Smooth vector field on $S^3$

Question

Consider the map $X:(x,y,z,w)=(-y,x,-w,z)\in\mathbb{R^4}$. What I want to show is that $X$ is a smooth vector field on $S^3$. I have no idea how to do this. I know that a vector field is smooth if its coefficients are smooth. However, I need to restrict this to $S^3$ and this can be done using an inclusion map. Any help is appreciated! Thank you

Answer 1

Since $\mathbb{S}^3$ is a submanifold of $\mathbb{R}^4$, the inclusion map $i\colon\mathbb{S}^3\rightarrow\mathbb{R}^4$ is smooth, therefore $X$ is smooth.

The only thing to check is that for all $x\in\mathbb{S}^3$, $X_x\in T_x\mathbb{S}^3$, which is obvious since $T_x\mathbb{S}^3=x^{\perp}$.

Answer 2

Here is how you can prove this generally. Suppose $N$ is a submanifold of $M$. Let $p\in N$ and $i:N\to M$ be the inclusion map which is smooth as $N$ is a submanifold of $M$. Then $(X|_N)(p)=(X\circ i)(p)=X_{i(p)}=X_p\in T_pN.$

Now consider the tangent bundle $\pi:TN\to N$. Observe that $(\pi\circ (X|_N))(p)=(\pi\circ X\circ i)(p)=\pi(X_p)=p\implies X|_N$ is a section of the tangent bundle. Finally, how can we show that this is smooth? Looks like you already know the answer. However, for completeness sake, since $X$ is smooth and $i$ is smooth $X|_N=X\circ i$ is smooth. Hence, $X|_N$ is smooth as a section of the tangent bundle $\pi:TN\to N$. Therefore, $X|_N$ is a smooth vector field on $N$ !