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Victor Protsak
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Looking at the images below, you recognize that the adjacency matrix of the graph $A_G$ splits up into three different colored submatrices, with $A_G=A_d+A_b+A_d$$A_G=A_r+A_b+A_d$ (where $d$ is dark, damn...). It's obvious that $A_k^2=1$. Now have a look a the right coloring: You'll see that $(A_dA_r)^2=(A_rA_b)^2=(A_bA_d)^2=1$ as well. Let's use $a,b,c$ instead. We summarize this as: $$ \langle a,b,c|a^2=b^2=c^2=(ab)^2=(bc)^2=(ca)^2=1\rangle =\Delta(2,2,2) $$ This is a presentation of a triangle group $\Delta(2,2,2)$, a special kind of Coxeter group. The left image resembles the situation: enter image description here

Now the right images deviates from the left at the inner square, where blue and dark are flipped; still a valid 3-edge coloring. It has Hamiltonian cycles (follow dark-red or red-blue edges), therefore its presentation is: $$ \langle a,b,c|a^2=b^2=c^2=(ab)^4=(bc)^4=(ca)^2=1 \rangle =\Delta(4,4,2) $$

So by flipping the inner cycle, we map $\Delta(2,2,2)\mapsto \Delta(4,4,2)$.

EDIT: This means that we change the classification from $\frac12+\frac12+\frac12=\frac32>1$, i.e. spherical, to $\frac14+\frac14+\frac12=1$, i.e. Euclidean.

Which branch of math deals with this mapping? What are good introductions to it?

Let's restrict to planar, bicubic graphs...

Looking at the images below, you recognize that the adjacency matrix of the graph $A_G$ splits up into three different colored submatrices, with $A_G=A_d+A_b+A_d$ (where $d$ is dark, damn...). It's obvious that $A_k^2=1$. Now have a look a the right coloring: You'll see that $(A_dA_r)^2=(A_rA_b)^2=(A_bA_d)^2=1$ as well. Let's use $a,b,c$ instead. We summarize this as: $$ \langle a,b,c|a^2=b^2=c^2=(ab)^2=(bc)^2=(ca)^2=1\rangle =\Delta(2,2,2) $$ This is a presentation of a triangle group $\Delta(2,2,2)$, a special kind of Coxeter group. The left image resembles the situation: enter image description here

Now the right images deviates from the left at the inner square, where blue and dark are flipped; still a valid 3-edge coloring. It has Hamiltonian cycles (follow dark-red or red-blue edges), therefore its presentation is: $$ \langle a,b,c|a^2=b^2=c^2=(ab)^4=(bc)^4=(ca)^2=1 \rangle =\Delta(4,4,2) $$

So by flipping the inner cycle, we map $\Delta(2,2,2)\mapsto \Delta(4,4,2)$.

EDIT: This means that we change the classification from $\frac12+\frac12+\frac12=\frac32>1$, i.e. spherical, to $\frac14+\frac14+\frac12=1$, i.e. Euclidean.

Which branch of math deals with this mapping? What are good introductions to it?

Let's restrict to planar, bicubic graphs...

Looking at the images below, you recognize that the adjacency matrix of the graph $A_G$ splits up into three different colored submatrices, with $A_G=A_r+A_b+A_d$ (where $d$ is dark, damn...). It's obvious that $A_k^2=1$. Now have a look a the right coloring: You'll see that $(A_dA_r)^2=(A_rA_b)^2=(A_bA_d)^2=1$ as well. Let's use $a,b,c$ instead. We summarize this as: $$ \langle a,b,c|a^2=b^2=c^2=(ab)^2=(bc)^2=(ca)^2=1\rangle =\Delta(2,2,2) $$ This is a presentation of a triangle group $\Delta(2,2,2)$, a special kind of Coxeter group. The left image resembles the situation: enter image description here

Now the right images deviates from the left at the inner square, where blue and dark are flipped; still a valid 3-edge coloring. It has Hamiltonian cycles (follow dark-red or red-blue edges), therefore its presentation is: $$ \langle a,b,c|a^2=b^2=c^2=(ab)^4=(bc)^4=(ca)^2=1 \rangle =\Delta(4,4,2) $$

So by flipping the inner cycle, we map $\Delta(2,2,2)\mapsto \Delta(4,4,2)$.

EDIT: This means that we change the classification from $\frac12+\frac12+\frac12=\frac32>1$, i.e. spherical, to $\frac14+\frac14+\frac12=1$, i.e. Euclidean.

Which branch of math deals with this mapping? What are good introductions to it?

Let's restrict to planar, bicubic graphs...

Looking at the images below, you recognize that the adajencyadjacency matrix of the graph $A_G$ splits up into three different colorcolored submatrices, with $A_G=A_d+A_b+A_d$ (where $d$ is dark, damn...). It's obvious that $A_k^2=1$. Now have a look a the right coloring: You'll see that $(A_dA_r)^2=(A_rA_b)^2=(A_bA_d)^2=1$ as well. Let's use $a,b,c$ instead. We summarize this as: $$ \langle a,b,c|a^2=b^2=c^2=(ab)^2=(bc)^2=(ca)^2=1\rangle =\Delta(2,2,2) $$ This is a presentation of a triangle group $\Delta(2,2,2)$, a special kind of Coxeter group. The left image resembles the situation:

$\hskip0.5in$  enter image description here

Now the right images deviates from the left at the inner square, where blue and dark are flipped; still a valid 3-edge coloring. It has hamiltonianHamiltonian cycles (follow dark-red or red-blue edges), therefore its presentation is: $$ \langle a,b,c|a^2=b^2=c^2=(ab)^4=(bc)^4=(ca)^2=1 \rangle =\Delta(4,4,2) $$

So by flipping the inner cycle, we map $\Delta(2,2,2)\mapsto \Delta(4,4,2)$.

EDIT: This means that we change the classification from $\frac12+\frac12+\frac12=\frac32>1$, i.e. spherical, to $\frac14+\frac14+\frac12=1$, i.e. euclideanEuclidean.

Which branch of math deals with this mapping? What are good introductionalsintroductions to it?

Let's restrict to planar, bicubic graphs...

Looking at the images below, you recognize that the adajency matrix of the graph $A_G$ splits up into three different color submatrices, with $A_G=A_d+A_b+A_d$ (where $d$ is dark, damn...). It's obvious that $A_k^2=1$. Now have a look a the right coloring: You'll see that $(A_dA_r)^2=(A_rA_b)^2=(A_bA_d)^2=1$ as well. Let's use $a,b,c$ instead. We summarize this as: $$ \langle a,b,c|a^2=b^2=c^2=(ab)^2=(bc)^2=(ca)^2=1\rangle =\Delta(2,2,2) $$ This is a presentation of a triangle group $\Delta(2,2,2)$, a special kind of Coxeter group. The left image resembles the situation:

$\hskip0.5in$enter image description here

Now the right images deviates from the left at the inner square, where blue and dark are flipped; still a valid 3-edge coloring. It has hamiltonian cycles (follow dark-red or red-blue edges), therefore its presentation is: $$ \langle a,b,c|a^2=b^2=c^2=(ab)^4=(bc)^4=(ca)^2=1 \rangle =\Delta(4,4,2) $$

So by flipping the inner cycle, we map $\Delta(2,2,2)\mapsto \Delta(4,4,2)$.

EDIT: This means that we change the classification from $\frac12+\frac12+\frac12=\frac32>1$, i.e. spherical, to $\frac14+\frac14+\frac12=1$, i.e. euclidean.

Which branch of math deals with this mapping? What are good introductionals to it?

Let's restrict to planar, bicubic graphs...

Looking at the images below, you recognize that the adjacency matrix of the graph $A_G$ splits up into three different colored submatrices, with $A_G=A_d+A_b+A_d$ (where $d$ is dark, damn...). It's obvious that $A_k^2=1$. Now have a look a the right coloring: You'll see that $(A_dA_r)^2=(A_rA_b)^2=(A_bA_d)^2=1$ as well. Let's use $a,b,c$ instead. We summarize this as: $$ \langle a,b,c|a^2=b^2=c^2=(ab)^2=(bc)^2=(ca)^2=1\rangle =\Delta(2,2,2) $$ This is a presentation of a triangle group $\Delta(2,2,2)$, a special kind of Coxeter group. The left image resembles the situation:  enter image description here

Now the right images deviates from the left at the inner square, where blue and dark are flipped; still a valid 3-edge coloring. It has Hamiltonian cycles (follow dark-red or red-blue edges), therefore its presentation is: $$ \langle a,b,c|a^2=b^2=c^2=(ab)^4=(bc)^4=(ca)^2=1 \rangle =\Delta(4,4,2) $$

So by flipping the inner cycle, we map $\Delta(2,2,2)\mapsto \Delta(4,4,2)$.

EDIT: This means that we change the classification from $\frac12+\frac12+\frac12=\frac32>1$, i.e. spherical, to $\frac14+\frac14+\frac12=1$, i.e. Euclidean.

Which branch of math deals with this mapping? What are good introductions to it?

Let's restrict to planar, bicubic graphs...

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