# An intuitive idea about fundamental group of $\mathbb{RP}^2$

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 $a$, $[2a] = [1]$"

• 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. May 6, 2013 at 17:28
• @K.Stm. tomorrow il'll try during class :)) May 6, 2013 at 18:03
• 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 $\mathbb{RP}^2$. May 6, 2013 at 22:09
• en.wikipedia.org/wiki/Plate_trick for some links to videos May 6, 2013 at 22:10

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.

• This is the first time I've seen the noun “homotopy” verbed. ☺ May 6, 2013 at 17:56
• I thought it was a quite common shorthand. I could say the same for the noun "verb" :D. May 6, 2013 at 18:03
• 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. May 6, 2013 at 20:08
• @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). Nov 21, 2014 at 13:09
• 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? Mar 28, 2017 at 3:52

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.

• Wow - I would not have thought that phillipine dance is connected to properties of electrons and algebraic topology..! +1 Jun 4, 2017 at 7:09

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).

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,

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!