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!

  • $\begingroup$ A generator is $\frac{\partial}{\partial \theta}$. Compare this with $V(S^2)$, not free. Hey, but $V(S^3)$ is again free. And $V(S^7)$. $\endgroup$ – Orest Bucicovschi Feb 15 '15 at 19:04
  • $\begingroup$ @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? $\endgroup$ – user113576 Feb 15 '15 at 19:19
  • 1
    $\begingroup$ @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. $\endgroup$ – 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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.