Prove that the $\lambda^2$ function of conformal mappings definition must be smooth

Question

Regarding this question (if a map is angle-preserving then it's conformal), I'd like to show that, if $<d\phi_p(v_1),d\phi_p(v_2)>=\lambda^2(p)<v_1,v_2>_p$ then $\lambda^2$ is differentiable. I suspect we don't even need to pass through the definition of the function, but maybe we do.

I could only think of evaluating (1) in the same $v_1,v_2$ to see what happens. I can't really think of how to differentiate with respect to $p$. Should I try it by definition?

Any hints are appreciated. Thanks!

Answer 1

Assuming that your main equality holds for all $p, v_1, v_2$, apply the equality to unit-length $v$s with $v_1 = v_2$. Then you have $$ \lambda^2(p) =\frac{<d\phi_p(v_1),d\phi_p(v_1)>}{<v_1,v_1>_p}\\ =\frac{<d\phi_p(v_1),d\phi_p(v_1)>}{1}\\ =<d\phi_p(v_1),d\phi_p(v_1)> $$ and as long as $\phi$ is sufficiently smoothly differentiable, and so is the metric, you're done.

Post-comment additions OP is apparently thinking of functions on some arbitrary manifold, $M$, so a little more detail is needed. What we're given is that for every $p \in M$ and every $v_1, v_2 \in T_p M$, $$ <d\phi_p(v_1),d\phi_p(v_2)>_{\phi(p)}=\lambda^2(p)<v_1,v_2>_p. $$ where $\phi$ is a differentiable map from $M$ to some manifold $K$, and the inner product on the left is at $\phi(p) \in K$, while the one on the right is at $p \in M$ (i.e., on the tangent space $T_p M$).

The key thing to observe is that $\lambda$ is assumed to be independent of $v_1, v_2$.

OP wants to know whether $\lambda^2$ is differentiable as a function of $p$. The most general answer is "no, at least not obviously, for $d\phi$ may not be differentiable."

But if we're willing to assume that $\phi$ is twice differentiable, then this is a purely local question: we need only show that for any point $p_0 \in M$, $\lambda^2$ is differentiable in a neighborhood of $p_0$.

Pick a local coordinate system $x : U \to \mathbb R^n$ in a neighborhood of $p_0$; then the vector field $v(p) = \frac{\partial}{\partial x_1} (p)$ is nonzero at $p_0$ and throughout $U$. Let $v_1(p) = v_2 (p) = v(p) / \| v(p) \|$ for $p \in U$. Then $v_1$ and $v_2$ are both vector fields on $U$, and unit length at every point, and their inner product, at every point, is 1. The assumed identity then tells us that $$ <d\phi_p(v_1),d\phi_p(v_2)>_{\phi(p)}=\lambda^2(p)<v_1,v_2>_p = \lambda^2(p). $$ If the inner product on $K$ is differentiable as a function of position, and if $d\phi$ is as well, then the chain rule allows us to compute the derivative of $\lambda^2$, hence $\lambda^2$ is differentiable.

Without the assumptions of a continuously varying metric, the theorem simply is not true. For instance, if we take the identity map on $\mathbb R^2$, where the domain has the usual metric, but on the codomain, the metric at $(a, b)$ is the usual metric for $a \le 0$, but twice the usual metric for $a > 0$, then the function $\lambda^2 (a, b)$ is $1$ for $a \le 0$ and $2$ for $a > 0$, which is evidently not even continuous.

If instead we pick the metric as the standard metric for $a \le 0$, but $(1+a)$ times the standard metric for $a > 0$, then the function $\lambda^2(a, b)$ is $1 $ for $a < 0$ and $1 + a$ for $a > 0$, which is continuous but not differentiable.

Hence the assumption of differentiability of the metric wrt position in the codomain is essential.

A similar example can be crafted to show that the assumption of differentiability of $d\phi$ is essential, I believe. I leave it to OP to give this a shot, following the model I've given.

* Further additions *

Let $$ f(x) = \begin{cases} x-x^2 & x \le 0 \\ x + x^2 & x > 0 \end{cases} $$ and $$ \phi(x, y) = (f(x), y). $$ Then $$ D\phi(x, y) = \begin{bmatrix} u(x) & 0 \\ 0 & 1\end{bmatrix}, $$ where $$ u(x) = \begin{cases} 1-2x & x \le 0 \\ 1x + 2x & x > 0 \end{cases} $$ so $\phi$ is clearly continuously differentiable, but not twice differentiable.

For $p = (-a, 0)$ and $v_1 = v_2 = [1, 0]^t$, we have $$ \newcommand{\ep}{\epsilon} d\phi_p (v_i) = \begin{bmatrix} 1-2a \\ 0 \end{bmatrix} $$ so $$ \lambda^2(-a, 0) = (1-2a)^2 $$ Similarly, $$ \lambda^2(a, 0) = (1 + 2a)^2 $$ The derivative of the first, with respect to $a$, is $2(1-2a)*(-2)$; for the second, it's $2(1+2a)2$. The first, evaluated at $0$, is $-4$; the second is $+4$. Hence the derivative of $\lambda^2(t, 0)$, as $t$ passes from negative to positive through $0$, is undefined.