2

Let $\mathbb{R}P^2$ denote the real projective plane and suppose that $a$ is a non-contractible loop in $\mathbb{R}P^2$. Since $\pi_1(\mathbb{R}P^2)=\mathbb{Z}_2$, $a^2$ is contractible. I am trying to figure out the following:

Let $i:\mathbb{R}P^2\to\mathbb{R}P^3$ be the inclusion map (e.g. we obtain $\mathbb{R}P^3$ by identifying antipodal points of $\mathbb{S}^3$, so we can obtain $\mathbb{R}P^2$ as quotient of the subspace of $\mathbb{S}^3$ where the $4th$ coordinate is constantly 0). If we now see $a$ as a loop in $\mathbb{R}P^3$, does it become contractible? or does it remain non-contractible? In other words, is $i_*[a]=0$, where $i_*$ is the induced group homomorphisms between the fundamental groups?

I have little understanding of the real projective space, I don't even know how to compute its fundamental group (although I'm trying for a Seifert-van Kampen application) and it is hard for me to conceptualize the loop $a$ in $\mathbb{R}P^3$. Any intuitive or explicit hints and answers to help me understand this are greatly appreciated.

Just dropped in
  • 16,664
  • If $\gamma$ is a curve on $S^n$ with $\gamma(1)=-\gamma(0)$ then as a closed loop on $S^n/\pm 1$ it is not contractible because $\gamma_t(1)=- \gamma_t(0)$ has to be preserved in an homotopy – reuns Jan 14 '20 at 02:04
  • @reuns Could you please elaborate on why this preservation does not allow contractability? – Just dropped in Jan 14 '20 at 02:11
  • 1
    The "lift" idea is just saying that the length of $\gamma_t$ on $S^n/\pm1$ is the same as the length on $S^n$, which is at least $\pi$ – reuns Jan 14 '20 at 02:16
  • Have you learned anything about covering space theory? Because if you have not, then my answer to your question would be "go learn covering space theory". With that, it is possible to write a more informative answer. – Lee Mosher Jan 14 '20 at 03:15
  • @LeeMosher just the basic theory – Just dropped in Jan 14 '20 at 03:24
  • Well, if you've ever seen the proof that $\pi_1(S^1)\approx\mathbb Z$ using the covering map $\mathbb R \mapsto S^1$, then you might understand the proof that $\pi_1(\mathbb RP^2) \approx \mathbb Z / 2 \mathbb Z$, and how to see your path $a$ as the generator, using the covering map $S^2 \mapsto \mathbb RP^2$. – Lee Mosher Jan 14 '20 at 03:42
  • And, actually, the answer to this question says pretty much everything that I would say along those lines. – Lee Mosher Jan 14 '20 at 03:45

1 Answers1

1

It is not contractible to see this consider the fibration $p_n:S^n\rightarrow \mathbb{R}P^n$,

A contractible loop $c:\rightarrow \mathbb{R}P^n$, lifts to a contractible loop of $S^n$ since for a Serre fibration homotopies can be lifted.

The non contractible loop of $\mathbb{R}P^n$ can be constructed by taking a path $c$ in the upper hemisphere of $S^n$ wich meets the equator in two points which are identified by $p_n$, the embedding $\mathbb{R}P^n\rightarrow \mathbb{R}P^{n+1}$ result from an embedding $i_n:S^n\rightarrow S^{n+1}$ and $i_n(c)$ is a non closed segment such that $p_n(i_{n+1}(c))$ is a loop, so it is not contractible since its lift is not a loop.

Tsemo Aristide
  • 89,587