Example of a space that has fundamental group with a specific representation

Question

I'm trying to solve a specific task of finding a space with $\pi_1(X) = <a,b,c | a^2cb, abc>$, but I'm interested if there is a way to think about this in general. I understand fairly intuitively the definition of a fundamental group and what a representation like this means algebraically. I'm completely lost when it comes to the geometry though and how one might go about constructing such a space. I'd be grateful if someone could do this or a similar example, or for any tips on the intuition behind such constructions.

Answer 1

If $\pi_1(X)$ is generated by $x_1,...,x_n$ and relations $R_1,....,R_m$. You can construct such a space as follows: consider a bouquet of $n$-circles, $1$ for each generator. For every relation you consider a $2$-disc that you circle the boundary along the circle to represent the relation.

Forexample for your case, you are going to construct a bouquet of three circles $a,b,c$ two discs $R_1,R_2$ you circle the boundary of $R_1$ twice qround $a$, once around $c$ and once around $b$. You circle the boundary of $R_2$ once around $a$, once around $b$ and once around $c$.

Answer 2

Any group which is finitely-presented, i.e. of the form $$ G = \langle S \mid R\rangle$$ for $S$ and $R$ finite, is the fundamental group of a $2$-complex.

To construct such a space, we start by taking the wedge product of $|S|$-many circles, say labelled by the generators (in your case, $a,b,c$) with a specific direction (say, "counter-clockwise"). This wedge will have fundamental group $\langle S \mid \emptyset \rangle$.

To get a relation $x_1 \cdots x_n =1$ where $x_k \in S^{\pm 1}$, we paste along the boundaries of these circles a disk whose boundary has been split up into $n$ segments corresponding to the letters $x_1,\ldots, x_n$ and glued along the corresponding circle (so an $s$ component of the boundary gets glued onto the $s$-labelled circle with the same orientation, while an $s^{-1}$ component of the boundary gets glued onto the $s$-labelled circle with the opposite orientation). The resulting fundamental group will be $\langle S \mid x_1\cdots x_n\rangle$.

We repeat this construction until we've added all of the relations. In your case, we glue first a square (with interior) whose sides going counter-clockwise around are labelled $a,a,c,b$ and then glue a triangle (with interior) whose sides going counter-clockwise around are labelled $a,b,c$.

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To show that this actually gives the proper fundamental group rigorously, Van Kampen's Theorem must be used.

First we show by induction that $\langle S \mid \emptyset\rangle$ is the fundamental group of a wedge of $|S|$-many circles. Let $|S|=n$. If $n=1$, then this is simply the fact that $\mathbb{Z} \cong \langle s_1 \rangle \cong \pi_1(S^1)$. Now suppose the result is true for $|S|=n$, and consider $S\sqcup \{s_{n+1}\}$. Consider the wedge $(S^1)^{\wedge (n+1)} = (S^1)^{\wedge n} \cup S^1$, where $(S^1)^{\wedge n} \cap S^1 = \{x_0\}$ (the base-point). Then $\pi_1((S^1)^{\wedge n}) = \langle s_1,\ldots, s_n\rangle$ by hypothesis and $\pi_1(S^1) = \langle s_{n+1}\rangle$ and $\pi_1(\{x_0\}) = 1$. Thus, we find that \begin{equation*} \pi_1((S^1)^{\wedge (n+1)}) \cong \langle s_1,\ldots, s_n,s_{n+1} \mid \emptyset \cup \emptyset \cup \{1 =1\} \rangle = \langle s_1, \ldots, s_n,s_{n+1}\rangle \end{equation*}

Now suppose we have a $2$-complex $Y_1$ with fundamental group $\langle S \mid R\rangle$. We show how to introduce a relation $r = t_1\cdots t_k$, where $t_i = s_j^{\pm 1}$ for each $i$. Let $Y_2$ be a regular $k$-gon (say, bounded within a disc of radius $1$), and label the edges counter-clockwise by $t_1,t_2,\ldots, t_k$. Then we define \begin{equation*} X = (Y_1 \cup Y_2)/\sim \end{equation*} where $\sim$ is an equivalence relation generated by identifying the edges $t_i =s_j^{\pm 1}$ with the loop $s_j$ (with the indicated orientation).

Write $X_1$ to be the image of a $Y_1 \cup Y_2 \setminus \overline{B_d(0,1/2)}$, i.e. $Y_1$ along with a band along the edge of $Y_2$ (how thin it needs to be doesn't matter, so long as it is non-trivial) and $X_2$ to be the image $B_d(0,3/4)$. Then let $X_0=X_1 \cap X_2$, which is the image of $B_d(0,3/4) \setminus \overline{B_d(0,1/2)}$.

No non-trivial identifications are made except among the boundary of $Y_2$. Thus, we see that $X_0 = B_d(0,3/4) \setminus \overline{B_d(0,1/2)}$ is homeomorphic to an annulus (which is homotopically equivalent to $S^1$) and so has fundamental group $\pi_1(X_0) \cong \mathbb{Z}\cong \langle s\rangle$. $X_1$ is homotopically equivalent to $Y_1$ (just shrink the edges down) so has fundamental group $\pi_1(X_1) \cong \langle S \mid R\rangle$. $X_2$ is contractible, so has fundamental group $\pi_1(X_2) \cong \langle \emptyset\rangle = \{1\}$.

Then Van Kampen's Theorem implies that \begin{equation*} \pi_1(X) \cong \langle S \mid R \cup \{\theta_1(s)=\theta_2(s)\} \rangle =\langle S \mid R \cup \{t_1t_2\cdots t_k = 1\}\rangle = \langle S \mid R\cup \{r\}\rangle \end{equation*}

Answer 3

Bouquets of circles give you a supply of spaces whose fundamental groups are free groups with $1$ generator for each circle in the bouquet. E.g., in your example the bouquet of three circles $A$, $B$ and $C$ has fundamental group the free group $F = <a. b, c>$. A relation word in $F$ of length $n$ (e.g., $abc$ with $n=3$) corresponds to a continuous mapping from the edges of an $n$-gon (e.g., a triangle) into the bouquet (that starts at the common point, then goes once round $A$, once round $B$ and once round $C$). If you form a new space by stitching on an $n$-gon to the bouquet for each relation word using these mappings to identify the edges of the $n$-gon with circles in the bouquet, you get a space whose fundamental group is as desired.