## The Fundamental Group of Euclidean n-Space.

### December 8, 2010

Sometimes, we can read a whole bunch of math and not get it until a simple picture is drawn.  For example, the notion of upper semi-continuous was not clear to me (at all) until a picture was drawn.  The "idea" of the mean value theorem is similar — drawing the tangent and parallel secant really shows what the theorem is trying to say.

This post is to say in words, and then show in pictures, why we should have the fundamental group about a basepoint $x_{0}$ of ${\mathbb R}^{n}$ be trivial.  In other words, we’d like to prove $\pi_{1}({\mathbb R}^{n}, x_{0}) = 0$

## With Words.

First, let’s think about this for a second.  What’s the definition of $\pi_{1}({\mathbb R}^{n}, x_{0})$?  Well, it’s the collection of equivalence classes of loops up to homotopy.  So, what does this mean?  Well, we take a loop and another loop.  Are they homotopic?  If so, they’re in the same class.  If not, they’re in different classes.  In this very specific example, we’ll show that every loop is the same as the constant loop.  And, in fact, there’s even a nice way to do this.

First, let’s just go ahead and suppose our basepoint $x_{0} = 0$.  Given some loop $\gamma:[0,1]\rightarrow {\mathbb R}^{n}$, we want to show that this is homotopy equivalent to the constant loop (which will act as the 0 element of our group).  Another way to say this is that $\gamma$ is nullhomotopic

(Note here that in many books we think of the loop as the physical image in Euclidean space, but it is actually just a map.  It is common to identify these notions, but just remember that there is a subtle difference between the map and the image.)

So, what is our loop going to look like?  Well, since it’s in Euclidean space, we can actually write down a function for it.

$\gamma(x) = (y_{1}(x), y_{2}(x), \dots, y_{n}(x))$

where each of these $y_{i}$‘s is continuous and is the restriction of our curve to the $i$-th dimension.  The following homotopy is called the "straight line homotopy" for reasons that will become clear in the pictures below.

$h(x,t) = \gamma(tx) = (ty_{1}(x), ty_{2}(x), \dots, ty_{n}(x))$

So, this homotopy tells us that, when $t = 0$ we have $h(x,0) = \gamma(0) = 0$ for every $x$ and when $t = 1$ then $h(x,1) = \gamma(x)$ for every $x$.  The "hard" part is to show this is actually continuous.  Because we’re working over Euclidean space with no holes and things, we can argue by basic calculus that the composition of continuous functions is continuous and the product is continuous if and only if the projection onto each coordinate is continuous.  Thus, this homotopy is continuous which implies $\gamma$ is homotopic to the loop which is always 0; this latter loop is called the constant loop.  We write $\gamma \simeq 0$ in this case.

Now, in the fundamental group $\pi_{1}({\mathbb R}^{n}, x_{0})$, we consider a basic element; say $[\gamma]$ where the brackets indicate it is actually an equivalence class of loops up to homotopy.  But what did we just learn about this loop?  Well, we have that $[0] = [\gamma]$ since they’re the same loop.  In particular, given any other loop, say, $\gamma_{0}:[0,1]\rightarrow {\mathbb R}^{n}$, we have (by the same arguments as above) that $[\gamma_{0}] = [0] = [\gamma]$Thus, in our fundamental group we have only one element, namely $[0]$.  Because we only have one element (the identity element) we say that the group is trivial and usually omit the bracket to write

$\pi_{1}({\mathbb R}^{n}, x_{0}) = 0$

which expresses exactly this idea: the only loop, up to homotopy, is the constant loop which just "stays at 0 the whole time."

## With Pictures.

First, a nice picture of ${\mathbb R}^{n}$

The astute reader will notice I have drawn this for $n = 3$, because my tablet won’t let me draw it for any higher $n$.  I think it may be a defect in the hardware.  Nonetheless, let’s proceed.

Here’s a loop, which we will identify with the "image" of a loop, just to be able to see a concrete thing.

Our loop looks a bit like a dinosaur.  Let’s call it $\gamma$ just to be able to talk about it.  So, what can we do to $\gamma$?  Well, let’s try to continuously take it back to the origin.  I’ll do this with a nicer looking loop, but the idea is exactly the same.

This picture shows our original loop (darkest, outermost) going through the homotopy.  I denoted a few points (one in green, one in light blue, one in purple) and showed where those go at each stage in the homotopy.  Note that this is a "continuous" thing, so it should actually have more than like, seven loops inside it, but it’s hard to draw infinitely many.

This is another kind of crappy picture showing that each of the points is going to travel over to 0.  In this one, I have not drawn any of the loops constituting the homotopy in, but just the general path that the point would probably take.  Note that with the "straight line homotopy" as we detailed above, all of these lines would be completely straight, as in:

There is even no problem if they "pass through" one another — you shouldn’t concern yourself too much with this homotopy, just prove it is continuous in this case and use it.

The fundamental group basically says what kind of loops we can have, and it turns out that if we have any loop, we can kind of squish it down in this way, so:

So each loop is really just the constant loop at 0.  This should help to see why $\pi_{1}({\mathbb R}^{n}) = 0$.