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For a harmonic function $\Phi$ on a simply connected subset $\Gamma$ of $\mathbb{R}^3$, define a guide curve $\gamma: I \mapsto \Gamma$ of $\Phi$ as a simple regular $C^1$ curve such that

  • all point in $\gamma(I)$ are critical points of $\Phi$, and
  • for all points $p$ in $\gamma(I)$ there exists a neighborhood $V$ of $p$ so that all critical points of $\Phi$ within $V$ are also in $\gamma(I)$.

My question is whether there are any such guide curves which do not have an analytic parametrization?

For a concrete example, consider $\Phi(x,y,z)=x\ y\ z$ for which any part of a coordinate axis not including the origin is a guide curve.

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  • $\begingroup$ Background: The question is relevant to networks of rf ion traps, where the trapping potential is the ponderomotive potential associated with an oscillating electric field. The local amplitude of electric potential oscillations is described by a harmonic function, and for practical reasons it is preferable to trap ions on critical points of this function so that transport would have to take place along guide curves as introduced above. The aim of my question is to establish what intersection topologies are possible for guide curves. $\endgroup$
    – Janus Wesenberg
    Commented Oct 21, 2010 at 7:28
  • $\begingroup$ You need to be a lot more careful with your quantifiers. Let $\Phi$ be a harmonic function without a critical point, say $\Phi(x,y,z) = x$. then trivially any curve $\gamma$ has the property that for every point $p\in \gamma$, for any neighborhood $V$ of $p$, all critical points of $\Phi$ in $V$ (which comprise the empty set) is in $\gamma$. Now, I am not sure what you mean by "analytical parametrization", but given that ALL curves $\gamma$ are allowed in the above example, if there exists curves that does not have analytical parametrization, then you can draw the obvious conclusion. $\endgroup$
    – Willie Wong
    Commented Oct 21, 2010 at 9:39
  • $\begingroup$ In analogy with the imaginary part of $(x + i y)^n$ being zero along $n$ lines through the origin in the plane, my guess is that one can arrange guide curves as lines through the origin in $R^3$ and the vertices of a regular polyhedron, not just the octahedron as you point out. Furthermore I thing harmonic polynomials suffice to do this. What seems less clear is larger numbers of points on the unit sphere. Note that for fixed constants $A,B,C$ the Laplacian commutes with $$ A \frac{\partial}{ \partial x} + B \frac{\partial}{ \partial y} + C \frac{\partial}{ \partial z} $$ $\endgroup$
    – Will Jagy
    Commented Oct 21, 2010 at 18:40
  • $\begingroup$ @Willie Wong: Thank you very much for pointing this one out. I have corrected my definition of guide curves so it now hopefully describes what I am looking for. $\endgroup$
    – Janus Wesenberg
    Commented Oct 22, 2010 at 7:00
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    $\begingroup$ Since the laplacian is elliptic with real-analytic coefficients, a harmonic function $f$ is real-analytic in its domain of definition. Hence the set $C$ of critical points of $f$ is a real-analytic subset of $\mathbb{R}^3$, and as such it admits a locally finite partition into real-analytic locally closed smooth submanifolds. Thus if $\dim C \leq 1$, it is locally a finite union of analytic open arcs and singular points (but the curves might not extend smoothly across those points). $\endgroup$
    – BS.
    Commented Oct 22, 2010 at 8:20

1 Answer 1

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[This is an answer made of two comments and an example]

Since the laplacian is elliptic with real-analytic coefficients, a harmonic function $f$ is real-analytic in its domain of definition. Hence the set $C$ of critical points of $f$ is a real-analytic subset of $R^3$, and as such it admits a locally finite partition into real-analytic locally closed smooth submanifolds. Thus if $\dim C≤1$, it is locally a finite union of analytic open arcs and singular points (but the curves might not extend smoothly across those points).

A reference on real analytic functions (reedited in 2002) might be

S. Krantz, H. Parks, A primer of real analytic functions. Birkhäuser Verlag, 1992.

But maybe the "curve selection lemma" in Milnor's "Singular points on complex hypersurfaces" would be enough . edit: it concerns real algebraic subsets.

As an example of a curve of critical points not extending throuh a singular point, take the harmonic polynomial $f(x,y,z)=y^3-3x^2y+y^3z-yz^3$, which has critical locus $y=0,z^3=-3x^2$. But of course you have the singular parametrization $x=3t^3, z=-3t^2$. I don't know if they exist in general.

Addendum : in fact the critical locus of a harmonic can polynomial can have an arbitrary (real) plane algebraic curve as a union of irreducible components. Let $P(x,y)$ be a real two variable polynomial, and define $$f(x,y,z)=\Sigma_k \frac{z^{2k+1}}{(2k+1)!}(-\Delta_{x,y})^k P(x,y) \; .$$ It is easy to check that $f$ is harmonic, and $df$ vanishes on $z=0$, $P(x,y)=0$.

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    $\begingroup$ Thank you very much, BS! Perhaps if I had not been so convinced that all curves of critical points would extend through the any singular points, I would have had better luck finding a counterexample myself :) Thus I learn again that (my) intuition and analysis should not be mixed. This is really amazing -- I have been pushing this problem to mathematician friends for a couple of years now with no progress, and then a few days after posting it to mathoverflow it is solved. The future is now (but then, it's 2010 -- so it'd better be) :) $\endgroup$
    – Janus Wesenberg
    Commented Oct 25, 2010 at 2:48
  • $\begingroup$ I will be writing up your answer as a short paper. Please get in touch if you would like to be listed as author, otherwise I will cite your answer here. You can reach me as janus@<domain listed in my profile>. $\endgroup$
    – Janus Wesenberg
    Commented Nov 15, 2023 at 10:46

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