旧博客原文
原题:Linear metric on F2, free group with two generator.

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 which we carefully constructed. Let we define the generators of free group
by
.
Step 1.
Constructed the linear normalized space . the space
was spanned by basis
,
are defined by look at the Cayley graph of
, 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
, because there is only countable many vertical vectors (for example,
are in the same vertical level,
are in the same vertical level,
are not in the same vertical level), we put a basis in
for every vertical level and claim we accomplished the construct of
, we do the same operation for
but only change the vertical level with horizontal level. Now we accomplished the construction of
, We spanned this with coefficient
and we get a linear space
. by Zorn’s lemma there exists a norm on the space, take one norm
we accomplished the construction of
.
Step 2:
Pullback the norm on
to the free group
. In fact there is a natural bijection
, which is given by following: On the Cayley graph (imaged it is embedding in
), identity
in the group
corresponding to the original, and more general every element in
exactly identify with a point in the Cayley graph, thanks to there is no relation between
. 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
with the point in Cayley graph with the shortest path. Now we could explain why the path lies
. 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
, the key point is there is only finite
. So we have defined the bijection
, and we could use the bijection to pullback the norm on
to a norm on
.
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 stay the structure in fact, the multiplier of elements
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 is a linear growth metric.
补充说明
以下是新整理的中文说明;上方旧博客原文保持不变。
自由群 $F_2=\langle a,b\rangle$ 的 Cayley graph 是一棵正则树。一个自然想法是:能否把这棵树嵌入某个线性赋范空间,然后把范数距离 pull back 回群上,得到一种由线性构造出来的 metric?

1. Cayley graph 图像
$F_2$ 的每个元素都是一个约化字。Cayley graph 的顶点是群元素,边对应左乘或右乘生成元。因为自由群没有非平凡关系,这个图没有环,是一棵树。
2. 从边方向构造线性空间
可以尝试给 Cayley tree 中不同“水平”的水平边、垂直边分配基向量。设这些基张成一个向量空间 $V$,再在 $V$ 上选择一个范数 $\|\cdot\|$。
每个群元素对应从单位元到该点的唯一 geodesic path,于是可把路径上的边向量相加,得到映射
$$\Phi:F_2\to V.$$
3. Pullback metric
定义
$$d(g,h)=\|\Phi(g)-\Phi(h)\|.$$
如果 $\Phi$ 是单射,这确实给出 metric。若范数选得合适,它可能与 word metric 有可比较关系;若选得太退化,则会丢失树的几何。
4. 需要注意的问题
真正困难在于,这种构造是否自然、是否左不变、是否 quasi-isometric 于 word metric。普通 word metric 满足
$$d_S(g,h)=|g^{-1}h|_S,$$
具有明显的左不变性;pullback metric 未必自动保留这个性质。
5. 几何意义
这个问题可以看作自由群嵌入 Banach space 的 toy model。它连接 Cayley graph、tree metric、coarse embedding 和 geometric group theory 中的线性化思想。
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