5
$\begingroup$
  • I would like to know how does one imagine/write-down the tangent bundle of the real/complex projective space. Is there something simplifying that happens especially for $\mathbb{RP}^1$? Isn't there some relationship of this tangent bundle to spheres and Möbius bands?

I can write down transition functions on these but that doesn't answer "what" is the tangent bundle.

  • I vaguely know that Möbius band can be imagined as the anti-podal identification of the normal bundle on the circle. What is the exact statement and how to see that?

Since projective spaces are easier to think of as quotients of simpler spaces like spheres I am motivated to ask the following question,

  • In general if a group action on a manifold is such that the quotient space is again a manifold then the same action will also do a quotient of the tangent bundle. Now is the quotient of the tangent bundle (will that be again a vector bundle?) anyway related to the tangent bundle of the quotient space?
$\endgroup$
1
  • 3
    $\begingroup$ About your 2nd point: I think you should try to come up with a precise statement yourself... That will do tons more for you than reading it from someone else. $\endgroup$
    – Mariano Suárez-Álvarez
    Sep 29, 2011 at 22:52

1 Answer 1

Reset to default
4
$\begingroup$

$\mathbf P^1(\mathbb R)$ is a closed curve, so its tangent bundle is trivial. There is not much to be imagined, really, in that case! In general, what kind of answer do you expect for your «what is the tangent bundle of $\mathbf P^n$?». It is the tangent bundle of $\mathbf P^n$...

Regarding your last question: if a group $G$ acts properly discontinuously and differentiably on a manifold $M$, then $M/G$ is a manifold, $G$ acts on $TM$ in a natural way (also properly discontinuously, and morover linearly on fibers), and $(TM)/G$ is "the same thing" as $T(M/G)$.

$\endgroup$
5
  • $\begingroup$ @Marino I didn't understand as to how you see that $\mathbb{RP}^1$ is a closed curve and hence its tangent bundle is trivial. I guess you are using some theorem which I am unaware of. How would you describe the tangent bundle for other $\mathbb{RP}^n$? May be you can give me some examples for some other n's? $\endgroup$
    – Student
    Sep 30, 2011 at 23:07
  • $\begingroup$ @Marino So I guess this case of getting $\mathbb{RP}^1$ from $S^1$ fits the bill for the case where you say that $TM/G = T(M/G)$? Can you may be give the idea of the proof or reference for it? $\endgroup$
    – Student
    Sep 30, 2011 at 23:09
  • $\begingroup$ Regarding your 2nd comment: the proof of that is a simple exercise in the definitions, really. As for $\mathbf P^1(\mathbb R)$, it is a compact manifold of dimension $1$, and there is exactly one of those up to diffeomorphism, namely $S^1$; you can probably prove that the tangent bundle to $S^1$ is trivial. One can conclude in many ways (not invoking the classification of one-dimensional manifolds) that $\mathbf P^1(\mathbb R)$ is diffeomorphic to $S^1$, really. $\endgroup$
    – Mariano Suárez-Álvarez
    Sep 30, 2011 at 23:54
  • $\begingroup$ Thanks for the reply. The triviality of the tangent bundle of $S^1$ is from the general reason for why it should be so for any Lie group. But though I knew that the real projective space is compact and 1 dimensional its bizarre that it never occurred to me that it then has to be the circle! $\endgroup$
    – Student
    Oct 1, 2011 at 20:40
  • 2
    $\begingroup$ Can you say more about tangent bundle of the higher real/complex projective spaces? Like $\mathbb{RP}^2$ or $\mathbb{RP}^3$ or $\mathbb{CP}^1(=S^2)$ OR $\mathbb{CP}^2$? Is there a systematic way to create/write down their tangent bundles? $\endgroup$
    – Student
    Oct 1, 2011 at 20:42

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .