INFINITE GROUPS WITH LARGE BALLS OF TORSION ELEMENTS AND SMALL ENTROPY

LAURENT BARTHOLDI, YVES DE CORNULIER

Abstract. We exhibit infinite, solvable, abelian-by-finite groups, with a fixed number of generators, with arbitrarily large balls consisting of torsion elements. We also provide a sequence of 3-generator non-nilpotent-by-finite polycyclic groups of algebraic entropy tending to zero. All these examples are obtained by taking appropriate quotients of finitely presented groups mapping onto the first Grigorchuk group.

The Burnside Problem asks whether a finitely generated group all of whose elements have finite order must be finite. We are interested in the following related question: fix n sufficiently large; given a group Γ, with a finite symmetric generating subset S such that every element in the n-ball is torsion, is Γ finite? Since the Burnside problem has a negative answer, a fortiori the answer to our question is negative in general. However, it is natural to ask for it in some classes of finitely generated groups for which the Burnside Problem has a positive answer, such as linear groups or solvable groups. This motivates the following proposition, which in particularly answers a question of Breuillard to the authors.

Proposition 1. For every n, there exists a group G, generated by a 3-element subset S consisting of elements of order 2, in which the n-ball consists of torsion elements, and satisfying one of the additional assumptions:

(1) G is solvable, virtually abelian, and infinite (more precisely, it has a free abelian normal subgroup of finite 2-power index); in particular it is linear.

(2) G is polycyclic, not virtually nilpotent.

(3) G is solvable, non-polycyclic.

Remark 2.

(1) Natural stronger hypotheses are the following: being linear in fixed dimension, being solvable of given solvability length. We have no answer in these cases. It is also natural to ask what happens it we fix a torsion exponent.

(2) By [Se, Corollaire 2, p.90], if G is a group and S is any finite generating subset for which the 2-ball of G consists of torsion elements, then G has Property (FA): every action of G on a tree has a fixed point. In particular, if G is infinite, then, by Stalling’s Theorem [St] it cannot be virtually free.

(3) For every sufficiently large prime p, and for all n, there exists an infinite, 2-generated word hyperbolic group in which the n-ball consists of elements of p-torsion [Ol].

Date: September 12, 2005.

2000 Mathematics Subject Classification. Primary 20F50; Secondary 20F16, 20F05.

Key words and phrases. Grigorchuk group; balls of torsion; algebraic entropy.

(4) We give in the sequel more quantitative statements: in (1), the free abelian subgroup can be chosen of index 2^{a_{n}}, where a_{n} ∼ 11n^{k} (that is, a_{n}/(11n^{k}) → 1), where k ≅ 6.60 is a constant (see Corollary 9).

With a similar construction, we obtain results on the growth exponent. Let G be generated by a finite symmetric set S, and let B_{n} be the n-ball. Then, by a standard argument, the limit h(G, S) = lim \frac{1}{n} log(#(B_{n})) exists. The (algebraic) entropy of G is defined as h(G) = inf_{S} h(G, S), where S ranges over all finite symmetric generating subsets of G. Osin has proved [Os1, Os2] that, for an elementary amenable finitely generated group, h(G) = 0 if and only if G is virtually nilpotent; on the other hand, J. Wilson [W] has constructed a finitely generated group with h(G) = 0 which is not virtually nilpotent; see [B2] for a simpler example. Relying on former work of Grigorchuk [G2], Osin observes in [Os2] that there exist elementary amenable groups (actually virtually solvable) with h > 0 arbitrary close to 0. This latter result can be improved as follows.

Proposition 3. For every ε > 0, there exists a polycyclic, virtually metabelian, 3-generated group G with 0 < h(G) < ε.

Propositions 1 and 3 are obtained by approximating the Grigorchuk group, first introduced in [G1], by finitely presented groups.

We recall below the definition of a family of 3-generated groups Γ_{n}, which are successive quotients (Γ_{n+1} is a quotient of Γ_{n} for all n). These are finitely presented groups obtained by truncating the presentation of Grigorchuk’s group. They are proved in [GH] to be virtually direct products of nonabelian free groups; they have larger and larger balls of torsion, and their entropy tends to zero. We get Propositions 1 and 3 by considering appropriate solvable quotients of the groups Γ_{n}.

Following Lysionok [L], the first Grigorchuk group is presented as follows. We start from the 3-generated group Γ_{−1} = ⟨a, b, c, d | a^{2} = b^{2} = c^{2} = d^{2} = bcd = 1⟩. Elements u_{n} and v_{n} of Γ_{−1} are defined below. Then, for 0 ≤ n ≤ ∞, Γ_{n} is defined as the quotient of Γ_{−1} by the relations u_{i} for i < n + 1 and v_{i} for i < n. The first Grigorchuk group Γ = Γ_{∞} has a wealth of remarkable properties. The most celebrated one is that Γ has non-polynomial subexponential growth [G2]. It is also a 2-group, i.e. a group in which every element is of finite 2-power order.

Let us define the relations u_{n} and v_{n}. Consider the substitution σ defined by σ(a) = aca, σ(b) = d, σ(c) = b, σ(d) = c; extend its definition to words in the natural way, and finally observe that it defines a group endomorphism of Γ_{−1}. Set u_{0} = (ad)^{4}, v_{0} = (adacac)^{4}, u_{n} = σ^{n}(u_{0}), v_{n} = σ^{n}(v_{0}).

For all n ≥ −1, the natural morphism Γ_{n} → Z/2Z sending b, c, d to 0 and a to 1 has kernel Ξ_{n} of index two.

We are going to focus on the finitely presented groups Γ_{n} rather than on Γ. Individually, and up to commensurability, the structure of these groups is not of special interest: Γ_{n} is commensurable to a direct product of 2^{n} non-abelian free groups [GH, Proposition 12]. However, since Γ_{∞} is torsion, for all n, there exists i(n) such that every element in the n-ball of Γ_{i(n)} is torsion. A quantative statement is given in Corollary 13: we can take i(n) = ⌊log_{α}(n) − 1⌋, with α ≅ 1.25 a root of the polynomial 2X^{3} − X^{2} − X − 1.

The following proposition specifies [GH, Proposition 12].

Proposition 4. For every n ≥ 0, Γ_{n} has a normal subgroup H_{n} of index 2^{α_{n}}, where α_{n} ≤ (11 · 4^{n} + 1)/3, and H_{n} is a subgroup of index 2^{β_{n}} in a finite direct product of 2^{n} nonabelian free groups of rank 3, where β_{n} ≤ (11 · 4^{n} − 8)/3 − 2^{n}.

Remark 5. The main difference with [GH, Proposition 12] is that the subgroup of finite index constructed there is not normal. Of course one could take a smaller normal subgroup of finite index, but this one need not a priori be of 2-power index, a fact we require to obtain solvable (and not only virtually solvable) groups in Propositions 1 and 3.

We use the following elementary lemma.

Lemma 6. Let G be a group, and let H be a proper subgroup of index 2^{a}, normalized by a subgroup of index two in G. Let N be the intersection of all conjugates of H. Then N has index 2^{b}, for some integer b ≤ 2a − 1.

Proof: If H is normal in G, the result is trivial. Otherwise, consider the unique conjugate H′ ≠ H of H, so that N = H ∩ H′. Taking the quotient by N, we can suppose that H ∩ H′ = {1} and we are reduced to proving that G is a 2-group of order d ≤ 2^{2a−1}. Let W be the normalizer of H. Since it has index 2 in G, it is normal in G, so that H′ ⊂ W. Since H and H′ are both normal subgroups of W and H ∩ H′ = {1}, [H, H′] = {1}. Accordingly, HH′ is a normal subgroup of G, contained in W, and is naturally the direct product of H and H′. The order of H is d/2^{a}, so that the order of HH′ is d^{2}/2^{2a}, and hence the index of HH′ in G is 2^{2a}/d. This proves that d is a power of 2, and d ≤ 2^{2a}; actually d ≤ 2^{2a−1} because HH′ ⊂ W, hence has index ≥ 2. ■

Remark 7. In Lemma 6, the assumption that the normalizer has index at most two is sharp: in the alternating group A_{4}, there are four subgroups of index 4, all conjugate; they have pairwise trivial intersection, hence of index 12, which is not a power of 2.

Recall that Ξ_{0} ⊂ Γ_{0} is a subgroup of index 2; it is generated by b, c, d, aba, aca, ada. By [GH, Proposition 1], the assignment i_{0}(b) = (a, c), i_{0}(c) = (a, d), i_{0}(d) = (1, b) extends to a unique group homomorphism i_{0} : Ξ_{0} → Γ_{−1} × Γ_{−1} such that, for all x, if i_{0}(x) = (x_{0}, x_{1}), then i_{0}(axa) = (x_{1}, x_{0}). By [GH, Proposition 10], this induces, for all n ≥ 1, an injective group homomorphism: i_{n} : Ξ_{n} → Γ_{n−1}×Γ_{n−1}.

Proof of Proposition 4. Let us proceed by induction on n. We start with the essential case when n = 0, worked out in [GH, Lemma 11]. Write Γ_{0} = ⟨a, b, d| a^{2} = b^{2} = d^{2} = (bd)^{2} = (ad)^{4} = 1⟩ (this is a Coxeter group). Let H_{0} be the normal subgroup generated by (ab)^{2}. Then, by an immediate verification, Γ_{0}/H_{0} is isomorphic to the direct product of a cyclic group of order 2 and a dihedral group of order 8.

We claim that H_{0} is free of rank 3. Let L be the normal subgroup of Γ_{0} generated by ab: then L has index 4 in Γ_{0}, contains H_{0}, and is shown, in the proof of [GH, Lemma 11], to be isomorphic to Z ∗ (Z/2Z).

Accordingly, by Kurosh’s Theorem, if H_{0} were not free, then it would contain a conjugate of ab, but this is not the case. Actually H_{0} is contained in a subgroup of index 8 in Γ_{0}, free of rank 2 (see the proof of [GH, Lemma 11]), hence has rank 3.

Now, for n ≥ 1, let us suppose that Γ_{n−1} has a normal subgroup H_{n−1} of index 2^{α_{n−1}}, which embeds as a subgroup of index 2^{β_{n−1}} in a direct product of 2^{n−1} nonabelian free groups of rank 3.

The homomorphism i_{n} described above embeds Ξ_{n} as a subgroup of index 8 in Γ_{n−1} × Γ_{n−1}. Define H′_{n} = i_{n}^{−1}(H_{n−1} × H_{n−1}): this a normal subgroup of index 2^{k} in Ξ_{n}, with k ≤ 2α_{n−1}.

Then, using Lemma 6, H_{n} = H′_{n} ∩ ^{a}H′_{n} has index 2^{α_{n}}, with α_{n} ≤ 4α_{n−1} − 1. Combining the inclusions H_{n} ⊂ H′_{n} ⊂ H_{n−1} × H_{n−1} ⊂ (F_{3})^{2^{n−1}} × (F_{3})^{2^{n−1}}, we obtain that H_{n} has index 2^{β_{n}} in (F_{3})^{2^{n}}, with β_{n} ≤ (2α_{n} − 1) + 8 + 2β_{n}.

Define α′_{n}11 · 4^{n} + 1)/3 and β′_{n} = (11 · 4^{n} − 8)/3 − 2^{n}. Then α′_{0} = 4, β′_{0} = 0, and they satisfy, for all n: α′_{n} = 4α′_{n−1} − 1 and β′_{n} = 2α′_{n} + 2β′_{n} + 2. Therefore, an immediate induction gives α_{n} ≤ α′_{n} and β_{n} ≤ β′_{n} for all n. ■

It is maybe worth restating the result avoiding reference to the particular sequence Γ_{n}.

Corollary 8. For every finitely presented group G mapping onto the first Grigorchuk group Γ, there exist normal subgroups N ⊂ H in G, with H of finite 2-power index, such that H/N is isomorphic to a finite index subgroup of a nontrivial direct product of free groups.

Proof: Let p : G → Γ be onto. Since G is finitely presented, p factors through Γ_{n} for some sufficiently large n, so that there exists a map p′ of G onto Γ_{n}. Then take N = Ker(p′) and H = p^{′−1}(H_{n}). ■

Combining with Corollary 13, we also obtain the following statement.

Corollary 9. In the group Γ_{i(n)}, the n-ball consists of 2^{i(n)+1}-torsion elements, and there exists a normal subgroup of index 2^{α_{i(n)}}, which embeds in a direct product of free groups, with α_{i(n)} ≤ (11 · n^{log_{α}(4)} + 1)/3, and log_{α}(4) ≅ 6.60.

Proof of Proposition 1. Take i sufficiently large so that the n-ball of Γ_{i} consists of torsion elements. Since H_{i} is a finite index subgroup in a nontrivial direct product of free groups (see Proposition 4), it has infinite abelianization. There is a short exact sequence

1 → H_{i}/[H_{i}, H_{i}] → Γ_{i}/[H_{i}, H_{i}] → Γ_{i}/H_{i} → 1 .

Accordingly, G = Γ_{i}/[H_{i}, H_{i}] is an infinite, virtually abelian group, in which the n-ball consists of torsion elements. Moreover, since Γ_{i}/H_{i} is a finite 2-group, G is also solvable.

For (3), take, instead, G = Γ_{i}/[[H_{i}, H_{i}], [H_{i}, H_{i}]]. Since H_{i} maps onto a nonabelian free group, its metabelianization is not virtually polycyclic, so that G is virtually metabelian, but not virtually polycyclic.

For (2), take a morphism of H_{i} onto, say, a polycyclic group W which is not virtually nilpotent, and let K be the kernel of this morphism. Since the normalizer of K has finite index in Γ_{i}, K has finitely many conjugates K_{1}, …, K_{l}. Set L = ⋂_{j=1}^{l} K_{i}. Then the diagonal map H_{i}/L → ∏_{j=1}^{l} H_{i}/K_{i} is injective, hence embeds H_{i}/L in W^{l}. On the other hand, observe that H_{i}/L projects onto W, so is not virtually nilpotent. It follows that G = Γ_{i}/L is polycyclic but not virtually nilpotent. If W has been chosen metabelian, then we also have that G is virtually metabelian.

Proof of Proposition 3. Keep the last construction in the previous proof. Then h(Γ_{i}/L) ≤ h(Γ_{i}). Moreover, h(Γ_{i}/L) > 0 since Γ_{i}/L is solvable but not virtually nilpotent [Os1]. On the other hand, it is proved in [GH] that h(Γ_{i}) → 0. Thus we can obtain h(G) arbitrary small. ■

Remark 10. Consider for every i an infinite quotient Q_{i} of Γ_{i}. In the topology of marked groups (defined in [G2]; see also, for instance, [C]), the sequence (Q_{i}) converges to the Grigorchuk group Γ. Indeed, otherwise it would have another cluster point, which would be a proper quotient of Γ, thus would be finite. This is a contradiction since the infinite groups form a closed subset in the topology of marked groups.

Appendix

We gather here the technical results, used in the proofs of the results above. They are slight modifications of results in the papers [GH], [B1].

Let α ≅ 1.25 be the real root of the polynomial 2X^{3} − X^{2} − X − 1.

We introduce on Γ_{−1} (hence on all its quotients) the metric | · | of [B1]: it is defined by attributing a suitable weight to each of the generators a, b, c, d: |a| = 2(α − 1) ≅ 0.47, |b| = 1 − |a| = α^{−3} ≅ 0.53, |c| = 2α^{2} − 3α + 1 ≅ 0.34, and |d| = −2α^{2} + α + 2 ≅ 0.19. Throughout this appendix, unless explicitly stated, the balls and the lengths are meant in the sense of this weighted metric.

To check that the length of a, b, c, d is exactly the weight we have imposed, it suffices to check this in the abelianization of Γ_{−1}, the F_{2}-vector space with basis (a, b, d) (which is also the abelianization of all Γ_{n}). There, it is a straightforward verification that the mapping |·| just defined extends to a length function by setting |aξ| = |a| + |ξ| for ξ ∈ {b, c, d}.

Observe that if ξ ∈ {b, c, d}, and i_{n}(ξ) = (ξ_{0}, ξ_{1}), we have

(1)  |ξ_{0}| + |ξ_{1}| = α^{−1}(|ξ| + |a|).

Lemma 11. Let x ∈ Γ_{0} be any element. Set x′ = x if x ∈ Ξ_{0} and x′ = xa otherwise; and set i_{0}(x′) = (x_{0}, x_{1}).

Then |x_{0}| + |x_{1}| ≤ α^{−1}(|x| + |a|).

Suppose moreover that x is of minimal length among its conjugates, and that x ∉ {b, c, d}. Then |x_{0}| + |x_{1}| ≤ α^{−1}|x|.

Proof: Fix x ∈ Γ_{0}, and let w be a word in the letters {a, b, c, d}, of minimal length^{1} representing x. Since every element in {b, c, d} is the product of the two others, w can be chosen without any two consecutive occurrences of letters in {b, c, d}.

Suppose now that x is of minimal length within its conjugacy class and that w is not a single letter. Maybe conjugating x by the last letter of w, we can suppose that w ends with the letter a. The minimality assumption then implies that w begins with a letter in {b, c, d}.

• First case: x ∈ Ξ_{0}. Write w = ξ^{1}(aξ^{2}a) … ξ^{2n−1}(aξ^{2n}a), where ξ^{i} ∈ {b, c, d} for i = 1, …, 2n. Write i(x) = (x_{0}, x_{1}) and i(ξ^{i}) = (ξ_{0}^{i}, ξ_{1}^{i}), so that i(ξ^{i}) =

(ξ_{1}^{i}, ξ_{0}^{i}). Then

|x_{0}| + |x_{1}| ≤ (|ξ_{0}^{1}| + |ξ_{1}^{2}| + ⋯ + |ξ_{0}^{2n−1}| + |ξ_{1}^{2n}|) + (|ξ_{1}^{1}| + |ξ_{0}^{2}| + ⋯ + |ξ_{1}^{2n−1}| + |ξ_{0}^{2n}|)

= (|ξ_{0}^{1}| + |ξ_{1}^{1}|) + (|ξ_{0}^{2}| + |ξ_{1}^{2}|) + ⋯ + (|ξ_{0}^{2n}| + |ξ_{1}^{2n}|).

1If w = u_{1} … u_{n}, the length of w is defined as |u_{1}| + ⋯ + |u_{n}|.

By (1), we get

|x_{0}| + |x_{1}| ≤ α^{−1} ∑_{i=1}^{2n}(|ξ^{i}| + |a|)

On the other hand, |x| = ∑_{i=1}^{2n}(|ξ^{i}| + |a|), so that finally |x_{0}| + |x_{1}| ≤ α^{−1}|x|.

• Second case: x ∉ Ξ_{0}, so that xa ∈ Ξ_{0}. Write w = ξ^{1}(aξ^{2}a) … ξ^{2n−1}(aξ^{2n}a)ξ^{2n+1}a, so that ξ^{1}(aξ^{2}a) … ξ^{2n−1}(aξ^{2n}a)ξ^{2n+1} represents xa in Γ_{0}. Write i_{0}(xa) =

(x_{0}, x_{1}), and i_{0}(ξ^{i}) = (ξ_{0}^{i}, ξ_{1}^{i}). Then

|x_{0}|+|x_{1}| ≤ (|ξ_{0}^{1}|+|ξ_{1}^{2}|+⋯+|ξ_{0}^{2n−1}|+|ξ_{1}^{2n}|+|ξ_{0}^{2n+1}|)+(|ξ_{1}^{1}|+|ξ_{0}^{2}|+⋯+|ξ_{1}^{2n−1}|+|ξ_{0}^{2n}|+|ξ_{1}^{2n+1}|)

= (|ξ_{0}^{1}| + |ξ_{1}^{1}|) + (|ξ_{0}^{2}| + |ξ_{1}^{2}|) + ⋯ + (|ξ_{0}^{2n+1}| + |ξ_{1}^{2n+1}|).

= α^{−1}(∑_{i=1}^{2n+1}(|ξ^{i}| + |a|)) again by (1).

Since |x| = ∑_{i=1}^{2n+1}(|ξ^{i}| + |a|), we get |x_{0}| + |x_{1}| ≤ α^{−1}|x|.

The other inequality |x_{0}| + |x_{1}| ≤ α^{−1}(|x| + |a|) is proved similarly: we must deal with the following cases:

• w begins and ends with the letter a: discussing whether or not x ∈ Ξ_{0}, in both cases we obtain the stronger inequality |x_{0}| + |x_{1}| ≤ α^{−1}(|x| − |a|).

• w begins and ends with letters in {b, c, d}: discussing whether or not x ∈ Ξ_{0}, in both cases we obtain the inequality |x_{0}| + |x_{1}| ≤ α^{−1}(|x| + |a|).

• w begins with the letter a and ends with a letter in {b, c, d}: in this case, replacing x and w by x^{−1} and w^{−1} (this is the word w read from the right to the left — recall that the generators are involutions), we reduce to the case, already carried out, when w begins with a letter in {b, c, d} and ends with the letter a, obtaining the inequality |x_{0}| + |x_{1}| ≤ α^{−1}|x|.

Since the verifications are similar to the computations above and since we do not use this case in the sequel, we omit the details. ■

Lemma 12. For every n ≥ −1, and every element in the open α^{n−1}-ball of Γ_{−1}, its image in Γ_{n} is of 2^{n+1}-torsion.

Proof: For n = −1, α^{−2} = |a| + |d| ≅ 0.66, and the elements in the open α^{−2}-ball are 1, a, b, c, and d, and are of 2-torsion in Γ_{−1}.

For n = 0, α^{−1} = |a| + |c| ≅ 0.81, and the elements in the open α^{−1}-ball are, besides the elements in the open α^{−2}-ball already quoted, ad and its inverse da, which are of 4-torsion in Γ_{0}.

We can start an induction, and suppose that, for some n ≥ 1, we have already proved that every element in the open α^{n−2}-ball of Γ_{−1}, its image in Γ_{n−1} is of 2^{n}- torsion. Pick x in the open α^{n−1}-ball of Γ_{−1}. We want to show that x^{2^{n+1}} = 1. We can suppose that x is of minimal length among its conjugates, and that x ∉ {b, c, d}. Define x′ as in Lemma 11, i.e. {x′} = {x, xa} ∩ Ξ_{−1}. Denote by [·] the projection of Γ_{−1} onto Γ_{0}. Set i_{0}([x′]) = (x_{0}, x_{1}).

• First case: x ∈ Ξ_{−1}, i.e. x = x′. By Lemma 11, for i = 0, 1, |x_{i}| ≤ |x_{0}| + |x_{1}| ≤ α^{−1}|x| ≤ α^{n−2}. By the induction hypothesis, x_{0} and x_{1} are of 2^{n}-torsion in Γ_{n−1}. Since i_{0} induces an injection of Ξ_{n} into Γ_{n−1} × Γ_{n−1}, this implies that x is of 2^{n}-torsion, hence of 2^{n+1}-torsion in Γ_{n}.

• Second case: x ∉ Ξ_{−1}, i.e. x′ = xa. Then x^{2} ∈ Ξ_{−1}, and
i_{0}([x^{2}]) = i_{0}([xaaxaa]) = i_{0}([xa])i_{0}([a(xa)a]) = (x_{0}x_{1}, x_{1}x_{0}),

which is conjugate to (x_{0}x_{1}, x_{0}x_{1}) in Γ_{−1} × Γ_{−1}. By Lemma 11, we have |x_{0}x_{1}| ≤ |x_{0}| + |x_{1}| ≤ α^{−1}|x| ≤ α^{n−2}. By the induction hypothesis, x_{0}x_{1} is of 2^{n}-torsion in Γ_{n−1}, and thus i_{0}([x^{2}]) is of 2^{n}-torsion in Γ_{n−1} × Γ_{n−1}. Since i_{0} induces an injection of Ξ_{n} into Γ_{n−1} × Γ_{n−1}, this implies that x^{2} is

of 2^{n}-torsion in Ξ_{n} ⊂ Γ_{n}, hence x is of 2^{n+1}-torsion in Γ_{n}. ■

Corollary 13. Set i(n) = ⌊log_{α}(n) − 1⌋. Then, in the n-ball of Γ_{i(n)} (for the word metric), every element is 2^{i(n)+1}-torsion.

Proof: Suppose that x has word length ≤ n. Then |x| ≤ |b|n = α^{−3}n = α^{log_{α}(n)−3} < α^{i(n)−1}. By Lemma 12, the image of x in Γ_{i(n)} is of 2^{i(n)+1}-torsion. ■

Acknowledgment. We thank Emmanuel Breuillard for encouragement and valuable discussions. We also thank Goulnara Arzhantseva and Pierre de la Harpe for useful remarks.

References

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Laurent Bartholdi

E-mail: laurent.bartholdi@epfl.ch

École Polytechnique Fédérale de Lausanne (EPFL) Institut de Mathématiques B (IMB) CH-1015 Lausanne, Switzerland Yves de Cornulier

E-mail: decornul@clipper.ens.fr

École Polytechnique Fédérale de Lausanne (EPFL) Institut de Géométrie, Algèbre et Topologie (IGAT) CH-1015 Lausanne, Switzerland