TITLE: Confusion about centraliser of $D_5$ in $GL_4(\mathbb{Z})$
QUESTION [8 upvotes]: I am trying to follow a derivation on a very old paper. My knowledge of group theory is limited, I have the basis but not much experience with advanced concepts.
We are working in 4 dimensions, so the paper quotes the 4D representations of the generators of $D_5$, $r$ (a rotation by $\pi/5$) and $p$ (the reflection $x \rightarrow x, y \rightarrow y$) as:
$$
\mathcal{R}(r) = \left (\begin{array}{ccc}
0 & 0 & 0 & -1  \\
1 & 0 & 0 & -1  \\
0 & 1 & 0 & -1  \\
0 & 0 & 1 & -1  \\
\end{array} \right ), 
$$
$$
\mathcal{R}(p) = \left (\begin{array}{ccc}
0 & 0 & 0 & 1  \\
0 & 0 & 1 & 0  \\
0 & 1 & 0 & 0  \\
1 & 0 & 0 & 0  \\
\end{array} \right ).
$$
Question 1: I thought all dihedral groups had irreducible representations of at most 2 dimensions, hence all >2 dimensional representations should be constructed from direct sums of these (so block diagonals)? Where do they get these expressions from?  
Then we calculate the centraliser of $D_5$ in $GL_4(\mathbb{Z})$, i.e. the largest subgroup of $GL_4(\mathbb{Z})$ which commutes with $D_5$. Unfortunately they skip directly to the answer because apparently "computing these groups in straightforward".  The generators of the centraliser $C$ are given by:
$$
\mathcal{R}(\delta) = \left (\begin{array}{ccc}
-1 & 1 & 0 & -1  \\
0 & 0 & 1 & -1  \\
-1 & 1 & 0 & 0  \\
-1 & 0 & 1 & -1  \\
\end{array} \right ), 
$$
$$
\mathcal{R}(\tau) = \left (\begin{array}{ccc}
-1 & 0 & 0 & 0  \\
0 & -1 & 0 & 0  \\
0 & 0 & -1 & 0  \\
0 & 0 & 0 & -1  \\
\end{array} \right ).
$$
Question 2: where do these come from? Is it there something very trivial that I'm not getting? Where would I even start?
My attempts:  
1)I started this by building a 4D representation of $D_5$ just by direct summing its 2D irreps. I am using $r_{2D}$ as the 2D rotation matrix and $p_{2D} = \left (\begin{array}{cc}
1 & 0   \\
0 & -1   \\
\end{array} \right ), 
$, to then make $
\mathcal{R}(r_{4D}) = \left (\begin{array}{cc}
r_{2D} & 0  \\
0 & r_{2D}  \\
\end{array} \right ), 
$
$
\mathcal{R}(p_{4D}) = \left (\begin{array}{ccc}
1 & 0 & 0 & 0  \\
0 & -1 & 0 & 0  \\
0 & 0 & 1 & 0  \\
0 & 0 & 0 & -1  \\
\end{array} \right ),
$
etc.
2) I used the generic element of $GL_4(\mathbb{Z})$ as $
\beta = \left (\begin{array}{ccc}
a & b & c & d  \\
e & f & g & h  \\
i & l & m & n  \\
q & s & t & u  \\
\end{array} \right ). 
$ 
Then I brute-forced computed the relationships between the entries by requiring $h$ to commute with every element $d$ of $D_5$: $h^{-1}dh = d$.  I found that the matrix above is reduced to $\beta = \left (\begin{array}{ccc}
a & 0 & c & 0  \\
0 & a & 0 & c  \\
s & 0 & u & 0  \\
0 & s & 0 & u  \\
\end{array} \right ). $ However this is where I stop, for I don't know how to compute the generator of the group of these matrices.

REPLY [2 votes]: $\def\glz{\operatorname{GL}_4(\mathbb Z)}$
$\def\inv{^{-1}}$
I'm posting some further observations, as if my previous gigantic post wasn't enough.
We have specific matrices $R  = \mathcal R(r)$ and $J = \mathcal R(p)$ in $\glz$ providing a representation of the dihedral group $D_5$.   It is required to find the centralizer of $D_5$ in $\glz$
We also have the matrix $M = R  + R\inv$, denoted $\mathcal R(\delta)$ in the original post, which is an element of $\glz$ commuting with $D_5$.
The first step is to find the commutant of $D_5$, that is the set of all matrices commuting with all elements of $D_5$.
Your original brute force method for computing the commutant of $R$ works fine.  I asked Mathematica to do it, and was told the solutions are of the form
$$\left(
\begin{array}{cccc}
 a & b & 0 & -b \\
 0 & a+b & b & -b \\
 -b & b & a+b & 0 \\
 -b & 0 & b & a \\
\end{array}
\right)$$
which by examination is 
$b M + (a + b) 1$.    So this gives back the solution for the commutant, that it is the set of linear combinations of $M$ and the identity, without any of the "fancy" arguments from my previous answer.
The determinant of $ a M + b 1$  is  $D(a, b) = (a^2 + ab - b^2)^2$.   So we have a Diophantine  problem to determine the integral solutions of $D(a, b) = 1$.  
I'm not going to provide a simpler solution to this problem than already appeared in my other answer (4th postlude), but I want to point out that  Fibonacci numbers pervade this problem.  
In fact, we can first observe that the sequence of powers of $M$ is a matrix Fibonacci sequence in the following sense.   Using the minimal polynomial of $M$, namely $x^2 + x - 1$ repeatedly, we find that, with $(f_k)_{k \ge 0}$ denoting the Fibonacci sequence,
$$
(-M)^k =  -f_k M + f_{k-1}  \quad \text{and} \quad M^{-k} = f_k M + f_{k+1} 1,
$$
for $k \ge 1$.
So what we have to show is that an integer pair $(a, b)$ satisfies $D(a, b) = 1$ if and only if 
$$(a, b) =  \pm (-f_k, f_{k-1})  \quad \text{or} \quad (a, b) = \pm(f_k, f_{k+1}),
$$
for some $k$.   (I am extending the Fibonacci sequence by taking $f_0 = 0$.)
This requires more or less the same sort of work as was contained in my previous 4th postlude, and I am going to skip the details.  (This is known: Jones, James P.
Diophantine representation of the Fibonacci numbers. 
Fibonacci Quart. 13 (1975), 84–88. )
In any case the OP is interested in generalizing these results to $D_n$, $n >5$, and all of these Fibonacci phenomena are very much tied to the case of $D_5$. 
Edit:  I am adding this in response to some questions posed to me by the OP.
$\def\glz{\operatorname{GL}_4(\mathbb Z)}
\def\inv{^{-1}}
\def\boldG{\mathbf G}
\def\boldn{\mathbf n}
\def\Z{\mathbb Z}
\def\R{\mathbb R}
\def\boldq{\mathbf q}
$
Let me start over with a few things. 
The starting point in the paper is the set of points $\boldG_j  = (\cos(2 \pi  j/ 5), \sin(2 \pi  j/5))$,  $0 \le j \le 4$ in the real vector space $\mathbb R^2$.  (It is also convenient to take the indices as representing residue classes mod 5.)   These $\boldG_j$ satisfy $\sum_{j = 0}^4 \boldG_j = 0$, but
$\{\boldG_j :  1 \le j \le 4\}$ is linearly independent over $\mathbb Z$, while of course linearly dependent over $\mathbb R$.  So the set $P$ of integer linear combinations of $\{\boldG_j :  1 \le j \le 4\}$ is a complicated configuration of points in the plane,  but is also   a free $\mathbb Z$--module of rank $4$.  The dihedral group $D_5$, generated by the geometric rotation $\tau$  by angle $2 \pi/5$ in the plane and the reflection through the $x$-axis $\sigma$ leaves invariant the set of 
$\{\boldG_j :  0 \le j \le 4\}$.  The group $D_5$ acts on $\mathbb R^2$ by real linear automorphisms,  and hence acts on $P$ by $\mathbb Z$--linear automorphisms.  The matrices of the generators  $\tau$, $\sigma$  with respect to the $\Z$--basis $\{\boldG_j :  1 \le j \le 4\}$ of $P$  are the matrices $M(\tau)$ and $M(\sigma)$ given in the paper. 
We have a $\Z$--module isomorphism  $\boldq : \Z^4 \to P$, $\boldq(\boldn) = \sum_{j = 1}^4 n_j \boldG_j$. This gives us a representation of $D_5$ on $\Z^4$ with the same representation matrices.
We can regard this representation as a representation on $\mathbb R^4$ or on $\mathbb C^4$, or whatever, by using the same representation matrices.
There is also an element $\delta = \tau + \tau\inv$. (It lives in the the group ring consisting of formal linear combinations of the group elements of $D^5$.).  It acts on $P$ or on $\Z^4$ by the matrix
$M(\delta) = M(\tau) + M(\tau)\inv$.  
All of these transformations $\tau, \sigma, \delta$ will act via the same matrices, either on $P$, with respect to the basis $\{\boldG_j :  1 \le j \le 4\}$ or on $\Z^4$ with respect to the standard basis.