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I am looking for a (already-studied or interesting) class of Matroids such that - Class of Gammoids are contained in it

One example would be Strongly-base-orderable Matroids. I would also be grateful if someone knows a class of Matroids such that

  • Class of Gammoids are contained in it AND
  • It is contained in the class of "Strongly-base-orderable" Matroids.

By the way, strongly-base-orderable is a property such that : GIven any two bases I,J of the Matroid, there exists a bijection \pi between I-J and J-I such that given any subset H of I-J, I- H +\pi(H) and J - \pi(H) + H is a base in the Matroid. (In Oxley's "Matroid theory" pp410 )

Motivation : Well, I have something I can show for Gammoids but cannot for Strongly-base-orderable Matroids, although computational evidence suggests that it holds for general Matroids.

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up vote 3 down vote accepted

The way I understand your question is as follows: you have a property which you believe holds for all matroids (these are rare); you can prove it for gammoids, but your proof does not extend to a certain bigger class of "strongly-base-orderable" (SBO) matroids. I take it you have an example of a SBO matroid where your proof breaks down because some intermediate lemma explicitly fails (even if the main property still holds). Without knowing if this is the right picture and what are all these properties and examples, it's hard to give you a good answer. However, if that is indeed your logic you might as well consider classes of matroids which contain gammoids but are not completely included in the class of SBO matroids. Say, Piff and Welsh proved in 1970 that every gammoid is representable over large enough field. So, how about the class of real-representable matroids? Not all of them are SBO, but so what - perhaps your "bad example" is not there...

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It is also a result of Piff and Welsh that gammoids are restrictions of transversal matroids right? –  Gjergji Zaimi Feb 18 '10 at 22:40
    
Yes, every gammoid is a contraction of a transversal matroid. Piff and Welsh proved this and concluded the representability result (for transversal matroids this is straightforward). –  Igor Pak Feb 19 '10 at 3:24
    
Thank you for the answer. Too bad it is hard to play around with algebraic matroids... –  Suho Oh Feb 19 '10 at 15:12
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