Last Saturday (March 23, 2013), the second Bourbaki seminar of this year took place at amphithéâtre Hermite of Institut Henri Poincaré (as usual), and the following topics were discussed:

- C. Moeglin talked about recent progresses (of J. Arthur) on the discrete spectrum of classical groups,
- G. Malle talked about Ore’s conjecture after Liebeck-O’Brien-Shalev-Tiep,
- J.-B. Gouéré talked about branching Brownian motion, and
- C. Lecuire talked about ending laminations after Brock-Canary-Minsky.

Once more the speakers did a great job in explaining these topics to an audience of non-experts, and, for this reason, I decided to make a post about one of these talk.

Contrary to last time, it was “easy” for me to choose which topic to pick: given my tastes, I had to choose between Ore’s conjecture and ending laminations, and I opted for Ore’s conjecture because the website Images de Mathématiques made available an excellent article (in French) by F. Guéritaud with a guided tour (with plenty of beautiful pictures!) around the works of Masur-Misnky and Brock-Canary-Minsky.

So, below I will transcript my notes of G. Malle’s talk about Ore’s conjecture. As usual, the eventual mistakes in what follows are my entire responsibility.

**1. Introduction **

Let be a group. Given , we denote by the commutator of and , and we define the *commutator subgroup* the subgroup of *generated* by the elements , for . In group theory, is used to define the derived series helping to distinguish certain “categories” of groups such as perfect groups (when ), solvable groups, etc.

By definition, , but, in general, this inclusion is *not* an equality: for example, there are two (non-isomorphic) groups of order such that the *set* of commutators is different from (cf. this article of R. Guralnick).

In a more positive tone, N. Ito and O. Ore (independently) showed that:

Theorem 1 (Ito, Ore (1951))For (symmetric group) and (alternating group), we have the equality .

Furthermore, O. Ore said in his paper that “*it is possible that a similar theorem holds for any simple group of finite order, but it seems that at present we do not have the necessary methods to investigate the question*”. For this reason, the question of determining whether the previous theorem extends to all finite simple groups became *Ore’s conjecture*.

In 2010, M. Liebeck, E. O’Brien, A. Shalev and P. Tiep showed in this paper here that Ore’s conjecture is true:

Theorem 2 (Liebeck-O’Brien-Shalev-Tiep (2010))If is a non-abelian finite simple group, then every is a commutator.

As G. Malle pointed out to us, the proof of Ore’s conjecture by LOST (Liebeck-O’Brien-Shalev-Tiep ) “confirms” O. Ore’s prediction that the tools to attack this problem were not available at the time he wrote his paper. Indeed, LOST’s proof of Ore’s conjecture uses:

- the classification of finite simple groups;
- Lusztig’s parametrization of irreducible characters of finite reductive groups (and, so, Weil conjectures);
- explicit (computer) calculations (approximately 3 years of computer calculations in CPU time)

Concerning the statement of Ore’s conjecture/LOST theorem, let us notice that the result is simply not true for general non-simple groups . For instance, R. Guralnick showed that if and are two finite groups such that is Abelian, , and , then the regular wreath product doesn’t satisfy the equality . (Here, we recall that the wreath product is formed by taking the direct product of several copies of indexed by the elements of and by observing that acts [regularly by left multiplication] on this direct product by permuting the indices, so that it makes sense to talk about the semi-direct product of and ). Also, it is known that the smallest example of a perfect group not fitting the conclusion of Ore’s conjecture/LOST theorem is an extension of (i.e., an elementary Abelian group of order ) by the alternating group .

A question related to Ore’s conjecture and normally attributed to J. G. Thompson is the following one:

**Conjecture (J. G. Thompson).** Let be a finite non-Abelian simple group. Then, there exists a conjugacy class such that equals to ().

Note that Thompson’s conjecture implies Ore’s conjecture. Indeed, if , then any has the form with . In particular, whenever , its inverse also belongs to : by hypothesis, we can write the identity element as a product for some , so that there is an element whose inverse also belongs to , and, *a fortiori*, the same is true for all elements of (as they are conjugated to ). Therefore, if we write as with , then, since (as we just saw), we know that there exists such that and, thus, .

Actually, prior to LOST theorem, the proofs of Ore’s conjecture for *certain families of groups* proceeded by showing that Thompson’s conjecture is true for those families (and this is why we’re mentioning Thompson’s conjecture here). However, this is *not* the way that LOST’s arguments will work and, in particular, Thompson’s conjecture is still *open* for the general finite simple group.

**2. Direct calculations **

The “essential” case in LOST’s proof is that of of Lie type, e.g.,

Indeed, the idea is that for the “remaining” groups (such as the 26 sporadic ones), one can verify Ore’s conjecture by “hands” (or, more precisely, using computers).

Before discussing the case of groups of Lie type, let us consider first the case of simple algebraic groups (that is, the “continuous” analogs of finite groups of Lie type). In this direction, one has the following theorem of S. Pasiencier and H. Wang (over the complex numbers) and R. Ree (over arbitrary algebraically closed fields):

Theorem 3 (Pasiencier-Wang, Ree (1964))Let be a semisimple linear algebraic group over an algebraically closed field. Then, every is a commutator.

*Proof:* Given , we have that belongs to some Borel subgroup (by a result of Borel). Denote by Levi’s decomposition of with the (unipotent) radical of and a maximal torus.

In this notation, the proof of the theorem is based on the following two facts:

- (1) For all , there exists a regular element (i.e., an element whose centralizer is ) and an element of the normalizer of such that .
- (2) if is regular, then is a single -conjugacy class.

The first item is due to Kostant and the second item is not hard to deduce (one shows by induction on the central series of that the map from to is bijective when is regular).

In any case, given with Jordan decomposition , and , we use item (1) to select regular and with . Since is regular, we have that is also regular, and, thus, by item (2), there exists such that . It follows that

so that the proof of the theorem is complete.

It is tempting to try to adapt this proof to the case of finite groups of Lie type, but, as it turns out, this is not so easy because it is not true that all elements lie in some Borel subgroup and the analog of item (1) above is simply *false*.

Nevertheless, E. Ellers and N. Gordeev obtained in these papers here the following result about Gauss decompositions of elements of finite simple groups of Lie type:

Theorem 4 (Ellers-Gordeev (1996))Let be a finite simple group of Lie type. Denote by a maximally split torus inside a Borel subgroup of with Levi decomposition . Consider the opposite Borel subgroup and let be its Levi decomposition.Fix . Then, for any , there exists such that with and .

The next two corollaries show why this theorem is helpful in our discussion:

Corollary 5If and denote the -conjugacy classes of two regular elements , then .

*Proof:* By Ellers-Gordeev theorem, if , then, by taking , we have

On the other hand, by applying item (2) in the proof of Theorem 3 ( is a single -conjugacy class when is regular), we deduce the existence of elements and such that

It follows that

so that the proof of the corollary is complete.

This corollary provides the following criterion for the validity of Thompson’s and Ore’s conjectures for finite simple groups of Lie type:

Corollary 6If there exists a regular element , then Thompson’s (and, a fortiori, Ore’s) conjecture is true.

*Proof:* By taking in the previous corollary, we have that any belongs to where is the conjugacy class of .

Using this corollary (among other arguments), E. Ellers and N. Gordeev showed in this paper here the following result:

Theorem 7 (Ellers-Gordeev (1998))Let be a finite simple group of Lie type over a finite field with . Then, Thompson’s (and Ore’s) conjecture is true.

In view of this theorem, Liebeck-O’Brien-Shalev-Tiep showed their theorem (Ore’s conjecture) by developing a character theory method for the remaining cases that we are going to discuss in the next section.

**3. Character theory **

Denote by the set of irreducible (complex) characters of . A classical lemma due to Frobenius gives the following criterion to recognize whether an element is a commutator in terms of the characters of :

Lemma 8 (Frobenius)Let be a finite group. Then, is a commutator if and only if

Note that this criterion is suitable for computer calculations: indeed, if we know the character table of , we can ask a computer to determine the sum in order to recognize what elements of are commutators. In particular, this approach works to verify Ore’s conjecture for the 26 *sporadic groups* (because their character tables are known).

LOST’s idea consists into applying Frobenius lemma in the following way. Firstly, we separate the contribution of the trivial character from other characters

Secondly, we use the classical bound (coming from Schur orthogonality relations) to see that

In particular, the desired sum is not zero if the “error term” in the right-hand side is small, that is, if the conjugacy class is small and/or the dimensions of non-trivial irreducible characters are large.

In order to make this approach work, LOST show the following “Cauchy-Schwarz inequality” controlling the contribution of (non-trivial) irreducible characters with “large” dimension :

Lemma 9Let be a finite group with conjugacy classes. Then, for all and ,

Of course, a preliminary question before applying this lemma is: what is the size of ? In general, if the finite simple group of Lie type has the form where denotes its rank, then it is known that is bounded by a polynomial in of degree . For example, for , J. Fulman and R. Guralnick proved that

Once we know how to control , the next preliminary question concerns lower bounds on the dimensions of non-trivial irreducible characters (because this gives an idea of the size of the integer for which we will apply the lemma). Here, the Lusztig’s classification of irreducible characters comes into play and one *often* founds that the following *gap* phenomenon: there are a few non-trivial irreducible characters close to the smallest possible dimensions and all other non-trivial irreducible characters have dimension at least the *square* of the smallest possible dimensions. For example, for the symplectic groups , it is possible to show that (cf. this paper of P. Tiep and A. Zalesski):

Lemma 10Let with odd and . Let be a non-trivial irreducible character of . Then:

- either and is one of the (four) so-called Weil characters of ;
- or .

*Proof:* The lower bound on is not hard to establish: we can embed into by considering as a -dimensional vector space over ; since has smallest degrees , we get the desired lower bound.

On the other hand, the “gap” result is hard to get: for symplectic groups, one can consult this paper for an elementary proof, but for other types one has to use the full Lusztig’s classification.

After answering the preliminary questions on the sizes of and , we are ready to get back to Ore’s conjecture. By using the information on and , LOST use the estimate on and the “gap” phenomenon above to safely concentrate on the few “Weil characters” with dimensions close to the lower bound. Then, they use *explicit* values of for a Weil character to show that their lemma works *when* the size of the -conjugacy class of is *small*.

In particular, the proof of Ore’s conjecture will be complete if one can show that an element with *large* conjugacy class is a commutator. In this direction, LOST introduce the concept of *breakable elements* and they show by induction that breakable elements are commutators. For example, they prove that:

Theorem 11Let be a breakable element, that is, with . Then, is a commutator.

Here, we said that the proof goes by induction because one can assume that this statement is true for and in order to show the statement for . However, one has to be careful here because some small cases such as require special attention.

Finally, it remains to deal with unbreakable elements with large conjugacy classes. As it is shown by LOST, an unbreakable element that is not covered in their “Cauchy-Schwarz inequality” argument has a *really* large conjugacy class: for instance, for , they show (sometimes by computing characters) that if is *unbreakable*, then is *small* as far as their arguments are concerned. Thus, there is a rather *small* number of such large conjugacy classes of unbreakable elements, so that one can hope to show that the elements in such a *small list* of conjugacy classes are commutators by simply exhibiting *random commutators* belonging to each of them, and this is precisely what LOST do to complete the proof of their theorem.

After presenting this beautiful sketch of proof of Ore’s conjecture/LOST theorem, G. Malle decided to comment on the relationship between Ore’s conjecture and *word maps* as the last topic of his talk.

**4. Word maps **

Let denote the free group on generators and let be any group. Given a word , , we define the *word map* as

We will denote the image of the word map by .

Example 1For the word , the surjectivity of the word map is equivalent to Ore’s conjecture for .

In 1983, A. Borel showed that the following “almost surjectivity” result for word maps on algebraic groups:

Theorem 12 (Borel (1983))Let be a semisimple linear algebraic group over an algebraically closed field. Then, for any non-trivial word , the word map is dominant (i.e., its image contains a Zariski open and dense subset of ).

*Proof:* Firstly, by taking the universal cover of , one can see that it suffices to prove the result when is simple. Secondly, by considering maximal torii, it is possible to further reduce to the case of of type (e.g., if , one reduces the result to the case of ). Finally, one does the type case directly.

Remark 1Word maps are not surjective in general: for example, in positive characteristic , the image of the word map associated does not contain regular unipotent elements.

This theorem allows to deduce the following result of M. Larsen, A. Shalev and P. Tiep:

Theorem 13 (Larsen-Shalev-Tiep (2011))Let be two non-trivial words. Then, there exists an integer such that for all finite non-Abelian simple groups with .

As a corollary of this result, we have:

Corollary 14If , then there exists such that for all (finite non-Abelian simple with) . In particular, .

A short sketch of proof of Theorem 13 goes as follows. We separate the discussion into two cases: groups of *bounded rank* and groups of *unbounded rank*.

For the bounded rank case, Larsen-Shalev-Tiep show that:

Theorem 15 (Larsen-Shalev-Tiep)Let and let be an infinite family of finite simple groups of fixed Lie type (e.g., with fixed). Then, there exists such that, for each , the set contains regular elements from any maximal torus of .

Then, they conclude the bounded rank case essentially by combining this theorem with the result of Ellers-Gordeev in Corollary 5 above.

Finally, for groups with unbounded rank, they use that for all finite simple groups of Lie type *except* type , there are pairs and of maximal torii such that if for some , then is the trivial character or the Steinberg character . At this point, G. Malle ran out of time and thus he decided to finish his talk here.

**5. Epilogue **

Closing this post, let me mention three question posed to G. Malle after the end of his talk:

- Y. de Cornulier asked about estimates on the
*number*of ways of writing as a commutator. Here, G. Malle mentioned that it is easy to produce lower bounds (essentially by changing the conjugacy classes of the elements you use to write ), but he was not sure about upper bounds. - J.-P. Serre asked about the nature of computer part of LOST’s proof (i.e., if there were “inner checks” for correctness, etc.) and G. Malle assured him that the computer programs used are not extremely sophisticated (they are based on GAP and Magma) and thus it is unlikely that there are errors coming from the corresponding codes. (Apparently J.-P. Serre was not completely satisfied with this answer as I could deduce from his facial expression…)
- I asked about whether one can write as the commutator of a pair of elements
*generating*(here I had in mind some potential links to origamis/square-tiled surfaces…), and G. Malle told me that, at least for certain families, this is likely to be known (and it is probably contained in the most recent LOST papers), but he was not aware of a reference were the general problem is treated.

The answer to your last question is “no”. I posed the same question some time ago to Shalev, and he said that it was shown already by BH Neumann,

e.g., even A_5 is a counter example and there are many others.

By:

Boris Kunyavskiion June 2, 2013at 12:30 pm

Thanks for your answer!

In fact, after writing this post, I took a look in the literature, and there are several papers on the question I posed and also on the related problem of understanding -systems in certain families of groups, e.g., as you pointed out, BH Newman, Higuchi and Miyamoto studied this for alternating groups, and, more recently, Evans, Guralnick, Pak, McCullough and Wanderley studied this for .

Furthermore, it seems that the presence of several -systems for certain groups has some interesting consequences in the application I had in mind (i.e., Lyapunov spectra of square-tiled surfaces [a.k.a. origamis]) and I hope to write more on this in the future…

By:

matheuscmsson June 2, 2013at 1:25 pm

There is a weaker question for which I do not know any answer.

Trying to make the subgroup in the representation of $g=[x,y]$

as large as possible, can one, say, bound its index

uniformly within some family of groups of Lie type

(fixing either q, or rank) ?

By:

Boris Kunyavskiion June 2, 2013at 1:29 pm