The gradient of a differentiable function $f:S\to R$ is a differentiable map grad $f:S\to R^3$ which assigns to each point $p\in S$ a vector grad $f(p)\in T_p(S)\subset R^3$ such that $$\langle \operatorname{grad}{f(p)}, v\rangle_p=df_p(v) \ \text{ for all } v\in T_p(S).$$
Show that
If $E,F,G$ are the coefficients of the first fundamental form in a parametrisation $\mathbf{x}:U\subset R^2\to S$, then grad $f$ on $\mathbf{x}(U)$ is given by $$\text{grad }f=\frac{f_uG-f_vF}{EG-F^2}\mathbf{x}_u+\frac{f_vE-f_uF}{EG-F^2}\mathbf{x}_v.$$
In particular, if $S=R^2$ with coordinates $x,y$, $$\operatorname{grad}{f}=f_xe_1+f_ye_2,$$ where $\{e_1,e_2\}$ is the canonical basis of $R_2$ (thus, the definition agrees with the usual definition of gradient in the plane).
The tools that I can think of right now are:
- The definition of $df_p(v)$, but the definition needs another map $\alpha:(-\epsilon,\epsilon)\to U$, and I think it is not necessary to introduce $\alpha$ in deriving $\operatorname{grad}{f}=\frac{f_uG-f_vF}{EG-F^2}\mathbf{x}_u+\frac{f_vE-f_uF}{EG-F^2}\mathbf{x}_v$.
- I also don't think we need to use the Jacobian in this case.
- Another doubt that I have is the expression that we are looking for $\operatorname{grad}{f}$ is inside the fundamental form $\langle \operatorname{grad}{f(p)}, v\rangle_p$ which we still need to find.
Can somebody give some clue on how to proceed?
You only need to check that $$\langle \operatorname{grad}{f(p)}, v\rangle_p=df_p(v) $$
is satisfied for $v = \mathbf{x}_u$ and $\mathbf{x}_v$, as these two vectors spans $T_pS$. Note that $df_p(\mathbf{x}_u) = f_u$ (Do you know why?). On the other hand,
$$\begin{split} \langle \text{grad}(f), \mathbf{x}_u\rangle &=\left\langle \frac{f_uG-f_vF}{EG-F^2}\mathbf{x}_u+\frac{f_vE-f_uF}{EG-F^2}\mathbf{x}_v, \mathbf{x}_u \right\rangle \\ &= \frac{f_uG-f_vF}{EG-F^2} E + \frac{f_vE-f_uF}{EG-F^2} F \\ &=\frac{1}{EG-F^2} \left(f_u EG - f_v FE + f_v EF - f_u F^2 \right)\\ &= f_u\\ &=df_p(\mathbf{x}_u). \end{split}$$
The checking for $\mathbf{x}_v$ is similar.