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I am interested in getting some geometrical or analytical perspective in studing the following complex pde. I would appreciate any help.

Consider $ (M,g)$ to be a 3 dimensional Riemannian manifold and consider solving the following pde near a surface $\Sigma$

$<d\Phi,d\Phi>_g =0$ where $\Phi$ is a complex valued function.

If the dimenion is 2 then any function of $z=x_1 + i x_2$ would do the job in isothermal coordinates.

I understand the question is a bit vague but I am interested in general observations about solvability of this pde. Thanks,

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  • $\begingroup$ Isothermal coordinates exist locally iff the manifold is conformally flat. This always happens in dimension two. In 3D this is equivalent with vanishing Cotton tensor and in higher dimensions Weyl tensor. That is, in general isothermal coordinates don't exist in higher dimensions. $\endgroup$
    – Joonas Ilmavirta
    Nov 28, 2014 at 20:21
  • $\begingroup$ you are completely right but this question is not necessarily asking to find isothermal coordinates in a 3-d manifold. I just put the word isothermal coordinates because in 2-d, solutions to the pde are related to isothermal coordinates. I have edited the title $\endgroup$
    – Ali
    Nov 28, 2014 at 22:08
  • $\begingroup$ I know the question is not about isothermal coordinates per se, but it sounded like you would have liked to have isothermal coordinates in higher dimensions as well. $\endgroup$
    – Joonas Ilmavirta
    Nov 29, 2014 at 8:21

2 Answers 2

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In the analytic category, there is a (local) existence theorem, but I don't know about the smooth category. (I rather suspect that there might be some obstructions in the smooth case.) Here is what you can say in the analytic case (and this works in all dimensions):

Suppose that $(M^n,g)$ is a real-analytic Riemannian manifold and that $\Sigma\subset M$ is an embedded real-analytic hypersurface with a choice of normal unit vector field $N$ along $\Sigma$. Denote the induced metric on $\Sigma$ by $\bar g$. Suppose that $f:\Sigma\to \mathbb{C}$ is a real-analytic function that satisfies $\langle \mathrm{d}f,\mathrm{d}f\rangle_{\bar g}= -h^2$ where $h:\Sigma\to\mathbb{C}$ is nonvanishing. Then there exists an open $\Sigma$-neighborhood $U\subset M$ and a function $\Phi:U\to\mathbb{C}$ that satisfies the differential equation $\langle \mathrm{d}\Phi,\mathrm{d}\Phi\rangle_{g}= 0$ and the initial conditions $\mathrm{d}\Phi(N) = h$ and $\Phi = f$ along $\Sigma$. Moreover, the solution is unique in the sense that any other solution $\Phi':U'\to\mathbb{C}$ of the differential equation with these initial conditions that is real-analytic satisfies $\Phi=\Phi'$ on some open $\Sigma$-neighborhood $U''\subset U\cap U'$.

All of this follows as a direct application of the Cartan-Kähler Theorem.

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Your solution for $2$ dimensions works for any dimension. You simply solve for co-ordinates $x^1, \dots, x^n$ such that the metric tensor $g_{ij}dx^idx^j$ satisfies $$ g_{11} = g_{22},\ g_{12} = 0 $$ and set $f = x^1$, $g = x^2$. This is effectively parameterized isothermal co-ordinates, so your title is justified. You should be able to get solutions near a point by fixing a foliation of $2$-dimensional surfaces and solving for isothermal co-ordinates along each surface in such a way that the solution depends smoothly on the other $n-2$ co-ordinates.

ADDED: Here's a more explicit description in dimension 3: Start with co-ordinates $y^1$, $y^2$, and $y^3$ (where the level sets of $y^3$ give the foliation.) and a metric $g = g_{ij}dy^idy^j$. We want to solve for functions $\phi^1$ and $\phi^2$ such that if $$ y^1 = \phi^1(x^1,x^2,x^3),\ y^2 = \phi^2(x^1,x^2,x^3),\ y^3 = x^3 $$ then the metric written with respect to $x^1, x^2, x^3$, $g = \hat{g}_{ij}dx^idx^j$ satisfies $$ \hat{g}_{11} = \hat{g}_{22} \text{ and }\hat{g}_{12} = 0. $$ A straightforward calculation shows that this is equivalent to $$ g_{ab}\partial_1\phi^a\partial_1\phi^b = g_{ab}\partial_1\phi^a\partial_1\phi^b\text{ and } g_{ab}\partial_1\phi^a\partial_2\phi^b = 0, $$ where we sum over $1 \le a, b \le 2$. This is the same system of PDE's corresponding to finding co-ordinates $x^1, x^2, x^3$ such that $x^1, x^2$ are isothermal co-ordinates on any level set of $x^3$. In particular, since it involves only differentiation in the $x^1$ and $x^2$ directions and is elliptic, it can be solved for each value of the "parameter" $x^3$ using the standard approach for constructing isothermal co-ordinates. You can then verify that $\Phi = dx^1 + i dx^2$ satisfies $\langle d\Phi,d\Phi\rangle_g = 0$.

CORRECTION: Robert Bryant points out that the wrong problem is being solved above. The correct PDE is $$ 0 = \langle\partial\Phi,\partial\Phi\rangle_g = g^{ij}\partial_i\Phi\partial_j\Phi. $$ Since $\Phi$ is complex-valued, this consists of 2 real equations for two real-valued functions. The linearization can be shown to be elliptic, if $\partial\Phi$ has maximal rank. Therefore, local solutions exist. However, the construction above of isothermal co-ordinates on each leaf of a foliation does not work.

CORRECTION to the CORRECTION: The linearized system is not elliptic. It is, however, generically hypoelliptic subelliptic. Given this, I believe that local solutions exist, but the only way I can see how to prove this is by proving smooth tame estimates for a solution to the hypoelliptic subelliptic PDE and using the Nash-Moser implicit function theorem.

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  • $\begingroup$ Thanks you for your response. Suppose as per your suggestion I pick a foliation of 2-dimensional surfaces $\Sigma_t$ and I pick $\Phi|_{\Sigma_0}$ to be isothermal coordinates on $\Sigma_0$. Then it is clear to see that we would have $\partial_\nu \Phi = 0$ where $\nu$ is the unit normal to the surface and that would ofcourse be problematic. so I am thinking you do need the condition that Robert mentioned in his response, i.e that h be non vanishing $\endgroup$
    – Ali
    Nov 29, 2014 at 21:47
  • $\begingroup$ Why is $\partial_\nu\Phi = 0$? $\endgroup$
    – Deane Yang
    Nov 29, 2014 at 22:18
  • $\begingroup$ Well we have $ <d\Phi,d\Phi>_g = (\partial_{\nu} \Phi)^2 + <d\Phi,d\Phi>_{g^0}$ where $g^0$ is the induced metric on the surface... $\endgroup$
    – Ali
    Nov 29, 2014 at 22:33
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    $\begingroup$ Deane, I think that there is a misunderstanding. The equations that say that $\Phi= x^1+i\,x^2$ satisfies $\langle\mathrm{d}\Phi,\mathrm{d}\Phi\rangle_g=0$ are not $g_{11}-g_{22}=g_{12}=0$. They are $g^{11}-g^{22}=g^{12}=0$. The former are two equations on the entire coordinate system, while the latter are two equations on the two functions $x^1$ and $x^2$, so they are different if $n>2$. In fact, if $\Phi = u + i\,v$, then $$0=\langle\mathrm{d}\Phi,\mathrm{d}\Phi\rangle_g= g(\nabla\Phi,\nabla\Phi) = \bigl(g(\nabla u,\nabla u){-}g(\nabla v,\nabla v)\bigr) + 2i\, g(\nabla u,\nabla v).$$ $\endgroup$
    – Robert Bryant
    Nov 30, 2014 at 2:49
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    $\begingroup$ @DeaneYang: I agree with your comments above about the hypoelliptic nature of the problem. However, I think that, at least in higher dimensions, the smooth tame estimates might be hard to come by because the linearization of the problem at a solution $\Phi = u+ i\,v$ will be hypoelliptic only if the $2$-plane field spanned by $\nabla u$ and $\nabla v$ is bracket-generating. This is generic in all dimensions, as you say, but as the dimension $n$ increases, it's higher and higher order and thus, presumably, more difficult to analyze. When $n=3$, though, it should be fairly straightforward. $\endgroup$
    – Robert Bryant
    Nov 30, 2014 at 11:44

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