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Jun 15, 2010 at 11:21 comment added Charles Matthews That's a negative result, as far as existence of common invariants is concerned. It removes one good reason why they might be there.
Jun 15, 2010 at 10:35 comment added Andi As far as I can see, the group is not finite (except in some very trivial examples). Does this mean that I should also consider alternatives to Galois theory? Also, I am very much dependent on algorithmic solutions to this problem, since the map $F$ is quite complicated, that is numerators and denominators are polynomials of total degree $n-1$ or $n$ in the $n$ variables. Hence, it stands out of the question to deal with these maps without using a CAS like Maple or Mathematica.
Jun 15, 2010 at 10:22 comment added Charles Matthews But it is possible that you have a group action (whatever your two involutions generate). If that group is finite, Galois theory might be more revealing. The formulation does suggest a slightly more algebraic approach to what is happening. It depends on the nature of F.
Jun 15, 2010 at 10:13 comment added Andi alright, thanks for the clarification.
Jun 15, 2010 at 10:12 comment added Charles Matthews I don't know how hard. The Galois theory isn't profound at all. I wanted to say that this is a recognisable type of problem.
Jun 15, 2010 at 10:10 history edited Charles Matthews CC BY-SA 2.5
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Jun 15, 2010 at 10:09 comment added Andi Thank you for your answer. I do however have no idea of Galois Theory (my traing has been mostly in integrable systems, mathematical physics, etc.) , so I do not even understand if - by your last two sentences - you mean whether this is a hard or easy problem...I guess I should start reading books on Galois theory :-)
Jun 15, 2010 at 10:04 history edited Charles Matthews CC BY-SA 2.5
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Jun 15, 2010 at 9:57 history answered Charles Matthews CC BY-SA 2.5