. 2
( 3)


so we choose » = f (w).
Now let W1 be the hyperbolic line v, w ’ »v , and let V1 = W1 , where orthog-

onality is de¬ned with respect to the form B. It is easily checked that V = V1 •W1 ,
and the restriction of the form to V1 is still non-degenerate. Now the existence of
the decomposition follows by induction.
The uniqueness of the decomposition will be proved later, as a consequence
of Witt™s Lemma (Theorem 3.15).
The number r of hyperbolic lines is called the polar rank of V , and (the iso-
morphism type of) U is called the germ of V .
To complete the classi¬cation of forms over a given ¬eld, it is necessary to
determine all the anisotropic spaces. In general, this is not possible; for exam-
ple, the study of positive de¬nite quadratic forms over the rational numbers leads
quickly into deep number-theoretic waters. I will consider the cases of the real
and complex numbers and ¬nite ¬elds.
First, though, the alternating case is trivial:
Proposition 3.7 The only anisotropic space carrying an alternating bilinear form
is the zero space.
In combination with Theorem 3.6, this shows that a space carrying a non-
degenerate alternating bilinear form is a direct sum of hyperbolic planes.
Over the real numbers, Sylvester™s theorem asserts that any quadratic form in
n variables is equivalent to the form
2 2 2 2
x1 + . . . + xr ’ xr+1 ’ . . . ’ xr+s ,
for some r, s with r + s ¤ n. If the form is non-singular, then r + s = n. If both r
and s are non-zero, there is a non-zero singular vector (with 1 in positions 1 and
r + 1, 0 elsewhere). So we have:

Proposition 3.8 If V is a real vector space of rank n, then an anisotropic form
on V is either positive de¬nite or negative de¬nite; there is a unique form of each
type up to invertible linear transformation, one the negative of the other.

The reals have no non-identity automorphisms, so Hermitian forms do not
Over the complex numbers, the following facts are easily shown:

(a) There is a unique non-singular quadratic form (up to equivalence) in n vari-
ables for any n. A space carrying such a form is anisotropic if and only if
n ¤ 1.

(b) If σ denotes complex conjugation, the situation for σ-Hermitian forms is the
same as for quadratic forms over the reals: anisotropic forms are positive or
negative de¬nite, and there is a unique form of each type, one the negative
of the other.

For ¬nite ¬elds, the position is as follows.

Theorem 3.9 (a) An anisotropic quadratic form in n variables over GF(q) ex-
ists if and only if n ¤ 2. There is a unique form for each n except when n = 1
and q is odd, in which case there are two forms, one a non-square multiple
of the other.

(b) Let q = r2 and let σ be the ¬eld automorphism ± ’ ±r . Then there is an
anisotropic σ-Hermitian form in n variables if and only if n ¤ 1. The form
is unique in each case.

Proof (a) Consider ¬rst the case where the characteristic is not 2. The multiplica-
tive group of GF(q) is cyclic of even order q ’ 1; so the squares form a subgroup
of index 2, and if · is a ¬xed non-square, then every non-square has the form ·±2
for some ±. It follows easily that any quadratic form in one variable is equivalent
to either x2 or ·x2 .
Next, consider non-singular forms in two variables. By completing the square,
such a form is equivalent to one of x2 + y2 , x2 + ·y2 , ·x2 + ·y2 .
Suppose ¬rst that q ≡ 1 (mod 4). Then ’1 is a square, say ’1 = β2 . (In
the multiplicative group, ’1 has order 2, so lies in the subgroup of even order
2 (q ’ 1) consisting of squares.) Thus x + y = (x + βy)(x ’ βy), and the ¬rst and
1 2 2

third forms are not anisotropic. Moreover, any form in 3 or more variables, when

converted to diagonal form, contains one of these two, and so is not anisotropic
Now consider the other case, q ≡ ’1 (mod 4). Then ’1 is a non-square
(since the group of squares has odd order), so the second form is (x + y)(x ’ y),
and is not anisotropic. Moreover, the set of squares is not closed under addition
(else it would be a subgroup of the additive group, but 1 (q + 1) doesn™t divide q);
so there exist two squares whose sum is a non-square. Multiplying by a suitable
square, there exist β, γ with β2 + γ2 = ’1. Then

’(x2 + y2 ) = (βx + γy)2 + (γx ’ βy)2 ,

and the ¬rst and third forms are equivalent. Moreover, a form in three variables
is certainly not anisotropic unless it is equivalent to x2 + y2 + z2 , and this form
vanishes at the vector (β, γ, 1); hence there is no anisotropic form in three or more
The characteristic 2 case is an exercise (see below).
(b) Now consider Hermitian forms. If σ is an automorphism of GF(q) of order
2, then q is a square, say q = r2 , and ±σ = ±r . We need the fact that every element
of Fix(σ) = GF(r) has the form ±±σ (see Exercise 3.3).
In one variable, we have f (x) = µxxσ for some non-zero µ ∈ Fix(σ); writing
µ = ±±σ and replacing x by ±x, we can assume that µ = 1.
In two variables, we can similarly take the form to be xxσ + yyσ . Now ’1 ∈
Fix(σ), so ’1 = »»σ ; then the form vanishes at (1, »). It follows that there is no
anisotropic form in any larger number of variables either.

Exercise 3.11 Prove that there is, up to equivalence, a unique non-degenerate al-
ternating bilinear form on a vector space of countably in¬nite dimension (a direct
sum of countably many isotropic planes).

Exercise 3.12 Let F be a ¬nite ¬eld of characteristic 2.

(a) Prove that every element of F has a unique square root.

(b) By considering the bilinear form obtained by polarisation, prove that a non-
singular form in 2 or 3 variables over F is equivalent to ±x2 + xy + βy2
or ±x2 + xy + βy2 + γz2 respectively. Prove that forms of the ¬rst shape
(with ±, β = 0) are all equivalent, while those of the second shape cannot be

3.4 Polar spaces
Polar spaces describe the geometry of vector spaces carrying a re¬‚exive sesquilin-
ear form or a quadratic form in much the same way as projective spaces describe
the geometry of vector spaces. We now embark on the study of these geometries;
the three preceding sections contain the prerequisite algebra.
First, some terminology. The polar spaces associated with the three types of
forms (alternating bilinear, Hermitian, and quadratic) are referred to by the same
names as the groups associated with them: symplectic, unitary, and orthogonal
respectively. Of what do these spaces consist?
Let V be a vector space carrying a form of one of our three types. Recall that
as well as a sesquilinear form b in two variables, we have a form f in one variable
” either f is de¬ned by f (x) = B(x, x), or b is obtained by polarising f ” and
we make use of both forms. A subspace of V on which B vanishes identically
is called a B-¬‚at subspace, and one on which f vanishes identically is called a
f -¬‚at subspace. (Note: these terms are not standard; in the literature, such spaces
are called totally isotropic (t.i.) and totally singular (t.s.) respectively.) The
unquali¬ed term ¬‚at subspace will mean a B-¬‚at subspace in the symplectic or
unitary case, and a q-¬‚at subspace in the orthogonal case.
The polar space associated with a vector space carrying a form is the geometry
whose ¬‚ats are the ¬‚at subspaces (in the above sense). Note that, if the form is
anisotropic, then the only member of the polar space is the zero subspace. The
polar rank of a classical polar space is the largest vector space rank of any ¬‚at
subspace; it is zero if and only if the form is anisotropic. Where there is no
confusion, polar rank will be called simply rank. (We will soon see that there is
no con¬‚ict with our earlier de¬nition of rank as the number of hyperbolic planes
in the decomposition of the space.) We use the terms point, line, plane, etc., just
as for projective spaces.
Polar spaces bear the same relation to formed spaces as projective spaces do
to vector spaces.
We now proceed to derive some properties of polar spaces. Let “ be a classical
polar space of polar rank r.

(P1) Any ¬‚at, together with the ¬‚ats it contains, is a projective space of dimen-
sion at most r ’ 1.

(P2) The intersection of any family of ¬‚ats is a ¬‚at.

(P3) If U is a ¬‚at of dimension r ’ 1 and p a point not in U, then the union of the

planes joining p to points of U is a ¬‚at W of dimension r ’ 1; and U ©W is
a hyperplane in both U and W .

(P4) There exist two disjoint ¬‚ats of dimension r ’ 1.
(P1) is clear since a subspace of a ¬‚at subspace is itself ¬‚at. (P2) is also clear.
To prove (P3), let p = y . The function x ’ B(x, y) on the vector space U is
linear; let K be its kernel, a hyperplane in U. Then the line (of the projective
space) joining p to a point q ∈ U is ¬‚at if and only if q ∈ K; and the union of all
such ¬‚at lines is a ¬‚at space W = K, y , such that W ©U = K, as required.
Finally, to prove (P4), we use the hyperbolic-anisotropic decomposition again.
If L1 , . . . , Lr are the hyperbolic planes, and xi , yi are the distinguished spanning
vectors in Li , then the required ¬‚ats are x1 , . . . , xr and y1 , . . . , yr .
The signi¬cance of the geometric properties (P1)“(P4) lies in the major result
of Veldkamp and Tits which determines all the geometries of rank at least 3 which
satisfy them. All these geometries are polar spaces (as we have de¬ned them) or
slight generalisations, together with a couple of exceptions of rank 3. In particular,
the following theorem holds:

Theorem 3.10 A ¬nite geometry satisfying (P1)“(P4) with r ≥ 3 is a polar space.

Exercise 3.13 Let P = PG(3, F) for some (not necessarily commutative) division
ring F. Construct a new geometry “ as follows:
(a) the ˜points™ of “ are the lines of P;

(b) the ˜lines™ of “ are the plane pencils in P (consisting of all lines lying in a
plane Π and containing a point p of Π);

(c) the ˜planes™ of “ are of two types: the pencils (consisting of all the lines
through a point) and the dual planes (consisting of all the lines in a plane).
Prove that “ satis¬es (P1)“(P4) with r = 3.
Prove that, if F is not isomorphic to its opposite, then “ contains non-isomorphic
(We will see later that, if F is commutative, then “ is an orthogonal polar

Exercise 3.14 Prove the Buekenhout“Shult property of the geometry of points
and lines in a polar space: if p is a point not lying on a line L, then p is collinear
with one or all points of L.

You should prove this both from the analytic description of polar spaces, and
using (P1)“(P4).

In a polar space “, given any set S of points, we let S⊥ denote the set of points
which are perpendicular to (that is, collinear with) every point of S. Polar spaces
have good inductive properties. Let G be a classical polar space. There are two
natural ways of producing a “smaller” polar space from G:
(a) Take a point x of G, and consider the quotient space x⊥ /x, the space whose
points, lines, . . . are the lines, planes, . . . of G containing x.

(b) Take two non-perpendicular points x and y, and consider {x, y}⊥ .
In each case, the space constructed is a classical polar space, having the same
germ as G but with polar rank one less than that of G. (Note that, in (b), the span
of x and y in the vector space is a hyperbolic plane.)

Exercise 3.15 Prove the above assertions.

There are more general versions. For example, if S is a ¬‚at of dimension d ’ 1,
then S⊥ /S is a polar space of rank r ’ d with the same germ as G. We will see
below how this inductive process can be used to obtain information about polar
We investigate just one type in more detail, the so-called hyperbolic quadric,
the orthogonal space which is a direct sum of hyperbolic planes (that is, having
germ 0). The quadratic form de¬ning this space can be taken to be x1 x2 + x3 x4 +
. . . + x2r’1 x2r .

Proposition 3.11 The maximal ¬‚ats of a hyperbolic quadric fall into two classes,
with the properties that the intersection of two maximal ¬‚ats has even codimension
in each if and only if they belong to the same class.

Proof First, note that the result holds when r = 1, since then the quadratic form is
x1 x2 and there are just two singular points, (1, 0) and (0, 1) . By the inductive
principle, it follows that any ¬‚at of dimension r ’ 2 is contained in exactly two
maximal ¬‚ats.
We take the (r ’1)-¬‚ats and (r ’2)-¬‚ats as the vertices and edges of a graph “,
that is, we join two (r ’ 1)-¬‚ats if their intersection is an (r ’ 2)-¬‚at. The theorem
will follow if we show that “ is connected and bipartite, and that the distance
between two vertices of “ is the codimension of their intersection. Clearly the

codimension of the intersection increases by at most one with every step in the
graph, so it is at most equal to the distance. We prove equality by induction.
Let U be a (r ’ 1)-¬‚at and K a (r ’ 2)-¬‚at. We claim that the two (r ’ 1)-
spaces W1 ,W2 containing K have different distances from U. Factoring out the
¬‚at subspace U © K and using induction, we may assume that U © K = 0. Then
U © K ⊥ is a point p, which lies in one but not the other of W1 ,W2 ; say p ∈ W1 . By
induction, the distance from U to W1 is r ’ 1; so the distance from U to W2 is at
most r, hence equal to r by the remark in the preceding paragraph.
This establishes the claim about the distance. The fact that “ is bipartite also
follows, since in any non-bipartite graph there exists an edge both of whose ver-
tices have the same distance from some third vertex, and the argument given shows
that this doesn™t happen in “.

In particular, the rank 2 hyperbolic quadric consists of two families of lines
forming a grid, as shown in Figure 1. This is the so-called “ruled quadric”, famil-
iar from models such as wastepaper baskets.

¥ ¢ e¡ t e7 f i  ¥
¡ e
ff ii d ¢ ¡ e t7 e f    7
d¥ i  ¡ ¥ ¢
¥ d¡  e t e  i  ¥ ¢¡
t ¢ f
t i d¥ ¢ d 7
f ¢
e 7
e ¡

d¡  7 e t e i7¥¡
fie¥ ¢ ¢
t d     fi
fte d 7 d e t e f¢
e¥ ¢ ¡ 
i 7 ¡¥
d     7¡i¢¥
7 et e
if¢ ¡ e   d
¥i¡ t 7 d   et e¥i f
f   7 ¡¢
¢7t e   d etfi
¥ d 7  ¡ ¢ ¥e
¢¥7 f e  e t
 i f e t e 7d d ¢ ¥efe i
 ¡ d tf
¢¥ it

¡   i f   t 7  d ¢ ¥ de
7 e

¢¥ i fe
 i   e 7  ¡ d ¥ dt
¡¢ e
   f te ¢
d if
¥ i f 7e  ¡e ¢ ¥

Figure 1: A ruled quadric

Exercise 3.16 Show that Proposition 3.11 can be proved using only properties
(P1)“(P4) of polar spaces together with the fact that an (r ’ 1)-¬‚at lies in exactly
two maximal ¬‚ats.

3.5 Finite polar spaces
The classi¬cation of ¬nite classical polar spaces was achieved by Theorem 3.6.
We subdivide these spaces into six families according to their germ, viz., one

symplectic, two unitary, and three orthogonal. (Forms which differ only by a
scalar factor obviously de¬ne the same polar space.) The following table gives
some information about them. In the table, r denotes the polar space rank, and δ
the vector space rank of the germ; the rank n of the space is given by n = 2r + δ.
The signi¬cance of the parameter µ will emerge shortly. This number, depending
only on the germ, carries numerical information about all spaces in the family.
Note that, in the unitary case, the order of the ¬nite ¬eld must be a square.

δ µ
Symplectic 0 0
Unitary 0
Unitary 1 2
Orthogonal 0
Orthogonal 1 0
Orthogonal 2 1
Table 1: Finite polar spaces

Theorem 3.12 The number of points in a ¬nite polar space of rank 1 is q1+µ + 1,
where µ is given in Table 1.

Proof Let V be a vector space carrying a form of rank 1 over GF(q). Then V
is the orthogonal direct sum of a hyperbolic line L and an anisotropic germ U of
dimension k (say). Let nk be the number of points.
Suppose that k > 0. If p is a point of the polar space, then p lies on the hyper-
plane p⊥ ; any other hyperplane containing p is non-degenerate with polar rank 1
and having germ of dimension k ’ 1. Consider a parallel class of hyperplanes in
the af¬ne space whose hyperplane at in¬nity is p⊥ . Each such hyperplane con-
tains nk’1 ’ 1 points, and the hyperplane at in¬nity contains just one, viz., p. So
we have
nk ’ 1 = q(nk’1 ’ 1),
from which it follows that nk = 1 + (n0 ’ 1)qk . So it is enough to prove the result
for the case k = 0, that is, for a hyperbolic line.
In the symplectic case, each of the q + 1 projective points on a line is isotropic.
Consider the unitary case. We can take the form to be

B((x1 , y1 ), (x2 , y2 )) = x1 y2 + y1 x2 ,

where x = xσ = xr , r2 = q. So the isotropic points satisfy xy + yx = 0, that is,
Tr(xy) = 0. How many pairs (x, y) satisfy this? If y = 0, then x is arbitrary. If
y = 0, then a ¬xed multiple of x is in the kernel of the trace map, a set of size q1/2
(since Tr is GF(q1/2 )-linear). So there are

q + (q ’ 1)q1/2 = 1 + (q ’ 1)(q1/2 + 1)

vectors, i.e., q1/2 + 1 projective points.
Finally, consider the orthogonal case. The quadratic form is equivalent to xy,
and has two singular points, (1, 0) and (1, 0) .

Theorem 3.13 In a ¬nite polar space of rank r, there are (qr ’ 1)(qr+µ + 1)/(q ’
1) points, of which q2r’1+µ are not perpendicular to a given point.

Proof We let F(r) be the number of points, and G(r) the number not perpen-
dicular to a given point. (We do not assume that G(r) is constant; this constancy
follows from the induction that proves the theorem.) We use the two inductive
principles described at the end of the last section.
Claim 1: G(r) = q2 G(r ’ 1).
Take a point x, and count pairs (y, z), where y ∈ x⊥ , z ∈ x⊥ , and z ∈ y⊥ . Choos-
ing z ¬rst, there are G(r) choices; then x, z is a hyperbolic line, and y is a point in
x, z ⊥ , so there are F(r ’ 1) choices for y. On the other hand, choosing y ¬rst, the
lines through y are the points of the rank r ’ 1 polar space x⊥ /x, and so there are
F(r ’ 1) of them, with q points different from x on each, giving qF(r ’ 1) choices
for y; then x, y and y, z are non-perpendicular lines in y⊥ , i.e., points of y⊥ /y,
so there are G(r ’ 1) choices for y, z , and so qG(r ’ 1) choices for y. thus

G(r) · F(r ’ 1) = qF(r ’ 1) · qG(r ’ 1),

from which the result follows.
Since G(1) = q1+µ , it follows immediately that G(r) = q2r’1+µ , as required.
Claim 2: F(r) = 1 + qF(r ’ 1) + G(r).
For this, simply observe (as above) that points perpendicular to x lie on lines
of x⊥ /x.
Now it is just a matter of calculation that the function (qr ’ 1)(qr+µ + 1)/(q ’
1) satis¬es the recurrence of Claim 2 and correctly reduces to q1+µ + 1 when
r = 1.

Theorem 3.14 The number of maximal ¬‚ats in a ¬nite polar space of rank r is
∏(1 + qi+µ).

Proof Let H(r) be this number. Count pairs (x,U), where U is a maximal ¬‚at
and x ∈ U. We ¬nd that
F(r) · H(r ’ 1) = H(r) · (qr ’ 1)/(q ’ 1),
H(r) = (1 + qr+µ )H(r ’ 1).
Now the result is immediate.
It should now be clear that any reasonable counting question about ¬nite polar
spaces can be answered in terms of q, r, µ. We will do this for the associated
classical groups at the end of the next section.

3.6 Witt™s Lemma
Let V be a formed space, with sesquilinear form B and (if appropriate) quadratic
form q. An isometry of V is a linear map g : V ’ V which satis¬es B(xg, yg) =
B(x, y) for all x, y ∈ V , and (if appropriate) q(xg) = q(x) for all x ∈ V . (Note that,
in the case of a quadratic form, the second condition implies the ¬rst.)
The set of all isometries of V forms a group, the isometry group of V . This
group is our object of study for the next few sections.
More generally, if V and W are formed spaces of the same type, an isometry
from V to W is a linear map from V to W satisfying the conditions listed above.
Exercise 3.17 Let V be a (not necessarily non-degenerate) formed space of sym-
plectic or Hermitian type, with radical V ⊥ . Prove that the natural map from V to
V /V ⊥ is an isometry.
The purpose of this subsection is to prove Witt™s Lemma, a transitivity assertion
about the isometry group of a formed space.
Theorem 3.15 Suppose that U1 and U2 are subspaces of the formed space V , and
h : U1 ’ U2 is an isometry. Then there is an isometry g of V which extends h if
and only if (U1 ©V ⊥ )h = U2 ©V ⊥ .
In particular, if V ⊥ = 0, then any isometry between subspaces of V extends to
an isometry of V .

Proof Assume that h : U1 ’ U2 is an isometry. Clearly, if h is the restriction of
an isometry g of V , then V ⊥ g = V ⊥ , and so

(U1 ©V ⊥ )h = (U1 ©V ⊥ )g = U1 g ©V ⊥ g = U2 ©V ⊥ .

We have to prove the converse.
First we show that we may assume that U1 and U2 contain V ⊥ . Suppose not.
Choose a subspace W of V ⊥ which is a complement to both U1 ©V ⊥ and U2 ©V ⊥
(see Exercise 3.18), and extend h to U1 • W by the identity map on W . This is
easily checked to be an isometry to U2 •W .
The proof is by induction on rk(U1 /V ⊥ ). If U1 = V⊥ = U2 , then choose any
complement W for V ⊥ in V and extend h by the identity on W . So the base step
of the induction is proved. Assume that the conclusion of Witt™s Lemma holds for
V , U1 , U2 , h whenever rk(U1 /(V )⊥ ) < rk(U1 /V ⊥ ).
Let H be a hyperplane of U1 containing V ⊥ . Then the restriction f of f to H
has an extension to an isometry g of V . Now it is enough to show that h(g )’1
extends to an isometry; in other words, we may assume that h is the identity on H.
Moreover, the conclusion is clear if h is the identity on U1 ; so suppose not. Then
ker(h ’ 1) = H, and so the image of h ’ 1 is a rank 1 subspace P of U1 .
Since h is an isometry, for all x, y ∈ U1 we have

B(xh, yh ’ y) = B(xh, yh) ’ B(xh, y)
= B(x, y) ’ B(xh, y)
= B(x ’ xh, y).

So, if y ∈ H, then any vector xh ’ x of P is orthogonal to y; that is, H ¤ P⊥ .
Now suppose that P ¤ U1 . Then U1 © P⊥ = U2 © P⊥ = H. If W is a comple-

ment to H in P⊥ , then we can extend h by the identity on W to obtain the required
isometry. So we may assume further that U1 ,U2 ¤ P⊥ . In particular, P ¤ P⊥ .
Next we show that we may assume that U1 = U2 = P⊥ . Suppose ¬rst that
U1 = U2 . If Ui = H, ui for i = 1, 2, let W0 be a complement for U1 + U2 in
P⊥ , and W = W0 , u1 + u2 ; then h can be extended by the identity on W to an
isometry on P⊥ . If U1 = U2 , take any complement W to U1 in P⊥ . In either case,
the extension is an isometry of P⊥ which acts as the identity on a hyperplane H
of P⊥ containing H. So we may replace U1 ,U2 , H by P⊥ , P⊥ , H .
Let P = x and let x = uh ’ u for some u ∈ U1 . We have B(x, x) = 0. In the
orthogonal case, we have

q(x) = q(uh ’ u) = q(uh) + q(u) ’ B(uh, u) = 2q(u) ’ B(u, u) = 0.

(We have B(uh, u) = B(u, u) because B(uh ’ u, u) = 0.) So P is ¬‚at, and there is a
hyperbolic plane u, v , with v ∈ P⊥ . Our job is to extend h to the vector v.
To achieve this, we show ¬rst that there is a vector v such that uh, v ⊥ =
u, v ⊥ . This holds because u, v ⊥ is a hyperplane in uh ⊥ not containing V ⊥ .
Next, we observe that uh, v is a hyperbolic plane, so we can choose a vector
v such that B(uh, v ) = 1 and (if relevant) Q(v ) = 0.
Finally, we observe that by extending h to map v to v we obtain the required
isometry of V .

Exercise 3.18 Let U1 and U2 be subspaces of a vector space V having the same
rank. Show that there is a subspace W of V which is a complement for both U1
and U2 .

Corollary 3.16 (a) The ranks of maximal ¬‚at subspaces of a formed space are
all equal.

(b) The Witt rank and isometry type of the germ of a non-degenerate formed
space are invariants.

Proof (a) Let U1 and U2 be maximal ¬‚at subspaces. Then both U1 and U2 con-
tains V ⊥ . If rk(U1 ) < rk(U2 ), there is an isometry h from U1 into U2 . If g is the
extension of h to V , then the image of U2 under g’1 is a ¬‚at subspace properly
containing U1 , contradicting maximality.
(b) The result is clear if V is anisotropic. Otherwise, let U1 and U2 be hyper-
bolic planes. Then U1 and U2 are isometric and are disjoint from V ⊥ . An isometry
⊥ ⊥
of V carrying U1 to U2 takes U1 to U2 . Then the result follows by induction.

Theorem 3.17 Let Vr be a non-degenerate formed space with polar rank r and
germ W over GF(q). Let Gr be the isometry group of Vr . Then

∏(qi ’ 1)(qi+µ + 1)q2i’1+µ
|Gr | = |G0 |
∏(qi ’ 1)(qi+µ + 1)
=q |G0 |,

where |G0 | is given by the following table:

δµ |G0 |
Symplectic 00 1
0 ’1
Unitary 1
q1/2 + 1
Unitary 12
0 ’1
Orthogonal 1
2 (q odd)
Orthogonal 1 0
1 (q even)
2(q + 1)
Orthogonal 2 1

Proof By Theorem 3.13, the number of choices of a vector x spanning a ¬‚at
subspace is (qr ’ 1)(qr+µ + 1). Then the number of choices of a vector y spanning
a ¬‚at subspace and having inner product 1 with x is q2r’1+µ . Then x and y span
a hyperbolic plane. Now Witt™s Lemma shows that Gr acts transitively on such
pairs, and the stabiliser of such a pair is Gr’1 , by the inductive principle.
In the cases where δ = 0, G0 is the trivial group on a vector space of rank 0.
In the unitary case with δ = 1, G0 preserves the Hermitian form xxq , so consists
of multiplication by (q1/2 + 1)st roots of unity. In the orthogonal case with δ =
1, G0 preserves the quadratic form x2 , and so consists of multiplication by ±1
only. Finally, consider the orthogonal case with δ = 2. Here we can represent
the quadratic form as the norm from GF(q2 ) to GF(q), that is, N(x) = xq+1 . The
GF(q)-linear maps which preserve this form a dihedral group of order 2(q + 1):
the cyclic group is generated by the (q + 1)st roots of unity in GF(q2 ), which is
inverted by the non-trivial ¬eld automorphism over GF(q) (since, if xq+1 = 1, then
xq = x’1 ).

4 Symplectic groups
In this and the next two sections, we begin the study of the groups preserving
re¬‚exive sesquilinear forms or quadratic forms. We begin with the symplectic
groups, associated with non-degenerate alternating bilinear forms.

4.1 The Pfaf¬an
The determinant of a skew-symmetric matrix is a square. This can be seen in
small cases by direct calculation:

0 a12
= a2 ,
det 12
’a12 0
« 
0 a12 a13 a14
¬ ’a12 a24 ·
0 a23
det ¬ · = (a12 a34 ’ a13 a24 + a14 a23 )2 .
 ’a13 a34 
’a23 0
’a14 ’a24 ’a34 0

Theorem 4.1 (a) The determinant of a skew-symmetric matrix of odd size is

(b) There is a unique polynomial Pf(A) in the indeterminates ai j for 1 ¤ i < j ¤
2n, having the properties

(i) if A is a skew-symmetric 2n — 2n matrix with (i, j) entry ai j for 1 ¤ i <
j ¤ 2n, then
det(A) = Pf(A)2 ;
(ii) Pf(A) contains the term a12 a34 · · · a2n’1 2n with coef¬cient +1.

Proof We begin by observing that, if A is a skew-symmetric matrix, then the
form B de¬ned by
B(x, y) = xAy
is an alternating bilinear form. Moreover, B is non-degenerate if and only if A is
non-singular: for xAy = 0 for all y if and only if xA = 0. We know that there is
no non-degenerate alternating bilinear form on a space of odd dimension; so (a)
is proved.

We know also that, if A is singular, then det(A) = 0, whereas if A is non-
singular, then there exists an invertible matrix P such that

01 01
PAP = diag ,..., ,
’1 0 ’1 0

so that det(A) = det(P)’2 . Thus, det(A) is a square in either case.
Now regard ai j as being indeterminates over the ¬eld F; that is, let K = F(ai j :
1 ¤ i < j ¤ 2n) be the ¬eld of fractions of the polynomial ring in n(2n ’ 1) vari-
ables over F. If A is the skew-symmetric matrix with entries ai j for 1 ¤ i <
j ¤ 2n, then as we have seen, det(A) is a square in K. It is actually the square
of a polynomial. (For the polynomial ring is a unique factorisation domain; if
det(A) = ( f /g)2 , where f and g are polynomials with no common factor, then
det(A)g2 = f 2 , and so f 2 divides det(A); this implies that g is a unit.) Now det(A)
contains a term
a2 a2 · · · a2 2n
12 34 2n’1
corresponding to the permutation

(1 2)(3 4) · · · (2n ’ 1 2n),

and so by choice of sign in the square root we may assume that (ii)(b) holds.
Clearly the polynomial Pf(A) is uniquely determined.
The result for arbitrary skew-symmetric matrices is now obtained by speciali-
sation (that is, substituting values from F for the indeterminates ai j ).

Theorem 4.2 If A is a skew-symmetric matrix and P any invertible matrix, then

Pf(PAP ) = det(P) · Pf(A).

Proof We have det(PAP ) = det(P)2 det(A), and taking the square root shows
that Pf(PAP ) = ± det(P) Pf(A); it is enough to justify the positive sign. For this,
it suf¬ces to consider the ˜standard™ skew-symmetric matrix

01 01
A = diag ,..., ,
’1 0 ’1 0

since all non-singular skew-symmetric matrices are equivalent. In this case, the
(2n ’ 1, 2n) entry in PAP contains the term p2n’1 2n’1 p2n 2n , so that Pf(PAP )
contains the diagonal entry of det(P) with sign +1.

Exercise 4.1 A one-factor on the set {1, 2, . . . , 2n} is a partition F of this set
into n subsets of size 2. We represent each 2-set{i, j} by the ordered pair (i, j)
with i < j. The crossing number χ(F) of the one-factor F is the number of pairs
{(i, j), (k, l)} of sets in F for which i < k < j < l.
(a) Let Fn be the set of one-factors on the set {1, 2, . . . , 2n}. What is |Fn |?

(b) Let A = (ai j ) be a skew-symmetric matrix of order 2n. Prove that

‘ (’1)χ(F) ∏
Pf(A) = ai j .
(i, j)∈F

4.2 The symplectic groups
The symplectic group Sp(2n, F) is the isometry group of a non-degenerate al-
ternating bilinear form on a vector space of rank 2n over F. (We have seen
that any two such forms are equivalent up to invertible linear transformation of
the variables; so we have de¬ned the symplectic group uniquely up to conju-
gacy in GL(2n, F).) Alternatively, it consists of the 2n — 2n matrices P satisfying
P AP = A, where A is a ¬xed invertible skew-symmetric matrix. If necessary, we
can take for de¬niteness either
On In
’In On
01 01
A = diag ,..., .
’1 0 ’1 0
The projective symplectic group PSp(2n, F) is the group induced on the set
of points of PG(2n ’ 1, F) by Sp(2n, F). It is isomorphic to the factor group
Sp(2n, F)/(Sp(2n, F) © Z), where Z is the group of non-zero scalar matrices.

Proposition 4.3 (a) Sp(2n, F) is a subgroup of SL(2n, F).

(b) PSp(2n, F) ∼ Sp(2n, F)/{±I}.

Proof (a) If P ∈ Sp(2n, F), then Pf(A) = Pf(PAP ) = det(P) Pf(A), so det(P) =
(b) If (cI)A(cI) = A, then c2 = 1, so c = ±1.

From Theorem 3.17, we have:

Proposition 4.4
n n
| Sp(2n, q)| = ∏(q ’ 1)q ∏(q2i ’ 1).
2i 2i’1
i=1 i=1

The next result shows that we get nothing new in the case 2n = 2.

Proposition 4.5 Sp(2, F) ∼ SL(2, F) and PSp(2, F) ∼ PSL(2, F).
= =

Proof We show that there is a non-degenerate bilinear form on F 2 preserved by
SL(2, F). The form B is given by

B(x, y) = det

for all x, y ∈ F 2 , where is the matrix with rows x and y. This is obviously a
symplectic form. For any linear map P : F 2 ’ F 2 , we have

xP x
= P,
yP y

B(xP, yP) = det = B(x, y) det(P),
and so all elements of SL(2, F) preserve B, as required.
The second assertion follows on factoring out the group of non-zero scalar
matrices of determinant 1, that is, {±I}.

In particular, PSp(2, F) is simple if and only if |F| > 3.
There is one further example of a non-simple symplectic group:

Proposition 4.6 PSp(4, 2) ∼ S6 .

Proof Let F = GF(2) and V = F 6 . On V de¬ne the “standard inner product”
x · y = ‘ xi yi

(evaluated in F). Let j denote the all-1 vector. Then

x·x = x· j

for all x ∈ X, so on the rank 5 subspace j⊥ , the inner product induces an alternating
bilinear form. This form is degenerate ” indeed, by de¬nition, its radical contains
j ” but it induces a non-degenerate symplectic form B on the rank 4 space j⊥ / j .
Clearly any permutation of the six coordinates induces an isometry of B. So S6 ¤
Sp(4, 2) = PSp(4, 2). Since

|S6 | = 6! = 15 · 8 · 3 · 2 = | Sp(4, 2)|,

the result is proved.

4.3 Generation and simplicity
This subsection follows the pattern used for PSL(n, F). We show that Sp(2n, F) is
generated by transvections, that it is equal to its derived group, and that PSp(2n, F)
is simple, for n ≥ 2, with the exception (noted above) of PSp(4, 2).
Let B be a symplectic form. Which transvections preserve B? Consider the
transvection x ’ x + (x f )a, where a ∈ V , f ∈ V — , and a f = 0. We have

B(x + (x f )a, y + (y f )a) = B(x, y) + (x f )B(a, y) ’ (y f )B(a, x).

So B is preserved if and only if (x f )B(a, y) = (y f )B(a, x) for all x, y ∈ V . We claim
that this entails x f = »B(a, x) for all x, for some scalar ». For we can choose x
with B(a, x) = 0, and de¬ne » = (x f )/B(a, x); then the above equation shows that
y f = »B(a, y) for all y.
Thus, a symplectic transvection (one which preserves the symplectic form)
can be written as
x ’ x + »B(x, a)a
for a ¬xed vector a ∈ V . Note that its centre and axis correspond under the sym-
plectic polarity; that is, its axis is a⊥ = {x : B(x, a) = 0}.

Lemma 4.7 For r ≥ 1, the group PSp(2r, F) acts primitively on the points of
PG(2r ’ 1, F).

Proof For r = 1 we know that the action is 2-transitive, and so is certainly prim-
itive. So suppose that r ≥ 2.

Every point of PG(2r ’ 1, F) is ¬‚at, so by Witt™s Lemma, the symplectic group
acts transitively. Moreover, any pair of distinct points spans either a ¬‚at subspace
or a hyperbolic plane. Again, Witt™s Lemma shows that the group is transitive on
the pairs of each type. (In other words G = PSp(2r, F) has three orbits on ordered
pairs of points, including the diagonal orbit

∆ = {(p, p) : p ∈ PG(2r ’ 1, F)};

we say that PSp(2r, F) is a rank 3 permutation group on PG(2r ’ 1, F).)
Now a non-trivial equivalence relation preserved by G would have to consist
of the diagonal and one other orbit. So to ¬nish the proof, we must show:

(a) if B(x, y) = 0, then there exists z such that B(x, z), B(y, z) = 0;

(b) if B(x, y) = 0, then there exists z such that B(x, z) = B(y, z) = 0.

This is a simple exercise.

Exercise 4.2 Prove (a) and (b) above.

Lemma 4.8 For r ≥ 1, the group Sp(2r, F) is generated by symplectic transvec-

Proof The proof is by induction by r, the case r = 1 having been settled earlier
(Theorem 2.6).
First we show that the group H generated by transvections is transitive on the
non-zero vectors. Let u, v = 0. If B(u, v) = 0, then the symplectic transvection

B(x, v ’ u)
x ’ x+ (v ’ u)
B(u, v)

carries u to v. If B(u, v) = 0, choose w such that B(u, w), B(v, w) = 0 (by (a) of the
preceding lemma) and map u to w to v in two steps.
Now it is enough to show that any symplectic transformation g ¬xing a non-
zero vector u is a product of symplectic transvections. By induction, since the
stabiliser of u is the symplectic group on u / u , we may assume that g acts
trivially on this quotient; but then g is itself a symplectic transvection.

Lemma 4.9 For r ≥ 3, and for r = 2 and F = GF(2), the group PSp(2r, F) is
equal to its derived group.

Proof If F = GF(2), GF(3), we know from Lemma 2.8 that any element induc-
ing a transvection on a hyperbolic plane and the identity on the complement is a
commutator, so the result follows. The same argument completes the proof pro-
vided that we can show that it holds for PSp(6, 2) and PSp(4, 3).
In order to handle these two groups, we ¬rst develop some notation which can
be more generally applied. For convenience we re-order the rows and columns of
the ˜standard skew-symmetric matrix™ so that it has the form

J= ,
’I O

where O and I are the r — r zero and identity matrices. (In other words, the ith
and (i + r)th basis vectors form a hyperbolic pair, for i = 1, . . . , r.) Now a matrix
C belongs to the symplectic group if and only if C JC = J. In particular, we ¬nd
(a) for all invertible r — r matrices A, we have

A’1 O
∈ Sp(2r, F);

(b) for all symmetric r — r matrices B, we have

∈ Sp(2r, F).

Now straightforward calculation shows that the commutator of the two matrices
in (a) and (b) is equal to
and it suf¬ces to choose A and B such that A is invertible, B is symmetric, and
B ’ ABA has rank 1.
The following choices work:
1 1 01
(a) r = 2, F = GF(3), A = ,B= ;
0 1 10
«  « 
1 10 1 0 1
(b) r = 3, F = GF(2), A =  0 0 1 , B =  0 1 .
1 00 1 1 1

Theorem 4.10 The group PSp(2r, F) is simple for r ≥ 1, except for the cases
(r, F) = (1, GF(2)), (1, GF(3)), and (2, GF(2)).

Proof We now have all the ingredients for Iwasawa™s Lemma (Theorem 2.7),
which immediately yields the conclusion.

As we have seen, the exceptions in the theorem are genuine.

Exercise 4.3 Show that PSp(4, 3) is a ¬nite simple group which has no 2-transitive
The only positive integers n such that n(n ’ 1) divides | PSp(4, 3)| are n =
2, 3, 4, 5, 6, 9, 10, 16, 81. It suf¬ces to show that the group has no 2-transitive action
of any of these degrees. Most are straightforward but n = 16 and n = 81 require
some effort.
(It is known that PSp(4, 3) is the smallest non-abelian ¬nite simple group with
this property.)

4.4 A technical result
The result in this section will be needed at one point in our discussion of the
unitary groups. It is a method of recognising the groups PSp(4, F) geometrically.
Consider the polar space associated with PSp(4, F). Its points are all the points
of the projective space PG(3, F), and its lines are the ¬‚at lines (those on which the
symplectic form vanishes). We call them F-lines for brevity. Note that the F-lines
through a point p of the porojective space form the plane pencil consisting of all
the lines through p in the plane p⊥ , while dually the F-lines in a plane Π are all
those lines of Π containing the point Π⊥ . Now two points are prthogonal if and
only if they line on an F-line.
The geometry of F-lines has the following property:

(a) Given an F-line L and a point p not on L, there is a unique point q ∈ L such
that pq is an F-line.

(The point q is p⊥ © L.) A geometry with this property (in which two points lie on
at most one line) is called a generalised quadrangle.

Exercise 4.4 Show that a geometry satisfying the polar space axioms with r = 2
is a generalised quadrangle, and conversely.

We wish to recognise, within the geometry, the remaining lines of the projec-
tive space. These correspond to hyperbolic planes in the vector space, so we will
call them H-lines. Note that the points of a H-line are pairwise non-orthogonal.
We observe that, given any two points p, q not lying on an F-line, the set

{r : pr and qr are F-lines}

is the set of points of {p, q}⊥ , and hence is the H-line containing p and q. This
de¬nition works in any generalized quadrangle, but in this case we have more:

(b) Any two points lie on either a unique F-line or a unique H-line.

(c) The F-lines and H-lines within a set p⊥ form a projective plane.

(d) Any three non-collinear points lie in a unique set p⊥ .

Exercise 4.5 Prove conditions (b)“(d).

Conditions (a)“(d) guarantee that the geometry of F-lines and H-lines is a pro-
jective space, hence is isomorphic to PG(3, F) for some (possibly non-commutative)
¬eld F. Then the correspondence p ” p⊥ is a polarity of the projective space,
such that each point is incident with the corresponding plane. By the Funda-
mental Theorem of Projective Geometry, this polarity is induced by a symplectic
form B on a vector space V of rank 4 over F (which is necessarily commutative).
Hence, again by the FTPG, the automorphism group of the geometry is in-
duced by the group of semilinear transformations of V which preserve the set of
pairs {(x, y) : B(x, y) = 0}. These transformations are composites of linear trans-
formations preserving B up to a scalar factor, and ¬eld automorphisms. It follwos
that, if F = GF(2), the automorphism group of the geometry has a unique minimal
normal subgroup, which is isomorphic to PSp(4, F).

5 Unitary groups
In this section we analyse the unitary groups in a similar way to the treatment
of the symplectic groups in the last section. Note that the treatment here applies
only to the isometry groups of Hermitian forms which are not anisotropic. So in
particular, the Lie groups SU(n) over the complex numbers are not included.
Let V be a vector space over F, σ an automorphism of F of order 2, and B a
non-degenerate σ-Hermitian form on V (that is, B(y, x) = B(x, y)σ for all x, y ∈ V ).
It is often convenient to denote cσ by c, for any element c ∈ F.
Let F0 denote the ¬xed ¬eld of F. There are two important maps from F to F0
associated with σ, the trace and norm maps, de¬ned by
Tr(c) = c + c,
N(c) = c · c.
Now Tr is an additive homomorphism (indeed, an F0 -linear map), and N is a
multiplicative homomorphism. As we have seen, the image of Tr is F0 ; the kernel
is the set of c such that cσ = ’c (which is equal to F0 if the characteristic is 2 but
not otherwise).
Suppose that F is ¬nite. Then the order of F is a square, say F = GF(q2 ),
and F0 = GF(q). Since the multiplicative group of F0 has order q ’ 1, a non-
zero element c ∈ F lies in F0 if and only if cq’1 = 1. This holds if and only if
c = aq+1 for some a ∈ F (as the multiplicative group of F is cyclic), in other
words, c = a · a = N(a). So the image of N is the multiplicative group of F0 , and
its kernel is the set of (q + 1)st roots of 1. Also, the kernel of Tr consists of zero
and the set of (q ’ 1)st roots of ’1, the latter being a coset of F0— in F — .
The Hermitian form on a hyperbolic plane has the form
B(x, y) = x1 y2 + y1 x2 .
An arbitrary Hermitian formed space is the orthogonal direct sum of r hyper-
bolic planes and an anisotropic space. We have seen that, up to scalar multiplica-
tion, the following hold:
(a) over C, an anisotropic space is positive de¬nite, and the form can be taken
to be
B(x, y) = x1 y1 + · · · + xs ys ;
(b) over a ¬nite ¬eld, an anisotropic space has dimension at most one; if non-
zero, the form can be taken to be
B(x, y) = xy.

5.1 The unitary groups
Let A be the matrix associated with a non-degenerate Hermitian form B. Then
A = A , and the isometry group of B (the unitary group U(V, B) consists of all
invertible matrices P which satisfy P AP = A.
Since A is invertible, we see that

N(det(P)) = det(P ) det(P) = 1.

So det(P) ∈ F0 . Moreover, a scalar matrix cI lies in the unitary group if and only
if N(c) = cc = 1.
The special unitary group SU(V, B) consists of all elements of the unitary
group which have determinant 1 (that is, SU(V, B) = U(V, B) © SL(V )), and the
projective special unitary group is the factor group SU(V, B)/SU(V, B) © Z, where
Z is the group of scalar matrices.
In the case where F = GF(q2 ) is ¬nite, we can unambiguously write SU(n, q)
and PSU(n, q), since up to scalar multiplication there is a unique Hermitian form
on GF(q2 )n (with rank n/2 and germ of dimension 0 or 1 according as n is even
or odd). (It would be more logical to write SU(n, q2 ) and PSU(n, q2 ) for these
groups; we have used the standard group-theoretic convention.

(a) | U(n, q) = qn(n’1)/2 ∏n (qi ’ (’1)i ).
Proposition 5.1 i=1

(b) | SU(n, q)| = | U(n, q)|/(q + 1).

(c) | PSU(n, q)| = | SU(n, q)|/d, where d = (n, q + 1).

Proof (a) We use Theorem 3.17, with either n = 2r, µ = ’ 1 , or n = 2r + 1, µ = 2 , 1
and with q replaced by q2 , noting that, in the latter case, |G0 | = q + 1. It happens
that both cases can be expressed by the same formula! On the same theme, note
that, if we replace (’1)i by 1 (and q + 1 by q ’ 1 in parts (b) and (c) of the
theorem), we obtain the orders of GL(n, q), SL(n, q), and PSL(n, q) instead.
(b) As we noted, det is a homomorphism from U(n, q) onto the group of (q +
1)st roots of unity in GF(q2 )— , whose kernel is SU(n, q).
(c) A scalar cI belongs to U(n, q) if cq+1 = 1, and to SL(n, q2 ) if cn = 1. So
|Z © SL(n, q2 )| = d, as required.

We conclude this section by considering unitary transvections, those which
preserve a Hermitian form. Accordingly, let T : x ’ x + (x f )a be a transvection,

where a f = 0. We have

B(xT, yT ) = B(x + (x f )a, y + (y f )a)
= B(x, y) + (x f )B(y, a) + (y f )B(x, a) + (x f )(y f )B(a, a).

So T is unitary if and only if the last three terms vanish for all x, y. Putting y = a we
see that (x f )B(a, a) = 0 for all x, whence (since f = 0) we must have B(a, a) = 0.
Now choosing y such that B(y, a) = 1 and setting » = (y f ), we have x f = »B(x, a)
for all x. So a unitary transvection has the form

x ’ x + »B(x, a)a,

where B(a, a) = 0. In particular, an anisotropic space admits no unitary transvec-
tions. Also, choosing x and y such that B(x, a) = B(y, a) = 1, we ¬nd that Tr(») =
0. Conversely, for any » ∈ ker(Tr) and any a with B(a, a) = 0, the above formula
de¬nes a unitary transvection.

5.2 Hyperbolic planes
In this section only, we use the convention that U(2, F0 ) means the unitary group
associated with a hyperbolic plane over F, and σ is the associated ¬eld automor-
phism, having ¬xed ¬eld F0 .

Theorem 5.2 SU(2, F0 ) ∼ SL(2, F0 ).

Proof We will show, moreover, that the actions of the unitary group on the polar
space and that of the special linear group on the projective space correspond, and
that unitary transvections correspond to transvections in SL(2, F0 ). Let K = {c ∈
F : c + c = 0} be the kernel of the trace map; recall that the image of the trace map
is F0 .
With the standard hyperbolic form, we ¬nd that a unitary matrix


must satisfy P AP = A, where

0 1
A= .
1 0

ac + ac = 0, bc + ad = 1, bd + bd = 0.
In addition, we require, that det(P) = 1, that is, ad ’ bc = 1.
From these equations we deduce that b + b = c + c = 0, that is, b, c ∈ K, while
a ’ a = d ’ d = 0, that is, a, d ∈ F0 .
Choose a ¬xed element u ∈ K. Then » ∈ K if and only if u» ∈ F0 . Also,
u’1 ∈ K. Hence the matrix
a ub
P† =
u’1 c d

belongs to SL(2, F0 ). Conversely, any matrix in SL(2, F0 ) gives rise to a matrix in
SU(2, F0 ) by the inverse map. So we have a bijection between the two groups. It
is now routine to check that the map is an isomorphism.
Represent the points of the projective line over F by F ∪ {∞} as usual. Recall
that ∞ is the point (rank 1 subspace) spanned by (0, 1), while c is the point spanned
by (1, c). We see that ∞ is ¬‚at, while c is ¬‚at if and only if c + c = 0, that is, c ∈ K.
So the map x ’ x takes the polar space for the unitary group onto the projective
line over F0 . It is readily checked that this map takes the action of the unitary
group to that of the special linear group.
By transitivity, it is enough to consider the unitary transvections x ’ x +
»B(x, a)a, where a = (0, 1). In matrix form, these are

P= ,
0 1

with » ∈ K. Then
1 u»
P† = ,
0 1
which is a transvection in SL(2, F0 ), as required.

In particular, we see that PSU(2, F0 ) is simple if |F0 | > 3.

5.3 Generation and simplicity
We follow the now-familiar pattern. First we treat two exceptional ¬nite groups,
then we show that unitary groups are generated by unitary transvections and that
most are simple. By the preceding section, we may assume that the rank is at
least 3.

The ¬nite unitary group PSU(3, q) is a 2-transitive group of permutations of
the q3 + 1 points of the corresponding polar space (since any two such points are
spanned by a hyperbolic pair) and has order (q3 + 1)q3 (q2 ’ 1)/d, where d =
(3, q + 1). Moreover, any two points span a line containing q + 1 points of the
polar space. The corresponding geometry is called a unital.
For q = 2, the group has order 72, and so is soluble. In fact, it is sharply
2-transitive: a unique group element carries any pair of points to any other.

Exercise 5.1 (a) Show that the unital associated with PSU(3, 2) is isomorphic
to the af¬ne plane over GF(3), de¬ned as follows: the points are the vectors
in a vector space V of rank 2 over GF(3), and the lines are the cosets of
rank 1 subspaces of V (which, over the ¬eld GF(3), means the triples of
vectors with sum 0).

(b) Show that the automorphism group of the unital has the structure 32 : GL(2, 3),
where 32 denotes an elementary abelian group of this order (the translation
group of V ) and : denotes semidirect product.

(c) Show that PSU(3, 2) is isomorphic to 32 : Q8 , where Q8 is the quaternion
group of order 8.

(d) Show that PSU(3, 2) is not generated by unitary transvections.

We next consider the group PSU(4, 2), and outline the proof of the following

Theorem 5.3 PSU(4, 2) ∼ PSp(4, 3).

Proof Observe ¬rst that both these groups have order 25 920. We will construct
a geometry for the group PSU(4, 2), and use the technical results of Section 4.4
to identify it with the generalised quadrangle for PSp(4, 3). Now it has index 2
in the full automorphism group of this geometry, as also does PSp(4, 3), which is
simple; so these two groups must coincide.
The geometry is constructed as follows. Let V be a vector space of rank 4 over
GF(4) carrying a Hermitian form of polar rank 2. The projective space PG(3, 4)
derived from V has (44 ’ 1)/(4 ’ 1) = 85 points, of which (42 ’ 1)(43/2 + 1)/(4 ’
1) = 45 are points of the polar space, and the remaining 40 are points on which
the form does not vanish (spanned by vectors x with B(x, x) = 1). Note that 40 =
(34 ’ 1)/(3 ’ 1) is equal to the number of points of the symplectic generalised
quadrangle over GF(3). Let „¦ denote this set of 40 points.

De¬ne an F-line to be a set of four points of „¦ spanned by the vectors of
an orthonormal basis for V (a set of four vectors x1 , x2 , x3 , x4 with B(xi , xi ) = 1
and B(xi , x j ) = 0 for i = j). Note that two orthogonal points p, q of „¦ span a
non-degenerate 2-space, which is a line containing ¬ve points of the projective
space of which three are ¬‚at and the other two belong to „¦. Then {p, q}⊥ is
also a non-degenerate 2-space containing two points of „¦, which complete {p, q}
to an F-line. Thus, two orthogonal points lie on a unique F-line, while two non-
orthogonal points lie on no F-line. It is readily checked that, if L = {p1 , p2 , p3 , p4 }
is an F-line and q is another point of „¦, then p has three non-zero coordinates in
the orthonormal basis corresponding to L, so q is orthogonal to a unique point of
L. Thus, the points of „¦ and the F-lines satisfy condition (a) of Section 4.4; that
is, they form a generalised quadrangle.
Now consider two points of „¦ which are not orthogonal. The 2-space they
span is degenerate, with a radical of rank 1. So of the ¬ve points of the corre-
sponding projective line, four lie in „¦ and one (the radical) is ¬‚at. Sets of four
points of this type (which are obviously determined by any two of their members)
will be the H-lines. It is readily checked that the H-lines do indeed arise in the
manner described in Section 4.4, that is, as the sets of points of „¦ orthogonal to
two given non-orthogonal points. So condition (b) holds.
Now a point p of „¦ lies in four F-lines, whose union consists of thirteen points.
If q and r are two of these points which do not lie on an F-line with p, then q and
r cannot be orthogonal, and so they lie in an H-line; since p and q are orthogonal
to p, so are the remaining points of the H-line containing them. Thus we have
condition (c). Now (d) is easily veri¬ed by counting, and the proof is complete.

Exercise 5.2 (a) Give a detailed proof of the above isomorphism.
(b) If you are familiar with a computer algebra package, verify computation-
ally that the above geometry for PSU(4, 2) is isomorphic to the symplectic
generalised quadrangle for PSp(4, 3).

In our generation and simplicity results we treat the rank 3 case separately. In
the rank 3 case, the unitary group is 2-transitive on the points of the unital.

Theorem 5.4 Let (V, B) be a unitary formed space of Witt rank 1, with rk(V ) = 3.
Assume that the ¬eld F is not GF(22 ).
(a) SU(V, B) is generated by unitary transvections.
(b) PSU(V, B) is simple.

Proof We exclude the case of PSU(3, 2) (with F = GF(22 ), considered earlier.
Replacing the form by a scalar multiple if necessary, we assume that the germ
contains vectors of norm 1. Take such a vector as second basis vector, where the
¬rst and third are a hyperbolic pair. That is, we assume that the form is
B((x1 , x2 , x3 ), (y1 , y2 , y3 )) = x1 y3 + x2 y2 + x3 y1 ,

so the isometry group is
{P : P AP = A}
« 
00 1
A = 0 1 0.
10 0
Now we check that the group
±« 

’a b
1 
Q = 0 1 a  : N(a) + Tr(b) = 0
 
0 01
is a subgroup of G = SU(V, B), and its derived group consists of unitary transvec-
tions (the elements with a = 0).
Next we show that the subgroup T of V generated by the transvections in G
is transitive on the set of vectors x such that B(x, x) = 1. Let x and y be two such
vectors. Suppose ¬rst that x, y is nondegenerate. Then it is a hyperbolic line,
and a calculation in SU(2, F0 ) gives the result. Otherwise, there exists z such that
x, z and y, z are nondegenerate, so we can get from x to y in two steps.
Now the stabiliser of such a vector in G is SU(x⊥ , B) = SU(2, F0 ), which is
generated by transvections; and every coset of this stabiliser contains a transvec-
tion. So G is generated by transvections.
Now it follows that the transvections lie in G , and Iwasawa™s Lemma (Theo-
rem 2.7) shows that PSL(V, B) is simple.

Exercise 5.3 Complete the details in the above proof by showing
(a) the group SU(2, F0 ) acts transitively on the set of vectors of norm 1 in the
hyperbolic plane;
(b) given two vectors x, y of norm 1 in a rank 3 unitary space as in the proof,
either x, y is a hyperbolic plane, or there exists z such that x, z and y, z
are hyperbolic planes.

Theorem 5.5 Let (V, B) be a unitary formed space with Witt rank at least 2. Then

(a) SU(V, B) is generated by unitary transvections.

(b) PSU(V, B) is simple.

Proof We follow the usual pattern. The argument in the preceding theorem
shows part (a) without change if F = GF(4). In the excluded case, we know
that PSU(4, 2) ∼ PSp(4, 3) is simple, and so is generated by any conjugacy class
(in particular, the images of the transvections of SU(4, 2)). Then induction shows
the result for higher rank spaces over GF(4). Again, the argument in 3 dimen-
sions shows that transvections are commutators; the action on the points of the
polar space is primitive; and so Iwasawa™s Lemma shows the simplicity.

6 Orthogonal groups
We now turn to the orthogonal groups. These are more dif¬cult, for two related
reasons. First, it is not always true that the group of isometries with determinant 1
is equal to its derived group (and simple modulo scalars). Secondly, in character-
istic different from 2, there are no transvections, and we have to use a different
class of elements.
We let O(Q) denote the isometry group of the non-degenerate quadratic form Q,
and SO(Q)the group of isometries with determinant 1. Further, PO(Q) and PSO(Q)
are the quotients of these groups by the scalars they contain. We de¬ne „¦(Q) to be
the derived subgroup of O(Q), and P„¦(Q) = „¦(Q)/(„¦(Q) © Z), where Z consists
of the scalar matrices. Sometimes „¦(Q) = SO(Q), and sometimes it is strictly
smaller; but our notation serves for both cases.
In the case where F is ¬nite, we have seen that for even n there is a unique
type of non-degenerate quadratic form up to scalar multiplication, while if n is
even there are two types, having germ of dimension 0 or 2 respectively, We
write O+ (n, q), O(n, q) and O’ (n, q) for the isometry group of a non-degenerate
quadratic form on GF(q)n with germ of rank 0, 1, 2 (and n even, odd, even respec-
tively). We use similar notation for SO, P„¦, and so on. Then we write Oµ (n, q) to
mean either O+ (n, q) or O’ (n, q). Note that, unfortunately, this convention (which
is standard notation) makes µ the negative of the µ appearing in our general order
formula (Theorem 3.17).
Now the order formula for the ¬nite orthogonal groups reads as follows.

| O(2m + 1, q)| = d ∏(q2i ’ 1)q2i’1
∏(q2i ’ 1),
= dq
∏(qi ’ 1)(qi’1 + 1)q2i’2
| O (2m, q)| =
(q ’ 1) ∏ (q2i ’ 1),
m(m’1) m
= 2q
| O (2m, q)| = 2(q + 1) ∏ (qi ’ 1)(qi+1 + 1)q2i


(q + 1) ∏ (q2i ’ 1),
m(m’1) m
= 2q
where d = (2, q ’ 1). Note that there is a single difference in sign between the
¬nal formulae for Oµ (2m, q) for µ = ±1; we can combine the two and write
| O (2m, q)| = 2(q ’ µ) ∏ (q2i ’ 1).
µ m
We have | SO(n, q)| = | O(n, q)|/d (except possibly if n is odd and q is even).
This is because, with this exclusion, the bilinear form B associated with Q is non-
degenerate; and the orthogonal group consists of matrices P satisfying P AP = A,
where A is the matrix of the bilinear form, so that det(P) = ±1. It is easy to show
that, for q odd, there are orthogonal transformations with determinant ’1. The
excluded case will be dealt with in Section 6.2. We see also that the only scalars
in O(Q) are ±I; and, in characteristic different from 2, we have ’I ∈ SO(Q) if
and only if the rank of the underlying vector space is even. Thus, for q odd, we
| SO(Q)| = | PO(Q)| = | O(Q)|/2,
| PSO(Q)| = | SO(Q)|/(n, 2).
For q and n even we have O(Q) = SO(Q) = PO(Q) = PSO(Q).
Exercise 6.1 Let Q be a non-degenerate quadratic form over a ¬eld of character-
istic different from 2. Show that O(Q) contains elements with determinant ’1.
[Hint: if Q(v) = 0, then take the transformation which takes v to ’v and extend it
by the identity on v⊥ (in other words, the re¬‚ection in the hyperplane v⊥ ).]

6.1 Some small-dimensional cases
We begin by considering some small cases. Let V be a vector space of rank n
carrying a quadratic form Q of Witt index r, where δ = n ’ 2r is the dimension of
the germ of Q. Let O(Q) denote the isometry group of Q, and SO(Q) the subgroup
of isometries of determinant 1.

Case n = 1, r = 0. In this case the quadratic form is a scalar multiple of x2 .
(Replacing q by a scalar multiple does not affect the isometry group.) Then
O(Q) = {±1}, a group of order 1 or 2 according as the characteristic is or is
not 2; and SO(Q) is the trivial group.

Case n = 2, r = 1. The quadratic form is Q(x1 , x2 ) = x1 x2 , and the isometry
group G = O(Q) is

» »
0 0
: » ∈ F— ,
0 »’1 »’1 0

a group with a subgroup H of index 2 isomorphic to F — , and such that an element
t ∈ G \ H satis¬es t 2 = 1 and t ’1 ht = h’1 for all h ∈ H. In other words, G is
a generalised dihedral group. If F = GF(q), then O(2, q) is a dihedral group of
order 2(q ’ 1). Note that H = SO(Q) if and only if the characteristic of F is not 2.

Case n = 2, r = 0. In this case, the quadratic form is

±x1 + βx1 x2 + γx2 ,
2 2

where q(x) = ±x2 + βx + γ is an irreducible quadratic over F. Let K be a splitting
¬eld for q over F, and assume that K is a Galois extension (in other words, that q
is separable over F: this includes the cases where either the characteristic is not 2
or the ¬eld F is ¬nite). Then, up to scalar multiplication, the form Q is equivalent
to the K/F norm on the F-vector space K. The orthogonal group is generated by
the multiplicative group of elements of norm 1 in K and the Galois automorphism
σ of K over F.
In the case F = GF(q), this group is dihedral of order 2(q + 1). In the case
F = R, the C/R norm is
z ’ zz = |z|2 ,
and so the orthogonal group is generated by multiplication by unit complex num-
bers and complex conjugation. In geometric terms, it is the group of rotations and
re¬‚ections about the origin in the Euclidean plane.
Again we see that SO(Q) has index 2 in O(F) if the characteristic of F is not 2.

Exercise 6.2 Prove that, if K is a Galois extension of F, then the determinant of
the F-linear map x ’ »x on K is equal to NK/F (»). [Hint: if » ∈ F, the eigenvalues
of this map are » and »σ .]

Case n = 3, r = 1. In this case and the next, we describe a group preserving a
quadratic form and claim without proof that it is the full orthogonal group. Also,
in this case, we assume that the characteristic is not equal to 2.


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