This question came up when I was trying to find out the details about the existence of isothermal coordinates for surfaces.

Given a surface in $\mathbb{R}^3$, at least $C^2$ for simplicity, at any point on the surface, it is known that in a suitable coordinates around that point, one can always write the first fundamental form

$ds^2=Edu^2+Fdudv+Gdv^2$ as $ds^2=\gamma^{2} (dx^{2}+dy^{2})$

where E,F,G are (at least) $C^1$ functions of u and v, E and G are strictly positive, (x,y) is a new local coordinate system for the surface and $\gamma$ is a $C^1$ function of x and y.

I am aware of a proof which reduces to solving a Beltrami equation. See the related Wikipedia entry, for example. A more detailed proof can be found in this Chern's paper. Their conditions on the coefficients are less demanding than what I have here, but the setups are basically the same. Apart from the setups, the systems we considered are quite different.

Here is what I propose to do. It is well-known that an orthogonal coordinate exists, as can be shown using the existence of real integrating factor, which I will not show here.

We may without loss of generality assume the first fundamental form is in the form


expand right hand side as

$(\sqrt{E}du+\sqrt{G}idv)(\sqrt{E}du-\sqrt{G} idv)$

we wish to find a complex integrating factor $\mu=p+qi$ of $\sqrt{E}du+\sqrt{G}idv$. In other words, we want to find p, q $C^1$ of u,v, which are not both zero at the point in question, such that $dU= \mu (\sqrt{E}du+\sqrt{G}idv)$ for some $U=x+yi$.

Before going to the actual system, let's see how this would solve our problem, assuming such $\mu$ exists.

Clearly, $d\bar{U}=\bar{\mu}(\sqrt{E}du-i\sqrt{G}dv)$.

So we can write

$ds^2=\frac{1}{\left| \mu\right|^{2}}dUd\bar{U}$.

We claim $(u,v) \rightarrow(x,y)$ is a local diffeomorphism.

Indeed, by comparing real and imaginary part of $dU$, we get $dx=p\sqrt{E}du-q\sqrt{G}dv$ and $dy=q\sqrt{E}du+p\sqrt{G}dv$


$\det{\begin{vmatrix}x_u&x_v\\y_u&y_v\end{vmatrix}}=\sqrt{EG}(p^2+q^2)$ which is non-vanishing near the point in question by hypothesis.

Therefore, this is a change of coordinates, and in x, y we can write

$ds^2=\frac{1}{\left| \mu\right|^{2}}(dx^2+dy^2)$. Since this is true for all points on surface, this proves existence of local isothermal coordinates.

There remains of question of existence of such $\mu$.

I propose to solve the following system in a small enough ball around the point in question, and apply the Poincare's lemma.

We wish to find a pair of p,q making both of the real and imaginary parts of the first factor of $ds^2$ closed. This is equivalent to

$\sqrt{E}p_v + p\sqrt{E}_v + \sqrt{G}q_u+q\sqrt{G}_u=0$ and $-\sqrt{E}q_v-q\sqrt{E}_v+\sqrt{G}p_u+p\sqrt{G}_u=0$

The only condition is that p,q are not both zero at the point in question. This is a first order linear homogeneous elliptic system of two unknowns in the plane. Elliptic here just means, upon dividing by $\sqrt{E}$ we have a system $\vec{X_v}+A \vec{X_u}+B\vec{X}=0$ Where $\vec{X}=\begin{bmatrix}p\\ q\end{bmatrix}$, A, and B some real matrices whose entries are $C^1$ functions in $(u,v)$ and A is a matrix with no real eigen-vectors.This terminology is standard, it is consistent with the definition in the book of M.A. Lavrent'ev.

For the purpose of geometry, we only require existence of p,q in a neighborhood (however small it maybe), not both zero at the point in question. This will suffice for our purpose because of Poincare's lemma

I am not very familiar with the theory of system of pde, apart from the very simple hyperbolic systems as introduced in Evan's pde.The system I arrived at looks more complicated than the Beltrami's equation, but I still want to see if it works. The interest in doing it this way is to generalize the method used in showing the existence of orthogonal coordinates, where a real integrating factor can be used. The pde here is more complicated than Beltrami's equation because of the zero order terms. The method for hyperbolic system which uses diagonal forms in an essential way does not seem applicable for the problem system I have in my hand.

One should have some faith in this approach because:

  1. in the case of real analytic metric, some variant of Cauchy-Kowalevski's theorem will show the existence of p-q pair. This was known to Gauss.
  2. The form of the equation is simple, it is linear, homogeneous, first order, so it does not seem to permit a Lewy-kind of counterexamples.
  3. The demand is modest:we don't need well-posedness, only existence of solution which is non-vanishing at a point.

A lot of existing results focus almost exclusively on boundary-value problems. I looked at Wendland's "Elliptic Systems in the Plane", and some books by Gilbert and Bers. Nothing there seems to be directly applicable to my system.

So my questions are:

  1. Is this system asking too much? Can one hope for existence?
  2. If one were to try to prove existence under such "modest conditions", what kind of techniques should one learn?


  • $\begingroup$ It is usually hard to prove the existence of a solution unless you've imposed enough conditions to make the solution unique. For a general elliptic system of PDEs, this is most easily done via a boundary value problem. So my suggestion is to restrict the PDE to a small ball and impose well-posed boundary conditions (probably something like $ap + bq = 1$, where $a$ and $b$ are appropriately chosen constants). There will be a unique solution, and since the boundary value is nonzero, so must be $(p,q)$ in the interior. $\endgroup$
    – Deane Yang
    Dec 7, 2018 at 17:25
  • 1
    $\begingroup$ I strongly recommend the paper Adrien Douady and X. Buff, Le théorème d’intégrabilité des structures presque complexes, The Mandelbrot set, theme and variations, London Math. Soc. Lecture Note Ser., vol. 274, Cambridge Univ. Press, Cambridge, 2000, pp. 307–324. MR 1765096, which provides a complete and elementary proof of the existence of isothermal coordinates, which is (for me) much clearer than Chern's proof. $\endgroup$
    – Ben McKay
    Jan 29, 2021 at 15:11

2 Answers 2


I have not received an answer yet. It is still not clear to me how one would tackle this system directly. But I did find a reference giving the result I need: recall I only wanted an integrating factor proof.

The paper I found is "On Integrating Factors and On Conformal Mappings" by Philip Hartman(1958).

My complex differential form has coefficients satisfying the codition of $(4)$ in his paper. He also reduced the problem to an elliptic system, but very cleverly avoided the introduction of zero order terms. So without solving my apparently more general system, he was able to prove the existence of integrating factor $\mu$ in my question. This fills in the last gap in my proof.

Also, because of the existence of complex integrating factor proved in the paper mentioned above, it follows that my seemingly "harder" system has at least one non-trivial solution with the properties I described in my question.

I am still interested in hearing more professional advices on how to go about solving systems with weak conditions.

  • 2
    $\begingroup$ I'm traveling right now and don't have access to the books, but I would suggest that you look at the books and papers of Michael E. Taylor. He has done a lot of work on partial differential equations in differential geometry, in particular, particularly complex structures with very low regularity. I think that he may have the best known current results along those lines. $\endgroup$ Jun 10, 2018 at 7:41
  • $\begingroup$ Thanks for the suggestion, I will look into it. Incidentally,I thought of this approach because of Taylor's book on pde. In the first volume of his pde series, he asked a question about existence of integrating factor in a very early chapter, under the section of differential forms. I would also like to clarify my follow-up question a little bit. When I said weak conditions, I meant conditions on the solutions, not on the equations. For example, existence of solutions without strong boundary conditions. But I am less interested in weakening the hypothesis on coefficients. $\endgroup$
    – Yujia Yin
    Jun 10, 2018 at 20:18

Let me first rephrase your question: Your complex-valued 1-form $$\sqrt{E}du+i\sqrt{G}dv$$ is a section of the canonical bundle of the (at least at the moment almost) complex structure induced by the metric (and the implicit choice of orientation given by the oriented chart $(u,v)$): The metric and orientation define the Hodge star operator $*$. Define the canonical bundle $$K=\{\omega\in T^*M\otimes\mathbb C\mid *\omega=-i\omega\}.$$ By construction, your complex-valued 1-form is a section of this bundle. As you have already observed it is enough to prove the existence of a closed 1-form in the canonical bundle around each point without zeros. There might be different possibilities to prove existence, e.g. by imposing boundary conditions, but the following is quite geometric: the metric induces a Levi-Civita connection, and its restriction to the canonical bundle satisfies $$(\nabla\omega)''=d\omega,$$ where you split the $K$-valued 1-form $\nabla\omega$ into $(1,0)$ and $(0,1)$ parts:$$\nabla\omega=(\nabla\omega)'+(\nabla\omega)''.$$ This follows from a short computation together with the fact that $\nabla$ is the Levi-Civita connection and therefore preserves the Hodge $*$. The connection $\nabla$ has some curvature $F^\nabla$. Add a $(1,0)$-form $\eta$ to $\nabla,$ i.e., $$\tilde\nabla=\nabla+\eta.$$ Then we have $$(\tilde\nabla)''\omega=(\nabla)''\omega=d\omega$$ and $$F^{\tilde\nabla}=F^\nabla+d\eta.$$ By standard elliptic PDE theory for the Laplacian $\Delta=* d* d$, there always exist a locally defined function $f\colon U\subset M$ with $\eta=df+ i*df\in\Gamma(U,K)$ and $$0=F^\nabla+d\eta=F^\nabla+id*df.$$ Hence, the connection can be made flat, and locally there exists nowhere vanishing parallel sections. These are automatically closed as 1-forms.

  • 4
    $\begingroup$ If we allow ourselves standard elliptic theory, then we might just remark that the original Beltrami equation has local solutions with arbitrary 1-jet at a point. $\endgroup$
    – Ben McKay
    Sep 1, 2020 at 15:44

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.