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An $n$ dimensional harmonic function is defined to be a real valued function $f$ in $\mathbb{R}^n$ such that $\nabla^2 f = 0 $. Equivalently, $f$ is the scalar potential of a conservative vector field $ F = \nabla f$ with zero divergence: $ \nabla F = 0 $.

The one dimensional harmonic functions are simply the linear functions. In two dimensions, the real and imaginary parts of a complex analytic function are harmonic, and so come in conjugate pairs. I have been trying for some time to generalise the notion of a harmonic conjugate, or possibly harmonic conjugates, plural, if higher dimensions require triples, quadruples, etc. of related harmonic functions rather that just a pair as in 2 dimensions, in a natural way. I haven't been able to come up with anything though, and a quick google search does not yield anything in that direction.

Taking inspiration from the fact that in 2-D harmonic conjugates have gradient vector fields that are orthogonal and of equal length, in fact, one vector field is simply the other one rotated by 90 degrees, I tried analogous ideas for 3 dimensions and higher, but to no avail. Simply rotating $F$ by 90 degrees does not generally result in another conservative vector field with no divergence. I tried a more general linear transformation than a rotation, namely $MF$ for some constant matrix $M$, but when I checked the conditions for $MF$ being a conservative vector field with zero divergence given that $F$ is, I got that $M$ must be a scalar matrix, which is of course trivial and uninteresting.

In another direction, considering how harmonic functions in 2 dimensions are real and imaginary parts of complex analytic functions, one could try a similar idea for higher dimensions. But of course, the Fundamental Theorem of Algebra famously states that there is no finite field extension of $\mathbb{R}$ other than $\mathbb{C}$, so I was thinking perhaps of dropping commutativity and looking for a skew field analogue, but again, could not find anything.

So is there a known natural generalisation? Or is the two dimensional case simply special? It is not uncommon in mathematics for particular low dimensional examples to be different, and the number 2 seems to be a special case in many other different ways as well. And if this is the case, does anyone have any insight as to why 2 should be singled out?

Thank you in advance.

Asier Calbet
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  • Hmm. Maybe because in $\mathbb {C}$, the concept of differentiation is different that in $\mathbb {R^2}$. – Martín Vacas Vignolo Mar 30 '16 at 12:35
  • I would say it is hard to make such a generalization, since $\Delta f=0$ is not conformal invariant when $n>2$, (by conformal invariance I mean $f$ is harmonic if and only if $f\circ \varphi$ is harmonic for all conformal map $\varphi$). –  Mar 30 '16 at 12:37

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