We have 3 different (but equivalent) definition of tangent space, one of them is by equivalent class of smooth path, i.e. let $c:(-1,1)\rightarrow M$ be a path on $M$ with $c(0)=p$, then $c_1\sim c_2$ if for every $f:M\rightarrow\mathbb{R}$, $(f\circ c_1)'(0)=(f\circ c_1)'(0)$.

However, when applying it to find the tangent space of north pole of $S^2$, I meet some trouble.

Consider path on $S^2$, $r(t)=(\text{sin}(\theta)\text{cos}(\varphi), \text{sin}(\theta)\text{sin}(\varphi), \text{cos}(\theta))$ where $\theta$ and $\varphi$ are smooth function of $t$.

Then for any $f:S^2\rightarrow \mathbb{R}$, we can regard $f$ as the function of the coordinates of the point on $S^2$,i.e. $f(P)=f(x,y,z)$, then we have $(f\circ r)'=f_xx'+f_yy'+f_zz'$ where $(x(t),y(t),z(t))=r(t)$.

So by definition, for two path to be equivalent, we must have their $x'$, $y'$, $z'$ are equal.

After calculation and set $\theta(0)=0$ (we can't set $\varphi(0)$ because it can be anything.), we get for $r_1$ and $r_2$ to be equivalent, we must have

$$ \text{cos}(\varphi_1(0))\theta_1'(0)=\text{cos}(\varphi_2(0))\theta_2'(0) $$ $$ \text{sin}(\varphi_1(0))\theta_1'(0)=\text{sin}(\varphi_2(0))\theta_2'(0) $$

From which we can get 2 solutions:

(1)$$\theta_1'(0)=\theta_2'(0)\neq 0$$$$\varphi_1(0)=\varphi_2(0)$$

(2)$$\theta_1'(0)=\theta_2'(0)= 0$$

I think the solution is quite strange, because it does not depend on $\varphi'(0)$ and the second condition doesn't depend on $\varphi(0)$. And if $\theta(t)=t^2$, then whatever $\varphi(t)$ is, those path are equivalent. It is obviously anti-intuition.

So, doesn't the tangent space at north pole be $\mathbb{R}^2$ like the intuition suggests?


The problem arises because the chart you have chosen for $S^2$ is not valid at the poles (There is no diffeomorphism between the a neighborhood of the north pole and a subset of $\mathbb{R}^2$ in those coordinates because you have a whole bunch of points mapping to the pole). If you change your chart (for example moving the north pole to $\phi = 0, \theta = \pi/2$, then you get the right result, that the tangent space at the north pole is isomorphic to $\mathbb{R}^2$.

  • $\begingroup$ Oh I see.. it is a stupid question..=.=.thx! $\endgroup$ – hxhxhx88 Oct 25 '12 at 12:10

A hint:

When you want to study various aspects of the tangent space at some point $p\in S^2$ you should avoid choosing a presentation of $S^2$ which has a singularity at $p$. You obtain healthy local coordinates at the north pole $(0,0,1)\in S^2$ using the presentation $${\bf r}:\quad (u,v)\mapsto\left\{\eqalign{x&=\sin u\cos v \cr y&= \sin v \cr z&=\cos u\cos v \cr}\right.\quad.$$ Then $|{\bf r}(u,v)|\equiv1$, ${\bf r}(0,0)=(0,0,1)$ and ${\bf r}_u\times{\bf r}_v\Bigr|_{(0,0)}=(0,0,1)$. This guarantees that ${\bf r}$ maps a suitable neighborhood of $(0,0)$ diffeomorphically onto a neighborhood of $p=(0,0,1)$ in the manifold $S^2$.

  • $\begingroup$ ...I see..I should know it..thx! $\endgroup$ – hxhxhx88 Oct 25 '12 at 12:10

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