Fundamental group

Revision as of 01:17, 13 December 2009 by Jam (talk | contribs)

Perhaps the simplest object of study in algebraic topology is the fundamental group.

Let $(X,x_0)$ be a based, topological space (that is, $X$ is a topological space, and $x_0\in X$ is some point in $X$). Note that some authors will require $X$ to be path-connected. Now consider all possible "loops" on $X$ that start and end at $x_0$, i.e. all continuous functions $f:[0,1]\to X$ with $f(0)=f(1)=x_0$. Call this collection $\Omega(X,x_0)$ (the loop space of $X$). Now define an equivalence relation $\sim$ on $\Omega(X,x_0)$ by saying that $f\sim g$ if there is a (based) homotopy between $f$ and $g$ (that is, if there is a continuous function $F:[0,1]\times[0,1]\to X$ with $F(a,0)=f(a)$, $F(a,1)=g(a)$, and $F(0,b)=F(1,b)=x_0$). Now let $\pi_1(X,x_0)=\Omega(X,x_0)/\sim$ be the set of equivalence classes of $\Omega(X,x_0)$ under $\sim$.

Now define a binary operation $\cdot$ (called concatenation) on $\Omega(X,x_0)$ by $(g\cdot h)(a)=\begin{cases} g(2a) & 0\le a\le 1/2, \\ h(2a-1) & 1/2\le a\le 1.\end{cases}$ One can check that if $f\sim f'$ and $g\sim g'$ then $f\cdot g\sim f'\cdot g'$, and so $\cdot$ induces a well-defined binary operation on $\pi_1(X,x_0)$.

One can now check that the operation $\cdot$ makes $\pi_1(X,x_0)$ into a group. The identity element is just the constant loop $e(a) = x_0$, and the inverse of a loop $f$ is just the loop $f$ traversed in the opposite direction (i.e. the loop $\bar f(a) = f(1-a)$). We call $\pi_1(X,x_0)$ the fundamental group of $X$.

Note that the fundamental group is not in general abelian. For example, the fundamental group of a figure eight is the free group on two generators, which is not abelian. However, the fundamental group of a circle is ${\mathbb{Z}}$, which is abelian.

More generally, if $X$ is an h-space, then $\pi_1(X)$ is abelian, for there is a second multiplication on $\pi_1(X)$ given by $(\alpha\beta)(t) = \alpha(t)\beta(t)$, which is "compatible" with the concatenation in the following respect:

We say that two binary operations $\circ, \cdot$ on a set $S$ are compatible if, for every $a,b,c,d \in S$, \[(a \circ b) \cdot (c \circ d) = (a \cdot c) \circ (b \cdot d).\]

If $\circ,\cdot$ share the same unit $e$ (such that $a \cdot e = e \cdot a = a \circ e = e \circ a = a$) then $\cdot = \circ$ and both are abelian.

Independence from base point

At this point, one might wonder how significant the choice of base point, $x_0$, was. As it turns out, as long as $X$ is path-connected, the choice of base point is irrelevant to the final group $\pi_1(X,x_0)$.

Indeed, pick consider any other base point $x_1$. As $X$ is path connected, we can find a path $\alpha$ from $x_0$ to $x_1$. Let $\bar\alpha(t) = \alpha(1-t)$ be the reverse path from $x_1$ to $x_0$. For any $f\in\Omega(X,x_0)$, define $\varphi_\alpha = \bar\alpha\cdot f\cdot\alpha \in\Omega(X,x_1)$ by \[\varphi_\alpha(f)(t) = (\bar\alpha\cdot f \cdot \alpha)(t) = \begin{cases} \bar\alpha(t) & 0\le t\le 1/3, \\  f(3a-1) & 1/3\le t\le 2/3,\\ \alpha(3t-2) & 2/3\le t\le 1. \end{cases}\] One can now easily check that $\varphi_\alpha$ is in fact a well-defined map $\pi_1(X,x_0)\to\pi_1(X,x_1)$, and furthermore, that it is a homomorphism. Now we may similarly define the map $\varphi_{\bar\alpha}:\pi_1(X,x_1)\to\pi_1(X,x_0)$ by $\varphi(g) = \alpha\cdot g\cdot\bar\alpha$. One can now easily verify that $\varphi_{\bar\alpha}$ is the inverse of $\varphi_\alpha$. Thus $\varphi_\alpha$ is an isomorphism, so $\pi_1(X,x_0)\cong \pi_1(X,x_1)$.

Therefore (up to isomorphism), the group $\pi_1(X,x_0)$ is independent of the choice of $x_0$. For this reason, we often just write $\pi_1(X)$ for the fundamental of $X$.

Invalid username
Login to AoPS