You might try the following:

Define a functional q mapping each member of F to a real number by defining q(f)=the supremum over x in Rn of ||f(x)-x||, where ||.|| is the Euclidean norm for Rn. You may then endow F with the unique coarsest topology such that q is continuous. Now define an equivalence relation on F such that f is equivalent to g if and only if q(f)=q(g). The equivalence classes are then topologically homeomorphic to the nonnegative reals with the usual topology.

Dear Kerry,

Thank you very much for your answer. It seems to me that defining q can be not so easy. Imagine F containing translations and rotations of the real plane. Then for rotations the value of q must be infinity. Do you have any ideas how to overcome this problem?

Best,

Alex

Yes, quite right. Let's change the definition of q. Let q(f)=the supremum over nonzero x in Rn of ||x||/(||x||+||f(x)||). Then we will always have 0

Rotations about the origin and translations will have value 1/2.

Dear Kerry,

Thank you for the answer. Now the problem is that q if considered as a metric on F does not preserve its topology. Ideally I would like to have 0

Happy to be of some help! Please let me know the eventual results, this is an interesting question.

Dear Kerry,

What I came to is to define the metric as an integral (or supremum) of ||f1(x)-f2(x)|| over a bounded domain D. Where D is such that for every pair f1, f2 not equal to each other the metric is non-zero. I guess that for every set of functions defined on Rn and having forming Lie group on Rn there must exist such D. But I haven't thought about the proof yet. What do you think?

Best,

Alex

Hmmm, I am thinking it might be possible to create an example of a set of such functions such that only an unbounded domain D will suffice to make the integral of ||f1(x)-f2(x)|| zero if and only if f1=f2. I'll try to come up with the details in the next day or so.

Kerry, I am looking forward to hearing from you. I was trying to build a counter example and to prove the statement, but I've succeeded in none so far.

Alex

I've attached a pdf of what I think is a counterexample. It's a group of mappings of the reals into the reals for which no bounded domain with the desired property can exist.

Thanks! But are you sure it is a group? I mean, it seems that the composition of fi and fj with non equal i and j does not belong to F.

No, G is much bigger than F, but contains F.

F itself is not a group;by adding all possible compositions of functions in F, we get the group G.

G is the smallest set containing F such that G is closed under composition.

Perfect. But can we be sure that it is a Lie group? If I remember correctly, the Lie group must be parametrizable by a finite-dimensional vector in such a way that it is smooth with respect to this vector. Do you think G possesses this property?

Yes, and I think f's must be smooth functions of their argument, but it does not change much here.

Dear Alexander,

Are you sure that this group must be a Lie group? What about a simple example: linear mappings given by powers of a non-degenerate matrix (simplest 2-dim. subcase: rotations, then we get either finite groups or a dense subgroup in so(2)).

Can you endow these groups with a structure of a Lie group?

Greetings,

Jan

Let n = 2k, identify R^n with the complex vectorspace C^k, and let F be the set of real-polynomial mappings from R^n to R^n that correspond to complex-polynomial automorphisms of C^k -- where by a complex-polynomial automorphism of C^k I mean a map f from C^k to C^k such that

(1) f is a complex polynomial map : each component f_1,..., f_k from C^k to C is a complex polynomial function in the standard coordinates z_1,...,z_k;

(2) f is a (set-theoretical) automorphism of C^k : it is one-to-one and onto, and therefore has a (set-theoretical) inverse map f^{-1}; and

(3) f^{-1} is also a complex polynomial map.

Examples of complex-polynomial automorphisms of C^k include all linear automorphisms of C^k, all translations of C^k, and all "generalized shears" of C^k, that is, all mappings of the form

(z_1,...,z_{k-1},z_k) |-> (z_1,...,z_{k-1},z_k+P(z_1,...,z_{k-1}),

for P(z_1,...,z_{k-1}) is a polynomial mapping.

Clearly the set of complex-polynomial automorphisms of C^k is a group, and therefore so is F. If k=1 then the group of complex-polynomial automorphism of C^k = C is the affine group {f : C -> C | for some complex numbers a and b, with a non-zero, f(z)=az+b}, so it is (in a natural way) a Lie group. If k=2, an old theorem of Jung (for a proof, see Abhyankar and Moh's paper "Embeddings of the Line in the Plane", where they give a simple derivation of Jung's theorem from their own Embedding Theorem, on embeddings of the line in the plane, proved in that paper; for a much shorter and [I think] much simpler proof of their Embedding Theorem, see my paper with the same title, also published in J. Reine Angew. Math.) says that the group of complex polynomial automorphisms of C^k = C^2 is generated by the linear automorphisms and translations (which together generate the affine group) along with the generalized shears. This group is certainly not a Lie group in the standard sense (which requires finite-dimensionality of the group space), because it contains as a subgroup the set (which is a group) of all generalized shears, and that group is isomorphic to the additive group of polynomial functions in two complex variables, which is infinite-dimensional in any reasonable topology. Nor is it, as it stands, an "infinite dimensional Lie group" in the fairly standard sense (which requires the group space to be a Banach manifold), because it is not complete (again, in any reasonable metric).

For k > 2 things are at least as bad (and perhaps much worse, since Jung's theorem doesn't generalize).

Dear Jan and Rufolf,

Thank you for your responses. I think because of the discussion we had with Kerry my original question was somehow lost. I assume that I KNOW that my functions form a Lie group, yet I do not know it's dimension, nor I know the corresponding manifold. I wonder how I can learn that. The first thing that comes into the mind is to define a metric on the functions (see Kerry's response). Yet it appears to be not so easy even in the plane when the group includes rotations.

Unfortunately I must say I do not understand your responses. Do you mean that not every set of functions form a Lie group? I agree.

Best,

Alex

"Do you mean that not every set of functions form a Lie group?"

Speaking only for myself, I do not mean that. Rather, I mean that not every group of functions is "small enough" to form a Lie group. For instance, a group G of functions (like the group of polynomial automorphisms of C^2) can contain a sequence of proper subgroups that are Lie groups the dimensions of which diverge to infinity--so G cannot itself be a Lie group in the usual sense (though it might be an "infinite dimensional Lie group" for which the underlying manifold is modeled on an infinite dimensional Hilbert space or Banach space). What is your reason for "guessing" (your original verb), or "assuming that you KNOW" (as you now write), that your set of functions isn't that big? What IS your set of functions? ... And, even more basic, why do you WANT your group of functions to be a Lie group? What interesting conclusions about, for instance, the context in which your group of functions arose can be shown to follow once you know that your group is a Lie group? For example, it's a theorem that the second homotopy group of (the component of the identity of) a Lie group is trivial: would it help you with your underlying problem to know that your group has a trivial second homotopy group?

Dear Lee,

Take for a example a group of rigid transformations of a plane, i.e. translations and rotations. Imagine now that you have a set of this functions, say, a finite set of functions defined by the image of a regular grid of points.. Now I would like to determine the dimensionality of this set of functions and to parametrize the set I have, so that the parametrization would preserve the topology. For that I would like to introduce a metric on this functions and then use dimension reduction techiques, like isomap, to find the dimension of the corresponding manifold. This can be easily done for translations, but becomes less trivial for rotations. One idea is to define a metric based on a mapping of a bounded domain in the plane, e.g. rho(f1, f2) = ||f2*f1^-1|| + ||f1*f2^-1|| and ||f|| = max_{x in X} ||x-f(x)||, where X is a bounded domain in plane.

The problem with this definition is that I cannot be sure that there always exist such X that for every two different mappings the metric defined this way will not vanish.

Best,

Alex

Hi Alex: I had another idea about your Lie group question. Its TeX expression follows:

Given a collection $C$ of functions $f:\mathbb{R}^n \rightarrow \mathbb{R}^n$ with an unknown Lie group structure, and a spanning collection $B$ of functions $B_i:\mathbb{R}^n \rightarrow \mathbb{R}^n$ orthonormal under some inner product $\left$, there is a natural mapping $\mu:C \times C\rightarrow \mathbb{R}^\infty$ given by $\mu\left(f,g\right)_j=\left$. This maps the collection $C$ into a smooth submanifold $S$ of $\mathbb{R}^\infty$ whose dimension should (I think) be the same as that of the unknown Lie group. Further, the topology induced on $S$ by the usual topology on $\mathbb{R}^\infty$ can be mapped back to $C$ by taking the smallest topology for which the function $\alpha:C \rightarrow \mathbb{R}^\infty$ given by $\alpha\left(f\right)_j=\left$ is continuous. In other words, the topology on $C$ generated by $\alpha^{-1}\left(U\right)$ as $U$ ranges over the open subsets of $S$.

Hi Kerry,

Thank you very much! Looks like it may work. I will looks more attentively into it.

Best,

Alex