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.

Cotangent Bundle $T^{*}(M)=\bigcup_{m\in M} M_m^{*}$ (disjoint union of contangent space) Could any one explain me how there is a natural projection from $T^{*}(M)\rightarrow M$ given by $\pi(f)=m$ if $f\in M^{*}_m$? I am not able to understand and feel the map naturally Please explain.

share|improve this question

1 Answer 1

For each point in $M$ you have a cotangent space $M_m^*$. So we now look at the disjoint union of all of them as a collection of fibers -- "above" each point in $M$ lies a cotangent space.

The canonical map just "projects" each fiber (end every point within it) onto the point above which it lies, just the same as with the tangent bundle.

share|improve this answer

Your Answer

 
discard

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.