Why is the fundamental group of the projective plane $C_{2}$?

Question

I just recently know that there are topologies with finite nontrivial fundamental groups (homotopy curve). I just can't wrap my mind around it at all.

If you have a curve, and somehow cannot shrink it to null, then there must be a hole blocking the curve from shrinking. And since there is a hole there, you can just wrap the curve around the hole as many times as you like, producing a more non-homotopic curve. So the fundamental group, if nontrivial, must contain a copy of $\mathbb{Z}$, right?

Apparently, that reasoning does not work, but I cannot wrap my mind about what went wrong here. So if it's not a hole that's blocking the shrinking, then what, exactly is that? What's a concrete way to visualize such a thing?

Answer 1

Here's an illustrated version of Sharkos' answer. We'll use the disk model of ℝP2: consider a disk with antipodal points on its boundary identified.

We can imagine a closed path that starts at the center, goes straight up (1st part in light blue) hits the topmost point of the disk's boundary - which is also the lowermost point of the disk's boundary - and then continues upwards (now from the bottom, 2nd part in dark blue) to reach the center again and close the loop:

We cannot shrink this loop to a point: we can modify it within the interior of the disk, but that doesn't get us very far. And if we try to change the loop at the point it crosses the boundary, we have to move the top and the bottom of the loop so that they cross at antipodal points, otherwise we brake the loop. For instance, doing the following breaks the loop:

We can slide both points so they stay opposite each other:

But that doesn't allow us to shrink this loop to a point.

Next we consider a loop that crosses the boundary twice:

The loops starts from the center, goes straight up to the top of the disk (1st part in light blue), comes up from the bottom to the center (2nd part in dark blue), continues up to the top of the disk (3rd part in pink) and finally comes up from the bottom of the disk back to the center to close the loop (4th part in red).

As in the previous example, if we move a point where the path crosses the boundary we have to move the opposite point in the opposite direction so they remain antipodal to one another. We can start to rotate the 1st and 2nd parts of the path counter-clockwise in a continuous way until it looks like this:

The loop is not broken in any way. We continue to rotate it:

Until we've rotated the first half of the path 180 degrees:

And this is where the magic happens: if we imagine a particle following along the loop, it travels down part 2 (in dark blue) and when it reaches the center it turns around and travels up part three (in pink). That means that we can pull the point where the end of part 2 and the beginning of part 3 meet upwards without breaking the loop:

We can now pull the 2-3 part of the loop past the boundary:

At which points it's evident that the loop is contractible.

As Sharkos said, it's sliding points around on the boundary that messes with our intuition: when there is only one point where the path crosses the boundary we cannot shrink the loop to a point because any time we move that point it must be "reflected" in the opposite side of the circle (this is more of an appeal to our intuition rather than a rigorous proof). When the path crosses the boundary twice we can slide one pair of opposite points around to effectively "untangle" the loop.

Answer 2

$\mathrm{SO}(3)$ (which is also $\mathbb{R}P^3$) is a famous 'basic' example for physicists. (See Fundamental group of $SO(3)$, Visualizing the fundamental group of SO(3), https://mathoverflow.net/questions/38219/intuition-on-finite-homotopy-groups and An intuitive idea about fundamental group of $\mathbb{RP}^2$ as linked to above - these last two tie in more with what I'm talking about). The Dirac belt or plate trick is one suggestive way of approaching the fundamental group for this manifold.

In order to make things concrete, we describe $\mathbb{R}P^2$ as a square as Joseph pointed out in the comments:

We would similarly describe $\mathrm{SO}(3)$ as being a solid ball with antipodal points identified. Then intuitively you can probably convince yourself quickly that

• any loop lying within the 'interior' of the square/ball (i.e. not reaching any antipodal points) is contractible.
• any loop which leaves the centre, goes 'over the boundary' once and continues back to the origin is not contractible.

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The trick, however, is in considering following this last loop twice. It turns out that by bending one of the crossings around you can cancel out the boundary-crossing.

I'm not going to open up Paint just now, but imagine a line which crosses from left to right twice very close by. (They should form a pair of parallel lines symmetrically placed around the middle.) Call the 'two' lines you draw $L,M$, going from $l_1 \to l_2$ and $m_1 \to m_2$ left to right respectively.

Now $l_2$ is constrained to lie antipodal to $m_1$ and similarly the other two points, but (as Stefan pointed out) there is freedom to move these pairs independently. Start sliding $l_2$ around the square anticlockwise. $m_1$ moves anticlockwise opposite it. This makes $L$ start to bend upwards on the right, and $M$ start to bend downwards on the left.

Continuing this, one brings $l_2$ all the way round to $m_2$ (and $m_1$ all the way to $l_2$). Now $L$ and $M$ just form a little loop over the boundary, and can be contracted away by bringing all four points together.

Just draw the intermediate stages and you might convince yourself!

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It's this freedom to slide stuff around which screws with your intuition. You really can't visualize projective planes as nice, oriented embedded surfaces, so your intuition of fundamental groups as being determined by genus is... horrifically broken to say the least.

Answer 3

Consider the map $z \mapsto z^2$ of the unit circle in the complex plane. This gives us a fibration $S^1 \to S^1$ of degree two. If you travel around the target circle once then this will take you to $-1$ in the source circle. You really need to run around the target circle twice to obtain an honest loop in the source circle.

Now consider the fibration $S^2 \to {\bf R}P^2$ given by identifying antipodal points. The intuition from the paragraph above applies again. You need to go around any non-trivial loop in ${\bf R}P^2$ twice to obtain an honest loop in $S^2$, which is then of course contractible because $S^2$ is simply connected. If you only travel once, you'll get a half-circle in $S^2$ connecting antipodal points and you can check yourself that there's no way to shrink this half-circle to a point.

Answer 4

Here's another way to look at why some loops in RP2 are contractible and some aren't, without referring to quotient spaces. In particular, we can visualize (without rigorous proof) a closed loop that is not contractible, but whose "double" is contractible.

Instead of considering quotient spaces, we can think of RP2 as the collection of lines through the origin in R3. We can represent these as the lines connecting antipodal points in the unit sphere. So in this view a point in RP2 is a line in the unit sphere through the origin:

We have to shift perspective here - this line (in the sphere) is a point in RP2. And a line in RP2 is a fan-like "sequence" of lines (in the sphere), all of which are "touching". For example, drawing a line in RP2 could look something like this:

It starts out as a single point in RP2 which is a line in the unit sphere, and then extends out to form a line in RP2.

One type of (contractible) loop looks like a cone:

Note that this is a closed loop because the very first point (the red line) of the path (the cone) is the same as the very last point (the same red line at the very end of the cone) of the path. You can see the loop in the more "traditional" viewpoint as the circle at the top of the cone, where the cone touches the unit sphere. In the square-with-opposite-sides-identified viewpoint this would be the equivalent of a simple loop drawn within the body of the square (does not cross the boundary at all). You can tell it's contractible because, well, it's easy to contract it to a point - we simply shrink the loop - that is "narrow" the cone - until it becomes just a single line (which of course in RP2 is a point):

So that's all well and good. Now let's visualize a non-contractible loop. First let's start with a "half-circle" in RP2: a curved path whose ends do not touch. This looks like a half-cone:

Notice how the very first red line represents the start point of this path in RP2, whereas the very last dark green line represents the end point of the path. They are very much not the same line, which is why this is not a closed path (not a full cone). And if you look on the surface of the sphere you can see a half-circle like path which is again not a closed loop.

Next, consider what happens when we "flatten" this cone:

The two end points (the red and green lines at the edges of the cone) get closer and closer together until they finally touch (when we form a "flat cone"). Since they're touching, this is now a closed loop. In the squares-with-opposite-sides-identified viewpoint this would be a loop that crosses the boundary once.

If you spend some time imagining how to contract this loop (i.e. cone in the unit sphere) to a point (i.e. a line), you may just convince yourself that this cannot be done. In particular, note that if we try to "shrink the cone" like we did for the full loops (full cones) above then the path breaks - we get a half-cone again which is not a closed loop:

So by trying to shrink the cone/loop we broke the loop. It's interesting to try to imagine what happens if you tried to "raise" this flat cone's edges towards the top of the sphere starting at one point (any point), and then propagating this change outward in two opposite directions along the edges of the flat cone:

It works great until you reach two antipodal points on the edge of the flat cone (our starting point the red line and our ending point the green line) at which point the "rising wave" from one side wants to twist the line in one sense (move the point in one direction), whereas the "rising wave" from the other side wants to twist the same line in another sense (move the same point in a different direction), resulting in a "tear" (the red line and the green line are no longer touching). Ultimately this feels like a problem of orientability: we have two different paths from where the wave starts to the red/green lines, which are the same when the cone is flat. But propagating that wave along the two paths results in two different directions of "up", causing the tear.

It's instructive to see a similar construction that starts with a full-cone / closed-loop. Flattening it gives us a similar "flat-cone":

Note however that each line/point in this flat-cone/closed-loop is represented twice! You can see it for example in how the dark red line of the cone meets the dark-green line opposite it when the cone flattens out completely. In the squares-with-opposite-sides-identified viewpoint this would be a loop that crosses the boundary twice.

Now if we just do the exact same process backwards we can see that this path can be shrunk to a point just fine, by narrowing the cone into a line:

The path/cone does not break when we do this. Unlike the flattened half-cone, since every point is accounted for twice in the loop, we can push one "up" and one "down" without tearing, giving us the two opposing sides of the cone.

I like this viewpoint because no points are singled out as "special", unlike the quotient space formulation where it feels like the boundary plays some mysterious role. And we see the issue of orientability more directly (as the animation of the wave propagating in two directions shows). To address your idea that there must be a hole: it's more like that there's a global twist in the space, that causes certain loops to have their ends touching, but in a "twisted" way. You cannot shrink it because the twist is still there and will ultimately cause a tear. But if you wrap it around twice the double "twist" cancels itself out, and shrinking becomes possible.