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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?

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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...

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  • $\begingroup$ 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

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