The Number of Sides of a Parallelogram

We deﬁne parallelograms of base a and b in a group. They appear as minimal relators in a presentation of a subgroup with generators a and b . In a Lie group they are realized as closed polygonal lines, with sides being orbits of left-invariant vector ﬁelds. We estimate the number of sides of parallelograms in a free nilpotent group and point out a relation to the rank of rational series.


Introduction
In IR 2 a parallelogram of base a and b can be defined as a closed polygon with the minimum number of sides parallel to a and b.In that paper we also consider parallelograms defined in more general groups.
In section 1. we first give some definitions and examples of parallelograms in Lie groups.These examples show the various complex situations occurring in the general case.In this paper we concentrate our attention on free nilpotent groups.This analysis will give universal properties for parallelograms.We obtain Theorem.The number of sides of a parallelogram on a free nilpotent group on two generators of order n is between n and n 2 .
We do not know what is the exact number of sides of parallelograms in a free nilpotent group neither how many non-equivalent parallelograms exist.We hope that an investigation of parallelograms might help understand general nilpotent groups.In particular it will be interesting to find presentations with relators of minimal size.
We have chosen in this paper to recall the basic properties and constructions of free Lie algebras in order to make it self-contained.That is done in section 2. In the last section we then introduce mth-order parallelograms and prove our result.A connection with rational series is pointed out at the end of the paper.
Our initial motivation to study parallelograms was the notion of curvature and holonomy of a connection for Riemannian manifolds and the generalization of those notions to sub-Riemannian geometry (see [FGR] and [BeR]).In classical differential geometry, curvature appears as the quadratic term in the asymptotic expansion of holonomy around short (four-sided) parallelograms, holonomy being the 1365-8050 c ✂ 1999 Maison de l'Informatique et des Mathématiques Discrètes (MIMD), Paris, France measure of the difference of the vector field by parallel translation around a closed loop.In the case of sub-Riemannian manifolds, the tangent space is naturally a nilpotent group ( [BeR]) and the holonomy associated to it will be calculated using parallelograms with many sides.The analog of sectional curvatures should be the holonomy associated to different parallelograms.
Another motivation is the approximation of a given element of the group by elements of a given subgroup.This occur for example in the search of symplectic integrators (see [K, Su]) that give numerical schemes for long-time integration of hamiltonian systems.Namely we try to approximate exp✄ x ☎ y✆ by a product of exp✄ x✆ and exp✄ y✆ .In this frame, minimal length of mth-order approximants are bounded by approximately 2 m .Example 2.2 Consider the Heisenberg group H 3 with Lie algebra generated by X ✡ Y ✡ Z, with ✘ X ✡ Y ✙ ✚✝ Z, all other brackets being null.One can verify, using the Campbell-Hausdorff formula that both are parallelograms.They are not equivalent as P 8 has at least one side of length two.On the other hand starting with X ✡ Z we get a parallelogram of 4 sides.
Example 2.3 Let L 4 be a free nilpotent group of order 4, generated by X and Y .We can verify that We thank the referee for pointing out the two following examples.
We could have given a more general definition of a parallelogram in an arbitrary group.Let a and b be two elements on a group G and G✧ a✡ b✪ be the subgroup generated by a✡ b.Consider the set of all relators, i. e., the set of words in a✡ b✡ a ✒ 1 ✡ b ✒ 1 which are the identity in G.One should consider only reduced words in the sense that if a is of order n and a n appears in a word, one should substitute the identity for a n .The same for b.A parallelogram of base a✡ b is a reduced relator (in the above sense) of minimal length with letters a✡ b✡ a ✒ 1 ✡ b ✒ 1 .Of course if G✧ a✡ b✪ is free in a✡ b there is no parallelogram.

Example 2.6 In the case of the symmetric group
one can verify that a minimal relator with base σ 1 ✡ σ 2 is ✄ σ 1 σ 2 ✆ 3 of length 6.On the other hand we have also that has a minimal relator of length 4.
In the case of Lie groups we would like to define infinitesimal parallelograms, that is parallelograms which remain the same in form when their sides are changed by a conformal factor.They will not exist in general but in the case of graded nilpotent groups their existence is assured.
Example 2.7 Consider the Lie group with Lie algebra generated by X ✡ Y with ✘ X ✡ Y ✙ ✯✝ X.Then we can construct a parallelogram which is not infinitesimal.Observe that That is a paralle- logram of length 5 with base ln 2 Y and X.It is clear that if we change the sides by a conformal factor this will no longer be a parallelogram.More generally, a polygon is a product The previous equation has clearly no integer solutions.
Example 2.8 Let us consider in IR 2 , X ✝ ∂ x and Y ✝ f ✄ x✆ ∂ y for a given analytic function f .The Lie algebra L✄ X ✡ Y ✆ is in general infinite dimensional as ✄ ad X ✆ n Y ✝ f ✲ n✳ ✄ x✆ ∂ y and is spanned, as vector space by X and ✌ ✴✄ ad X ✆ n Y ✎ .By noticing that exp✄ λad X ✆ Y ✝ f ✄ x ☎ λ✆ ∂ y , we deduce that This gives a parallelogram of length 8.

Magnus Groups and Algebras
Let us first introduce some notations and recall some results about free groups, free associative algebras and free Lie algebras.All these results can be found in ( [B, La, R]).
Let X be a set (alphabet).We denote by X ✷ the free monoid generated by X, that is, the set of words including the empty word denoted by 1, with concatenation as a product.X ✷ is totally ordered by the lexicographic order.The free magma M ✄ X ✆ is the set of words with parentheses, generated by X and A✄ X ✆ denotes the free associative algebra, that is to say the Q-algebra of X ✷ .An element P in A✄ X ✆ will be written ∑ w✸ X ✹ ✄ P✡ w✆ w.
We denote by L✄ X ✆ the free Lie algebra on A. It is the quotient of the Q-algebra of M ✄ X ✆ by the ideal generated by the elements ✄ u✡ u✆ and ✄ u✡ ✕✄ v✡ w✆ ✖✆ ✵☎ ✺✄ v✡ ✗✄ w✡ u✆ ✖✆ ✑☎ ✻✄ w✡ ✕✄ u✡ v✆ ✖✆ .The associative algebra A✄ X ✆ may be identified to the enveloping algebra of L✄ X ✆ by considering ✘ v✡ w✙ ✯✝ vw ✜ wv.We denote by ad x the map y ✼ ✽ ✾✘ x✡ y✙ .
The free group generated by X is denoted by F ✄ X ✆ .

Gradations
The sets L✄ X ✆ ✩✡ F ✄ X ✆ so as A✄ X ✆ are graded by -the length (the unique homomorphism that extends the function x ✼ ✽ 1 on X).
) is the submodule generated by monomials of length n.
-the multi-degree which is the unique homomorphism from X ✷ (resp. ) denotes the submodule generated by monomials of degree α.
As a consequence, we have

Formal series
We define L✄ X ✆ and Â✄ X ✆ as L✄ X ✆ ✫✝ We will write x ✏ L✄ X ✆ (resp.Â✄ X ✆ ✖✆ as a series ∑ n ❀ 0 x n .L✄ X ✆ so as Â✄ X ✆ are algebras with multiplications law We will also use It is a subgroup of the invertible elements of Â✄ X ✆ .One defines the exponential and the logarithm as They are mutually reciprocal functions and we have (see [B, Ch.II, ❊ 5]) the Theorem 3.1 (Campbell-Hausdorff) Definition 3.3 Let us consider the map µ✢ : F ✄ X ✆ ✯✽ Γ✄ X ✆ as the unique group homomorphism that extends x ✼ ✽ exp✄ x✆ , for x ✏ X.We set D✢ ❀ n ✄ X ✆ ✛✝ µ✢ ✒ 1 ✄ 1 ☎ Â❀ n ✄ X ✆ ✮✆ .This defines central filtrations of F ✄ X ✆ .We have clearly that F❀ n ✄ X ✆ ✛❂ D❀ n ✄ X ✆ and F❀ n ✄ X ✆ ✭❂ D✢ ❀ n ✄ X ✆ .
In fact Magnus proved a stronger result (see [B]) We will use the following corollary to establish the lower bound to the number of sides of parallelogram on the free nilpotent group.

Corollary 3.2 The projection of g in
In fact we need only the if part of the corollary for the lower bound, that is not dependent on Magnus result but on the inclusion F❀ n ✄ X ✆ ❃❂ D❀ n ✄ X ✆ .

mth-order parallelograms
Definition 4.1 The order of g in F ✄ X ✆ is the biggest integer k such that g ✏ F❀ k ✄ X ✆ .An element of order k will be called kth-order polygon.
Using proposition (3.1), a mth-order polygon g satisfies Here none of a✢ i s nor b✢ i s is 0.
we sill say that it is a p-sided polygon.For example xyx ✒ 1 y ✒ 1 is a 4-sided second-order parallelogram of length 4. In formula (4), we have l ✄ g✆ ✞✝ ∑ n i❅ 1 ✄ ✕✠ a i ✠ ✮☎ ✻✠ b i ✠ ✆ .We thus deduce that for any The inequality is strict only if terms of g 1 cancel terms of g 2 .
Definition 4.3 For m ✏ IN, we define l m as the lowest length of mth-order polygons.A mth-order parallelogram will be a mth-order polygon of minimal length.
Before discussing the lower and upper bounds for the length and the number of factors of mth-order parallelograms, let us show some transformations that preserve polygons.Proposition 4.1 Let αβ be a mth-order polygon then so is βα.
Proof.-The proposition comes from the fact that F ❄ F❀ m ✄ X ✆ is abelian.Let us suppose that g ✝ x a 1 y b 1 ✰ ✖✰ ✕✰ y b p x a p✓ 1 is a mth-order polygon.Then has smaller length as ✠ a 1 ☎ a p ✣ 1 ✠ ❑✟ ▲✠ a 1 ✠ ✮☎ ▼✠ a p ✣ 1 ✠ and is also a mth-order polygon. ◆ We can now suppose that for any integer m, an mth-order parallelogram has an even number of factors.We will now discuss lower and upper bound of l m .where z ✏ L❀ m ✄ X ✆ and none of the a✢ i s nor b i 's is 0. Considering the word w ✝ ✍✄ xy✆ n , we have

Lower bound
and so m ✟ 2n ✟ l m .In fact the number of sides itself is bigger than m.◆

Upper bound
First of all, let us show some small-order parallelograms.
which is a consequence of the following Lemma 4.1 For any m 2, l m is even.
Proof.-This is a consequence of We have seen g 2 as the commutator of two first-order polygons.We will now build a sequence g m of mth-order polygons, each g m being constructed as commutator of g p and g m✒ p for some p.We first use the following lemma Lemma 4.2 Let g p and g q be two polygons of order p and q respectively, then ✄ g p ✡ g q ✆ has order at least p ☎ q and has length at most 2✄ l ✄ g p ✆ ✚☎ l ✄ g q ✆ ✖✆ .
Remark.-This is also a consequence of the fact that ✄ F❀ n ✄ X ✆ ✕✆ n is a central filtration but we will show it by using the Hausdorff series.
Proof.-Let us write then we have order polygon and has length 2✄ l ✄ g p ✆ ❙☎ l ✄ g q ✆ ✖✆ .In order to be sure to obtain a ✄ p ☎ q✆ -th order polygon let us show that Lemma 4.
If x p and y q are not proportional, then ✄ α✡ β✆ has order exactly p ☎ q.

Rational series
In fact there is a strong connection with the rank of rational series.The set Â✄ X ✆ is usually denoted by Q✧ ❵✧ X ✪ ❛✪ and is called the set of formal series.
Consider the following operation of X ✷ on Â✄ X ✆ ; for u ✏ X ✷ , let We extend it by linearity to obtain Â✄ X ✆ as a right module over A✄ X ✆ .A combinatorial interpretation of that operation in the case where S ✝ v is a single word says that u ✒ 1 v vanishes , unless v starts with u, that is, v ✝ uv✢ , and in that case u ✒ 1 v ✝ v✢ .

Definition 4.4 A formal series is rational if it is an element of the closure of A✄ X ✆
A fundamental theorem due to M.-P.Schützenberger assures that the orbits of the action of A✄ X ✆ are finite dimensional over Q on rational series.We may then state the following Definition 4.5 The rank of a rational series S is the dimension of the space S ❜ A✄ X ✆ .
We state now corollary 3.6 of [BR].

Proposition 4 . 2
For any m ✏ IN we have m ✟ l m .Proof.-Let us consider the following equality exp✄ a 1 x✆ exp✄ b 1 y✆ ✴✰ ✖✰ ✕✰ exp✄ a n x✆ exp✄ b n y✆ ✞✝ exp✄ z✆ ✩✔ (8) If the group generated by exp✄ X ✆ and exp✄ Y ✆ is free, then there is no parallelogram of base X and Y .