Let $\mathbb{R}P^m$ be the $m$-dimensional real projective space and let $\mathbb{R}P^m\setminus\{*\}$ be the punctured space. I observe:

  1. $\mathbb{R}P^2\setminus\{*\}$ is homeomorphic to a (open) Mobius strip, that is, the total space of a non-trivial line bundle over $\mathbb{R}P^1$.

  2. $\mathbb{R}P^1\setminus\{*\}$ is homeomorphic to $\mathbb{R}$, which can be regarded as $\mathbb{R}P^0\times\mathbb{R}$.

Could I generalize (1) and (2) to the following?

(i). If $m$ is odd, then $\mathbb{R}P^{m+1}\setminus\{*\}$ is homeomorphic to the total space of a non-trivial line bundle over $\mathbb{R}P^{m}$.

(ii). If $m$ is even, then $\mathbb{R}P^{m+1}\setminus\{*\}$ is homeomorphic to $\mathbb{R}P^{m}\times\mathbb{R}$.

Question: Whether are (i) and (ii) true or false? Are there any related references?

  • 1
    $\begingroup$ You might want to read about Thom spaces. $\endgroup$ – Todd Trimble May 23 '16 at 11:35
  • 2
    $\begingroup$ Thom spaces? All we do is remove two antipodal points from the round sphere and take the quotient by the antipodal map. The projection to the equator is an $\mathbb R$-bundle. It is nontrivial in the quotient: try to smoothly push $RP^m$ off itself. Then there would be $\mathbb Z_2$-invariant push off upstairs, and visibly it does not exist. Alternatively, you could note that the antipodal map of $S^n$ changes orientation iff $n$ is even, so $RP^{m+1}\setminus\{*\}$ and $RP^m$ cannot be both orientable, hence the normal bundle is non-orientable, and in particular non-trivial. $\endgroup$ – Igor Belegradek May 23 '16 at 11:58
  • 1
    $\begingroup$ @IgorBelegradek The OP was asking about related references, and so I'm just mentioning that the Thom space of the tautological line bundle over $\mathbb{RP}^m$ is $\mathbb{RP}^{m+1}$. $\endgroup$ – Todd Trimble May 23 '16 at 13:31
  • 1
    $\begingroup$ Alternatively one could say that the line bundle over ${\mathbb {RP}}^{n-1}$ is nontrivial because its restriction over ${\mathbb {RP}}^1$ is nontrivial. $\endgroup$ – Allen Hatcher May 23 '16 at 17:52

(i) is true and (ii) is false. In fact, (i) is true for all $m\ge1$. More precisely, dropping a coordinate yields a map from $\mathbb RP^{m+1}\setminus\{*\}\to\mathbb RP^m$ which is a locally trivial bundle with fiber $\mathbb R$. This bundle is not trivial for $m\ge1$ because, e.g., $\mathbb R P^m$ is orientable if and only if $m=0$ or $m$ is odd.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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