This question was asked in my assignment and I am having trouble solving it. So, I need help.
On $S^{1} \subseteq \mathbb{R}^2$ , $X_{(x,y)} = y \frac{\partial}{\partial x} - x\frac {\partial } {\partial y}$ Prove that X is a vector field on $S^{1}$.
I need help in proving that $X_{(x,y)} \in T_{(x,y)} S^1$.
Attempt: Clearly $ T_P (S^1) \subseteq T_P\mathbb{R}^2 = \mathbb{R}^2$, I have proved it earlier $T_P=${$ v\in \mathbb{R}^2 : <v,p>=0$ } for a unit sphere. But I am not able to move forward.
A hint is given: Prove that $X_{(x,y)} =(y,-x)$ but I couldn't prove it.
Using it or otherwise can you please outline a proof.
The statement $X_{(x,y)} =(y,-x)$ seems like a tautology, and then your result can be applied as $\langle(x,y),(y,-x)\rangle=0$, and you are done.
Alternatively, another way to see it is you have a path $\phi\colon [0,1]\to S^1$ mapping $t\mapsto (\cos(-t+c),\sin(-t+c))$.
For any function $f$ defined on a neighbourhood of $S^1$, we have: \begin{eqnarray*}\frac{df(\phi(t))}{dt}&=&\frac{d\cos(-t+c)}{dt}\frac{\partial f}{\partial x}+\frac{d\sin(-t+c)}{dt}\frac{\partial f}{\partial y}\\ &=&\left(y\frac\partial{\partial x}-x\frac\partial{\partial y}\right)f\end{eqnarray*}
Thus the derivation associated to $y\frac\partial{\partial x}-x\frac\partial{\partial y}$ is just differentiation along the path $\phi$ which lies in $S^1$.