# 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.

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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 $c_0$.