MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $R$ be a ring generated by $k$ rational functions in the variables $x_1,...,x_n$ over the real numbers.

Is there an algorithm that computes a set of rational functions $f_1,...,f_l \in R$ which generate the subring $Q$ of $R$ consisting of all those rational functions in $R$ which are invariant under the transformation $(x_1,...,x_n) \rightarrow (-x_1,...,-x_n)$?

Background of the question: I have a birational map $F$ on n dimensional real affine space and I would like to find conserved quantities of the map, that is a function H such that $H \circ f = H$. Now, the map f is the composition of two involutions $I_1$, $I_2$, where $I_1$ is simply defined by $x \rightarrow -x$. The involution $I_2$ is a quite complicated map, but I am nevertheless able to find a set of rational functions generating the ring of all conserved quantities of $I_2$. Hence, I would like to know whether $I_1$ and $I_2$ have common conserved quantities, or in other words whether $I_2$ has conserved quantities which are invariant under $x \rightarrow -x$.

share|cite|improve this question
Are you doing this for arbitrary $n$ or for a fixed $n$? Also, does the other involution act linearly? – Charles Siegel Jun 15 '10 at 9:47
Arbitrary $n$ would be great, but most of my examples are for $n=3,4,5,6$. The other involution does not act linearly. It is birational. – Andi Jun 15 '10 at 9:53

Let $S$ be the rational functions in variables $x_i^2$. They are all in your invariants. The point is is that $R>S$ is a galois extensions with a group $C_2^n$. In particular, $R$ is a $2^n$-dimensional vector space over $S$ with basis $x_{a_1} \ldots x_{a_k}$ with $a_1 < \ldots < a_k$.

The field of invariants you are interested in is a $2^{n-1}$-dimensional subspace spanned by all monomials with even $k$. Proofs are easy exercises in Galois theory as Charles has pointed out.

share|cite|improve this answer

So at the level of function fields, you have a group of order 2 acting on L, the field of fractions of R, and there is a subfield K fixed by it. The extension L/K will be of degree 2, so quadratic. In other words L is generated by adjoining a square root of a function. All that is happening at the field level is that this square root is being replaced by its negative. (All this is standard Galois theory.)

So you want to chase this down to the level of R. By analogy with rings of algebraic integers, how hard will it be? In effect you are asking how to compute the field trace in a quadratic extension? To clarify the problem, taking any one of the variables shows how to adjoin a square root (the square of any one of the x's is in K).

share|cite|improve this answer
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 :-) – Andi Jun 15 '10 at 10:09
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. – Charles Matthews Jun 15 '10 at 10:12
alright, thanks for the clarification. – Andi Jun 15 '10 at 10:13
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. – Charles Matthews Jun 15 '10 at 10:22
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. – Andi Jun 15 '10 at 10:35

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.