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What is the genus of the complete $4-$partite graph $K_{4,2,2,2}$?

What i know: Since $K_{4,4,2}$ is a subgraph of $K_{4,2,2,2}$, and genus of $K_{4,4,2}$ is 2, $K_{4,2,2,2}$ has genus greater than or equal to 2. Also $K_{4,2,2,2}$ is a subgraph of $K_{10}$, so it genus is less than or equal to 4. My conjecture is that it is 2, so i try to embed on double torus, but i failed.

I have tried to find the genus even by using sage. But my computer is slow. Any one could help me to find the genus using sage programming. One could find the genus using the following program.

sage: g = Graph([(0,4),(0,5),(0,6),(0,7),(0,8),(0,9),(1,4),(1,5),(1,6),(1,7),(1,8),(1,9),(2,4),(2,5),(2,6),(2,7),(2,8),(2,9),(3,4),(3,5),(3,6),(3,7),(3,8),(3,9),(4,6),(4,7),(4,8),(4,9),(5,6),(5,7),(5,8),(5,9),(6,8),(6,9),(7,8),(7,9)])

sage: g.genus()

Thanks in advance.

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1 Answers1

2

Here is a proof only using plain Python that the genus of $K_{4,2,2,2}$ is $3$. We use the following adjacency list representation of the graph, where the partition is $\{0,1,2,3\},\{4,5\},\{6,7\},\{8,9\}$:

0: 4 5 6 7 8 9
1: 4 5 6 7 8 9
2: 4 5 6 7 8 9
3: 4 5 6 7 8 9
4: 0 1 2 3 6 7 8 9
5: 0 1 2 3 6 7 8 9
6: 0 1 2 3 4 5 8 9
7: 0 1 2 3 4 5 8 9
8: 0 1 2 3 4 5 6 7
9: 0 1 2 3 4 5 6 7

Suppose the graph has genus $2$, then by Euler's formula it would triangulate the surface of genus $2$ when embedded. The rotation system corresponding to such an embedding – which is like the above adjacency list but with the neighbour list for each vertex ordered as a cyclic word – must then satisfy rule R*:

If $vwx$ appears in the neighbour list for $u$, $xuv$ appears in the neighbour list for $w$.

Consider the neighbour lists for vertices $0$ to $3$. Clearly $4$ cannot follow $5$ in any of these lists since $4$ and $5$ are not adjacent, and the followers of $4$ across the lists must all be distinct by rule R*. Similar logic applies for vertices $5$ to $9$, leading to a very strong restriction on what the first four lists can be: they are directed Hamiltonian cycles that together use all $24$ directed arcs of the octahedron formed by vertices $4$ to $9$ exactly once.

The following Python code determines all admissible neighbour lists up to vertex relabelling:

#!/usr/bin/env python3
from itertools import permutations, combinations, product
from functools import reduce
from collections import Counter

cycles = [s[:-1] for p in permutations("56789") if all(e not in (s := "4"+"".join(p)+"4") for e in ["45", "54", "67", "76", "89", "98"])]
d = {c: [c[i-1]+c[i] for i in range(6)] for c in cycles}
cands = {cmb for cmb in combinations(d, 4) if len(Counter([e for c in cmb for e in d[c]])) == 24}


def r4(s):
    return s[(i := s.index("4")):] + s[:i]
def permute_cmb(cmb, perm):
    return tuple(sorted(r4(c.translate(str.maketrans("456789", perm))) for c in cmb))
def cform_under_perms(cmb, gens, ords):
    fchains = [[gen for (gen, o) in zip(gens, exponents) for _ in range(o)] for exponents in product(*(range(o) for o in ords))]
    return min(reduce(permute_cmb, fchain, cmb) for fchain in fchains)


cands = {cform_under_perms(cmb, ["894567", "458967", "546789", "457689", "456798"], [3, 2, 2, 2, 2]) for cmb in cands}
print(cands)

We find that there are exactly two: $$\begin{array}{c|cccccc} 0&4&6&5&8&7&9\\ 1&4&7&5&9&6&8\\ 2&4&8&5&6&9&7\\ 3&4&9&5&7&8&6 \end{array}\qquad\begin{array}{c|cccccc} 0&4&6&5&8&7&9\\ 1&4&7&5&9&6&8\\ 2&4&8&6&9&5&7\\ 3&4&9&7&8&5&6 \end{array}$$ In either case, however, rule R* forces vertex $4$'s neighbour list to have the contiguous sublists $609,718,827,936$, which cannot be overlapped to form a single cycle of length $8$ as the embedding requires. We arrive at a contradiction, so the genus of $K_{4,2,2,2}$ is at least $3$.

For the upper bound, note that $K_{4,2,2,2}$ is a subgraph of $K_{10}-K_3$, which has a triangular embedding into the genus-$3$ surface:

0: 1 7 6 2 8 5 4 9 3
1: 2 7 0 3 8 6 5 9 4
2: 3 7 1 4 8 0 6 9 5
3: 4 7 2 5 8 1 0 9 6
4: 5 7 3 6 8 2 1 9 0
5: 6 7 4 0 8 3 2 9 1
6: 0 7 5 1 8 4 3 9 2
7: 0 1 2 3 4 5 6
8: 0 2 4 6 1 3 5
9: 0 4 1 5 2 6 3

So $K_{4,2,2,2}$'s genus is at most $3$, and thus exactly $3$.

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