Directional gradient on sphere

Question

We consider the following function

$$f:(\mathbb S^n)^N \rightarrow \mathbb R^{n+1} \text{ such that } f(x_1,...,x_N)= \sum_{i=1}^N x_i.$$

This function can be written in Cartesian coordinates as $f(x)=(f_1(x),..,f_{n+1}(x))$ and I would like to know if one can find a simple expression for the derivative

$$\nabla_{x_1} \left(\frac{f_1(x)}{\Vert f(x) \Vert_{\mathbb R^{n+1}}}\right)$$ where $\nabla_{x_1}$ is the gradient on $\mathbb S^n$ with respect to $x_1.$

Can one somehow carry out this differentiation? I am a bit struggeling with computing $\nabla_{x_1} f_1(x)$ here.

Answer 1

$\newcommand\R{\mathbb R}\newcommand\S{\mathbb S}$Take any $(x_1,x_2,\dots,x_N)\in(\S^n)^N$. Let $(-1,1)\ni t\mapsto X_1(t)\in\S^n$ be any smooth curve such that $$X_1(0)=x_1.$$ For any $t\in(-1,1)$, let $$X(t):=(X_1(t),x_2,\dots,x_N)$$ and $$S(t):=f(X(t))=X_1(t)+x_2+\dots+x_N[\in\R^{n+1}],$$ so that $X(0)=(x_1,x_2,\dots,x_N)$ and $$S(0)=s:=x_1+x_2+\dots+x_N.$$ Let $v:=X'_1(0)$, so that $S'(0)=v$. Let $$S_1(t):=f_1(X(t))=e_1\cdot S(t),$$ where $\cdot$ denotes the dot product and $e_1$ is the first vector of the standard basis of $\R^{n+1}$. So, $S'_1(0)=e_1\cdot v$. So, for $$r(t):=\frac{f_1(X(t))}{\|f(X(t))\|}=\frac{S_1(t)}{\|S(t)\|}$$ we have $$r'(0)=\frac1{\|S(0)\|^2}\Big(\|S(0)\|S'_1(0)-S_1(0)\frac{S(0)}{\|S(0)\|}\cdot S'(0)\Big) \\ =\frac1{\|s\|^2}\Big(\|s\|e_1\cdot v-(e_1\cdot s)\frac{s}{\|s\|}\cdot v\Big).$$ Thus, the gradient in question is $g-(g\cdot x_1)x_1$, where $$ g:=\frac1{\|s\|^2}\Big(\|s\|e_1-(e_1\cdot s)\frac{s}{\|s\|}\Big).$$ $\big($Note: $g-(g\cdot x_1)x_1$ is the orthogonal projection of the vector $g$ onto the tangent hyperplane to the unit sphere $\S^n$ at point $x_1$.$\big)$