Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I have question about vector fields:(pls help me)

Consider 2-sphere $$\mathbb{S}^2= \left\{(x_1,x_2,x_3) \in R^3: x_1^2+x_2^2+x_3^2=1\right\}$$

Let $P \in \mathbb{S}^2$ has spherical coordinates$(1,\phi,\theta)$.Give a base of tangent space $T_p{S}^2$.For which values ​​of theta doesnot work?

Give some example of vector fields on ${\Bbb S}^2$ using above (who so ever touching on ${\Bbb S}^2$)Count for each example the number of points where the vector field equals $0$.

share|improve this question

closed as off-topic by Weapon of Choice, Tunk-Fey, studiosus, Claude Leibovici, JimmyK4542 Sep 6 at 4:39

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Weapon of Choice, Tunk-Fey, studiosus, Claude Leibovici, JimmyK4542
If this question can be reworded to fit the rules in the help center, please edit the question.

Could you make the question more clear. –  viplov_jain Feb 18 at 20:25

2 Answers 2

up vote 0 down vote accepted

The parametrizarion is $$x(\theta,\phi)=(\cos\theta\sin\phi,\sin\theta\sin\phi,\cos\phi)$$ and so the basis of the tangent space are $$x_\theta(\theta,\phi)=(-\sin\theta\sin\phi,\cos\theta\sin\phi,0) \\ x_\phi(\theta,\phi)=(\cos\theta\cos\phi,\sin\theta\cos\phi,-\sin\phi)$$ For example, at $(\pi/4,\pi/4)$ we $$x_\theta(\pi/4,\pi/4)=(-1/2,1/2,0) \\ x_\phi(\pi/4,\pi/4)=(1/2,1/2,-1/\sqrt2)$$

share|improve this answer
but for which theta doesnot work? –  moji Feb 19 at 21:49
And how can I construct a vector field on $\mathbb{S}^2$ with just one zero –  moji Feb 19 at 21:51

The tangent space of $S^2$ at $p\in S^2$ is nothing but the normal hyperspace to the vector $p$, i.e. $$ p^\perp=\{x\in\mathbb R^3 : (x,p)=0\}. $$

share|improve this answer

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