Understanding Tangent Space of S2n+1 in Cn+1

ozlem
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1. Let p be an arbitrary point on the unit sphere S2n+1 of Cn+1=R2n+2. Determine the tangent space TpS2n+1 and show that it contains an n-dimensional complex subspace of Cn+1

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3. It is easy to find tangent space of S1; it is only tangent vector field of S1. But what must do for higher dimension and how can I show it contains an n-dimensional subspace of Cn+1. Thanks for your helps.
 
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What is the definition of ##S^{2n+1}##? Given a point on the sphere, what infinitesimal Steps can you take from that point and still be on the sphere?
 
ozlem said:
It is easy to find tangent space of S1; it is only tangent vector field of S1.
By the way, this does not really answer the question. You are essentially saying "the tangent space is the tangent space". You should specify exactly what type of vectors are in this space.
 
Orodruin said:
By the way, this does not really answer the question. You are essentially saying "the tangent space is the tangent space". You should specify exactly what type of vectors are in this space.
I think that tangent vector field must be
X=-x2d/dx1+x1d/dx2 for any P(x1,x2) point on the C1. d/dx stand for partial derivative.
 
ozlem said:
I think that tangent vector field must be
X=-x2d/dx1+x1d/dx2 for any P(x1,x2) point on the C1. d/dx stand for partial derivative.
And how did you figure this out?

Edit: Also note that any vector proportional to that one will do.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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