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Someone can explain me with an example, what is the meaning of $\pi(\mathbb{RP}^2,x_0) \cong \mathbb{Z}_2$?

We consider the real projective plane as a quotient of the disk.

I didn't receive an exhaustive answer to this question from my teacher, in fact he said that the loop $2a$ with base point $P$ is homotopically equivalent to the "constant loop" with base point $P$. but this doesn't solve my doubts.

Obviously I can calculate it, so the problem is NOT how to calculate it using Van Kampen theorem, but I need to get an idea of "why for every loop $a$, $[2a] = [1]$"

Balarka Sen
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Riccardo
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    There was this one guy from Vienna who took off his belt during a algebraic topology tutorial, twistet it twice and somehow managed to explain how this represents the fundamental group of $ℝP²$, but I really couldn’t concentrate much for I had to laugh so hard. – k.stm May 06 '13 at 17:28
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    @K.Stm. tomorrow il'll try during class :)) – Riccardo May 06 '13 at 18:03
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    The demonstration that you witnessed is usually called the Dirac belt trick or the Feynman plate trick. It shows that $\pi_1(SO(3)) = \mathbb{Z}_2$, which is equivalent since $SO(3)$ is homeomorphic to $\mathbb{RP}^3$, whose $2$-skeleton agrees with that of $\mathbb{RP}^2$. – Sammy Black May 06 '13 at 22:09
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    http://en.wikipedia.org/wiki/Plate_trick for some links to videos – Sammy Black May 06 '13 at 22:10

4 Answers4

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The following argument is essentially an application of the path lifting property for covering spaces.

Let's think about $\mathbb{R}P^2$ as being the quotient space you get by identifying antipodal points on the sphere $S^2$. That is, let $x\sim -x$, let $\mathbb{R}P^2=S^2/\sim$ and let $p\colon S^2\rightarrow\mathbb{R}P^2$ be the quotient map. Let $z$ be the base point of $S^2$ and $y$ be the base point of $\mathbb{R}P^2$.

Now, consider a non-trvial loop $\gamma\colon[0,1]\rightarrow\mathbb{R}P^2$ based at the point $y\in\mathbb{R}P^2$ (so $\gamma$ can not be homotoped to a constant loop). Note that the preimage of $y$ under $p$ is exactly two points in $S^2$ which are $z$ and $-z$. If we lift the loop $\gamma$ up to $S^2$ via the lift $\tilde{p}$, the end points of the lifted path $\tilde{\gamma}\colon[0,1]\rightarrow S^2$ will either both be at $z$, or $\tilde{\gamma}(0)=z$ and $\tilde{\gamma}(1)=-z$.

But note that if both end points are at $z$, then $\tilde{\gamma}$ is a loop and we know that $S^2$ is simply connected so such a loop can be homotoped to a constant loop. Such a homotopy induces a similar homotopy in the loop $\gamma$ and so $\gamma$ must be trivial. This is a contradiction as we asked for $\gamma$ to be non-trivial. So, $\tilde{\gamma}(0)=z$ and $\tilde{\gamma}(1)=-z$.

Now, in this case, the path $\tilde{\gamma}$ can not be homotoped to a constant loop without moving the fixed ends of the path but if we consider the lift of the path $2\gamma$ via $\tilde{p}$, then the lifted path $\tilde{2\gamma}$ is a loop in $S^2$. Again, $S^2$ is simply connected and so such a loop can be homotoped to a constant loop and such a homotopy induces a similar homotopy in the loop $2\gamma$ and so $2\gamma$ is a trivial loop.

Dan Rust
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    This is the first time I've seen the noun “homotopy” verbed. ☺ – Harald Hanche-Olsen May 06 '13 at 17:56
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    I thought it was a quite common shorthand. I could say the same for the noun "verb" :D. – Dan Rust May 06 '13 at 18:03
  • Ok got it. have to draw an example to master the idea but ok, it is clear now! never thought about using lifting but i think it is pretty natural in this case :) – Riccardo May 06 '13 at 18:23
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    Yeah, verbing “verb” was a deliberate joke for the occasion. Anyhow, the universal cover by a sphere is definitely the right way to see it, mathematically – in any dimension $\ge2$. Ergo, +1. – Harald Hanche-Olsen May 06 '13 at 20:08
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    @DanielRust This answer is very helpful. I just want to confirm one thing. Your reasoning works because $p:S^2\to \mathbf RP^2$ is not just a quotient map but also a covering map right? (I am very new to algebraic topology). – caffeinemachine Nov 21 '14 at 12:09
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    @caffeinemachine Sure, you need the quotient map to be a covering map in order to use the path lifting property (a general quotient map won't have unique lifts of paths). – Dan Rust Nov 21 '14 at 13:09
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    @DanielRust Thank you. From your comment I infer that arbitrary quotient maps admit, possibly non-unique, lifts of paths. Is this true? Further, can you clarify what you mean by $\tilde p$ in your answer (it appears in the last paragraph)? As far as I know the overhead tilde denotes the lift of a path. But I don't think lifting $p$ makes sense, hence the confusion. Lastly, when you write $2\gamma$, do you means $\gamma*\gamma$? Thanks again. – caffeinemachine Nov 21 '14 at 13:55
  • @caffeinemachine That's a good question. I think it's likely that all quotient maps have non-unique path lifting in the sense that if $q\colon X\to Y$ is a quotient map and $\gamma\colon[0,1]\to Y$ is a path, then there exists a path $\tilde{\gamma}\colon[0,1]\to X$ such that $q\circ \tilde{\gamma}=\gamma$. I'd need to think about it. With regard to the notation: $\tilde{p}$ is sometimes used in a handwavey way to mean the map which sends a path $\gamma$ to its lifted path $\tilde{\gamma}$ according to the covering map $p$ and a prescribed basepoint in the total space. $2\gamma=\gamma*\gamma$ – Dan Rust Nov 21 '14 at 14:07
  • @DanielRust That was very helpful. Thank you. – caffeinemachine Nov 21 '14 at 14:09
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    How do we know that $2\gamma$ is a loop when lifted to $S^2$ and doesn't also start at one point and end at its antipode? – QCD_IS_GOOD Mar 28 '17 at 03:52
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    @JoshuaLin it comes from general covering space theory and the idea that lifting multiples of paths corresponds to applying the same permutation of the fiber. The fiber is only two points, so if the original lift of $\gamma$ swapped the two points in the fiber, then the lift of $2\gamma$ must act trivially on the fiber, hence the new path both begins and ends at the base point. – Dan Rust Mar 28 '17 at 09:24
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    Sorry for reviving this. This proves that every element of the fundamental group has order $2$, but I think we'd need a bit more to prove it's not something like $\mathbb{Z}_2\times\mathbb{Z}_2$. I think it's not a lot of work, if we take two of such loops which respective liftings end at $-z$, we can take the lifting of their product which is homotopic to the trivial loop, so any two of such loops are homotopic inverses. – LeviathanTheEsper Apr 04 '18 at 04:03
  • I think LeviathanTheEsper is right. Or actually I think it proves only that every non-trivial element is of order two, if it exists. – Bipolar Minds Jan 26 '22 at 20:30
  • Question 1: How do we know that $2\gamma$ is a loop when lifted? Question 2: How do we know that such a non-trivial $\gamma$ exists? – mathematics-and-caffeine Jul 16 '23 at 16:47
  • @mathematics-and-caffeine Q1. The end points of $\gamma$ are at $z$ and $-z$ and so when you lift $2\gamma$, we have $\overline{2\gamma}(0)=z, \overline{2\gamma}(1/2)=-z, \overline{2\gamma}(1)=z$ (because the 'second' $\gamma$ has to have two different end points just like the first $\gamma$ did). Q2. Our quotient map is a covering space projection. Any non-trivial covering implies that the codomain space has non-trivial fundamental group through the correspondence theorem for covering spaces (I guess for this question, that's one place where 'intuition' plays a role). – Dan Rust Jul 17 '23 at 09:35
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Try watching Your palm is a spinor on youtube. This move is part of a traditional phillipine dance – watch about 40 seconds into the clip.

As you go from the performer's more or less stationary shoulder to the hand that holds the glass, you are in fact following a homotopy from the trivial loop to the loop that rotates 720 degrees around a vertical axis.

The move is not hard to learn. But try it with an empty glass at first.

Edit: I neglected to add that this is really about $\mathrm{SO}(3)\simeq\mathbb{R}P^3$, not $\mathbb{R}P^2$. It's the same sort of thing going on, really. To see that $\mathrm{SO}(3)\simeq\mathbb{R}P^3$, think of a rotation as specified by a vector $x\in\mathbb{R}$ with $\|x\|\le\pi$, the direction giving the axis and the length the angle of rotation in the positive direction, as seen from the positive end of the axis. This identifies antipodal points on the sphere of radius $\pi$, thus turning the closed ball into a projective 3-space.

Harald Hanche-Olsen
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  • Wow - I would not have thought that phillipine dance is connected to properties of electrons and algebraic topology..! +1 – exchange Jun 04 '17 at 07:09
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Here is a bit more algebraic perspective: if $X$ and $Y$ are path connected and $p: Y\rightarrow X$ is a covering map, then the number of sheets is equal to index of the subgroup $p_*(\pi_1(Y))$ in $\pi_1(X)$. Since $S^2 \rightarrow \mathbb{R}P^2$ is a 2-sheeted universal cover, it follows that $\pi_1(\mathbb{R}P^2)$ has 2 elements (trivial subgroup has index 2).

lanskey
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You can see another set of related pictures here, which gives the script for this video Pivoted Lines and the Mobius Band (1.47MB).

The term "Pivoted Lines" is intended to be a non technical reference to the fact that we are discussing rotations, and their representations. The video shows the "identification" of the Projective Plane as a Mobius Band and a disk, the identification being shown by a point moving from one to the other. Then the point makes a loop twice round the Mobius Band, move

as in the above, and this loop moves off the Band onto the disk and so to a point. Thus we are representing motion of motions!

Community
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Ronnie Brown
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