Want to find the action of the group $D_4$ on the set of $2$-colorings of square, as per the source. It states, in section 4.1, on page #12.
Kindly vet the action of symmetry elements on the $2^4=16$ members of the set $X,$ and need help with doubts.
\begin{array}{c|c|c|c|c|c|c|c|c|c|c|} \text{S. No.} & e & r= (1234) & r^2= (13)(24) & r^3=(1432) & f=(12) (34)& rf= (13) & r^2f= (14)(23)& r^3f=(24) &\mathcal O & Stab(x)\\ \hline 0& 0000=0_{10} & 0000=0_{10} & 0000=0_{10} & 0000=0_{10} & 0000=0_{10} &0000=0_{10} & 0000=0_{10} & 0000=0_{10} & \{0_{10}\}=\mathcal O_0 &Stab(0000)= D_4 \\ \hline 1& 0001=1_{10} & 0010 = 2_{10} & 0100=4_{10} & 1000=8_{10} & 0010 =2_{10}& 0100 =4_{10}& 1000 = 8_{10}& 0001= 1_{10} & \{1,2,4,8\}=\mathcal O_1&Stab(0001)= \{e,r^3f\} \\ \hline 2& 0010= 2_{10} & 0100 = 4_{10} & 1000=8_{10} & 0001=1_{10} & 0001 =1_{10}& 0010 =2_{10}& 0100 = 4_{10}& 1000= 8_{10} & \{1,2,4,8\}=\mathcal O_1&Stab(0010)=\{e, rf\} \\ \hline 3& 0011= 3_{10} & 0110 = 6_{10} & 1100=12_{10} & 1001=9_{10} & 0011 =3_{10}& 0110 =6_{10}& 1100 = 12_{10}& 1001= 9_{10} & \{3,6,9,12\} =\mathcal O_2 &Stab(0011)=\{e, f\} \\ \hline 4 & 0100=4_{10} & 1000=8_{10} & 0001=1_{10} & 0010=2_{10} & 1000=8_{10}& 0001 =1_{10}& 0010 = 2_{10}& 0100= 4_{10} & \{1,2,4,8\} =\mathcal O_1 &Stab(0100)=\{e, r^3f\} \\ \hline 5 & 0101=5_{10} & 1010=10_{10} & 0101=5_{10} & 1010=10_{10} & 1010=10_{10}& 0101 =5_{10}& 1010 = 10_{10}& 0101= 5_{10} & \{5,10\} =\mathcal O_3 &Stab(0101)=\{e, r^2, rf, r^3f\} \\ \hline 6 & 0110=6_{10} & 1100=12_{10} & 1001=9_{10} & 0011=3_{10} & 1001=9_{10}& 0011 =3_{10}& 0110 = 6_{10}& 1100= 12_{10} & \{3,6,9,12\} =\mathcal O_2&Stab(0110)=\{e, r^2f\} \\ \hline 7 & 0111=7_{10} & 1110=14_{10} & 1101=13_{10} & 1011=11_{10} & 1011=11_{10}& 0111= 7_{10}& 1110 = 14_{10}& 1101= 13_{10} & \{7,11,13,14\} =\mathcal O_4&Stab(0111)=\{e, rf\} \\ \hline 8 & 1000=8_{10} & 0001=1_{10} & 0010=2_{10} & 0100=4_{10} & 0100=4_{10}& 1000= 8_{10}& 0001 = 1_{10}& 0010= 2_{10} & \{1,2,4,8\} =\mathcal O_1&Stab(1000)=\{e, rf\} \\ \hline 9 & 1001=9_{10} & 0011=3_{10} & 0110=6_{10} & 1100=12_{10} & 0110=6_{10}& 1100= 12_{10}& 1001 = 9_{10}& 0011= 3_{10} & \{3,6,9,12\} =\mathcal O_2&Stab(1001)=\{e, r^2f\} \\ \hline 10 & 1010=10_{10} & 0101=5_{10} & 1010=10_{10} & 0101=5_{10} & 0101=5_{10}& 1010= 10_{10}& 0101 = 5_{10}& 1010= 10_{10} & \{5,10\} =\mathcal O_3&Stab(1010)=\{e, r^2,rf,r^3f\} \\ \hline 11 & 1011=11_{10} & 0111=7_{10} & 1110=14_{10} & 1101=13_{10} & 0111=7_{10}& 1110= 14_{10}& 1101 = 13_{10}& 1011= 11_{10} & \{7,11,13,14\} =\mathcal O_4&Stab(1011)=\{e, r^3f\} \\ \hline 12 & 1100=12_{10} & 1001=9_{10} & 0011=3_{10} & 0110=6_{10} & 1100=12_{10}& 1001= 9_{10}& 0011 = 3_{10}& 0110= 6_{10} & \{3,6,9,12\} =\mathcal O_2&Stab(1100)=\{e, f\} \\ \hline 13 & 1101=13_{10} & 1011=11_{10} & 0111=7_{10} & 1110=14_{10} & 1110=14_{10}& 1101= 13_{10}& 1011 = 11_{10}& 0111= 7_{10} & \{7,11,13,14\} =\mathcal O_4&Stab(1101)=\{e, rf\}\\ \hline 14 & 1110=14_{10} & 1101=13_{10} & 1011=11_{10} & 0111=7_{10} & 1101=13_{10}& 1011= 11_{10}& 0111 = 7_{10}& 1110= 14_{10} & \{7,11,13,14\} =\mathcal O_4&Stab(1110)=\{e, r^3f\} \\ \hline 15 & 1111=15_{10} & 1111=15_{10} & 1111=15_{10} & 1111=15_{10} & 1111=15_{10}& 1111=15_{10}& 1111=15_{10}& 1111=15_{10} & \{15\} =\mathcal O_5&Stab(1111)=D_4\\ \hline Fix(g) & X & \{0,15\} & \{0,5,10,15\} &\{0,15\}& \{0,3,12,15\} & \{0,2,5,7,8,10,13,15\}&\{0,6,9,15\}&\{0,1,4,5,10,11,14,15\}& & \\ \end{array}
Hence, a total of six orbits are there. We have $\sum_{x\in X}|\mathrm{Stab}(x)|=8+2\cdot 12+4\cdot 2+8 = 48.$
To compute the fixed points, need take $g$ as constant, while $x$ varies. $sum= 16+2+4+2+4+8+4+8= 16+2\cdot 2+3\cdot 4+2\cdot 8= 16+4+12+16=48.$
Also, on page #13, the link states the same values of fixed points.
Doubts:
#1. Why at the bottom of page #12, it states that the possible actions of $(12)(34),$ and $(14)(23)$ on the set $X$ of $2$-colorings of square $=4$ each.
Or, it could be found that what is the relevance of the same.
At the top of page #14, the author states:
The six colorings:
- $0000,$ all of color $X,$
- $1111,$ all of color $Y,$
- $0101=5, 1010=10,$
- $0011=3, 1100=12,0110=6$
- $0111=7, 1011=11, 1101=13, 1110=15,$
- $0001=1, 0010=2, 0100=4, 1000=8,$
But, that does mean that he equates colorings with orbits.
But, the coloring produced by the symmetry $(12)(34)$ are: $1,2,4,5,6,7,8,10,11,13.$ The rest four colorings are fixed.
Just stating that the colorings produced are (i.e, outside the fixed set) lying in the four classes: #3, #4, #5, #6; doesn't ring any significance still to me.
#2. Why each column has each member of the set $X,$ occuring once only; while the row can have repetition?
The images for the above link's relevant page #12, #13, #14 are attached.
