Linear metric on F2, free group with two generator.

img_0515.jpg

I may have made a stupid mistake, but if not, we could construct a metric by pullback a metric on a suitable linear normalized space H which we carefully constructed. Let we define the generators of free group F_2 by a,b.

Step 1.

Constructed the linear normalized space H. the space H was spanned by basis \Lambda=\Lambda_a \coprod \Lambda_b, \Lambda_a, \Lambda_b are defined by look at the Cayley graph of F_2, there is a lot of vertical vector and horizontal vector in the Cayley graph, for every level set of vertical vector we put a basis in \Lambda_a, because there is only countable many vertical vectors (for example, a,a^2,a^{-5} are in the same vertical level, bab^{-1}, ba^{10}b^{-1} are in the same vertical level, bab^{-1},a are not in the same vertical level), we put a basis in \Lambda_a for every vertical level and claim we accomplished the construct of \Lambda_a, we do the same operation for \Lambda_b but only change the vertical level with horizontal level. Now we accomplished the construction of \Lambda, We spanned this with coefficient \mathbb Z and we get a linear space V. by Zorn’s lemma there exists a norm on the space, take one norm \|\cdot\| we accomplished the construction of H=(V,\|\cdot\|).

Step 2:

Pullback the norm \|\cdot\| on H to the free group F_2. In fact there is a natural bijection T: F_2\to H, which is given by following: On the Cayley graph (imaged it is embedding in \mathbb R^2), identity 1 in the group F_2 corresponding to the original, and more general every element in F_2 exactly identify with a point in the Cayley graph, thanks to there is no relation between a,b. And then there is of course infinity many of path from original to the point, but there is only one shortest path , thanks to there is no loop in the Cayley graph. We identify the elements in F_2 with the point in Cayley graph with the shortest path. Now we could explain why the path lies H. This path only across to finite vertical level and horizontal level and on every level it only pass finite step, this already given a representation \sum_{e_i\in \Lambda}c_i\cdot e_i, c_i\in \mathbb Z, the key point is there is only finite c_i\neq 0. So we have defined the bijection T:F_2\to H, and we could use the bijection to pullback the norm on H to a norm on F_2.

Step 3:

Now we begin to proof the norm we get by pullback satisfied the condition we need. We need only to proof the condition of linear growth and triangle inequality. The conjugation invariance is automatically by linear growth by the comments of Tobias Fritz. The triangle inequality is automatically, due to the bijection T stay the structure in fact, the multiplier of elements x_1,x_2 \in_2 could be view as put the two path together but this  is not true… merely because of the addition operation is not commutative.

The space we should consider is the path space equipped with the composition operation. I image there exists a “big space” such that the natural metric on the “big space” restrict on the embedding image of \mathbb F_2 is a linear growth metric.

广告

How to compute the Gromov-Hausdorff distance between spheres $latex S_n$ and $latex S_m$?

There is the question, because when we consider the Gromov-Hausdorff distance, we must fix the metric, so we use the natural metric induced from the embedding \mathbb{S}_n \to \mathbb{R}^{n+1}. Is it possible for us to compute the Gromov-Hausdorff distance d_{G-H}(\mathbb{S}_n,\mathbb{S}_m) for two different spheres \mathbb{S}_n and \mathbb{S}_m, m\neq n?

For example if we want to calculate d_{G-H}(\mathbb{S}_2,\mathbb{S}_3)=\inf_{M,f,g}d_{M}(\mathbb{S}_2,\mathbb{S}_3), where M ranges over all possible metric space and f:\mathbb{S}_2\to M and g:\mathbb{S}_3\to M range over all possible isometric (distance-preserving) embeddings.

At least we can embed \mathbb{S}_2,\mathbb{S}_3 into \mathbb{R}^3 in a canonical way. This will lead to a upper bound: d_{G-H}(\mathbb{S}_2,\mathbb{S}_3)\leq \sqrt{2}. And in general case we have d_{G-H}(\mathbb{S}_m,\mathbb{S}_n)\leq d_{G-H}(point,S_m)+d_{G-H}(point,S_n)\leq 2,\forall 0\leq n\leq m. But it is difficult to get a lower bound control for me. Because we need to take the inf in all possible metric spaces M. Especially I conjecture d_{G-H}(\mathbb{S}_m,\mathbb{S}_n)\geq \lambda_{m,n}\frac{m-n}{m},\forall 0\leq n\leq m, where \liminf_{m,n\to \infty}\lambda_{m,n}>0.

I only know the knowledge of Gromov-Hausdorff from Peterson’s Riemann Geometry. Unfortunately there is not enough information to compute the Gromov-Hausdorff distance, so this problem may be very stupid, I will appreciate any pointer.

 

 

And we know for the case S_n,S_m, if n,m is very near to each other,then the two space should be more near, and there is a canonical embed S_0\subset S_1 \subset S_2 ....\subset S_n \subset .... So it is natural to conjecture if m,n is very near then the distance d_{G-H}(S_n,S_m) is very small. I have a very rough strategy to prove the conjecture, that is inspired by the Nash embedding theorem. I just mean if we consider the problem in this frame d_{G-H}(S_n,S_m)=\inf_{M,g,f}(d_M(f(S_n),g(S_m))) then the difficult is the deformation space of M,g,f is too large. so the first step is to establish a regular lemma, to prove the function d_M(f(S_n),g(S_m)) is continues under the small perbutation of M and reduced to the situation of space $M,g,f$ with very nice regularity. the second part is to embed M to a big euclid space R^N as subspace, and the embedding stay the length of geodesic.locally this is determine by a group of pde:u_i(x)u_j(x)=g_{ij}(x),at least in the cut locus.but there should be some critical point,and I do not know how to deal with them.the third,i.e. the last step is to calculate d_{G-H}(S_n,S_m) in the very some deformation space M,f,g.

@Mark Sapir,Appreciate for help!I am reading the article you point out,it seems this article mainly focus on investigating the Gromov-Hausdorff limit space of a sequence of hyperbolic group equipped with modified G-H metric defined in 2.A with some special condition to ensure the limit space exists.and take a sequences corvarage to the limit space,the hyperbolic property and some other thing is stayed by the process of take limit.
@Mark Sapir,So it is natural for us to investigate the original space by some information from the limit space.there is a series of bi-product state in 3.B.but I do not see where the author exactly calculate some groom-hausdorff distance of two different space,may you point out it?appreciate again!
@MarianoSuárez-Álvarez,Corrected, thanks.

Y:
I fixed numerous typos. In particular, you should use spacing after each punctuation mark; capitals to begin sentences and names.

H:
Thank you very much for helping me to correct the mistakes! I will know how to write in a correct style.

Y:
23.1k
Your conjecture would imply that the GH distance is unbounded. But it’s clearly bounded, since the GH distance of any sphere to a point is equal to 2 (when the sphere is endowed with the restriction of Euclidean distance, as you seem to assume, or \pi when endowed with geodesic distance) and hence the GH distance between any two spheres is \le 4.

H:
73
You are right,In fact if we use the canonical embed, then we can get d_{G-H}(S_n,S_m)\leq 2 by another equivalent definition of GH distance.I confuse the geometry picture of the pairs T_n,S_n with the pairs S_n,S_m,for S_n,S_m case,I thick the seems correct conjecture will be d_{G-H}(S_n,S_m)\sim \frac{m-n}{m},0\leq n\leq m,m,n\to \infty.

Y:
16:37
Clearly from standard embeddings we get d_{GH}(S_n,S_m)\le\sqrt{2} for all n,m\ge 0. Would it be reasonable to simply conjecture that it’s an equality whenever n\neq m?

H:
73
Yeah, you are right,d_{G-H}(S_n,S_m)\leq \sqrt{2} for all n,m\geq 0.I find the interesting problem when I want to find a toy model of a kind of problem,roughly speaking is to investigate a map f:X\to Y from low-dimensions space X to high-dimension space Y stay some affine structure of the low-dimension space X. This structure could have some control by the distance function on the low-dimension space, so if we can get some control on the variation of the Energy of distance function, this will share some line on the original problem I consider.
And we know for the case S_n,S_m, if n,m is very near to each other,then the two space should be more near, and there is a canonical embed S_0\subset S_1 \subset S_2 ....\subset S_n \subset .... So it is natural to conjecture if m,n is very near then the distance d_{G-H}(S_n,S_m) is very small. .
I have a very rough strategy to prove the conjecture, that is inspired by the Nash embedding theorem. I just mean if we consider the problem in this frame d_{G-H}(S_n,S_m)=\inf_{M,g,f}(d_M(f(S_n),g(S_m))) then the difficult is the deformation space of M,g,f is too large. so the first step is to establish a regular lemma, to prove the function d_M(f(S_n),g(S_m)) is continues under the small perbutation of M and reduced to the situation of space M,g,f with very nice regularity.
The second part is to embed M to a big euclid space R^N as subspace, and the embedding stay the length of geodesic.locally this is determine by a group of pde:u_i(x)u_j(x)=g_{ij}(x),at least in the cut locus.but there should be some critical point,and I do not know how to deal with them.the third,i.e. the last step is to calculate d_{G-H}(S_n,S_m) in the very some deformation space M,f,g.
I need come back to explain why we expect the groom-hausdorff distance d_{G-H}(S_d,S_m),0\leq n\leq m should be much small than \sqrt 2 when frac{n}{m} is small.
Let consider a toy model of the problem,in a graph model,i.e. now we do not consider to take the Infimum in all space but in discrete space endow with metric. this can be view as a complete graph equipped metric, i.e. M=\{(G,d_G)\}. So there is also some space very like S_n,$S_m$ in the Euclid space, Let remark them as G_{S_n},G_{S_m}.
, oberseve that (G,d_G)\in M then (G,\hat d_{G})\in M,\hat d_{G} is a scaling of d_Gso it is natural to consider a cut off of M,called M_{\lambda} which is just a subset of M and satisfied if (G,d_G)\in M_{\lambda},then \inf_{x\neq y}d_{G}(x,y)\geq d.
Now,in the space M_{\lambda} let us consider a Distance distribution:\mu_G((a,b))=\frac{\#\{x,y\in G|a<d_G(x,y)<b\}}{\#G\times G}. Then this distribution will give us some information of the distance of the two different set G_1,G_2 in M_{\lambda}.
(removed)
Now,in the space M_{\lambda} let us consider a Distance distribution:\mu_G((a,b))=\frac{\#\{x,y\in G|a<d_G(x,y)<b\}}{\#G\times G}. Then this distribution will give us some information of the distance of the two different set G_1,G_2 in M_{\lambda}.
and obviously we will see that if n,m is close,then the distribution of fuzzy approximation G_{S_n},G_{S_m} is near, and the reverse is also true. I think this can explain why the conjecture d_{G-H}(S_n,S_m) =O(\frac{m-n}{m}) may be right.

 

by the way,it is a very good exercise to proof d_{G-H}(S_n,S_0)=1,\forall n\in N^*.