7
$\begingroup$

I'm trying to compute the fundamental group of $\mathbb{RP}^2$ minus 2 points. I'm using the presentation $\langle a\mid a^2\rangle$. Meaning that I'm taking the disk and identifying the two sides.

I'm conjecturing that it is $\mathbb{Z}\times\mathbb{Z}$. I'm trying to use Seifert -Van Kampen but I can't get my open sets to work nicely. The reason why I think it is $\mathbb{Z}\times\mathbb{Z}$ is because clearly a loop around each hole would give you two distinct loops, say $a$ and $b$. But then it seems that $ab$ is homotopic to $ba$. This relation leads me to think that it might be $\mathbb{Z}\times\mathbb{Z}$.

Any hint would be appreciated.

$\endgroup$
  • $\begingroup$ The fundamental group is $\mathbb{Z}$, not $\mathbb{Z}\times \mathbb{Z}$. $\endgroup$ – Bombyx mori Jan 7 '15 at 0:32
  • 3
    $\begingroup$ The projective plane minus a point/small disc is homeomorphic to a Mobius band. Show this, and proceed from there. $\endgroup$ – user98602 Jan 7 '15 at 0:33
  • $\begingroup$ @Bombyxmori: OP said minus two points, so is a Möbius band without a point $\endgroup$ – janmarqz Jan 7 '15 at 2:52
5
$\begingroup$

Here is the standard way of tackling such problems:

RP2

Think of $\mathbb{RP}^2$ as as the disc $D^2$, with boundary identifiead in the oposite direction. Then after removing $2$ points the space retracts onto the space $X$. Now you can easily see that $X$ is nothing but $S^1\vee S^1$. So you have, $\pi_1(\mathbb{RP}^2\setminus\{\text{two points}\})\cong\pi_1(X)\cong\mathbb{Z}*\mathbb{Z}$

$\endgroup$
  • $\begingroup$ The picture makes so much sense. $\endgroup$ – Phanu9000 Jan 7 '15 at 16:20
  • $\begingroup$ Would be able to elaborate on this retraction? $\endgroup$ – user475040 Jan 9 at 21:37
  • $\begingroup$ @crispypizza The (deformation) retraction is simply "radially" pushing everything away from the two puncture points. You'll end up with the space $X$. If you're asking for an explicit map, that would be quite messy! $\endgroup$ – ChesterX Jan 10 at 5:06
  • $\begingroup$ Yes definitely not, I guess I am not sure where the segment connecting $u$ is coming from. We are getting another generator $b$? $\endgroup$ – user475040 Jan 10 at 20:56
  • $\begingroup$ @crispypizza It is one of the generators. If you consider the horizontal segment before performing the retraction, it is a non trivial loop. $\endgroup$ – ChesterX Jan 11 at 6:23
4
$\begingroup$

This is akin to the approach taken by Bombyx mori. Let $\mathbb{RP}^2$ be identified with $S^2$ under the quotient of antipodal points. Thus we are removing two pairs of antipodal points and then quotienting. Let those points be $$(\sqrt{2}/2, \sqrt{2}/2,0), (-\sqrt{2}/2, \sqrt{2}/2,0), (\sqrt{2}/2, -\sqrt{2}/2,0), (-\sqrt{2}/2,-\sqrt{2}/2,0)$$ i.e. evenly spaced around the equator. Then there is a deformation retract of $S^2 - \{ \text{points}\}$ onto the union of great circles $$A = \{(x,y,z) \in S^2 | x = 0\} \cup \{ (x,y,z) \in S^2 | y = 0\}$$ Furthermore, we can arrange it so this deformation respects the antipodal quotient. But this descends then to a deformation retract $$ \mathbb{RP}^2 \to A/\{\pm\} \cong S^1 \vee S^1$$ And this has fundamental group $\mathbb{Z} * \mathbb{Z}$.

$\endgroup$
4
$\begingroup$

Here might be an unconventional way to see this to get around the Mobius band. Let $\mathbb{RP}^{2}$ to be $\mathbb{S}^{2}$ with antipodal points identified. Now removing four points, with two pair of them anti-podal then take the quotient is the same as removing two points from $\mathbb{RP}^{2}$. But the former one is easy to describe - it is $\mathbb{S}^{1}\vee \mathbb{S}^{1}\vee \mathbb{S}^{1}$. Now by taking the quotient we identify two such circles into one circle. So in the end we get $\mathbb{S}^{1}\vee \mathbb{S}^{1}$, and the fundamental group is $\mathbb{Z}*\mathbb{Z}$.

In this light my earlier comment (as janmarqz pointed out) was wrong, as the comments showed I could not see this clearly until thinking more carefully about it. In fact the Mobius band picture is equally clear, and I was confused with the extra $\mathbb{S}^{1}$ coming from the shrinking part of the Mobius band.

$\endgroup$
  • $\begingroup$ excuse me, man, if I was too rude, but you already see, (+1) for all $\endgroup$ – janmarqz Jan 7 '15 at 4:27
  • $\begingroup$ @janmarqz: Yes, by the point you posted your comment I knew it was $\mathbb{Z}*\mathbb{Z}$ already. So I want some reasoning from others to convince me I was false by assuming it should be a circle. Since I got zero feedback I have to think more carefully. It is okay. $\endgroup$ – Bombyx mori Jan 7 '15 at 4:30
  • $\begingroup$ @janmarqz: Forgot to say that thanks for pointing out. $\endgroup$ – Bombyx mori Jan 7 '15 at 4:31
  • $\begingroup$ you're welcome. $\endgroup$ – janmarqz Jan 7 '15 at 4:33

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.