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Q&A Is there a $(n_3)$ configuration which is not self-dual?

1 answer  ·  posted 4mo ago by WheatWizard‭  ·  last activity 4mo ago by Peter Taylor‭

Question combinatorics
#3: Post edited by user avatar WheatWizard‭ · 2024-01-04T15:22:35Z (4 months ago)
  • 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*.
  • 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*.
#2: Post edited by user avatar WheatWizard‭ · 2024-01-04T04:26:42Z (4 months ago)
A little about the automorphism group.
  • 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.
  • However for 3-configurations it is not as apparent.
  • Is there a $(n_3)$ configuration which is not self-dual?
  • ### References
  • * Coxeter, (1950) *Self-dual configurations and regular graphs*.
  • 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: Initial revision by user avatar WheatWizard‭ · 2024-01-04T03:47:14Z (4 months ago)
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.

However for 3-configurations it is not as apparent.

Is there a $(n_3)$ configuration which is not self-dual?

### References

* Coxeter, (1950) *Self-dual configurations and regular graphs*.