How do I prove that the second derivative of a function $f:M\to\mathbb{R}$ defined on a surface $M\subset\mathbb{R}^n$ is well defined?

Solution 1:

Below is an example showing that the second derivative is not well-defined in the sense you are asking about. The example is followed by a short theoretical discussion.

Example: Let $S^1$ denote the unit circle in $\mathbb{R}^2$. Consider the following two different local parametrizations of $S^1$ around the point $p=\left(\sqrt{\frac{1}{2}},\sqrt{\frac{1}{2}}\right)$: $$\varphi:(-1,1)\to S^1,\qquad t\mapsto\left(\sqrt{1-t^2},t\right)$$ and $$\psi:\left(-\frac{\pi}{2\sqrt{2}},\frac{\pi}{2\sqrt{2}}\right)\to S^1,\qquad s\mapsto\left(\cos\left(\sqrt{2}s\right),\sin\left(\sqrt{2}s\right)\right).$$ Consider the tangent vector $v\in T_pS^1$ given by $$v=-\frac{\partial}{\partial x}+\frac{\partial}{\partial y}=\frac{d\varphi}{dt}\left(\sqrt{\frac{1}{2}}\right)=\frac{d\psi}{dt}\left(\frac{\pi}{4\sqrt{2}}\right).$$ Let $f:S^1\to\mathbb{R}$ be given by $(x,y)\mapsto y$. Then the composition $f\circ\varphi$ is given by $t\mapsto t$, whereas the composition $f\circ\psi$ is given by $s\mapsto\sin \left(\sqrt{2}s\right).$ Let us compute the second derivative of $f$ with respect to both parametrizations: $$\frac{d^2(f\circ\varphi)}{dt^2}\left(\sqrt{\frac{1}{2}}\right)=0,\qquad \frac{d^2(f\circ\psi)}{ds^2}\left(\frac{\pi}{4\sqrt{2}}\right)=-2\sin\left(\sqrt{2}\frac{\pi}{4\sqrt{2}}\right)=-\sqrt{2}.$$ Hence, each parametrization leads to a different result, implying that the second derivative $d^2f(v,v)$ is not well-defined.

Discussion: The second derivative is, in fact, well-defined in the following sense. If $X$ and $Y$ are two vector fields on a smooth manifold $M$, then for a smooth $f:M\to\mathbb{R}$ the second derivative $X(Y(f))$ is a perfectly legit, coordinate free expression. Indeed, the first derivative $Y(f)$ is well-defined at every point and thus defines a new function. Then, the derivative $X(Y(f))$ is well-defined as the (first) derivative of a well-defined function. The question at hand could be posed as follows:

Question: Let $X,X',Y,Y'$ be vector fields on a smooth manifold $M$, and let $p\in M$ such that $X(p)=X'(p)$ and $Y(p)=Y'(p)$. Let $f:M\to\mathbb{R}$ be smooth (or at least twice differentiable). Do we have $$X(Y(f))(p)=X'(Y'(f))(p)?$$

The answer to the question is no, but let us see why. First, note that replacing $X$ by $X'$ does not affect the result. Indeed, the first derivative $Y(f)$ is a well-defined function, and so, the second derivative $X(Y(f))(p)$ only depends on the value of $X$ at $p$. On the other hand, replacing $Y$ by $Y'$ does affect the result. Let us calculate: \begin{align}X(Y(f))(p)-X(Y'(f))(p)&=X\left(Y(f)-Y'(f)\right)(p)\\&=X((Y-Y')(f))(p).\end{align} Now, since $Y(p)=Y'(p)$, we have $(Y-Y')(f)(p)=0$. But, in general, we have $(Y-Y')(f)(q)\ne0$ for $q\ne p$. Hence, the derivative $X((Y-Y')(f))(p)$ is just the directional derivative of a function which vanishes at $p$. There is no reason to expect such a derivative to vanish as well.