Hello,
I've two questions given:
(1)Let $X:S^2 \to \Bbb R^3$ be a vectorfield with $X\begin{pmatrix}x \\ y\\z\end{pmatrix}= \begin{pmatrix}-y \\ x\\0\end{pmatrix}$
Show: there's no smooth function $f:S^2 \to \Bbb R$ such that $\nabla f=\begin{pmatrix}-y \\ x\\0\end{pmatrix}$
(2)Let $f\in \mathcal C^\infty(S^n)$ be a function in the stereographic projection given chart$(S^n\setminus\{{S\}},p_S)$.How does the gradient look like?
My attempt:
(1)Maybe it's naive, but I said that $f$ must look like $f(x,y,z)=\frac{1}{2}(x^2-y^2)-c$ and than you could define the set $S=\{(x,y,z) \in \Bbb R^3 : x^2-y^2=2c \}$ and the set looks like an hyperbola for $c=\frac{1}{2}$. And its graph is disconnected , so $f$ isn't smooth.
(2)Here I'm completely unsure. Here's what I know: $p_S:S^n\setminus \{S\} \to \Bbb R^n$ with $p_S(q)=\frac{(q_1,...,q_n)}{1+q_{n+1}} $ and $S=(0,0,-1)$. Let $f(q_1,..,q_n)=q_1^2+...+q_n^2-1$ so $\nabla f=(2q_1,....2q_n)$
Thanks for corrections/hints.