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I am wondering if there is a known example of a pair of non-isomorphic graphs $G$ and $H$ that are both Cayley graphs for $\mathbb{Z}_2^n$ (for some $n$) and are both distance regular and have the same intersection array. Or is it known that this is not possible?

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1 Answer 1

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The answer is yes.

There are many difference sets on the group $\mathbb Z_2^6$, which can be found in the La Jolla repository.

Below is the SageMath code to generate two nonisomorphic $\mathbb Z_2^6$-cayley graphs:

a1=[(0,0,0,0,0,0),(1,0,0,0,0,0),(0,1,0,0,0,0),(0,0,1,0,0,0),(0,0,0,1,0,0),(0,0,0,0,1,0),(0,0,0,0,0,1),(1,1,0,0,0,0),(1,0,1,0,0,0),(1,0,0,1,0,0),(1,0,0,0,1,0),(0,1,0,0,0,1),(0,0,1,0,0,1),(0,0,0,1,0,1),(0,0,0,0,1,1),(1,1,1,0,0,1),(1,1,0,1,0,1),(1,1,0,0,1,1),(1,0,1,1,0,1),(1,0,1,0,1,1),(1,0,0,1,1,1),(0,1,1,1,1,0),(1,1,1,1,1,0),(1,1,1,1,0,1),(1,1,1,0,1,1),(1,1,0,1,1,1),(1,0,1,1,1,1),(0,1,1,1,1,1)];

a2=[(0,0,0,0,0,0),(1,0,0,0,0,0),(0,1,0,0,0,0),(0,0,1,0,0,0),(0,0,0,1,0,0),(0,0,0,0,1,0),(0,0,0,0,0,1),(1,1,0,0,0,0),(1,0,1,0,0,0),(1,0,0,1,0,0),(1,0,0,0,1,0),(0,1,1,0,0,0),(0,1,0,0,0,1),(0,0,1,0,1,0),(0,0,0,1,0,1),(0,0,0,0,1,1),(1,1,0,1,0,0),(1,0,1,0,0,1),(0,0,1,1,1,0),(0,0,1,1,0,1),(1,1,0,1,1,0),(1,1,0,0,1,1),(1,0,0,1,1,1),(0,1,1,1,0,1),(0,1,1,0,1,1),(1,1,1,1,1,0),(0,1,1,1,1,1),(1,1,1,1,1,1)];

def gen_cayley_graph(diff_set):
    g=Graph()
    X=GF(2)
    for x1 in X:
        for x2 in X:
            for x3 in X:
                for x4 in X:
                    for x5 in X:
                        for x6 in X:
                            for r in range(1,len(diff_set)):
                                b=diff_set[r];
                                g.add_edge((x1,x2,x3,x4,x5,x6),(x1+b[0],x2+b[1],x3+b[2],x4+b[3],x5+b[4],x6+b[5]))
    return g

g1=gen_cayley_graph(a1);
g2=gen_cayley_graph(a2);

It's evident from the code that the function gen_cayley_graph generates a Cayley graph on $\mathbb Z_2^6$ by the difference set diff_set.

The graphs g1 and g2 are both strongly-regular graphs with parameters (64, 27, 10, 12), but they are nonisomorphic:

sage: g1.is_strongly_regular(parameters=True)
(64, 27, 10, 12)
sage: g2.is_strongly_regular(parameters=True)
(64, 27, 10, 12)
sage: g1.is_isomorphic(g2)
False
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  • $\begingroup$ Thanks, this was very helpful! $\endgroup$ Commented Mar 28, 2023 at 10:29

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