Tangent Bundles, T(MxN) is Diffeomorphic to TM x TN

  • Thread starter Thread starter BrainHurts
  • Start date Start date
  • Tags Tags
    Bundles Tangent
BrainHurts
Messages
100
Reaction score
0

Homework Statement



If M and N are smooth manifolds, then T(MxN) is diffeomorphic to TM x TN

Homework Equations





The Attempt at a Solution



So I'm here

let ((p,q),v) \in T(MxN)

then p \in M and q \in N and v \in T(p,q)(MxN).

so T(p,q)(MxN) v = \sum_{i=1}^{m+n} v_{i}\frac{\partial}{∂x_{i}}|_{(p,q)}

= \sum_{i=1}^{m} v_{i}\frac{\partial}{∂x_{i}}|_{p} + \sum_{i=}^{m+1} v_{i}\frac{\partial}{∂x_{i}}|_{q}

not sure if this is it?
 
Physics news on Phys.org
You kind of assume that \frac{\partial}{\partial x^i} are vector fields on both M\times N and N. I think you should be a little more careful than this...
 
Hi Micro, I'm not sure I understand what you mean, am am assuming however that

\sum_{i=1}^{m} v_{i}\frac{\partial}{∂x_{i}}|_{p} and \sum_{i=}^{m+1} v_{i}\frac{\partial}{∂x_{i}}|_{q}

are bases for TM and TN respectively and I'm not sure if that's good enough.
 
Well, you assume \frac{\partial}{\partial x_i}\vert_p is both in T_pM and T_pN. You can't do that.
 
v = \sum_{i=1}^{m} v_{i}\frac{\partial}{∂x_{i}}|_{(p,q)} + \sum_{i=}^{m+1} v_{i}\frac{\partial}{∂x_{i}}|_{(p,q)}

so i need to find a map that takes v \rightarrow (w,y)

where w\inTpM and y\inTpN

and (w,y) \in TpM x TqN

does this sound better? I was thinking about this problem a lot. I'm not sure if that's a diffeo right off the get.
 
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...
Back
Top