Is there a $(n_3)$ configuration which is not self-dual?
Coxeter points out that for a self-dual configuration $(m_c,n_d)$ it must be that $m=n$ and $c=d$, so we may abbreviate it $(m_c)$.
However I'm interested in the other direction of this implication, i.e. is there a configuration $(m_c,m_c)$ which is not self-dual? For $c=2$ there is none. All polygons are combinatorially self-dual.
It's also clear that if the automorphism group of the configuration is flag-transitive, then it is of the form $_n\{p\}_n$ and must be self dual. (All polygons fall into this category.)
However this still leaves a lot of open cases. In particular I am interested in 3-configurations.
Is there a $(n_3)$ configuration which is not self-dual?
References
- Coxeter, (1950) Self-dual configurations and regular graphs.
1 answer
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| User | Comment | Date |
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| WheatWizard | (no comment) | Jan 5, 2024 at 02:01 |
OEIS has
- A001403 Number of combinatorial configurations of type (n_3).
- A100001 Number of self-dual combinatorial configurations of type (n_3).
They first differ at $n=11$. An example of a non-self-dual configuration of type $(11_3)$ has points $0$ to $10$ and lines $[0, 1, 2]$, $[0, 3, 4]$, $[0, 5, 6]$, $[1, 3, 7]$, $[1, 5, 8]$, $[4, 5, 9]$, $[3, 8, 10]$, $[6, 8, 9]$, $[2, 9, 10]$, $[6, 7, 10]$, $[2, 4, 7]$.
This was constructed by using nauty to generate regular bipartite graphs, filtering to configurations, and testing the automorphism groups. (And, for what it's worth, I found the OEIS entries on the basis of counts for small $n$ produced during the search).
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