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Given a smooth and closed toroidal surface (Cliffor torus, non-axysimmetric torus, torus knot...) $S$ parameterized by coordinates $u,v$ such that its metric can be written in the form $ ds^2 = \phi(u,v)^2(du^2 + dv^2)$ implies that the coordinates $u,v$ are isothermal coordinates, satisfying $\nabla_S^2 u = \nabla_S^2 v = 0$, where $\nabla_S^2$ is the Laplace-Beltrami operator.

Say $f = f(u,v)$ is a new function and we want to find $f$ such that $\nabla_S^2 f = 0$. The latter implies $ \Big ( \frac{\partial^2}{\partial u^2} + \frac{\partial^2}{\partial v^2} \Big ) f = 0$. The previous equation could be solved by separation of variables and a unique solution could be found with given boundary conditions.

The latter means there are infinitely many linearly-independent solutions to the equation $\nabla_S^2 f = 0$? I am just trying to make some sense on this since seems like it contradicts the literature stating that there are only two "harmonic solutions" for toroidal surfaces.

Ted Shifrin
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Francisco Sáenz
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  • The existence of isothermal coordinates is a local statement: For any point $p$ on a Riemannian surface $(\Sigma, g)$, there is an open subset $U \subset \Sigma$ with coordinates $(u, v) : U \to \Bbb R^2$. For a closed surface, $U$ cannot be all of $\Sigma$ (otherwise $U$ would be compact). There are infinitely many linearly independent solutions to $\Delta f = 0$ on $U$, but they need not all extend to harmonic functions on $\Sigma$. – Travis Willse Jun 02 '25 at 20:11
  • From what I understand, the $ \nabla_S \cdot \nabla_S = \nabla_S^2$, and this would be regardless of the coordinate system used for the parameterization of $\Sigma$. Particularly, if we choose $u,v$ as our coordinate system then the Laplace-Beltrami is simplified as $\nabla_S^2 = (\partial_u^2 + \partial_v^2)$.

    How is it that a solution to $\nabla_S^2 f = 0$ does not extend to $\Sigma$?

     – Francisco Sáenz Jun 02 '25 at 20:50
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    Computations done in coordinates $u, v$ only apply to the patch of $S$ where the coordinates are defined, not all of $S$. (Isn't the Laplace-Beltrami operator $\phi^{-2} (\partial_u^2 + \partial_v^2)$, by the way?) Consider, for example, the flat torus $T$ in $\Bbb C^2$ (the product of the unit circles) and the chart with inverse the parameterization $r : (0, 2 \pi) \times (0, 2 \pi) \to T \subset \Bbb C^2$, $r : (u, v) \mapsto (e^{iu}, e^{iv})$ with the induced metric . Now, the coordinate functions $u, v$ are both harmonic, but they don't even extend to continuous functions $T \to \Bbb R$. – Travis Willse Jun 02 '25 at 22:11
  • Indeed $u, v$ are harmonic ($\nabla_S^2 u = \nabla_S^2 v = 0$) but not periodic (not continuous functions). This is the case for all isothermal coordinates of any smooth and continuous toroidal surface. Going back to the question of $\nabla_S^2 f = 0$ having infinitely many independent solutions... you are saying that there are multiple independent solutions but they are also discontinuous? – Francisco Sáenz Jun 03 '25 at 00:14
  • Even if you extend the example functions to all of $T$, the resulting functions will not be harmonic along the discontinuity. On the coordinate patch, you can find infinitely many linearly independent functions. The global situation is different: The only harmonic functions on any compact, connected manifold are the constants. – Travis Willse Jun 03 '25 at 01:11
  • Thanks for the clarification, this is really helpful. Maybe just a final observation.

    If we now work on the patch $U$ and try to find solutions to $\nabla_S^2 f = 0$ ($f$ not continuous), then $f$ could be described as $f(u,v) = Au + Bv + \sum C_i f_i (u,v)$, where $f_i$ corresponds to the solutions obtained by assuming separation of variables $f_i = H_i(u)J_i(v)$ ? Are the $f_i$ solutions linearly independent from $u$ and $v$?

     – Francisco Sáenz Jun 03 '25 at 17:02
  • You'd need to allow the sum to be a series, but even then there might be some further subtlety here. I think even a simple example like $f(u, v) = u^2 - v^2$ needs infinitely many terms to express in terms of separable solutions, so depending on what you're trying to do a series representation might not be the most helpful one. – Travis Willse Jun 03 '25 at 17:23
  • Indeed, the sum is a series since that's the natural solution to the equation $\partial_u^2 f + \partial_v^2 f = 0$. The solutions I am looking for are functions whose derivatives are continuous at the boundaries of the patch. Thus, solutions of the form $f(u,v) = u^2 - v^2$ are not possible.

    Thanks again for your help.

     – Francisco Sáenz Jun 03 '25 at 18:56

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On any compact, connected Riemannian manifold $(M, g)$, the Maximum Principle implies that the only harmonic functions, i.e., solutions $f : M \to \Bbb R$ of Laplace's equation, $\nabla_g^2 f = 0$, are the constant functions; see this answer for a clever proof that uses Stokes' Theorem. In particular that's true for a smooth torus (of any dimension) irrespective of the metric $g$.

For local solutions, the story is entirely different. For any surface and around any point there are coordinates $(u, v)$ on a such that $g$ has coordinate representation $$g(u, v) = \phi(u, v)^2 (du^2 + dv^2)$$ for some function $\phi$ (such coordinates are called isothermal coordinates). In these coordinates the Laplacian operator is $\nabla_g^2 = \phi^{-2}\bar\nabla^2,$ where $\bar\nabla$ is the Levi-Civita connection of the standard metric $du^2 + dv^2$. In particular, any harmonic function in $u, v$ is harmonic with respect to $g$, and you can find an infinite set of independent harmonic functions in $(u, v)$, e.g., $$\{1, u, v, u^2 - v^2, 2 u v, \ldots\} .$$

I suspect that the statement "there are only two 'harmonic solutions' for toroidal surfaces" refers to the fact that the space of harmonic $1$-forms on a $2$-torus is $2$-dimensional, hence a basis of such functions has $2$ elements: On $S^1 \subset \Bbb C$ the pullback of the $1$-form $-y \,dx + x \,dy$ is harmonic (with respect to the round metric $g$), and the pullbacks of this form by the $2$ projections $S^1 \times S^1 \to S^1$ comprise a basis of $1$-forms harmonic with respect to the flat metric $g \oplus g$.

Travis Willse
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