$ be the factorisation mentioned in the question: $\pi_1$ maps polynomials to their restrictions to the unit sphere $\mathbb S^{n-1}$, and $\pi_2$ extends that harmonically to the unit ball $\mathbb B^n$. Then $\pi_2$ clearly extends to the usual extension from $L^2(\mathbb S^{n-1})$ to the harmonic Hardy space $\mathscr H^2(\mathbb B^n)$ in the unit ball, given by the Poisson integral. And for $\pi_1$ all that we need is to be able to restrict our function to the unit sphere and get something square-integrable (so for example the Sobolev space $H^{1/2}(\mathbb B^n)$ will do). However, if we require our projections to be in $\mathscr H(\mathbb R^n)$, the class of entire harmonic functions, then their power series converge everywhere, which is a severe restriction. I am not aware of any intrinsic characterisation of the inverse image of $\mathscr H(\mathbb R^n)$ through (the extension of) $\pi_1$, let alone $\pi = \pi_2 \circ \pi_1$.

When it comes to the main question, I have troubles understanding the proposed construction. Of course any (reasonable — say, integrable with respect to the surface measure) boundary values on $T$ correspond to a harmonic function $h$ in the complement of $V$, again given by a Poisson integral (with kernel which is no longer known explicitly). This $h$ is given uniquely if we assume, say, that $h$ is bounded at infinity. If we are lucky, this function $h$ might extend to the complement of $K$, but I am not aware of any reasonable conditions for such an extension to exist, even in the simplest possible setting when $K$ were a point and $T$ a sphere (this is then essentially what I was trying to describe in the first part of this comment, after a Kelvin transform).

So it looks like I got something completely wrong...

  • $\begingroup$ You’re right about how the Kelvin transform of a polynomial isn’t necessarily a polynomial. I had the description of the theory wrong, and I’ll edit my post. However, my question on how harmonic function theory translates to a knot complement is still unanswered. $\endgroup$
    – Max Lipton
    Oct 5 '20 at 1:57
  • $\begingroup$ I am afraid I still do not understand the question. (1) The "canonical projection" now maps polynomials to harmonic functions in $\mathbb R^n \setminus \{0\}$. (2) Solutions to the Dirichlet problem give a similar "projection" onto the class of harmonic functions in the complement of $V$ for whatever (reasonable) set $V$. (3) Extension to the complement of $K$ is problematic, I believe. $\endgroup$ Oct 5 '20 at 7:42
  • $\begingroup$ Damn, this clearly did not deserve the bounty... $\endgroup$ Oct 12 '20 at 21:04
  • $\begingroup$ Do not be discouraged! They are merely points. As you know, we academics can be busy people with multiple forces pulling on us. I'm still formulating a response to your question. I've been thinking about my problem, and I didn't want to say anything until I was ready. $\endgroup$
    – Max Lipton
    Oct 15 '20 at 16:04

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.