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Last year, Emmanuel Breuillard, Ben Green, Bob Guralnick, and I wrote a paper entitled “Strongly dense free subgroups of semisimple Lie groups“. The main theorem in that paper asserted that given any semisimple algebraic group ${G(k)}$ over an uncountable algebraically closed field ${k}$, there existed a free subgroup ${\Gamma}$ which was strongly dense in the sense that any non-abelian subgroup of ${\Gamma}$ was Zariski dense in ${G(k)}$. This type of result is useful for establishing expansion in finite simple groups of Lie type, as we will discuss in a subsequent paper.

An essentially equivalent formulation of the main result is that if ${w_1, w_2 \in F_2}$ are two non-commuting elements of the free group ${F_2}$ on two generators, and ${(a, b)}$ is a generic pair of elements in ${G(k) \times G(k)}$, then ${w_1(a,b)}$ and ${w_2(a,b)}$ are not contained in any proper closed algebraic subgroup ${H}$ of ${G(k)}$. Here, “generic” means “outside of at most countably many proper subvarieties”. In most cases, one expects that if ${(a, b)}$ are generically drawn from ${G(k) \times G(k)}$, then ${(w_1(a,b), w_2(a,b))}$ will also be generically drawn from ${G(k) \times G(k)}$, but this is not always the case, which is a key source of difficulty in the paper. For instance, if ${w_2}$ is conjugate to ${w_1}$ in ${F_2}$, then ${w_1(a,b)}$ and ${w_2(a,b)}$ must be conjugate in ${G(k)}$ and so the pair ${(w_1(a,b), w_2(a,b))}$ lie in a proper subvariety of ${G(k) \times G(k)}$. It is currently an open question to determine all the pairs ${w_1, w_2}$ of words for which ${(w_1(a,b), w_2(a,b))}$ is not generic for generic ${a,b}$ (or equivalently, the double word map ${(a,b) \mapsto (w_1(a,b),w_2(a,b))}$ is not dominant).

The main strategy of proof was as follows. It is not difficult to reduce to the case when ${G}$ is simple. Suppose for contradiction that we could find two non-commuting words ${w_1, w_2}$ such that ${w_1(a,b), w_2(a,b)}$ were generically trapped in a proper closed algebraic subgroup. As it turns out, there are only finitely many conjugacy classes of such groups, and so one can assume that ${w_1(a,b), w_2(a,b)}$ were generically trapped in a conjugate ${H^g}$ of a fixed proper closed algebraic subgroup ${H}$. One can show that ${w_1(a,b)}$, ${w_2(a,b)}$, and ${[w_1(a,b),w_2(a,b)]}$ are generically regular semisimple, which implies that ${H}$ is a maximal rank semisimple subgroup. The key step was then to find another proper semisimple subgroup ${H'}$ of ${G}$ which was not a degeneration of ${H}$, by which we mean that there did not exist a pair ${(x,y)}$ in the Zariski closure ${\overline{\bigcup_{g \in G} H^g \times H^g}}$ of the products of conjugates of ${H}$, such that ${x, y}$ generated a Zariski-dense subgroup of ${H'}$. This is enough to establish the theorem, because we could use an induction hypothesis to find ${a,b}$ in ${H'}$ (and hence in ${G(k)}$ such that ${w_1(a,b), w_2(a,b)}$ generated a Zariski-dense subgroup of ${H'}$, which contradicts the hypothesis that ${(w_1(a,b),w_2(a,b))}$ was trapped in ${\bigcup_{g \in G} H^g \times H^g}$ for generic ${(a,b)}$ (and hence in ${\overline{\bigcup_{g \in G} H^g \times H^g}}$ for all ${(a,b)}$.

To illustrate the concept of a degeneration, take ${G(k) = SO(5)}$ and let ${H = SO(3) \times SO(2)}$ be the stabiliser of a non-degenerate ${2}$-space in ${k^5}$. All other stabilisers of non-degenerate ${2}$-spaces are conjugate to ${H}$. However, stabilisers of degenerate ${2}$-spaces are not conjugate to ${H}$, but are still degenerations of ${H}$. For instance, the stabiliser of a totally singular ${2}$-space (which is isomorphic to the affine group on ${k^2}$, extended by ${k}$) is a degeneration of ${H}$.

A significant portion of the paper was then devoted to verifying that for each simple algebraic group ${G}$, and each maximal rank proper semisimple subgroup ${H}$ of ${G}$, one could find another proper semisimple subgroup ${H'}$ which was not a degeneration of ${H}$; roughly speaking, this means that ${H'}$ is so “different” from ${H}$ that no conjugate of ${H}$ can come close to covering ${H'}$. This required using the standard classification of algebraic groups via Dynkin diagrams, and knowledge of the various semisimple subgroups of these groups and their representations (as we used the latter as obstructions to degeneration, for instance one can show that a reducible representation cannot degenerate to an irreducible one).

During the refereeing process for this paper, we discovered that there was precisely one family of simple algebraic groups for which this strategy did not actually work, namely the group ${G = Sp(4) = Spin(5)}$ (or the group ${SO(5)}$ that is double-covered by this group) in characteristic ${3}$. This group (which has Dynkin diagram ${B_2=C_2}$, as discussed in this previous post) has one maximal rank proper semisimple subgroup up to conjugacy, namely ${SO(4)}$, which is the stabiliser of a line in ${k^5}$. To find a proper semisimple group ${H'}$ that is not a degeneration of this group, we basically need to find a subgroup ${H'}$ that does not stabilise any line in ${k^5}$. In characteristic larger than three (or characteristic zero), one can proceed by using the action of ${SL_2(k)}$ on the five-dimensional space ${\hbox{Sym}^4(k^2)}$ of homogeneous degree four polynomials on ${k^2}$, which preserves a non-degenerate symmetric form (the four-fold tensor power of the area form on ${k^2}$) and thus embeds into ${SO(5)}$; as no polynomial is fixed by all of ${SL_2(k)}$, we see that this copy of ${SL_2(k)}$ is not a degeneration of ${H}$.

Unfortunately, in characteristics two and three, the symmetric form on ${\hbox{Sym}^4(k^2)}$ degenerates, and this embedding is lost. In the characteristic two case, one can proceed by using the characteristic ${2}$ fact that ${SO(5)}$ is isomorphic to ${Sp(4)}$ (because in characteristic two, the space of null vectors is a hyperplane, and the symmetric form becomes symplectic on this hyperplane), and thus has an additional maximal rank proper semisimple subgroup ${Sp(2) \times Sp(2)}$ which is not conjugate to the ${SO(4)}$ subgroup. But in characteristic three, it turns out that there are no further semisimple subgroups of ${SO(5)}$ that are not already contained in a conjugate of the ${SO(4)}$. (This is not a difficulty for larger groups such as ${SO(6)}$ or ${SO(7)}$, where there are plenty of other semisimple groups to utilise; it is only this smallish group ${SO(5)}$ that has the misfortune of having exactly one maximal rank proper semisimple group to play with, and not enough other semisimples lying around in characteristic three.)

As a consequence of this issue, our argument does not actually work in the case when the characteristic is three and the semisimple group ${G}$ contains a copy of ${SO(5)}$ (or ${Sp(4)}$), and we have had to modify our paper to delete this case from our results. We believe that such groups still do contain strongly dense free subgroups, but this appears to be just out of reach of our current method.

One thing that this experience has taught me is that algebraic groups behave somewhat pathologically in low characteristic; in particular, intuition coming from the characteristic zero case can become unreliable in characteristic two or three.

Emmanuel Breuillard, Ben Green, Robert Guralnick, and I have just uploaded to the arxiv our paper “Strongly dense free subgroups of semisimple algebraic groups“, submitted to Israel J. Math.. This paper was originally motivated by (and provides a key technical tool for) another forthcoming paper of ours, on expander Cayley graphs in finite simple groups of Lie type, but also has some independent interest due to connections with other topics, such as the Banach-Tarski paradox.

Recall that one of the basic facts underlying the Banach-Tarski paradox is that the rotation group $O(3)$ contains a copy of the free non-abelian group $F_2$ on two generators; thus there exists $a, b \in O(3)$ such that $a,b$ obey no nontrivial word identities.  In fact, using basic algebraic geometry, one can then deduce that a generic pair $(a,b)$ of group elements $a, b \in O(3)$ has this property, where for the purposes of this paper “generic” means “outside of at most countably many algebraic subvarieties of strictly smaller dimension”.    (In particular, using Haar measure on $O(3)$, almost every pair has this property.)  In fact one has a stronger property, given any non-trivial word $w \in F_2$, the associated word map $(a,b) \mapsto w(a,b)$ from $O(3) \times O(3)$ to $O(3)$ is a dominant map, which means that its image is Zariski-dense.  More succinctly, if $(a,b)$ is generic, then $w(a,b)$ is generic also.

In contrast, if one were working in a solvable, nilpotent, or abelian group (such as $O(2)$), then this property would not hold, since every subgroup of a solvable group is still solvable and thus not free (and similarly for nilpotent or abelian groups).  (This already goes a long way to explain why the Banach-Tarski paradox holds in three or more dimensions, but not in two or fewer.)  On the other hand, a famous result of Borel asserts that for any semisimple Lie group $G$ (over an algebraically closed field), and any nontrivial word $w \in F_2$, the word map $w: G \times G \to G$ is dominant, thus generalising the preceding discussion for $O(3)$.  (There is also the even more famous Tits alternative, that asserts that any linear group that is not (virtually) solvable will contain a copy of the free group $F_2$; as pointed out to me by Michael Cowling, this already shows that generic pairs of generators will generate a free group, and with a little more effort one can even show that it generates a Zariski-dense free group.)

Now suppose we take two words $w_1, w_2 \in F_2$, and look at the double word map $(w_1,w_2): G \times G \to G \times G$ on a semisimple Lie group $G$.  If $w_1, w_2$ are non-trivial, then Borel’s theorem tells us that each component of this map is dominant, but this does not mean that the entire map is dominant, because there could be constraints between $w_1(a,b)$ and $w_2(a,b)$.  For instance, if the two words $w_1, w_2$ commute, then $w_1(a,b), w_2(a,b)$ must also commute and so the image of the double word map is not Zariski-dense.  But there are also non-commuting examples of non-trivial constraints: for instance, if $w_1, w_2$ are conjugate, then $w_1(a,b), w_2(a,b)$ must also be conjugate, which is also a constraint that obstructs dominance.

It is still not clear exactly what pairs of words $w_1, w_2$ have the dominance property.  However, we are able to establish that all pairs of non-commuting words have a weaker property than dominance:

Theorem. Let $w_1, w_2 \in F_2$ be non-commuting words, and let $a, b$ be generic elements of a semisimple Lie group $G$ over an algebraically closed field.  Then $w_1(a,b), w_2(a,b)$ generate a Zariski-dense subgroup of $G$.

To put it another way, $G$ not only contains free subgroups, but contains what we call strongly dense free subgroups: free subgroups such that any two non-commuting elements generate a Zariski-dense subgroup.

Our initial motivation for this theorem is its implications for finite simple groups $G$ of Lie type.  Roughly speaking, one can use this theorem to show that a generic random walk in such a group cannot be trapped in a (bounded complexity) proper algebraic subgroup $H$ of $G$, and this “escape from subgroups” fact is a key ingredient in our forthcoming paper in which we demonstrate that random Cayley graphs in such groups are expander graphs.

It also has implications for results of Banach-Tarski type; it shows that for any semisimple Lie group G, and for generic $a, b \in G$, one can use $a, b$ to create Banach-Tarski paradoxical decompositions for all homogeneous spaces of $G$.  In particular there is one pair of $a,b$ that gives paradoxical decompositions for all homogeneous spaces simultaneously.

Our argument is based on a concept that we call “degeneration”.  Let $a, b$ be generic elements of $G$, and suppose for contradiction that $w_1(a,b), w_2(a,b)$ generically generated a group whose algebraic closure was conjugate to a proper algebraic subgroup $H$ of $G$.  Borel’s theorem lets us show that $w_1(a,b), w_2(a,b)$, and latex [w_1(a,b), w_2(a,b)]\$ each generate maximal tori of $G$, which by basic algebraic group theory can be used to show that $H$ must be a proper semisimple subgroup of $G$ of maximal rank.  If we were in the model case $G = SL_n$, then we would already be done, as there are no such maximal rank semisimple subgroups; but in the other groups, such proper maximal semisimple groups unfortunately exist.  Fortunately, they have been completely classified, and we take advantage of this classification in our argument.

The degeneration argument comes in as follows.  Let $(a,b)$ be a non-generic pair in $G \times G$.  Then $(a,b)$ lies in the Zariski closure of the generic pairs, which means that $(w_1(a,b), w_2(a,b))$ lies in the Zariski closure of the set formed by $H \times H$ and its conjugates.  In particular, if the non-generic pair is such that $w_1(a,b), w_2(a,b)$ generates a group that is dense in some proper algebraic subgroup $H'$, then $H' \times H'$ is in the Zariski closure of the union of the conjugates of $H \times H$.  When this happens, we say that $H'$ is a degeneration of $H$.  (For instance, $H$ could be the stabiliser of some non-degenerate quadratic form, and $H'$ could be the stabiliser of a degenerate limit of that form.)

The key fact we need (that relies on the classification, and a certain amount of representation theory) is:

Proposition. Given any proper semisimple maximal rank subgroup $H$ of $G$, there exists another proper semisimple subgroup $H'$ that is not a degeneration of $H$.

Using an induction hypothesis, we can find pairs $(a,b)$ such that $w_1(a,b), w_2(a,b)$ generate a dense subgroup of $H'$, which together with the preceding discussion contradicts the proposition.

The proposition is currently proven by using some known facts about certain representation-theoretic invariants of all the semisimple subgroups of the classical and exceptional simple Lie groups.  While the proof is of finite length, it is not particularly elegant, ultimately relying on the numerical value of one or more  invariants of $H$ being sufficiently different from their counterparts for $H'$ that one can prevent the latter being a degeneration of the former.  Perhaps there is another way to proceed here that is not based so heavily on classification.