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.