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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 $\mathbb{R}^2$.

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  • $\begingroup$ 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 $U$. $\endgroup$ Dec 7, 2020 at 4:55
  • $\begingroup$ 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. $\endgroup$
    – freakish
    Dec 7, 2020 at 8:26
  • $\begingroup$ 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. $\endgroup$
    – freakish
    Dec 7, 2020 at 8:51
  • $\begingroup$ @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! $\endgroup$ Dec 7, 2020 at 8:52
  • $\begingroup$ @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. $\endgroup$
    – freakish
    Dec 7, 2020 at 11:08

1 Answer 1

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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.

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