What distinguishes the elements of the fundamental group of the circle?

Question

I made a search for similar questions on the site but have found no intuitive answer. If $m$ and $n$ are distinct integers, what would make a loop going around the circle $n$ times different from one going around the circle $m$ times?

Answer 1

Both look the same if you consider the geometric image of course, which is the whole circle in the case where, e.g, $m$ and $n$ are positive. But they are not homotopic as maps.

An intuitive way to see it is if you have a shoe lace that goes 3 times around a pole, and another one that goes 5 times, you will never be able to put the first one in the position of the second without cutting/opening it.

Another intuitive way (which actually begins to look like a formal proof) is to see that if your loop is nice enough in the sense that all points of the circle are visited finitely many times, then if you consider those points where the loop never "stops" to start going backwards, all these points are visited the exact same number$^\star$ of times, and this number is $m$ (respectively, $n$). And this property never changes as you move the loop using a homotopy: it's what we call a homotopy invariant.

$^\star$ count $+1$ each time the loop visits a point while running counterclockwise, and $-1$ each time it visits the point clockwise.

Answer 2

Probably the clearest way to see what's going on is via covering spaces. In particular, $\Bbb R$ is the covering space of $S^1$ (the circle) with covering map $p:\Bbb R \to \theta$ given by $p(\theta) = (\cos \theta, \sin \theta)$ (where I have identified $S^1$ with the unit circle in $\Bbb R^2$). If $\gamma: [0,1] \to S^1$ satisfies $\gamma(0) = \gamma(1)$ (i.e. $\gamma$ is a loop in the circle), then by the lifting property there exists a unique $\Gamma:[0,1] \to \Bbb R$ (the "lift" of $\gamma$) such that

• $\Gamma(0) \in [0,2\pi)$ satisfies $p(\Gamma(0)) = \gamma(0)$,
• $p \circ \Gamma = \gamma$.

Suppose now that we have a homotopy of loops in $S^1$. That is, we have a continuous map $\gamma(x,t):[0,1] \times [0,1] \to S^1$ such that for any $t \in [0,1]$, $\gamma_t := \gamma(\theta,t):[0,1] \to S^1$ defines a loop, which is to say that $\gamma_t(0) = \gamma_t(1)$. We can "lift" these maps to a corresponding homotopy $\Gamma:[0,1] \times [0,1] \to \Bbb R$ of paths in $\Bbb R$. Note that each $t$, $\Gamma_t := \Gamma(\cdot,t)$ is a lift of the loop $\gamma_t$. It will not necessarily hold that $\Gamma_t(0) = \Gamma_t(1)$.

Consider the map $\phi:[0,1] \to \Bbb R$ given by $\phi(t) = \Gamma_t(1) - \Gamma_t(0)$. Because $\gamma_t(0) = \gamma_t(1)$, $\Gamma_t(1) - \Gamma_t(0)$ must be an integer multiple of $2\pi$. However, $\phi$ is continuous, so in fact we find that $\phi(t)$ must be constant. Now, we note that $\frac 1{2 \pi}\phi(t)$ is the number of times that $\gamma_t$ winds around the circle. So, $\gamma_0$ and $\gamma_1$ must wind around the same number of times. That is, if we can continuously deform $\gamma_0$ to $\gamma_1$, they must wind around the same number of times.

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Here is a slightly more intuitive and less formal break-down of this proof. For any loop $\gamma:[0,1] \to S^1$, we can keep track of the angle $\Gamma(s)$ that $\gamma(s)$ makes with the $x$-axis, where we define the angle in such a way that $\gamma(0)$ is in $[0,2\pi)$. Intuitively, it should be clear that the "angle function" is completely determined by $\gamma$, and that any continuous angle-function corresponds to some path in $S^1$.

We want to define this angle-function $\Gamma:[0,1]$ to be continuous, so in some cases $\Gamma(s)$ is forced to leave the interval $[0,2\pi)$. For example, if $\gamma$ starts at $(0,0)$ and wraps around the circle counterclockwise twice at a constant rate, then the only angle-function that works is $\Gamma(s) = 4 \pi s$. Of course, $\Gamma(1) = 4 \pi$ does not lie in our starting-set of angles $[0,2\pi)$.

Looking at these phase functions, we notice that $\frac{1}{2 \pi}(\Gamma(1) - \Gamma(0))$ (which measures how far we have wandered from our initial starting angle) keeps track of how many times we have wrapped around the circle. If we continuously deform $\gamma_0$ to $\gamma_1$, then we have correpondingly deformed $\Gamma_0$ to $\Gamma_1$. Because deformations are continuous, $\frac{1}{2 \pi}(\Gamma(1) - \Gamma(0))$ is not allowed to suddenly "switch" from one multiple of $2\pi$ to another, which means that it must stay constant. So, $\Gamma_0$ and $\Gamma_1$ must traverse the same angle-difference, and correspondingly $\gamma_0$ and $\gamma_1$ must wrap around the same number of times.