arXiv:1606.03863v1 [] 13 Jun 2016

Some geometric properties of metric ultraproducts
of finite simple groups
Andreas Thom and John Wilson
Abstract. In this article we prove some previously announced results about metric ultraproducts of finite simple groups. We show that any non-discrete metric ultraproduct of alternating or special linear groups is a geodesic metric space. For more general non-discrete metric
ultraproducts of finite simple groups, we are able to establish path-connectedness. As expected, these global properties reflect asymptotic properties of various families of finite simple
groups.

1. Introduction
The study of metric ultraproducts of finite groups that are equipped with natural geometrically or combinatorially defined bi-invariant metrics has attracted some interest recently.
First of all, Elek and Szabó [4] established that a discrete group is sofic if and only it can
be embedded in a metric ultraproduct of alternating groups. Secondly, the model theory of
metric structures (as developed in [2]) provides a natural setting in which to study asymptotic metric properties of families of metric spaces. Families of finite simple groups serve
as interesting examples in this context. Some results about the metric geometry of such ultraproducts were announced in [10]. The aim of this article is to give complete proofs of some
of these results. We refer for the basic terminology to [10] and [9].
Throughout the paper, ω is a fixed non-principal ultrafilter on N. Let (Gi , di )i∈N be
a sequence of groups with normalized bi-invariant metrics. Then the metric ultraproduct
Q
(G, d) := i→ω (Gi , di ) is again a metric space, with metric induced by the pseudometric
d((gi ), (hi )) := lim di (gi , hi )
i→ω

Q

on the abstract ultraproduct i→ω Gi . The metric ultraproduct is the quotient of the abstract
ultraproduct that makes this pseudo-metric a metric; see [9]. It is clear that the metric
geometry of G reflects asymptotic features of the behaviour of the metrics di . When the groups
Gi are finite simple groups, the most natural choices of bi-invariant metrics are asymptotically
equivalent [9] and so the topology on G, unlike the metric itself, is independent of the choices.
1

2

ANDREAS THOM AND JOHN WILSON

In our context, the alternating groups An come equipped with the normalized Hamming
distance
µ −1
d(σ, τ ) = σ τ , for all σ, τ ∈ An ,
n
where for each permutation σ we write µσ ∈ {0, . . . , n} for the number of elements moved
by σ. The groups of Lie type are usually considered with the normalized rank metric or
projective rank metric, coming from the natural embedding into either GLn (F ) or PGLn (F )
for a field F and integer n; the normalized rank metric on GLn (F ) is defined by
rk(x − y)
, for all x, y ∈ GLn (F ),
n
and the projective rank metric on PGLn (F ) by
d(x, y) =

d(x̄, ȳ) = inf

λ∈F ×

rk(x − λy)
,
n

for all x, y ∈ PGLn (F ),

where x, y ∈ GLn (F ) denote arbitrary lifts of x̄, ȳ ∈ PGLn (F ). All of these metrics are
equivalent to the more intrinsically defined conjugacy metrics. See [9] for details.
The following result, which is essentially known, clarifies the implications of the classification of finite simple groups for the structure of metric ultraproducts of finite simple groups.
Theorem 1.1. Let (Gi )i∈N be a sequence of (non-abelian) finite simple groups with biQ
invariant metrics and let G be the metric ultraproduct i→ω (Gi , di ). If G is infinite then one
of the following holds:
(i) G is a Chevalley group (possibly twisted) over an ultraproduct of finite fields and the
metric on G is discrete;
(ii) G is a metric ultraproduct of alternating groups of ranks ni with ni →ω ∞, or
(iii) a metric ultraproduct of finite simple classical groups all of the same type (linear,
symplectic, unitary or orthogonal) over finite fields and of ranks ni with ni →ω ∞.
Proof. With finitely many exceptions each (non-abelian) finite simple groups is isomorphic to an alternating group or a (possibly twisted) group of Lie type j Xn (K) for a
suitable finite field K, where j Xn belongs to the list
An , Bn , Cn , Dn , 2 An , 2 Dn

(∗)

or to the list
E6 , E7 , E8 , F4 , G2 , 2 B2 , 3 D4 , 2 G2 , 2 E6 , 2 F4 .
Since an ultrafilter always selects one from a finite number of possibilities, it follows that G
is isomorphic to one of the following: (a) an ultraproduct of isomorphic simple groups, (b)
a group as in (ii) in the statement of the theorem, (c) an ultraproduct of groups j Xn with
X, j fixed. The groups in (a) are finite, and if in (c) the integers n are unbounded on each
set in U then G is as in (iii). If in case (c) the integers n are bounded on a set in U , then
they can be assumed constant, and then a theorem of F. Point [8] asserts that G is itself

METRIC ULTRAPRODUCTS OF FINITE SIMPLE GROUPS

3

a group j Xn (F ) for some field F ; being simple it coincides with the metric ultraproduct.
From [6, Corollary 1.9] there is a constant c such that each Gi has the property that for all
non-trivial g1 , g2 ∈ Gi the element g2 is a product of at most c conjugates of g1 . So therefore
does G. Choose h ∈ G \ {1} and let d(h, 1) = κ. If g ∈ G \ {1} then since h is a product
of at most c conjugates of g we have κ 6 cd(1, g). Therefore d(1, g) > κ/c and the result
follows.

Remark 1.2. If in (ii), (iii) above the group G is simple then it follows from a result of
[7] that the topology on G is induced by the conjugacy metrics on the groups Gi , and then
from results of [9] that the topology is induced from the Hamming metrics in case (ii) and
from the projective rank metrics in case (iii).
It follows from more general facts that the non-discrete simple groups in Theorem 1.1
are complete; see [11, p. 147]. Here we shall show that they are path-connected. We shall
deduce this as a consequence of a stronger result. We recall that a geodesic from one point
x of a metric space (X, d) to another point y is an isometric embedding p : [0, d(x, y)] → X
such that p(0) = x and p(d(x, y)) = y, and that X is called a geodesic space if all pairs of
points are connected by geodesics. A space X is called star-shaped if it has a point x0 that is
connected to each element of X by a geodesic.
Our main result is the following theorem.
Theorem 1.3. The non-discrete metric ultraproducts of sequences of alternating groups
equipped with the normalized Hamming metrics or projective linear groups equipped with the
projective rank metrics are complete geodesic metric spaces.
Moreover, every non-discrete simple metric ultraproduct G of finite simple groups has
geodesic subgroups H1 , H2 , H3 and a star-shaped subspace H4 such that G = H1 H2 H1 H3 H4 .
If follows that the groups in the first part above are path-connected. Let G and the
subspaces Hi be as in the second part. Given x, y ∈ G we write x−1 y = h1 . . . h5 with factors
hj in the appropriate subsets Hi . Concatenating translates of geodesics from 1 to the elements
hj we obtain a path from x to y. In particular, all non-discrete metric ultraproducts of finite
simple groups are path-connected.
2. Metric ultraproducts of groups An and PSLn
To prove that certain groups are geodesic metric spaces we use the following known result.
For the reader’s convenience we include a proof.
Proposition 2.1. Let (Xi , di ) be a sequence of metric spaces, and (εi ) a sequence of
positive reals with εi →ω 0. Suppose that for each i, and for all xi , yi in Xi there exists
zi ∈ Xi with di (xi , zi ) 6 21 di (xi , yi ) + εi and di (zi , yi ) 6 12 di (xi , yi ) + εi . Then the metric
Q
ultraproduct X = i→ω (Xi , di ) is a geodesic space.

4

ANDREAS THOM AND JOHN WILSON

Proof. Let x, y be elements of X, represented by sequences (xi )i∈N and (yi )i∈N . Let
d(x, y) = ℓ and let D be the set of dyadic rationals in the closed unit interval [0, 1]. We define
p(λℓ) for all λ ∈ D recursively as follows. Set pi (0) = xi and pi (ℓ) = yi . If λ is of form 2−k m
with 0 < k ∈ N and m an odd integer, define pi (λℓ) to be a point zi ∈ Xi satisfying
1
di (pi ((λ + j2−k )ℓ), zi ) 6 di (pi ((λ − 2−k )ℓ), pi ((λ + 2−k )ℓ) + εi
2

for j = ±1.

′

An easy induction on k + k′ shows that if λ = 2−k m and λ = 2−k m′ with k, k′ > 0 and
m, m′ odd then
di (pi (λ), pi (λ′ )) − |λ − λ′ | 6 (k + k′ )εi .
Therefore the sequence (pi )i∈N induces an isometry p from {λℓ | λ ∈ D} to X with p(0) = x
and p(ℓ) = y. Since the metric ultraproduct is complete, the map p admits a unique extension
to an isometric embedding from [0, ℓ] to X, and this shows that the metric ultraproduct is a
geodesic space.

Lemma 2.2. Let t ∈ [0, 1] and let g be an even permutation. Then there are even permutations h, k with g = hk such that |µh − tµg | 6 32 and |µk − (1 − t)µg | 6 32 .
Proof. It is well known that for each cycle π and integer l with 1 < l < µπ we can write
π = λρ with λ, ρ of lengths l, µπ − l + 1.
Consider the disjoint cycle decomposition of g; let h1 be a product of as many of the cycles
as possible subject to being even and satisfying µh1 6 tµg , and k1 a product of as many cycles
of h−1
1 g as possible with k1 even and µk1 6 (1 − t)µg . We may suppose that g 6= h1 k1 since
otherwise the conclusion is clear. Set r1 = tµg − µh1 and r2 = (1 − t)µg − µk1 . If d is a cycle
of odd length or a product of two cycles of even length in the cycle decomposition of k1−1 h−1
1 g
−1 −1
then µd > max(r1 , r2 ). Hence k1 h1 g = d and µd = µg − µh1 − µk1 = r1 + r2 .
First suppose that d is a cycle. If r1 6 32 take h = h1 ; if r1 > 32 and r2 6 23 take h = h1 d.
Otherwise µd > 5. Let l be the odd integer satisfying r1 − 12 6 l < r1 + 23 . Then 3 6 l 6 µd − 2
and r2 − 21 < µd − l + 1 6 r2 + 23 . Write d as a product of cycles h2 , k2 of lengths l and
µd − l + 1. It follows easily that the even permutations h = h1 h2 and k = k2 k1 have the
required property.
The procedure if d is a product of two cycles of even length is similar; one of the cycles of
d has length l 6 max(tµg − µh1 , (1 − t)µg − µk1 ) and we take it as one of the cycles of h2 or
k2 . The other cycle of length µd − l can be written as a product of two cycles one of which
has odd length in any prescribed interval of length 2 in [0, µd − l], and the result follows. 
Lemma 2.3. Let g be an element of GLn (F ) where F is a field. Suppose that g − λ1 has
rank r where λ 6= 0, and let ϕ1 , ϕ2 be non-negative real numbers with sum r. Then
(a) there exist elements h, k ∈ GLn (F ) with g = hk such that
|rk(h − λ1) − ϕ1 | 6 2

and

|rk(k − 1) − ϕ2 | 6 2.

(∗)

METRIC ULTRAPRODUCTS OF FINITE SIMPLE GROUPS

5

(b) If in addition g ∈ SLn (F ) then there exist h, k ∈ SLn (F ) with g = hk such that
|rk(h − λ1) − ϕ1 | 6 3

and

|rk(k − 1) − ϕ2 | 6 3.

Proof. (a) The element g1 = λ−1 g satisfies the hypothesis with λ = 1. If g1 = h1 k and
h1 − 1 and k − 1 have ranks satisfying the equivalents for g1 , h1 , k of the inequalities in (∗)
then h := λh1 and k have the correct property. Therefore we can suppose that λ = 1. We
argue by induction on n. Since result is clear for n = 1 we assume that n > 2.
First suppose that the natural module V for GLn (F ) is a cyclic (right) F hgi-module. Let
e1 be a module generator and define ei+1 = e1 (g − 1)i for i = 1, . . . , n − 1. Thus e1 , . . . , en is
a basis for V , and r = rk(g − 1) is n or n − 1. Write s = ⌈ϕ1 ⌉. If s = n we take h = g, k = 1;
if s = 0 we take h = 1, k = g. Otherwise, let h be the invertible map defined by ei 7→ ei + ei+1
for i 6 s, es+1 7→ e1 and ei 7→ ei for i > s + 1. So rk(h − 1) = s + 1. Moreover k := h−1 g fixes
e2 , . . . , es+1 , and so rk(k − 1) 6 n − s 6 n − ϕ1 6 1 + ϕ2 . Finally, since g − 1 = h(k − 1) + h − 1
we have
rk(k − 1) > rk(g − 1) − rk(h − 1) > r − s − 1 > r − ϕ1 − 2 = ϕ2 − 2.
The inequalities in (∗) follow.
Since in general V is a direct sum of cyclic F hgi-modules, we can now assume that V is a
direct sum U ⊕ W of non-zero F hgi-submodules. First suppose that r1 = rk(g − 1)|U satisfies
r1 6 ϕ1 . Then rk(g − 1)|W = r − r1 , and by induction applied to W and the numbers ϕ1 − r1 ,
ϕ2 we can find h2 , k2 ∈ GL(W ) with g|W = h2 k2 and
|rk(h2 − 1) − (ϕ1 − r1 )| 6 1 and

|rk(k2 − 1) − ϕ2 | 6 1.

Define h, k to be the maps acting as h2 , k2 on W and acting respectively as g and the identity
on U . It is clear that h, k have the required properties.
Similarly the result follows if r2 = rk(g−1)|W satisfies r2 6 ϕ2 . Since r1 +r2 = r = ϕ1 +ϕ2
one of the inequalities r1 6 ϕ1 , r2 6 ϕ2 must hold, and (a) follows.
(b) Let g ∈ SLn (F ). By (a) we can write g = h′ k′ with
|rk(h′ − λ1) − ϕ1 | 6 2 and

|rk(k′ − 1) − ϕ2 | 6 2.

Let d be the diagonal matrix with first diagonal entry det k′ and remaining diagonal entries
1, and define h = h′ d, k = d−1 k′ . Then h, k ∈ SLn (F ) and g = hk. Since h − λ1 =
(h′ −λ1)d+λ(d−1) we have |rk(h−λ1)−rk(h′ −λ1)| 6 1. Similarly |rk(k−1)−rk(k′ −1)| 6 1.
The assertion now follows.

Q
Proof of Theorem 1.3. First let G = i→ω (Ani , di ) be a metric ultraproduct of groups
Ani with the Hamming metric, and with ni →ω ∞. Define εi = 3/2ni for each i; thus εi →ω 0.
Given xi , yi ∈ Ani , set g = x−1
i yi and in the notation of Lemma 2.2 find h, k ∈ Ani with
g = hk, and
µh 6 21 µg + 32 , µk 6 12 µg + 32 .

6

ANDREAS THOM AND JOHN WILSON

Thus di (1, h) 6 12 di (1, g) + εi , di (1, k) 6 12 di (1, g) + εi . Hence di (xi , xi h) 6 12 di (xi , yi ) + εi ,
and
di (xi h, yi ) = di (xi g, yi k) = di (yi , yi k) = di (1, k) 6 21 di (xi , yi ) + εi .
Therefore the assertion of (a) for alternating groups follows from Proposition 2.1.
Q
The argument for ultraproducts G =
i→ω (PSLni (Fi ), di ) of (not necessarily finite)
groups of type PSL with the rank metric is similar. Define εi = 3/ni . Let x̄i , ȳi ∈ PSLni (Fi )
and let xi , yi be preimages in SLni (Fi ). Set g = x−1
i yi , choose λ ∈ F \ {0} with r := rk(g − λ1)
as small as possible, and using Lemma 2.3 find h, k ∈ SLni (Fi ) with rk(h − λ1) 6 21 r + 3,
rk(k − 1) 6 12 r + 3. Let h̄i be the image of h in PSLni (Fi ). Then di (x̄i , ȳi ) = r and
di (x̄i , x̄i h̄i ) 6 21 r + εi ,

di (x̄i h̄, ȳi ) 6 21 r + εi .

Hence the result follows from Proposition 2.1.



3. Geodesic and star-shaped subspaces
Our treatment of metric ultraproducts of the other classical groups depends basically
on the fact that these groups have a BN-pair. As preparation, we identify some geodesic
Q
subgroups of metric ultraproducts i→ω (PSLni (Fi ), di ). For the rest of the paper (with a
brief exception in the proof of Lemma 3.2 (b)), all metrics are defined in terms of projective
rank and they are denoted simply by d. For the classical groups under consideration, we do
not require the fields to be finite. We begin with a technical calculation.
In the next two lemmas, for n > 2 and an element λ of a field F , we write mλ,n for the
diagonal matrix with first diagonal entry λ−(n−1) and remaining diagonal entries λ.
Lemma 3.1. Let n > 2, and F be a field. Let E be the subgroup {mλ,n | λ 6= 0} of SLn (F ).
Let X be the set of matrices a ∈ GLn (F ) such that
dim ker(a − λ1n ) 6 dim ker(a − 1n ) + 2

for all λ 6= 0.

(∗∗∗)

Then GLn (F ) = EX.
Proof. Let a ∈ GLn (F ), and choose λ 6= 0 with s = dim ker(a−λ1n ) as large as possible.
Set x = mλ−1 ,n a. Since (mλ−1 ,n − λ−1 1n )a has rank at most 1, we have
dim ker(x − 1n ) = dim ker((mλ−1 ,n − λ−1 1n )a + λ−1 (a − λ1n )) > s − 1,
while for each µ 6= 0 we have
dim ker(x − µ1n ) = dim ker((mλ−1 ,n − λ−1 1n )a + λ−1 (a − λµ1n )) 6 s + 1.
Thus x ∈ X and the result follows.
Q
Lemma 3.2. Let G = i→ω (PSL2ni (Fi ), d), where ni →ω ∞.




(a) Let Ui , Vi be respectively the subgroup of SL2ni (Fi ) of all block matrices 10 k1 with

k skew-symmetric with zero diagonal and the subgroup of all matrices 10 k1 with k

METRIC ULTRAPRODUCTS OF FINITE SIMPLE GROUPS

7

symmetric, and let UiT , ViT be the images of Ui , Vi under the transposition map T .
Q
Q
Q
Q
Then the images in G of i→ω (Ui , d), i→ω (UiT , d), i→ω (Vi , d) and i→ω (ViT , d)
are geodesic subgroups.
(b) Suppose that ¯ is an automorphism of each Fi of order at most 2; write ¯ also for
the map induced on GLni (Fi ). For each i set
(
!
)
!
)
(
a−1 0
mλ−1 ,ni
0
Hi =
| λ 6= 0
| a ∈ GLni (Fi ) , Ei =
0 āT
0
mλ̄,ni

−1
and let Yi be the set of elements a 0 ā0T in Hi with a satisfying the condition (∗∗∗).
Q
Q
Q
Then the images H, E, Y of i→ω (Hi ∩ SL2ni (Fi ), d), i→ω (Ei , d), i→ω (Yi ∩
SL2ni (Fi ), d) satisfy H = EY ; the elements in E can be joined to the identity by a
path of length 1 and Y a star-shaped space.
Q
Proof. (a) Let U be the image in G of i→ω (Ui , d) and let u ∈ U be represented by the

sequence (ui ) with ui ∈ Ui for each i. Let ui = 10 k1i . Then there is a matrix ci ∈ GLni (Fi )
such that ki = ci di cTi , where di is the block diagonal matrix with si diagonal 2 × 2 entries

0 1
−1 0 and ni − 2si scalar entries 0; so rk(ui − 1) = 2si , see [1] for this and similar facts
especially in characteristic 2. For t ∈ [0, 1] let di (t) be the matrix having the same first 2⌊si t⌋
rows as di and remaining rows zero, and let ki (t) = ci di (t)cTi . Then
!
!
1 ki (t)
rk
− 1 = 2⌊si t⌋,
0
1
which differs from t rk(ui − 1) by at most 2. On the other hand, for λ 6= 0, 1 we have
!
!
1 ki (t)
rk
− λ1 = 2n.
0
1

So the composite of the function t 7→ 10 ki1(t) and the quotient map from the product of
the groups Ui to U is a geodesic in U from 1 to u.
Q
The proof for the image of i→ω (Vi , d) is similar but slightly more complicated. For ki
symmetric we can find ci ∈ GLni (Fi ) with ki = ci di cTi and with di a block diagonal matrix

with si entries 01 10 , then s′i non-zero diagonal entries, and finally ni − 2si − s′i zero entries;
here, si , s′i satisfy 2si + s′i = rk(ki ). We take di (t) to be the matrix having the same first
2⌊(si + 12 s′i )t⌋ rows as di and remaining rows zero, and argue as in the above paragraph. The
same arguments give the result for the transposed subgroups.
(b) It follows easily from Lemma 3.1 that Hi ∩ SL2ni (Fi ) = Ei (Yi ∩ SL2ni (Fi )) for each i
and hence H = EY . An argument very similar to the proof in (a) shows that every element
of E can be joined to the identity element by a path in H of length at most 1. We now
prove that for each y ∈ Y there is a geodesic in Y connecting 1 to
ℓ = d(1, y) be the
 y. Let

projective distance to the identity matrix and let the sequence

a−1
0
i
0 āT
i

represent y. So,

8

ANDREAS THOM AND JOHN WILSON

writing ri := rk(ai − 1) for each i, we can assume that rk(ai − λ1) > ri − 2 for each λ 6= 0. It
follows that
min(rk(ai − λ−1 1) + rk(ai − λ̄1)) > 2ri − 4.
λ6=0

n−1
i ri

Hence
→ω ℓ.
1ni −ri vi 
;
For each i there is a matrix xi ∈ GLni (Fi ) such that x−1
i ai xi has the form
0
wi
′
here wi is invertible and does not have 1 as an eigenvalue. Let wi be the result of multiplying
wi by a diagonal matrix with one diagonal entry (det wi )−1 and all other diagonal entries 1.
Thus wi′ ∈ SLri (Fi ) and dim rk(wi′ − 1ri ) 6 1. By Lemma 2.1 and Lemma 2.3, the metric
ultraproduct of the groups SLri (Fi ) with the ordinary rank metric is a geodesic space. In this
space the image w of the sequence (wi′ ) has length 1. Let (qi (t)) represent a geodesic from 1
to w in this space. Thus for all t1 , t2 ∈ [0, 1] we have r1i rk(qi (t1 )−1 qi (t2 ) − 1) → |t1 − t2 |.
For each t ∈ [0, ℓ] we now define
!
!
′ (t)−1
1
0
q
0
ni −ri
i
,
x−1
and pi (t) =
qi′ (t) = xi
i
qi′ (t)T
0
0
qi (ℓ−1 t)
and let (pi (t)) represent p(t) in G. Thus p(0) = 1, p(ℓ) = y. Moreover for t1 , t2 ∈ [0, ℓ] we
have
!
!
(qi′ (t1 )−1 qi′ (t2 ))−1
0
1
− λ1
min rk
T
2ni λ6=0
qi′ (t1 )−1 qi′ (t2 ))
0
6


T
1
rk((qi′ (t1 )−1 qi′ (t2 ))−1 − 1) + rk(qi′ (t1 )−1 qi′ (t2 )) − 1)
2ni

ri 1
rk((qi (λ−1 t1 )−1 qi (λ−1 t2 ))−1 − 1)
n i ri
→ |t1 − t2 | as i →ω ∞.
=

Therefore d(p(t1 ), p(t2 )) 6 |t1 − t2 | for all t1 , t2 and since d(p(0), p(ℓ)) = ℓ it follows that p is
a geodesic from 1 to y, as required.


4. Metric ultraproducts of symplectic, unitary and orthogonal groups
For the properties of symplectic, unitary and orthogonal groups that we shall use, we refer
the reader to [5, Chapter 2] and [3, Chapter 1 and §11.3].
Lemma 4.1. Let G = Sp2n (F ), and regard G as the subgroup of SL2n (F ) of block matrices

with adT − bcT = 1n and abT , cdT symmetric. Then G = U T U U T H where
(
!
)
(
!
)
1 k
1
0
U=
| k symmetric , U T =
| k symmetric
0 1
k 1

a b
c d

METRIC ULTRAPRODUCTS OF FINITE SIMPLE GROUPS

and
H=

(

a−1 0
0 aT

!

9

)

| a ∈ GLn (F ) .


Proof. Clearly U , U T , and H are subgroups of G. Let g = ac db ∈ G. There are

matrices y1 , y2 ∈ GLn (F ) withy1 ay2 = 10r 00 where r = rk(a). Thus we can find x1 , x2 ∈ H


with P an r × r
such that x1 gx2 has the form ac11 db11 with a1 = 10r 00 . Write b1 = PR Q
S

matrix. Since b1 aT1 is symmetric we have R = 0, and hence S is invertible since the matrix


0
(a1 b1 ) must have rank n. Let x3 = k1 01 , where k = 00 1n−r
; then x1 gx2 x3 has the form



a2 b2
1Q
where a2 is the invertible matrix 0 S . Adjusting the matrix x1 ∈ H to effect
c2 d2
further row operations in the matrix (a2 b2 ), we can now assume that x1 gx2 x3 has the form


1 b2
T
T
1 −b2 we find
c2 d2 . Now b2 , c2 are symmetric and d2 − b2 c2 = 1; and defining x4 = 0 1



that c12 db22 x4 = c12 01 . It follows that G = HU T U U T H. However H normalizes U and
U T , and so G = U T U U T H, as required.



The conclusion of Theorem 1.3 for metric ultraproducts of symplectic groups (over not
necessarily finite fields) follows immediately from Lemmas 3.2 and 4.1.
Modifications to the proof of Lemma 4.1, together with Lemma 3.2, give the conclusion
of the main theorem for metric ultraproducts of some other classical groups.
Lemma 4.2. Let G = Ω+
2n (F ) or G = SU2n (F ) with n > 3. In the latter case, there is an
involution ¯ on F ; in the former case we define ¯ to be the identity map on F . We regard G

as the subgroup of block matrices in SL2n (F ) of the form ac db with
ad¯T + bc̄T = 1n ,

ab̄T + bāT = 0

Then G = V V T V (H ∩ G) where
(
!
)
1 k
V =
| k̄T = −k ,
0 1
and
H=

(

a−1 0
0 āT

VT =
!

and

(

cd¯T + dc̄T = 0.

1 0
k 1

!

| k̄T = −k

(∗∗)

)

)

| a ∈ GLn (F ) .

Proof. Clearly V , V T are subgroups of G. We have H 6 G if and only if det a is fixed
by the automorphism ¯ for each a ∈ GLn (F ). In particular this holds if ¯ is the identity map
on F , and in this case the conclusion takes the form G = V V T V H.

Let g = ac db ∈ G. Arguing as in Lemma 4.1 we find x1 , x2 ∈ H such that x1 gx2 has the



, where P is an r × r matrix and S is invertible.
form ac11 db11 with a1 = 10r 00 and b1 = P0 Q
S
Changing x1 ∈ H to effect further row operations on (a1 b1 ), we can assume that S = 1n−r
and Q = 0.

10

ANDREAS THOM AND JOHN WILSON
T

We claim that there is an (n − r) × (n − r) matrix k1 satisfying k1 + k1 = 0 and with
rank n − r except possibly in the case when n − r is odd, char F 6= 2, and ¯ = idF ; and that
T
in this latter case there is an (n − r) × (n − r) matrix k1 satisfying k1 + k1 = 0 and with rank
at least n − r − 1. We begin with a matrix of block diagonal form with ⌊ 12 (n − r)⌋ diagonal

0 1 . If n − r is odd, we must add a final diagonal entry λ satisfying λ̄ = −λ. If
entries −1
0
char F = 2 we may take λ = 1. If char F 6= 2 and ¯ 6= idF then F is obtained from the fixed
field of ¯ by adjoining a square root, and we may take λ to be this square root. Otherwise,
we take λ = 0. Our claim follows.


Let k be the n×n block matrix 00 k01 and x3 the element k1 01 of V T . An easy calculation

shows that x1 gx2 x3 has the form ac22 db22 where rk(a2 ) > rk(k1 ).
Suppose that a2 is not invertible; so n is odd, char F 6= 2 and ¯ is the identity. In
particular, H 6 G. Adjusting x1 as above, we can assume that the top half of x1 gx2 x3 has

0P 0
form 1n−1
, with P an (n − 1) × (n − 1) matrix and each 0 a row or column vector.
0 0 0 1
From (∗∗) we have P + P̄ T = 0, and so y ∈ V T where


1n−1 0
0
0
 0
1
0
0


y=
.
 −P 0 1n−1 0
0

0

0

Therefore z := x1 gx2 x3 y ∈ G, and det z = 1. Now z

1n−1 0
0
 0
0
0


 c11 c12 d11
c21 c22 d21

and row expansion yields

1

has the form

0
1 

,
d12 
d22

c12 d11
det z = − det
c22 d21

!

.

However from (∗∗) we have
1n−1 0
0
0

!

d11 d21
d12 d22

!

+

!
!
c11 c21
0 0
= 1n ,
c12 c22
0 1


, and this is
and hence d11 = 1, d21 = 0, c12 = 0, c22 = 1. Therefore det z = − det 01 1n−1
0
equal to −1 because n is odd. Since char F 6= 2 this is a contradiction.

−1
It follows that a2 is invertible. Replacing x1 by a2 0T x1 we can assume that a2 = 1.
0 a2

The equations (∗∗) show that b2 + b̄T2 = 0, and setting x4 = 10 −b1 2 ∈ V we find that
x1 gx2 x3 x4 ∈ V T . Hence g ∈ HV T V V T H and g ∈ V T V V T H as H normalizes V , V T . Our
conclusion now follows.


METRIC ULTRAPRODUCTS OF FINITE SIMPLE GROUPS

11

The above lemma shows that the conclusion of Theorem 1.3 holds for metric ultraproducts
of (not necessarily finite) groups of type PSU2n (F ) and of type PΩ+
2n (F ). To complete the
proof of Theorem 1.3 it suffices to prove that all remaining metric ultraproducts of finite
simple groups that arise are isomorphic to such ultraproducts.
Q
Proposition 4.3.
(a) Every metric ultraproduct i→ω (PSUni (Fi ), d) with ni →ω ∞
is isomorphic to such an ultraproduct with all ni even.
(b) Every metric ultraproduct of finite simple orthogonal groups of ω-unbounded degree
is isomorphic to an ultraproduct of groups PΩ+
2mi (Fi ) with mi →ω ∞.
For the proof we need the following lemmas.
Lemma 4.4. Let V = L ⊕ S be an orthogonal decomposition of a space V with a nonsingular sesquilinear or quadratic form and let g : V → V be a form-preserving isomorphism.
Then there is a form-preserving isomorphism h : L → L with rk(g|L − h) 6 3 dim S.
Proof. We recall that every subspace W of V contains a subspace D with V = D ⊕ D ⊥
and dim(W/D) 6 dim(V /W ). If dim W 6 21 dim V this is clear; if not, then W has a nonzero subspace C on which the form is non-singular, and the assertion follows by induction on
consideration of C ⊥ and its subspace C ⊥ ∩ W .
Let M = g−1 (L); thus dim(V /(L ∩ M )) 6 2 dim S, and there is a subspace U 6 L ∩ M
with V = U ⊕ U ⊥ and with dim(V /U ) 6 4 dim S. The map g|U is a form-preserving map
to a subspace of L; since the form restricted to L is non-singular, by Witt’s Lemma g|U
can be extended to a form-preserving isomorphism h : L → L. Thus h|U = g|U and so
rk(g|L − h) 6 dim(L/U ) 6 3 dim S.

Lemma 4.5. For each i ∈ N let Vi be a vector space of finite dimension ni having a nonsingular sesquilinear or quadratic form fi , let Vi be the orthogonal direct sum of subspaces
Li , Si and let G′i , Hi′ be the derived groups of the groups Gi , Hi of form-preserving maps on
Vi , Li . Suppose that the sequence (dim Si ) is bounded. Then the maps Hi → Gi induce an
isomorphism of the metric ultraproducts H, G of the quotients Hi′ /Z(Hi′ ) and the quotients
G′i /Z(G′i ).
Proof. Suppose that dim Si 6 κ for each i. We may regard Hi as the subgroup of Gi
fixing Si pointwise. We may also assume that all quotients Hi /Z(Hi ), Gi /Z(Gi ) are simple;
thus the inclusion maps Hi′ → G′i induce maps Hi′ /Z(Hi′ ) → G′i /Z(G′i ) and induce a map
H → G that is evidently injective. To prove that it is surjective, it will suffice to prove that
it contains the commutator of any pair g(1) , g(2) of elements of the simple group G.
(1)
(2)
Let the sequences (gi ), (gi ) be preimages in the product of the groups (G′i ) of g(1) , g(2) .
(1) (2)
Using Lemma 4.4, for each i we construct hi , hi ∈ GL(Li ) preserving the form, with
(k)

rk(g(k) |Li − hi ) 6 3κ

for k = 1, 2.

12

ANDREAS THOM AND JOHN WILSON
(1)

(2)

It follows that for all non-zero λi , λi
(k)

|rk(gi

(k)

in the field we have
(k)

− λi id) − rk(hi

(k)

− λi id)| 6 6κ
(k)

for k = 1, 2,

(k)

and hence for k = 1, 2 the two sequences (hi ), (gi ) map to the same element of G. The
result now follows easily.

Proof of Proposition 4.3. The simple unitary and orthogonal groups are the quotients modulo the centre of derived groups of isometry groups of spaces with a non-singular
unitary form and spaces with a non-singular quadratic form. Every finite-dimensional space
with a non-singular unitary form has an orthonormal basis (see for example [5, Proposition
2.3.1]); and so it can be written as an orthogonal direct sum L ⊕ S with dim L even and
dim S 6 1.
Let V be a finite space with a non-singular quadratic form Q and associated bilinear form
ϕ. Then V has a basis of one of the following types:
(i) {e1 , . . . , em , f1 . . . , fm } with Q(ei ) = Q(fi ) = 0 and ϕ(ei , fj ) = δij for all i, j;
(ii) {e1 , . . . , em , f1 . . . , fm , x, y} with Q(ei ) = Q(fi ) = 0 and ϕ(ei , fj ) = δij for all i, j
and with the subspace spanned by x, y an orthogonal direct summand;
(iii) {e1 , . . . , em , f1 . . . , fm , x} with Q(ei ) = Q(fi ) = 0 and ϕ(ei , fj ) = δij for all i, j and
with the subspace spanned by x an orthogonal direct summand.
(See [5, Proposition 2.5.3].) Take L to be the space spanned by the elements ei , fj and S to
be the space spanned by the remaining basis elements. In each case we obtain an orthogonal
decomposition V = L ⊕ S with dim S 6 2 and with the form induced on L having isometry
group with derived group Ω+
2m . Therefore the above lemmas can be applied and Proposition
4.3 follows.

Acknowledgments
The research leading to these results has received funding from the European Research
Council under the European Union’s Seventh Framework Programme – ERC Starting Grant
n◦ 277728. This paper was written while the second author was Leibniz Professor at the
University of Leipzig, and he thanks this university for its hospitality.
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A.T., Institut für Geometrie, TU Dresden, Germany
E-mail address: andreas.thom@tu-dresden.de
J.W., Mathematical Institute, Oxford, United Kingdom
E-mail address: john.wilson@maths.ox.ac.uk

