Furthermore, if E is finite-dimensional, then dim(U ) = dim(U ) and in a suitable basis, the
matrix representing ϕ is of the form
0
A 0
A
0
0
0
0 B
We say that ϕ is a neutral form if it is nondegenerate, E is finite-dimensional, and if W = (0).
⊥
Sometimes, we use the notation U1 ⊕ U2 to indicate that in a direct sum U1 ⊕ U2,
the subspaces U1 and U2 are orthogonal. Then, in Definition 14.14, we can write that
⊥
E = (U ⊕ U ) ⊕ W .
As a warm up for Proposition 14.27, we prove an analog of Proposition 14.22 in the case
of a symmetric bilinear form.
Proposition 14.26. Let ϕ : E ×E → K be a nondegenerate symmetric bilinear form with K
a field of characteristic different from 2. For any nonzero isotropic vector u, there is another
nonzero isotropic vector v such that ϕ(u, v) = 2, and u and v are linearly independent. In
the basis (u, v/2), the restriction of ϕ to the plane spanned by u and v/2 is of the form
0 1 .
1 0
Proof. Since ϕ is nondegenerate, there is some nonzero vector z such that (rescaling z if
necessary) ϕ(u, z) = 1. If
v = 2z − ϕ(z, z)u,
then since ϕ(u, u) = 0 and ϕ(u, z) = 1, note that
ϕ(u, v) = ϕ(u, 2z − ϕ(z, z)u) = 2ϕ(u, z) − ϕ(z, z)ϕ(u, u) = 2,
and
ϕ(v, v) = ϕ(2z − ϕ(z, z)u, 2z − ϕ(z, z)u)
= 4ϕ(z, z) − 4ϕ(z, z)ϕ(u, z) + ϕ(z, z)2ϕ(u, u)
= 4ϕ(z, z) − 4ϕ(z, z) = 0.
If u and z were linearly dependent, as u, z = 0, we could write z = µu for some µ = 0, but
then, we would have
ϕ(u, z) = ϕ(u, µu) = µϕ(u, u) = 0,
contradicting the fact that ϕ(u, z) = 0. Then u and v = 2z − ϕ(z, z)u are also linearly
independent, since otherwise z could be expressed as a multiple of u. The rest is obvious.
14.6. TOTALLY ISOTROPIC SUBSPACES. WITT DECOMPOSITION
379
Proposition 14.26 yields a plane spanned by two vectors u1, v1 such that ϕ(u1, u1) =
ϕ(v1, v1) = 0 and ϕ(u1, v1) = 1. Such a plane is called an Artinian plane. Proposition 14.26
also shows that nonzero isotropic vectors come in pair.
Remark: Some authors refer to the above plane as a hyperbolic plane. Berger (and others)
point out that this terminology is undesirable because the notion of hyperbolic plane already
exists in differential geometry and refers to a very different object.
We leave it as an exercice to figure out that the group of isometries of the Artinian plane,
the set of all 2 × 2 matrices A such that
0 1
0 1
A
A =
,
1 0
1 0
consists of all matrices of the form
λ
0
0
λ , λ ∈ K − {0}.
0 λ−1
or
λ−1 0
In particular, if K = R, then this group denoted O(1, 1) has four connected components.
The first step in showing the existence of a Witt decomposition is this.
Proposition 14.27. Let ϕ be an -Hermitian form on E, assume that ϕ is nondegenerate
and satisfies property (T), and let U be any totally isotropic subspace of E of finite dimension
dim(U ) = r.
(1) If U is any totally isotropic subspace of dimension r and if U ∩ U⊥ = (0), then U ⊕ U
is nondegenerate, and for any basis (u1, . . . , ur) of U , there is a basis (u1, . . . , ur) of U
such that ϕ(ui, uj) = δij, for all i, j = 1, . . . , r.
(2) If W is any totally isotropic subspace of dimension at most r and if W ∩ U⊥ = (0),
then there exists a totally isotropic subspace U with dim(U ) = r such that W ⊆ U
and U ∩ U⊥ = (0).
Proof. (1) Let ϕ be the restriction of ϕ to U × U . Since U ∩ U⊥ = (0), for any v ∈ U ,
if ϕ(u, v) = 0 for all u ∈ U, then v = 0. Thus, ϕ is nondegenerate (we only have to check
on the left since ϕ is -Hermitian). Then, the assertion about bases follows from the version
of Proposition 14.3 for sesquilinear forms. Since U is totally isotropic, U ⊆ U⊥, and since
U ∩ U⊥ = (0), we must have U ∩ U = (0), which show that we have a direct sum U ⊕ U .
It remains to prove that U + U is nondegenerate. Observe that
H = (U + U ) ∩ (U + U )⊥ = (U + U ) ∩ U⊥ ∩ U ⊥.
Since U is totally isotropic, U ⊆ U⊥, and since U ∩ U⊥ = (0), we have
(U + U ) ∩ U⊥ = (U ∩ U⊥) + (U ∩ U⊥) = U + (0) = U,
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CHAPTER 14. BILINEAR FORMS AND THEIR GEOMETRIES
thus H = U ∩ U ⊥. Since ϕ is nondegenerate, U ∩ U ⊥ = (0), so H = (0) and U + U is
nondegenerate.
(2) We proceed by descending induction on s = dim(W ). The base case s = r is trivial.
For the induction step, it suffices to prove that if s < r, then there is a totally isotropic
subspace W containing W such that dim(W ) = s + 1 and W ∩ U⊥ = (0).
Since s = dim(W ) < dim(U ), the restriction of ϕ to U × W is degenerate. Since
W ∩ U⊥ = (0), we must have U ∩ W ⊥ = (0). We claim that
W ⊥ ⊆ W + U⊥.
If we had
W ⊥ ⊆ W + U⊥,
then because U and W are finite-dimensional and ϕ is nondegenerate, by Proposition 14.12,
U ⊥⊥ = U and W ⊥⊥ = W , so by taking orthogonals, W ⊥ ⊆ W + U⊥ would yield
(W + U ⊥)⊥ ⊆ W ⊥⊥,
that is,
W ⊥ ∩ U ⊆ W,
thus W ⊥ ∩ U ⊆ W ∩ U, and since U is totally isotropic, U ⊆ U⊥, which yields
W ⊥ ∩ U ⊆ W ∩ U ⊆ W ∩ U⊥ = (0),
contradicting the fact that U ∩ W ⊥ = (0).
Therefore, there is some u ∈ W ⊥ such that u /
∈ W + U⊥. Since U ⊆ U⊥, we can add to u
any vector z ∈ W ⊥ ∩ U ⊆ U⊥ so that u + z ∈ W ⊥ and u + z /
∈ W + U⊥ (if u + z ∈ W + U⊥,
since z ∈ U⊥, then u ∈ W + U⊥, a contradiction). Since W ⊥ ∩ U = (0) is totally isotropic
and u /
∈ W + U⊥ = (W ⊥ ∩ U)⊥, we can invoke Lemma 14.25 to find a z ∈ W ⊥ ∩ U such that
ϕ(u + z, u + z) = 0. If we write x = u + z, then x /
∈ W + U⊥, so W = W + Kx is a totally
isotopic subspace of dimension s + 1. Furthermore, we claim that W ∩ U⊥ = 0.
Otherwise, we would have y = w + λx ∈ U⊥, for some w ∈ W and some λ ∈ K, and
then we would have λx = −w + y ∈ W + U⊥. If λ = 0, then x ∈ W + U⊥, a contradiction.
Therefore, λ = 0, y = w, and since y ∈ U⊥ and w ∈ W , we have y ∈ W ∩ U⊥ = (0), which
means that y = 0. Therefore, W is the required subspace and this completes the proof.
Here are some consequences of Proposition 14.27. If we set W = (0) in Proposition
14.27(2), then we get:
Proposition 14.28. Let ϕ be an -Hermitian form on E which is nondegenerate and sat-
isfies property (T). For any totally isotropic subspace U of E of finite dimension r, there
exists a totally isotropic subspace U of dimension r such that U ∩ U = (0) and U ⊕ U is
nondegenerate.
14.6. TOTALLY ISOTROPIC SUBSPACES. WITT DECOMPOSITION
381
Proposition 14.29. Any two -Hermitian neutral forms satisfying property (T) defined on
spaces of the same dimension are equivalent.
Note that under the conditions of Proposition 14.28, (U, U , (U ⊕ U )⊥) is a Witt de-
composition for E. By Proposition 14.27(1), the block A in the matrix of ϕ is the identity
matrix.
The following proposition shows that every subspace U of E can be embedded into a
nondegenerate subspace.
Proposition 14.30. Let ϕ be an -Hermitian form on E which is nondegenerate and satisfies
property (T). For any subspace U of E of finite dimension, if we write
⊥
U = V ⊕ W,
for some orthogonal complement W of V = rad(U ), and if we let r = dim(rad(U )), then
there exists a totally isotropic subspace V of dimension r such that V ∩ V = (0), and
⊥
(V ⊕ V ) ⊕ W = V ⊕ U is nondegenerate. Furthermore, any isometry f from U into
another space (E , ϕ ) where ϕ is an -Hermitian form satisfying the same assumptions as
⊥
ϕ can be extended to an isometry on (V ⊕ V ) ⊕ W .
Proof. Since W is nondegenerate, W ⊥ is also nondegenerate, and V ⊆ W ⊥. Therefore, we
can apply Proposition 14.28 to the restriction of ϕ to W ⊥ and to V to obtain the required V .
We know that V ⊕ V is nondegenerate and orthogonal to W , which is also nondegenerate,
⊥
so (V ⊕ V ) ⊕ W = V ⊕ U is nondegenerate.
We leave the second statement about extending f as an exercise (use the fact that f (U ) =
⊥
f (V ) ⊕ f(W ), where V1 = f(V ) is totally isotropic of dimension r, to find another totally
isotropic susbpace V1 of dimension r such that V1 ∩ V1 = (0) and V1 ⊕ V1 is orthogonal to
f (W )).
⊥
The subspace (V ⊕ V ) ⊕ W = V ⊕ U is often called a nondegenerate completion of U.
The subspace V ⊕ V is called an Artinian space. Proposition 14.27 show that V ⊕ V has
a basis (u1, v1, . . . , ur, vr) consisting of vectors ui ∈ V and vj ∈ V such that ϕ(ui, uj) = δij.
The subspace spanned by (ui, vi) is an Artinian plane, so V ⊕ V it is the orthogonal direct
sum of r Artinian planes. Such a space is often denoted by Ar2r.
We now sharpen Proposition 14.27.
Theorem 14.31. Let ϕ be an -Hermitian form on E which is nondegenerate and satisfies
property (T). Let U1 and U2 be two totally isotropic maximal subspaces of E, with U1 or U2
of finite dimension. Write U = U1 ∩ U2, let S1 be a supplement of U in U1 and S2 be a
supplement of U in U2 (so that U1 = U ⊕ S1, U2 = U ⊕ S2), and let S = S1 + S2. Then,
there exist two subspaces W and D of E such that:
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CHAPTER 14. BILINEAR FORMS AND THEIR GEOMETRIES
(a) The subspaces S, U + W , and D are nondegenerate and pairwise orthogonal.
⊥
⊥
(b) We have a direct sum E = S ⊕ (U ⊕ W ) ⊕ D.
(c) The subspace D contains no nonzero isotropic vector (D is anisotropic).
(d) The subspace W is totally isotropic.
Furthermore, U1 and U2 are both finite dimensional, and we have dim(U1) = dim(U2),
dim(W ) = dim(U ), dim(S1) = dim(S2), and codim(D) = 2 dim(F1).
Proof. First observe that if X is a totally isotropic maximal subspace of E, then any isotropic
vector x ∈ E orthogonal to X must belong to X, since otherwise, X + Kx would be a
totally isotropic subspace strictly containing X, contradicting the maximality of X. As a
consequence, if xi is any isotropic vector such that xi ∈ U⊥
i
(for i = 1, 2), then xi ∈ Ui.
We claim that
S1 ∩ S⊥
2 = (0)
and S2 ∩ S⊥
1 = (0).
Assume that y ∈ S1 is orthogonal to S2. Since U1 = U ⊕ S1 and U1 is totally isotropic, y is
orthogonal to U1, and thus orthogonal to U, so that y is orthogonal to U2 = U ⊕ S2. Since
S1 ⊆ U1 and U1 is totally isotropic, y is an isotropic vector orthogonal to U2, which by a
previous remark implies that y ∈ U2. Then, since S1 ⊆ U1 and U ⊕ S1 is a direct sum, we
have
y ∈ S1 ∩ U2 = S1 ∩ U1 ∩ U2 = S1 ∩ U = (0).
Therefore S1 ∩ S⊥
2 = (0). A similar proof show that S2 ∩ S⊥
1 = (0). If U1 is finite-dimensional
(the case where U2 is finite-dimensional is similar), then S1 is finite-dimensional, so by
Proposition 14.12, S⊥
1 has finite codimension. Since S2 ∩ S⊥
1 = (0), and since any supplement
of S⊥
1 has finite dimension, we must have
dim(S2) ≤ codim(S⊥
1 ) = dim(S1).
By a similar argument, dim(S1) ≤ dim(S2), so we have
dim(S1) = dim(S2).
By Proposition 14.27(1), we conclude that S = S1 + S2 is nondegenerate.
By Proposition 14.20, the subspace N = S⊥ = (S1 + S2)⊥ is nondegenerate. Since
U1 = U ⊕ S1, U2 = U ⊕ S2, and U1, U2 are totally isotropic, U is orthogonal to S1 and to
S2, so U ⊆ N. Since U is totally isotropic, by Proposition 14.28 applied to N, there is a
totally isotropic subspace W of N such that dim(W ) = dim(U ), U ∩ W = (0), and U + W
is nondegenerate. Consequently, (d) is satisfied by W .
To satisfy (a) and (b), we pick D to be the orthogonal of U ⊕ W in N. Then, N =
⊥
⊥
⊥
⊥
(U ⊕ W ) ⊕ D and E = S ⊕ N, so E = S ⊕ (U ⊕ W ) ⊕ D.
14.6. TOTALLY ISOTROPIC SUBSPACES. WITT DECOMPOSITION
383
As to (c), since D is orthogonal U ⊕ W , D is orthogonal to U, and since D ⊆ N and N
is orthogonal to S1 + S2, D is orthogonal to S1, so D is orthogonal to U1 = U ⊕ S1. If y ∈ D
is any isotropic vector, since y ∈ U⊥
1 , by a previous remark, y ∈ U1, so y ∈ D ∩ U1. But,
D ⊆ N with N ∩ (S1 + S2) = (0), and D ∩ (U + W ) = (0), so D ∩ (U + S1) = D ∩ U1 = (0),
which yields y = 0. The statements about dimensions are easily obtained.
We obtain the following corollaries.
Theorem 14.32. Let ϕ be an -Hermitian form on E which is nondegenerate and satisfies
property (T).
(1) Any two totally isotropic maximal spaces of finite dimension have the same dimension.
(2) For any totally isotropic maximal subspace U of finite dimension r, there is another
totally isotropic maximal subspace U of dimension r such that U ∩ U = (0), and
U ⊕ U is nondegenerate. Furthermore, if D = (U ⊕ U )⊥, then (U, U , D) is a Witt
decomposition of E, and there are no nonzero isotropic vectors in D (D is anisotropic).
(3) If E has finite dimension n ≥ 1, then E has a Witt decomposition (U, U , D) as in (2).
There is a basis of E such that
(a) if ϕ is alternating ( = −1 and λ = λ for all λ ∈ K), then n = 2m and ϕ is
represented by a matrix of the form
0
Im
−Im
0
(b) if ϕ is symmetric ( = +1 and λ = λ for all λ ∈ K), then ϕ is represented by a
matrix of the form
0 I
r
0
I
,
r
0
0
0
0 P
where either n = 2r and P does not occur, or n > 2r and P is a definite symmetric
matrix.
(c) if ϕ is -Hermitian (the involutive automorphism λ → λ is not the identity), then
ϕ is represented by a matrix of the form
0
I
r
0
I
r
0
0 ,
0
0 P
where either n = 2r and P does not occur, or n > 2r and P is a definite matrix
such that P ∗ = P .
384
CHAPTER 14. BILINEAR FORMS AND THEIR GEOMETRIES
Proof. Part (1) follows from Theorem 14.31. By Proposition 14.28, we obtain a totally
isotropic subspace U of dimension r such that U ∩ U = (0). By applying Theorem 14.31
to U1 = U and U2 = U , we get U = W = (0), which proves (2). Part (3) is an immediate
consequence of (2).
As a consequence of Theorem 14.32, we make the following definition.
Definition 14.15. Let E be a vector space of finite dimension n, and let ϕ be an -Hermitian
form on E which is nondegenerate and satisfies property (T). The index (or Witt index ) ν
of ϕ, is the common dimension of all totally isotropic maximal subspaces of E. We have
2ν ≤ n.
Neutral forms only exist if n is even, in which case, ν = n/2. Forms of index ν = 0
have no nonzero isotropic vectors. When K = R, this is satisfied by positive definite or
negative definite symmetric forms. When K = C, this is satisfied by positive definite or
negative definite Hermitian forms. The vector space of a neutral Hermitian form ( = +1) is
an Artinian space, and the vector space of a neutral alternating form is a hyperbolic space.
If the field K is algebraically closed, we can describe all nondegenerate quadratic forms.
Proposition 14.33. If K is algebraically closed and E has dimension n, then for every
nondegenerate quadratic form Φ, there is a basis (e1, . . . , en) such that Φ is given by
n
m
x
Φ
x
i=1
ixm+i
if n = 2m
iei
=
m
x
i−1
i=1
ixm+i + x2
2m+1
if n = 2m + 1.
Proof. We work with the polar form ϕ of Φ. Let U1 and U2 be some totally isotropic
subspaces such that U1 ∩ U2 = (0) given by Theorem 14.32, and let q be their common
dimension. Then, W = U = (0). Since we can pick bases (e1, . . . eq) in U1 and (eq+1, . . . , e2q)
in U2 such that ϕ(ei, ei+q) = 0, for i, j = 1, . . . , q, it suffices to proves that dim(D) ≤ 1. If
x, y ∈ D with x = 0, from the identity
Φ(y − λx) = Φ(y) − λϕ(x, y) + λ2Φ(x)
and the fact that Φ(x) = 0 since x ∈ D and x = 0, we see that the equation Φ(y − λy) = 0
has at least one solution. Since Φ(z) = 0 for every nonzero z ∈ D, we get y = λx, and thus
dim(D) ≤ 1, as claimed.
We also have the following proposition which has applications in number theory.
Proposition 14.34. If Φ is any nondegenerate quadratic form such that there is some
nonzero vector x ∈ E with Φ(x) = 0, then for every α ∈ K, there is some y ∈ E such that
Φ(y) = α.
The proof is left as an exercise. We now turn to the Witt extension theorem.
14.7. WITT’S THEOREM
385
14.7
Witt’s Theorem
Witt’s theorem was referred to as a “scandal” by Emil Artin. What he meant by this is
that one had to wait until 1936 (Witt [110]) to formulate and prove a theorem at once so
simple in its statement and underlying concepts, and so useful in various domains (geometry,
arithmetic of quadratic forms).1
Besides Witt’s original proof (Witt [110]), Chevalley’s proof [20] seems to be the “best”
proof that applies to the symmetric as well as the skew-symmetric case. The proof in
Bourbaki [11] is based on Chevalley’s proof, and so are a number of other proofs. This is
the one we follow (slightly reorganized). In the symmetric case, Serre’s exposition is hard to
beat (see Serre [93], Chapter IV).
Theorem 14.35. (Witt, 1936) Let E and E be two finite-dimensional spaces respectively
equipped with two nondegenerate -Hermitan forms ϕ and ϕ satisfying condition (T), and
assume that there is an isometry between (E, ϕ) and (E , ϕ ). For any subspace U of E,
every injective metric linear map f from U into E extends to an isometry from E to E .
Proof. Since (E, ϕ) and (E , ϕ ) are isometric, we may assume that E = E and ϕ = ϕ (if
h : E → E is an isometry, then h−1 ◦ f is an injective metric map from U into E. The
details are left to the reader). We begin with the following observation. If U1 and U2 are
two subspaces of E such that U1 ∩ U2 = (0) and if we have metric linear maps f1 : U1 → E
and f2 : U2 → E such that
ϕ(f1(u1), f2(u2)) = ϕ(u1, u2) for ui ∈ Ui (i = 1, 2),
(∗)
then the linear map f : U1 ⊕ U2 → E given by f(u1 + u2) = f1(u1) + f2(u2) extends f1 and
f2 and is metric. Indeed, since f1 and f2 are metric and using (∗), we have
ϕ(f1(u1) + f2(u2), f1(v1) + f2(v2)) = ϕ(f1(u1), f1(v1)) + ϕ(f1(u1), f2(v2))
+ ϕ(f2(u2), f1(v1)) + ϕ(f2(u2), f2(v2))
= ϕ(u1, v1) + ϕ(u1, v2) + ϕ(u2, v1) + ϕ(u2, v2)
= ϕ(u1 + u2, v2 + v2).
Furthermore, if f1 and f2 are injective, then so if f .
We now proceed by induction on the dimension r of U . The case r = 0 is trivial. For
the induction step, r ≥ 1 so U = (0), and let H be any hyperplane in U. Let f : U → E
be an injective metric linear map. By the induction hypothesis, the restriction f0 of f to H
extends to an isometry g0 of E. If g0 extends f , we are done. Otherwise, H is the subspace
of elements of U left fixed by g−1
0
◦ f. If the theorem holds in this situation, namely the
1Curiously, some references to Witt’s paper claim its date of publication to be 1936, but others say 1937.
The answer to this mystery is that Volume 176 of Crelle Journal was published in four issues. The cover
page of volume 176 mentions the year 1937, but Witt’s paper is dated May 1936. This is not the only paper
of Witt appearing in this volume!
386
CHAPTER 14. BILINEAR FORMS AND THEIR GEOMETRIES
subspace of U left fixed by f is a hyperplane H in U , then we have an isometry g1 of E
extending g−1
0
◦ f, and g0 ◦ g1 is an isometry of E extending f. Therefore, we are reduced to
the following situation:
Case (H). The subspace of U left fixed by f is a hyperplane H in U .
In this case, the set D = {f(u) − u | u ∈ U} is a line in U (a one-dimensional subspace).
For all u, v ∈ U, we have
ϕ(f (u), f (v) − v) = ϕ(f(u), f(v)) − ϕ(f(u), v) = ϕ(u, v) − ϕ(f(u), v) = ϕ(u − f(u), v),
that is
ϕ(f (u), f (v) − v) = ϕ(u − f(u), v) for all u, v ∈ U,
(∗∗)
and if u ∈ H, which means that f(u) = u, we get u ∈ D⊥. Therefore, H ⊆ D⊥. Since ϕ is
nondegenerate, we have dim(D) + dim(D⊥) = dim(E), and since dim(D) = 1, the subspace
D⊥ is a hyperplane in E.
Hypothesis (V). We can find a subspace V of E orthogonal to D and such that
V ∩ U = V ∩ f(U) = (0).
Then, we have
ϕ(f (u), v) = ϕ(u, v) for all u ∈ U and all v ∈ V ,
since ϕ(f (u), v) − ϕ(u, v) = ϕ(f(u) − u, v) = 0, with f(u) − u ∈ D and v ∈ V orthogonal to
D. By the remark at the beginning of the proof, with f1 = f and f2 the inclusion of V into
E, we can extend f to an injective metric map on U ⊕ V leaving all vectors in V fixed. In
this case, the set {f(w) − w | w ∈ U ⊕ V } is still the line D. We show below that the fact
that f can be extended to U ⊕ V implies that f can be extended to the whole of E.
We are reduced to proving that a subspace V as above exists. We distinguish between
two cases.
Case (a). U ⊆ D⊥.
In this case, formula (∗∗) show that f(U) is not contained in D⊥ (check this!). Conse-
quently,
U ∩ D⊥ = f(U) ∩ D⊥ = H.
We can pick V to be any supplement of H in D⊥, and the above formula shows that V ∩U =
V ∩ f(U) = (0). Since U ⊕ V contains the hyperplane D⊥ (since D⊥ = H ⊕ V and H ⊆ U),
and U ⊕ V = D⊥ (since U is not contained in D⊥ and V ⊆ D⊥), we must have E = U ⊕ V ,
and as we showed as a consequence of hypothesis (V), f can be extended to an isometry of
U ⊕ V = E.
Case (b). U ⊆ D⊥.
In this case, formula (∗∗) shows that f(U) ⊆ D⊥ so U + f(U) ⊆ D⊥, and since D =
{f(u) − u | u ∈ U}, we have D ⊆ D⊥; that is, the line D is isotropic.
14.7. WITT’S THEOREM
387
We show that case (b) can be reduced to the situation where U = D⊥ and f is an
isometry of U . For this, we show that there exists a subspace V of D⊥, such that
D⊥ = U ⊕ V = f(U) ⊕ V.
This is obvious if U = f (U ). Otherwise, let x ∈ U with x /
∈ H, and let y ∈ f(U) with y /
∈ H.
Since f (H) = H (pointwise), f is injective, and H is a hyperplane in U , we have
U = H ⊕ Kx, f(U) = H ⊕ Ky.
We claim that x + y /
∈ U. Otherwise, since y = x + y − x, with x + y, x ∈ U and since
y ∈ f(U), we would have y ∈ U ∩ f(U) = H, a contradiction. Similarly, x + y /
∈ f(U). It
follows that
U + f (U ) = U ⊕ K(x + y) = f(U) ⊕ K(x + y).
Now, pick W to be any supplement of U + f (U ) in D⊥ so that D⊥ = (U + f (U )) ⊕ W , and
let
V = K(x + y) + W.
Then, since x ∈ U, y ∈ f(U), W ⊆ D⊥, and U + f(U) ⊆ D⊥, we have V ⊆ D⊥. We also
have
U ⊕ V = U ⊕ K(x + y) ⊕ W = (U + f(U)) ⊕ W = D⊥
and
f (U ) ⊕ V = f(U) ⊕ K(x + y) ⊕ W = (U + f(U)) ⊕ W = D⊥,
so as we showed as a consequence of hypothesis (V), f can be extended to an isometry of
the hyperplane D⊥, and D is still the line {f(w) − w | w ∈ U ⊕ V }.
The above argument shows that we are reduced to the situation where U = D⊥ is a
hyperplane in E and f is an isometry of U . If we pick any v /