# Constructing a Hausdorff space with a given fundamental group via a matrix space

Consider the topological $$\mathbb{R}$$-vector space $$V$$ of $$2 \times \infty$$ matrices $$A$$ of the form $$A=\left [ \begin{matrix} a_1 & a_2 & a_3 & \cdots \\ b_1 & b_2 & b_3 & \cdots \\ \end{matrix} \right ]$$

where all but finitely many entries are zero. The topology is induced by a metric $$d$$ defined by $$d(A,B)=\max_{ij}(|A_{ij}-B_{ij}|)$$.

We consider in particular the subspace (topological subspace, not a vector subspace) $$U$$ of $$V$$ consisting of matrices whose columns span $$\mathbb{R}^2$$.

We want to prove two things. First, that $$U$$ is contractible (as a topological space). Second, and this is really the hard part, we want to use $$U$$ to construct, for any finite subgroup $$G$$ of $$GL_{2}(\mathbb{R})$$, a Hausdorff space $$X_{G}$$ whose fundamental group is $$G$$.

I'm already stuck on contractibility. A very naive approach is trying to contract things to $$0$$, but this doesn't work since the matrix with all entries $$0$$ does not have columns that span $$\mathbb{R}^2$$.

• Try to contract to the matrix that starts with the $2 \times 2$ identity matrix and is zeroes afterwards. That at least is actually a matrix in $U$. Dec 7, 2020 at 4:55
• For the second question I assume we take the standard action of $G$ on $U$ and consider the orbit space $U/G$. There are some theorems about the fundamental group of an orbit space, but I'm not sure about the details. Dec 7, 2020 at 8:26
• Here's an idea for the first question. Not sure if it works though. Try to make the identity homotopic to the shift map (i.e. insert zero column at the begining of $A$). Just linear homotopy. Hopefuly it works. With that the identity is homotopic to double shift map. And then you have zero $2\times 2$ matrix at the begining which is homotopic to the $\left [ \begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix} \right ]$ matrix and the rest can be contracted to zero. Dec 7, 2020 at 8:51
• @freakish I agree that the orbit space approach is promising and natural, but haven't been able to put together the details. Thanks for the insightful comments! Dec 7, 2020 at 8:52
• @TheDayBeforeDawn the second question can be answered by Allen Hatcher's "Algebraic Topology" Proposition 1.40. Since $G$ acts on $U$ via covering space action, then $\pi_1(U/G)$ is isomorphic to $G$ if contractible, path connected and locally path connected. Dec 7, 2020 at 11:08

I will be using the notation $$(u_1, u_2,u_3, \ldots ) := \pmatrix{a_1 & a_2 & a_3 & \ldots \cr b_1 & b_2 & b_3 & \ldots }$$ where each $$u_j$$ is the $$j^{\text{th}}$$ column of the matrix on the right hand side.

Here is how to construct a homotopy from the identity map on $$U$$ to the shift map $$S:(u_1, u_2,u_3, \ldots ) \mapsto (0, u_1, u_2,\ldots ).$$

Define $$H: (t, u) \in [0,1]\times U\mapsto v\in U,$$ where $$v_j = \left\{\matrix{ (1-t)u_1, & \text{if } j=1, \cr (1-t)u_j + t u_{j-1}, & \text{if } j>1. } \right.$$

In other words, the $$1^{\text{st}}$$ coordinate of $$H(t, u)$$ consists of a slowly vanishing $$1^{\text{st}}$$ coordinate of $$u$$, while the $$j^{\text{th}}$$ coordinate of $$H(t, u)$$ runs along te segment joininig the $$j^{\text{th}}$$ and $$(j-1)^{\text{st}}$$ coordinates of $$u$$.

It is evident that $$H(0,\cdot)$$ is the identity map, that $$H(1,\cdot)$$ is the shift, so it is enough to check that $$H$$ is well defined in the sense that $$H(t,u)$$ indeed lies in $$U$$ for every $$(t,u)\in [0,1]\times U$$. In order to do this we will verify that when $$v=H(t,u),$$ the span of the columns of $$v$$, here denoted $$\text{span}(v)$$, contains $$\text{span}(u)$$.

When $$t=1$$ this is obvious, so we assume that $$t\in [0,1)$$. Since $$v_1$$ is a nonzero multiple of $$u_1$$, we have that $$u_1\in \text{span}(v)$$. By induction, assuming that $$u_{j-1}$$ lies in $$\text{span}(v)$$, notice that since $$v_j=(1-t)u_j + t u_{j-1},$$ we have that $$u_j = \frac{v_j- t u_{j-1}}{1-t} \in \text{span}(v).$$

The rest of the argument to prove that $$U$$ is contractible has been explained in the comments by @freakish, and the last part of the question follows from Allen Hatcher's "Algebraic Topology" book, also pointed out by @freakish.

Incidentally, the argument above is essentially also due to @freakish even though they claim in a comment that it does not work.