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I just watched a video lecture which proved that the fundamental group of the $1-$sphere $S^1$ is the integers $\mathbb Z$ under addition. While I followed most of the proof, it was very long and technical (the video was over one hour), and though the end result was interesting, the proof itself was not particularly motivating to me.

Are there some simpler examples of non-trivial fundamental groups of (path-connected) topological spaces, to help motivate the definition of fundamental group and also to see how one might go about computing the fundamental group of a space?

Math1000
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    The projective plane? – Angina Seng Dec 18 '19 at 19:40
  • @LordSharktheUnknown Feel free to elaborate and add as an answer; I am not familiar with this notion. – Math1000 Dec 18 '19 at 19:42
  • Clearly the video wasted more time than necessary. – cardinalRed Dec 18 '19 at 19:47
  • I don't know if it's quite what you're looking for, but once you know the fundamental group of $S^1$ you can do some computations for yourself (if you know a few properties of $\pi_1$). For example, the torus $S^1 \times S^1$, a solid torus $D^2 \times S^1$, and the Mobius strip all have nontrivial fundamental groups. – kamills Dec 18 '19 at 19:48
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    I think the proof that the fundamental group is $\Bbb Z$ is not very enlightening. If you understand why the result makes sense in terms of "number of oriented loops around a circle," that is the important part. – pancini Dec 18 '19 at 19:49
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    @ElliotG The proof teaches you how to use the universal covering. If that isn't enlightening, then the proof that you saw didn't highlight what was important. – cardinalRed Dec 18 '19 at 19:52
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    Your video proof probably used a covering map $\mathbb{R} \to S^1$. It turns out that this kind of approach works for other spaces, too, and is pretty beautiful. So the video you watched actually does show you one way fundamental groups can get computed! Maybe it just didn't tell you this explicitly... – kamills Dec 18 '19 at 19:53
  • The idea of the proof is nice, but the details are usually tedious, depending on what you take as given. – pancini Dec 18 '19 at 19:56
  • Yes, the professor did mention that he was proving a stronger result than was necessary. I believe that is the covering map/universal covering mentioned in the comments. @ElliotG The "number of oriented loops around a circle" is intuitive, and I got that immediately. But the sheer amount of notation and intermediate results of the proof got me a bit confused. – Math1000 Dec 18 '19 at 20:55

2 Answers2

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There does not exist any simpler example. Indeed, proving that any space has nontrivial fundamental group is at least as hard as proving that $S^1$ has nontrivial fundamental group, because if any space has nontrivial fundamental group it follows immediately that $S^1$ must have nontrivial fundamental group. To see this, suppose $\pi_1(S^1,x_0)$ is trivial. Then in particular the identity map of pointed spaces $(S^1,x_0)\to (S^1,x_0)$ is nullhomotopic. This then gives a nullhomotopy of any pointed map $(S^1,x_0)\to (Y,y_0)$ for any pointed space $(Y,y_0)$, by simply composing with a nullhomotopy of the identity map on $(S^1,x_0)$, and so $\pi_1(Y,y_0)$ is trivial.

Eric Wofsey
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  • I'm afraid I do not follow your argument. I am only beginning to study algebraic topology. Could you perhaps "dumb it down" a little for me? – Math1000 Dec 18 '19 at 21:14
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    Dumbed down explanation: The fundamental group is all about loops. If THE loop $S^1$ is trivial, then ALL loops are trivial. – Lee Mosher Dec 18 '19 at 21:54
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    More explicitly: as Eric says, if $\pi_1(S^1, x_0) = 0$ then in particular there is a (pointed) homotopy $H\colon (S^1, x_0) \times I \to (S^1, x_0)$ such that $H(x, 0) = x$ and $H(x, 1) = x_0$ for all $x\in S^1$. Then if $\gamma\colon (S^1, x_0) \to (Y, y_0)$ is any loop the composition $(S^1, x_0) \times I \stackrel{H}{\to} (S^1, x_0) \stackrel{\gamma}{\to} (Y, y_0)$ is a homotopy between $\gamma$ and the constant loop at $y_0$. Therefore $\pi_1(Y, y_0) = 0$ for every $Y$. – William Dec 18 '19 at 23:35
  • I am confused. I thought a loop was a map $\gamma : [0,1]\to X$ such that $\gamma(0)=\gamma(1)$. What is the meaning of a "loop" between two different spaces? – Math1000 Dec 18 '19 at 23:46
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    @Math1000: The quotient space of $[0,1]$ where you identify $0$ and $1$ is homeomorphic to $S^1$, so a loop in $X$ is equivalently a map $S^1\to X$. – Eric Wofsey Dec 18 '19 at 23:47
  • The fundamental group of the figure eight is the free group on two letters is another nontrivial example; I suppose whether it is "simpler" is a matter of opinion though. – Math1000 Dec 22 '19 at 07:55
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This is a somewhat incomplete answer, but by and large the computation of $\pi_1(X)$ for a space $X$ is just as difficult as computing $\pi_1(S^1)$, and often actually uses the fact that $\pi_1(S^1) \cong \mathbb{Z}$.

$S^1$ has the most elementary (non-trivial) fundamental group computation: there is a direct hands-on argument (see for example Massey's "A Basic Course in Algebraic Topology", Chapter II $\S 5$). The gist is that you divide $S^1$ into two halves, consider a loop $\gamma$ in $S^1$, and analyze the way $\gamma$ travels between the two halves, and describe how to homotope $\gamma$ into a loop that goes directly around the circle an integer number of times. The argument is elementary, but it's extremely difficult to do something similar for other spaces.

For a more general technique that can be used on other spaces, there are a few options. There's the Seifert-van Kampen theorem, where you write your space as a union of two open sets (with some assumptions) and your fundamental group is described as an "amalgamated product" of the fundamental groups of the two open sets and their intersection. A typical application of this theorem is to compute the fundamental group of any closed surface in terms of its fundamental polygon, and the usual argument uses the fact that $\pi_1(S^1) \cong \mathbb{Z}$. Unfortunately this theorem does NOT apply to compute $\pi_1 (S^1)$, but there is a more sophisticated version for "fundamental groupoids" which does apply (it may seem very daunting with all the abstraction, see for example this question, but as Eric Wofsey commented it actually essentially boils down to the argument I sketched above).

Another general tool for computing fundamental groups is the theory of Covering Spaces, which also isn't really elementary, though the arguments tend to be very nice. One outcome is that if $G$ is a discrete group acting on a simply-connected space $X$ (in a nice enough way) then the quotient space $X/G$ has fundamental group $G$. This theory does apply to computing $\pi_1(S^1) \cong \mathbb{Z}$, since $\mathbb{Z}$ acts on $\mathbb{R}$ by translation and $\mathbb{R}/\mathbb{Z} \cong S^1$, but the hands-on argument above is more elementary. Another example is the projective space $\mathbb{R}P^n$ for $n \geq 2$: the group $\mathbb{Z}/2$ acts on $S^n$, where the non-trivial element sends a vector $v\in S^n$ to $-v$ and the quotient by this action is by definition the projective space, so $\pi_1(\mathbb{R}P^n)\cong \mathbb{Z}/2$. (More generally there is a theory of Fibrations which can be used to compute $\pi_1$ in some nice cases, but covering spaces are deeply linked to the fundamental group.)

William
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    I think it's a little misleading to say that using groupoid van Kampen to compute $\pi_1(S^1)$ is hardly elementary. Of course if you state everything in terms of universal properties for groupoids it gets pretty hairy with the abstraction. But what the proof of van Kampen gives you is not the universal property but an explicit description (which you can then prove is equivalent to the universal property). If you just use that explicit description directly for $S^1$, this is essentially the same as the elementary "hands-on" argument you describe first. – Eric Wofsey Dec 18 '19 at 23:45
  • Good point, I will edit my answer – William Dec 18 '19 at 23:46