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.

Let $M$ be a manifold and let $U_{\alpha}$ and $U_{\beta}$ be coordinate charts with coordinates $x^{\alpha}$ and $x^{\beta}$, respectively.

How to show that $f_{\alpha} : U_{\alpha}\times\mathbb{R}^n\to TM_{|U_{\alpha}}$, $(p,a^{\alpha})\to\sum_{i} a^\alpha_i\frac{\partial}{\partial x^\alpha_i}|_{p}$ is a bundle isomorphism?

share|improve this question
What is your question? –  Eric O. Korman Oct 22 '12 at 16:07
What is the relation of $U_\beta$? With what are you having trouble? –  levap Oct 23 '12 at 11:01
add comment

1 Answer

Checking that $f_{\alpha}$ is a bundle isomorphism is the same as checking two things

  • $f_{\alpha}$ is a diffeomorphism ;

  • $f_{\alpha}$ commutes with the projection maps $\pi_{\alpha} : \mathrm{U}_{\alpha} \times \mathbb R^n \to \mathrm{U}_{\alpha}$ and $\pi'_{\alpha} : \mathrm{TM}_{|\mathrm U_{\alpha}} \to \mathrm{U}_{\alpha}$ i.e you have to check that $f_{\alpha} \circ \pi_{\alpha} = \pi'_{\alpha} \circ f_{\alpha}$.

By definition of the tangent bundle $$\mathrm{TM}_{|\mathrm U_\alpha} = \bigcup_{x \in \mathrm U_{\alpha}} \{x\} \times \mathbb R^n$$

We deduce that $f_{\alpha}$ is a diffeomorphism. The only thing left to check is the compatibility with the projection maps. But this comes immediately.

Hence $f_{\alpha}$ is a bundle isomorphism.

share|improve this answer
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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