# Harmonic functions without critical points (global isothermal coordinates)

Let $$(M,g)$$ be a compact, orientable Riemannian surface with non-empty boundary $$\partial M$$.

Question. Does there always exist a smooth function $$f:M\rightarrow \mathbb{R}$$ with $$\Delta_gf=0$$ in the interior of $$M$$ and $$d_pf\neq 0$$ for all $$p\in M$$?

1. For $$M\subset \mathbb{R}^2$$, equipped with the Euclidean metric, this is clear (just take an affine linear map).
2. Given a finite subset $$P\subset M$$, one can always find a smooth function $$f:M\rightarrow \mathbb{R}$$ such that, for all $$p\in P$$ we have $$\Delta f = 0$$ near $$p$$ and $$d_pf\neq 0$$. (This follows from a well known trick used to construct isothermal coordinates, see below.)
3. If $$M$$ is contractible, this is equivalent to the existence of global isothermal coordinates on $$(M,g)$$.

Proof of 2)

Extend $$M$$ to a closed surface $$(N,g)$$ and write $$R:H_\perp^3(N)\rightarrow \mathbb{R}^{\vert P \vert}$$ for the map that sends $$f$$ to the list $$(\vert d_p f\vert: p\in P)\in \mathbb{R}^{\vert P \vert}$$. Here $$H^s_\perp(N)$$ consists of Sobolev-functions of regularity $$s$$ with zero mean (i.e. $$\perp\{\mathrm{constants}\}$$). Further write $$F:H_\perp^1(N)\rightarrow H_\perp^3(N)$$ for the map that sends $$h$$ to the solution $$f$$ of $$\Delta f = h$$. Now the point is that the set $$D_P=\{h\in H_\perp^1(N)\cap C^\infty(N):h \text{ vanishes near } P\}$$ is dense in $$H_\perp^1(N)$$, which implies that $$RF(D_P)\subset \mathbb{R}^{\vert P\vert}$$ is dense (as both $$R$$ and $$F$$ are continuous and surjective). In particular there is a vector in $$RF(D_P)\subset\mathbb{R}^{\vert P\vert}$$ with all coordinates non-zero and an $$R$$-preimage $$f\in F(D_P)$$ satisfies the desired requirements.

Let $$k\ge 1$$ be the number of boundary components of $$M$$ and $$\gamma$$ the genus of the closed surface $$\Sigma =\Sigma_M$$ obtained by gluing a disk to each boundary component.
Lemma. Suppose $$\gamma =0$$ and $$k\ge 1$$. Then there is a conformal diffeomorphism $$F:M\rightarrow F(M)\subset \mathbb{R}^2$$ onto a a smooth domain $$F(M)$$. For $$k=1$$ this can be chosen to be the unit-disk.
Proof. Extend $$g$$ arbitrarily to all of $$\Sigma$$ and denote the extension also with $$g$$. We now use the fact that all Riemannian metrics on $$S^2$$ are conformally equivalent (or equivalently there is only one complex structure) to obtain a conformal diffeomorphism $$f:(\Sigma,g)\rightarrow (S^2,g_0)$$ (with $$g_0$$ the standard round metric). Without loss of generality we may assume that the north-pole $$p$$ does not lie in the image $$f(M)$$ and we consider the stereographic projection $$P:(S^2\backslash p,g_0)\xrightarrow{\sim} \mathbb{C}$$, which is well known to be conformal. In particular, $$F=P\circ f\vert_{M}:(M,g)\rightarrow \mathbb{C}$$ is a conformal diffeomorphism onto the image $$F(M)\subset \mathbb{C}\equiv \mathbb{R}^2$$. The assertion about $$k=0$$ is just the Riemann mapping theorem. q.e.d.
This answeres the question for $$\gamma =0$$: Just take any harmonic map $$f_0:F(M)\rightarrow \mathbb{R}$$ without critical points and pull it back via $$F$$. As $$F$$ is a conformal map between surfaces, $$f=F^*f_0$$ is harmonic as well (Section 2.3 in these notes) and it clearly has no critical points.
For $$\gamma >0$$ I actually suspect that the question has to be answered negatively, but I don't know how to prove it yet. I think that in that case $$(M,g)$$ necessarily has a closed geodesic $$c:S^1\rightarrow M$$ and one might get a contradiction from looking at $$f\circ c$$, where $$f$$ is the supposed harmonic map.