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K2,2

According to Ringel, the optimal genus for $K_{2,2}$ is $g=0$ . We were able to obtain the same genus using a Cayley Map with group $Z_4$ and rotation $\rho=(1$ $3).$

C_M(Z_4,(1 — $C_M(Z_4,(1$ $3))$
The face set is $\mathcal{F}=\{(3),(1)\}$ , where $F_0$ and $F_1$ are both 4-gons. Using the Euler Characteristic Formula, $\chi=2$ , so $g=0$ .

Cayley Map $C_M(Z_4,(1$ $3))$ embedded into a sphere ( $g=0$ ). The blue square uses permutation $\lambda_0=(3)$ and forms the top half of the sphere, and the yellow square uses permutation $\lambda_1=(1)$ and forms the bottom half.

C_M(Z_4,(1 — Cayley Map $C_M(Z_4,(1$ $3))$ embedded into a sphere ( $g=0$ ). The blue square uses permutation $\lambda_0=(3)$ and forms the top half of the sphere, and the yellow square uses permutation $\lambda_1=(1)$ and forms the bottom half.

Using the Dihedral Group $D_2$ yielded the same genus, but with one distinct face appearing twice. The reason for this difference is that each element in the rotation using $Z_4$ is followed by its inverse, whereas in the rotation for $D_2$ , each element is its own inverse.

Cayley Map $C_M(D_2,(f$ \hspace{1} $rf))$ .
The face set is $\mathcal{F}=\{(f$ \hspace{1} $rf)\}$ , where $F_0$ is two 4-gons. By the Euler Characteristic Formula, $\chi=2$ , so $g=0$ .

C_M(D_2,(f — Cayley Map $C_M(D_2,(f$ \hspace{1} $rf))$ .
The face set is $\mathcal{F}=\{(f$ \hspace{1} $rf)\}$ , where $F_0$ is two 4-gons. By the Euler Characteristic Formula, $\chi=2$ , so $g=0$ .

Cayley Map $C_M(D_2,(f$ $rf))$ embedded into a sphere ( $g=0$ ). The blue and green squares use permutation $(f$ $rf)$ , and each face forms half the sphere.

C_M(D_2,(f — Cayley Map $C_M(D_2,(f$ $rf))$ embedded into a sphere ( $g=0$ ). The blue and green squares use permutation $(f$ $rf)$ , and each face forms half the sphere.