Replacing entries of dice by average of it neighbours

I am interested in Representation Theory. I came across the following answer while reading this question on Mathoverflow.

An example from Kirillov’s book on representation theory: write numbers 1,2,3,4,5,6 on the faces of a cube, and keep replacing (simultaneously) each number by the average of its neighbours. Describe (approximately) the numbers on the faces after many iterations.

However I do not understand how to approach this problem via representation theory.

Solutions Collecting From Web of "Replacing entries of dice by average of it neighbours"

Let the value on each of the six sides of the dice be $x_1, x_2, … , x_6$

You can create equations for each new value. So for example, the value $x_1$ will be replaced by the average of $x_2, x_3, x_4$ and $x_5$

We can say that $x_{1,n+1} = \frac 1 4 x_{2,n} + \frac 1 4 x_{3,n}+ \frac 1 4 x_{4,n}+ \frac 1 4 x_{5,n} $

You now need to write down the other five equations.

The equation I have given you can be represented by the first row of the matrix below:

$\begin{bmatrix} x_{1,n+1}\\ x_{2,n+1} \\ x_{3,n+1}\\x_{4,n+1}\\x_{5,n+1}\\x_{6,n+1} \end{bmatrix} = \begin{bmatrix} 0 & 0.25 &0.25&0.25 & 0.25 & 0 \\ .&.&.&.&.&. \\.&.&.&.&.&.\\.&.&.&.&.&.\\.&.&.&.&.&.\\.&.&.&.&.&. \end{bmatrix}\begin{bmatrix} x_{1,n}\\ x_{2,n} \\ x_{3,n}\\x_{4,n}\\x_{5,n}\\x_{6,n} \end{bmatrix}$

You can then find out what happens to the matrix when you put it to a large power…

I don’t know how to bring representation theory into the game. But anyway: Here is a solution proposal that simplifies and generalizes this problem at the same time.

Write any real or complex numbers you like on the faces of the cube $[{-1},1]^3$. Denote the average of the two numbers on the opposite faces $x_i=\pm 1$ by $z_i$, whereby $i$ is taken modulo $3$. Already after one step the numbers on opposite faces are equal. It therefore suffices to consider the iteration map
$$T: \quad z\mapsto z’$$
defined by
$$z_i’:={1\over2}(z_{i-1}+z_{i+1})\ .$$
We now look what happens to the discrete Fourier transform of $z$ under $T$. Put $\omega:=e^{2\pi i/3}$ and
$$c_j:={1\over3}\sum_k z_k\omega^{-jk}\ .$$
Then
$$c_j’={1\over3}\sum_k z_k’\omega^{-jk}={1\over3}\sum_k {z_{k+1}+z_{k-1}\over2}\omega^{-jk}={\omega^j+\omega^{-j}\over2}\>c_j\ .$$
It follows that
$$c_0’=c_0,\qquad c_j’=-{1\over2}c_j \quad(j=\pm1)\ .$$
This shows that the $c_j$ with $j=\pm1$ (these encode the “oscillation” inherent in $z$) converge to $0$ under iteration, whereas the average $c_0={1\over3}(z_1+z_2+z_3)$ stays fixed. Therefore the limiting number on all faces of the cube is $c_0$.