TITLE: A conformal map whose Jacobian vanishes at a point is constant?
QUESTION [13 upvotes]: Let $f:M \to N$ be a smooth weakly conformal map between connected $d$-dimensional Riemannian manifolds, i.e. $f$ satisfies $df^Tdf =(\det df)^{\frac{2}{d}} \, \text{Id}_{TM}$. 

Assume $d \ge 3$ and that $df=0$ at some point.  Is it true that $f$ is constant?

A proof for the Euclidean case, can be found in "Geometric Function Theory and Non-linear Analysis", by Iwaniec and Martin. Their proof uses the fact both $M,N$ are Euclidean.
If I am not mistaken, a conformal map (in $d\ge3$) is determined by its 2-jet at a point, so it suffices to prove all the second derivatives at the point where $df$ vanishes are zero. (But maybe this wrong; it's possible that the "2-jet determination" only holds for conformal maps whose differentials are everywhere non-zero. I am not sure.)
Edit:
David E Speyer proved below that if $f$ is conformal and $df_{x_0}=0$ then all the derivatives of $f$ vanish at $x_0$ (for all orders). Thus, if the above statement about the two-jet-determination is true, the only solutions are indeed constants.
Does anybody know whether or not this "two-jet-determination" hold in the case where we allow the Jacobian to vanish?

REPLY [2 votes]: In a different direction, there are solutions where $f$ is $C^1$ and the metric $g_M$ is $C^0$. Let $\phi : \mathbb{R}_{>0} \to \mathbb{R}_{>0}$ be a $C^1$ function such that $\lim_{s \to 0^+} \phi(s) = \lim_{s \to 0^+} s \tfrac{\phi'(s)}{\phi(s)} = 0$. An example is $\phi(s) = \tfrac{1}{-\log s}$. (Or rather, that function near $0$ smoothed out somehow to not blow up at $s=1$.)
Write $s$ for the function $\sum x_i^2$ on $\mathbb{R}^n$. Let $M=N=\mathbb{R}^n$ and define $f:M \to N$ by $f(\vec{x}) = \phi(s)\ \vec{x}$. Let $g_N$ be the standard metric and define $g_M = \tfrac{1}{\phi(s)^2} f^{\ast} g_N$ on $M \setminus \{ 0 \}$. I claim that $g_M$ extends continuously to $0$. 
We compute: The Jacobian of $f$ at the point $\vec{x}$ is
$$J = \phi(s) \mathrm{Id} + \phi'(s)\  \vec{x} \ \vec{x}^T.$$
Using the identity $\vec{x}^T \vec{x}=s$, we compute:
$$g_M = \tfrac{1}{\phi(s)^2} J J^T = \mathrm{Id} + \left(2 \tfrac{\phi'(s)}{\phi(s)} + \tfrac{s \phi'(s)^2}{\phi(s)^2} \right) \vec{x} \ \vec{x}^T .$$
We will show that the second term zpproaches $0$ as $\vec{x} \to 0$. Note that each entry of $\vec{x}^T \vec{x}$ is $O(|\vec{x}|^2) = O(s)$. So each entry in the second term is bounded by 
$$2 \tfrac{s \phi'(s)}{\phi(s)} + \left( \tfrac{s \phi'(s)}{\phi(s)} \right)^2$$
and we assumed $\tfrac{s \phi'(s)}{\phi(s)} \to 0$.
Unfortunately, the hypothesis that $\tfrac{s \phi'(s)}{\phi(s)} \to 0$ gives $\tfrac{\phi'}{\phi} = o(s^{-1})$ so $\log |\phi| = o(\log s)$ and $\phi$ must approach $0$ more slowly than any $s^{\epsilon}$. So $f(x,0,0,\ldots,0)$ has derivative $0$ at $0$ but is larger than any $x^{1+\epsilon}$. So this method can't be pushed to $C^2$ functions.