# Prove that the set of all vector fields $V(S^1)$ is a free $C^{\infty}(S^1)$-module

I need to prove that the set of all vector fields, $X:S^1\to TS^1$ name it: $V(S^1)$, is a free $C^{\infty}(S^1)$-module. So i need a basis $\frac {d}{dx_1},...,\frac {d}{dx_n}$ for $V(S^1)$.It's easy to show that $V(S^1)$ is an $C^{\infty}(S^1)$-module. The difficulty lies in the basis.

For example given a vector field $X\in V(S^1)$, then around $p_1\in S^1$ there is a map $(U,\phi=(x_1,...,x_n))$ and $f_i:U\to \mathbb R$ which are $C^{\infty}$ such that $X(q)=\sum f_i(q)\frac {d}{dx_i}|_q$ in $U$. We write $X=\sum f_i\frac {d}{dx_i}$.

If $Y=\sum g_i\frac {d}{dy_i}$ is another vector field around $p_2$ with map $(W,\psi)$, in order to write $Y$ in terms of $\frac {d}{dx_i}$ we need to change the bases of the maps, and thus we need $p_1,p_2\in W\cap U$ in order to have $\frac {d}{dy_j}=\sum_i \frac {dx_i}{dy_j}\frac {d}{x_i}$ in $W\cap U$.

The problem is that if i take $p_1$ the north pole and $p_2$ the south and $U=S^1\setminus \{N\}$ and $W=S^1\setminus \{S\}$, then $p_1,p_2\not \in U\cap W$.

Any thoughts? I can know homotopy,homology,diff. geometry so feel free to write. Although i prefer a diff. geometry approach.

Thank you!

• A generator is $\frac{\partial}{\partial \theta}$. Compare this with $V(S^2)$, not free. Hey, but $V(S^3)$ is again free. And $V(S^7)$. – Orest Bucicovschi Feb 15 '15 at 19:04
• @orangeskid, the $\frac {d}{d\theta}$ is from the "angles" map from the $x'x$ axis?if it is this i can se it:) also, which is the generator in the $V(S^3)$ case? – user113576 Feb 15 '15 at 19:19
• @user113576 think about how to generalize this to vector bundles. the point is that a vector bundle of rank $n$ is trivial iff it has $n$ ptwise linearly indepdent sections. – Mister Benjamin Dover Feb 15 '15 at 19:40

@user113576: Totally!. Now, there are some vector fields on $\mathbb{R}^2$ that restrict to this one on the circle ( yep, they are tangent to it). You can take $-y \frac{\partial }{\partial x} + x \frac{\partial}{ \partial y}$. It is obtained from the $(x,y)\simeq x \frac{\partial }{\partial x} + y \frac{\partial}{ \partial y}$ by rotating with $\frac{\pi}{2}$, that is, multiply $(x,y) \simeq x + iy$ by $i$.
For $S^3 \subset \mathbb{R}^4$ start with $(x_0, x_1, x_2, x_3)\simeq x_0 \frac{\partial }{\partial x_0}+x_1 \frac{\partial }{\partial x_1}+x_2 \frac{\partial }{\partial x_2}+x_3 \frac{\partial }{\partial x_3}$. Now $\mathbb{R}^4 = \mathbb{H}$, the quaternions. See if you can get three other vector fields perpendicular to this.