Quadratic forms

Abstract

Quaternion algebras, as algebras equipped with a standard involution, are intrinsically related to quadratic forms. We develop this connection in the next two chapters.

Quaternion algebras, as algebras equipped with a standard involution, are intrinsically related to quadratic forms. We develop this connection in the next two chapters.

1 $\triangleright$ Reduced norm as quadratic form

Let F be a field with ${{\,\mathrm{char}\,}}F \ne 2$ and let $B=({a,b} \mid {F})$ be a quaternion algebra over F. We have seen (3.2.9) that B has a unique standard involution and consequently a reduced norm map, with

$$\begin{aligned} {{\,\mathrm{nrd}\,}}(t+xi+yj+zij)=t^2-ax^2-by^2+abz^2 \end{aligned}$$

(4.1.1)

for $t,x,y,z \in F$. The reduced norm therefore defines a quadratic form, a homogeneous polynomial of degree 2 in F[t, x, y, z] (thought of as a function of the coefficients of an element respect to the basis $1,i,j,k$). It should come as no surprise, then, that the structure of the quaternion algebra B is related to properties of the quadratic form ${{\,\mathrm{nrd}\,}}$.

Let $Q :V \rightarrow F$ be a quadratic form. Then Q can be diagonalized by a change of variables: there is a basis $e_1,\dots ,e_n$ of V such that

$$\begin{aligned} Q(x_1e_1+\dots +x_ne_n)=Q(x_1,\dots ,x_n) = a_1x_1^2 + \dots + a_nx_n^2 \end{aligned}$$

with $a_i \in F$. We define the discriminant  of Q to be the (well-defined) product ${{\,\mathrm{disc}\,}}(Q) :=a_1\cdots a_n/2^n \in F / F^{\times 2}$. (The factor $2^n$ is for consistency with more general notions; it is harmless if a bit annoying.) We say that a quadratic form is nondegenerate if its discriminant is nonzero. The reduced norm quadratic form (4.1.1) is already diagonal in the basis $1,i,j,k$, and it is nondegenerate because $a,b \ne 0$.

A similarity from Q to another quadratic form $Q' :V' \rightarrow F$ is a pair (f, u) where $f :V \xrightarrow {\sim } V'$ is an F-linear isomorphism and $u \in F^\times$ satisfy $Q'(f(x))=uQ(x)$ for all $x \in V$. An isometry is a similarity with $u=1$. The orthogonal group of Q is the group of self-isometries of Q, i.e.,

$$\begin{aligned} {{\,\mathrm{O}\,}}(Q)(F) :=\{f \in {{\,\mathrm{Aut}}}_F(V) : Q(f(x))=Q(x) \text { for all }x \in V\}. \end{aligned}$$

An isometry $f \in {{\,\mathrm{O}\,}}(Q)(F)$ is special if $\det f$, and the special orthogonal group of Q is the group of special isometries of Q.

More generally, we have seen that any algebra with a standard involution has a quadratic form ${{\,\mathrm{nrd}\,}}$. We say that the standard involution is nondegenerate whenever the quadratic form ${{\,\mathrm{nrd}\,}}$ is so. Generalizing Theorem 3.1.1, we prove the following (see Main Theorem 4.4.1 for the proof).

Main Theorem 4.1.2

Let B be an F-algebra. Then B has a nondegenerate standard involution if and only if one of the following holds:

• (i) $B=F$;

• (ii) $B=K$ has $\dim_F K=2$ and either $K \simeq F \times F$ or K is a field; or

• (iii) B is a quaternion algebra over F.

This theorem gives another way of characterizing quaternion algebras: they are noncommutative algebras with a nondegenerate standard involution.

In Section 2.4, we saw that the unit Hamiltonians $\mathbb H ^1$ act on the pure Hamiltonians $\mathbb H ^0$ (Section 2.4) by rotations: the standard Euclidean quadratic form (sum of squares) is preserved by conjugation. This generalizes in a natural way to an arbitrary field, and so we can understand the group of linear transformations that preserve a ternary (or quaternary) form in terms of the unit group of a quaternion algebra B (Proposition 4.5.10): there is an exact sequence

$$\begin{aligned} 1 \rightarrow F^\times \rightarrow B^\times \rightarrow {{\,\mathrm{SO}\,}}({{\,\mathrm{nrd}\,}}|_{B^0})(F) \rightarrow 1 \end{aligned}$$

where ${{\,\mathrm{SO}\,}}(Q)(F)$ is the group of special (or oriented) isometries of the quadratic form Q.

2 Basic definitions

In this section, we summarize basic definitions and notation for quadratic forms over fields. The “Bible for all quadratic form practitioners” (according to the MathSciNet review by K. Szymiczek) is the book by Lam [Lam2005]; in particular, Lam gives a very readable account of the relationship between quadratic forms and quaternion algebras over F when ${{\,\mathrm{char}\,}}F \ne 2$ [Lam2005, Sections III.1–III.2] and many other topics in the algebraic theory of quadratic forms. Also recommended are the books by Cassels [Cas78], O’Meara [O’Me73], and Scharlau [Scha85], as well as the book by Grove [Grov2002], who treats quadratic forms from a geometric point of view in terms of the orthogonal group. For reference and further inspiration, see also the hugely influential book by Eichler [Eic53].

Let F be a field. (For now, we allow ${{\,\mathrm{char}\,}}F$ to be arbitrary.)

Definition 4.2.1

A quadratic form Q is a map $Q:V\rightarrow F$ on an F-vector space V satisfying:

• (i) $Q(ax)=a^2Q(x)$ for all $a \in F$ and $x \in V$; and

• (ii) The map $T :V \times V \rightarrow F$ defined by

  $$\begin{aligned} T(x,y)=Q(x+y)-Q(x)-Q(y) \end{aligned}$$

  is F-bilinear.

We call the pair (V, Q) a quadratic space and T the associated bilinear form.

We will often abbreviate a quadratic space (V, Q) by simply V. If Q is a quadratic form then the associated bilinear form T is symmetric, satisfying $T(x,y)=T(y,x)$ for all $x,y \in V$; in particular, $T(x,x)=2Q(x)$ for all $x \in V$, so when ${{\,\mathrm{char}\,}}F \ne 2$ we recover the quadratic form from the symmetric bilinear form.

For the remainder of this section, let $Q:V \rightarrow F$ be a quadratic form with associated bilinear form T.

4.2.2

Suppose $\dim_F V=n<\infty$. Let $e_1,\dots ,e_n$ be a basis for V, giving an isomorphism $V \simeq F^n$. Then Q can be written

$$\begin{aligned} Q(x_1e_1+\dots +x_ne_n)=\sum_i Q(e_i)x_i^2 + \sum_{i<j} T(e_i,e_j)x_ix_j \in F[x_1,\dots ,x_n] \end{aligned}$$

as a homogeneous polynomial of degree 2.

The Gram matrix of Q in the basis $e_i$ is the (symmetric) matrix

$$\begin{aligned}{}[T] :=(T(e_i,e_j))_{i,j} \in {{\,\mathrm{M}}}_n(F). \end{aligned}$$

We then have $T(x,y)=x^{\textsf {t} }[T] y$ for $x,y \in V \simeq F^n$ as column vectors. Under a change of basis $A \in {{\,\mathrm{GL}}}_n(F)$ with $e_i'=Ae_i$, the Gram matrix $[T]'$ in the basis $e_i'$ has

$$\begin{aligned}{}[T]'=A^{\textsf {t} }[T]A. \end{aligned}$$

(4.2.3)

Definition 4.2.4

A similarity of quadratic forms from $Q:V \rightarrow F$ to $Q':V' \rightarrow F$ is a pair (f, u) where is an F-linear isomorphism and $u \in F^\times$ satisfy $Q'(f(x))=uQ(x)$ for all $x \in V$, i.e., such that the diagram

(4.2.5)

commutes. In a similarity (f, u), the scalar u is called the similitude factor  of the similarity. An isometry of quadratic forms (or isomorphism of quadratic spaces) is a similarity with similitude factor $u=1$; we write in this case $Q \simeq Q'$.

Definition 4.2.6

The general orthogonal group (or similarity group) of the quadratic form Q is the group of self-similarities of Q under composition

$$\begin{aligned} {{\,\mathrm{GO}\,}}(Q)(F) :=\{(f,u) \in {{\,\mathrm{Aut}}}_F(V) \times F^\times : Q(f(x))=uQ(x) \text { for all }x \in V\}; \end{aligned}$$

the orthogonal group of Q is the group of self-isometries of Q, i.e.,

$$\begin{aligned} {{\,\mathrm{O}\,}}(Q)(F) :=\{f \in {{\,\mathrm{Aut}}}_F(V) : Q(f(x))=Q(x) \text { for all }x \in V\}. \end{aligned}$$

Remark 4.2.7. A similarity allows isomorphisms of the target F (as a one-dimensional F-vector space). The notion of isometry comes from the connection with measuring lengths, when working with the usual Euclidean norm form on a vector space over $\mathbb R$: similarity allows these lengths to scale uniformly (e.g., similar triangles).

There is a canonical exact sequence

$$\begin{aligned} \begin{aligned} 1 \rightarrow {{\,\mathrm{O}\,}}(Q)(F) \rightarrow {{\,\mathrm{GO}\,}}(Q)(F)&\rightarrow F^\times \\ (f,u)&\mapsto u \end{aligned} \end{aligned}$$

(4.2.8)

realizing ${{\,\mathrm{O}\,}}(Q)(F) \le {{\,\mathrm{GO}\,}}(Q)(F)$ as the subgroup of self-similarities with similitude factor $u=1$.

4.2.9

Returning to 4.2.2, suppose $\dim_F V=n<\infty$ and ${{\,\mathrm{char}\,}}F \ne 2$. Then one can understand the orthogonal group of Q quite concretely in matrix terms as follows. Choose a basis $e_1,\dots ,e_n$ for V and let [T] be the Gram matrix of Q with respect to this basis, so that $2Q(x)=x^{\textsf {t} }[T] x$ for all $x \in V \simeq F^n$. Then ${{\,\mathrm{Aut}}}_F(V) \simeq {{\,\mathrm{GL}}}_n(F)$ and $A \in {{\,\mathrm{GL}}}_n(F)$ belongs to ${{\,\mathrm{O}\,}}(Q)$ if and only if

$$\begin{aligned} (Ax)^{\textsf {t} }[T] (Ax) = x^{\textsf {t} }(A^{\textsf {t} }[T] A)x = x^{\textsf {t} }[T] x \end{aligned}$$

for all $x \in V$, and therefore

$$\begin{aligned} {{\,\mathrm{O}\,}}(Q)(F) = \{A \in {{\,\mathrm{GL}}}_n(F) : A^{\textsf {t} }[T] A = [T]\} \end{aligned}$$

(4.2.10)

and

$$\begin{aligned} {{\,\mathrm{GO}\,}}(Q)(F) = \{(A,u) \in {{\,\mathrm{GL}}}_n(F) \times F^\times : A^{\textsf {t} }[T] A = u[T]\} \end{aligned}$$

(4.2.11)

From now on, let $Q:V \rightarrow F$ be a quadratic form and let $T:V \times V \rightarrow F$ be the symmetric bilinear form associated to Q.

Definition 4.2.12

Let $x,y \in V$. We say that x is orthogonal  to y (with respect to Q) if $T(x,y)=0$.

Since T is symmetric, x is orthogonal to y if and only if y is orthogonal to x for $x,y \in V$, and so we simply say x, y are orthogonal. If $S \subseteq V$ is a subset, we write

$$\begin{aligned} S^\perp :=\{x \in V : T(v,x)=0\text { for all }v \in S\} \end{aligned}$$

for the subspace of V which is orthogonal to (the span of) S.

4.2.13

Let B be an algebra over F with a standard involution. Then ${{\,\mathrm{nrd}\,}}:B \rightarrow F$ is a quadratic form on B. Indeed, ${{\,\mathrm{nrd}\,}}(a\alpha )=a^2 \alpha$ for all $\alpha \in B$, and the map T given by

$$\begin{aligned} T(\alpha ,\beta ) = (\alpha +\beta )\overline{(\alpha +\beta )}-\alpha \overline{\alpha }-\beta \overline{\beta }= \alpha \overline{\beta }+\beta \overline{\alpha }=\alpha \overline{\beta }+\overline{\alpha \overline{\beta }}= {{\,\mathrm{trd}\,}}(\alpha \overline{\beta }) \end{aligned}$$

(4.2.14)

for $\alpha ,\beta \in B$ is bilinear, and

$$\begin{aligned} T(\alpha ,\beta )={{\,\mathrm{trd}\,}}(\alpha \overline{\beta })={{\,\mathrm{trd}\,}}(\alpha ({{\,\mathrm{trd}\,}}(\beta )-\beta ))={{\,\mathrm{trd}\,}}(\alpha ){{\,\mathrm{trd}\,}}(\beta )-{{\,\mathrm{trd}\,}}(\alpha \beta ). \end{aligned}$$

(4.2.15)

So $\alpha ,\beta \in B$ are orthogonal with respect to ${{\,\mathrm{nrd}\,}}$ if and only if

$$\begin{aligned} {{\,\mathrm{trd}\,}}(\alpha \overline{\beta })=\alpha \overline{\beta }+\beta \overline{\alpha }=0 \end{aligned}$$

if and only if

$$\begin{aligned} {{\,\mathrm{trd}\,}}(\alpha \beta )={{\,\mathrm{trd}\,}}(\alpha ){{\,\mathrm{trd}\,}}(\beta ). \end{aligned}$$

Thus 1 and $\alpha \in B$ are orthogonal if and only if ${{\,\mathrm{trd}\,}}(\alpha )=0$ if and only if $\alpha ^2=-{{\,\mathrm{nrd}\,}}(\alpha )$. Moreover, rearranging (4.2.14),

$$\begin{aligned} \begin{aligned} \alpha \beta +\beta \alpha&={{\,\mathrm{trd}\,}}(\beta )\alpha +{{\,\mathrm{trd}\,}}(\alpha )\beta -T(\alpha ,\beta ). \end{aligned} \end{aligned}$$

(4.2.16)

In particular, if $1,\alpha ,\beta \in B$ are linearly independent over F, then by (4.2.16) they are pairwise orthogonal if and only if $\beta \alpha =-\alpha \beta$.

In this way, we see that the multiplication law in B is governed in a fundamental way by the reduced norm quadratic form.

Definition 4.2.17

Let $Q:V \rightarrow F$ be a quadratic form. We say that Q represents an element $a \in F$ if there exists $x \in V$ such that $Q(x)=a$. A quadratic form is universal if it represents every element of F.

Definition 4.2.18

A quadratic form Q (or a quadratic space V) is isotropic if Q represents 0 nontrivially (there exists $0 \ne x \in V$ such that $Q(x)=0$) and otherwise Q is anisotropic.

Remark 4.2.19. The terminology isotropic is as least as old as Eichler [Eic53, p. 3], and perhaps it goes back to Witt. The word can be used to mean “having properties that are identical in all directions”, and so the motivation for this language possibly comes from physics: the second fundamental form associated to a parametrized surface $z=f(x,y)$ in $\mathbb R ^3$ is a quadratic form, and (roughly speaking) this quadratic form defines the curvature at a given point. In this sense, if the quadratic form vanishes, then the curvature is zero, and things look the same in all directions.

4.2.20

Let $Q':V' \rightarrow F$ be another quadratic form. We define the orthogonal direct sum

$$\begin{aligned} Q \boxplus Q':V \oplus V'&\rightarrow F \\ (Q \boxplus Q')(x+x')&=Q(x)+Q(x') \end{aligned}$$

where $x \in V$ and $x' \in V'$; the associated bilinear form $T \boxplus T'$ has

$$\begin{aligned} (T \boxplus T')(x+x',y+y') =T(x,y) + T(x',y') \end{aligned}$$

for all $x,y \in V$ and $x',y'\in V'$. By definition, under the natural inclusion of $V,V' \subseteq V \oplus V'$, we have $V' \subseteq V^\perp$ (and $V \subseteq (V')^\perp$).

4.2.21

For $a \in F$, we write $\langle a \rangle$ for the quadratic form $ax^2$ on F. More generally, for $a_1,\dots ,a_n \in F$, we write

$$\begin{aligned} \langle a_1 \rangle \boxplus \dots \boxplus \langle a_n \rangle :=\langle a_1,\dots ,a_n \rangle \end{aligned}$$

for the quadratic form on $F^n$ defined by $Q(x_1,\dots ,x_n)=a_1x_1^2+\dots +a_nx_n^2$.

To conclude this introduction, we state an important result due originally to Witt which governs the decomposition of quadratic spaces into orthogonal sums up to isometry.

Theorem 4.2.22

Let $V \simeq V'$ be isometric quadratic spaces with orthogonal decompositions $V \simeq W_1 \boxplus W_2$ and $V' \simeq W_1' \boxplus W_2'$.

• (a) If $W_1 \simeq W_1'$, then $W_2 \simeq W_2'$.

• (b) If is an isometry, then there exists an isometry such that $f|_{W_1}=g$ and $f(W_2)=W_2'$.

Proof. The proof is requested in Exercise 4.16. For a proof and the equivalence between Witt cancellation (part (a)) and Witt extension (part (b)), see Lam [Lam2005, Proof of Theorem I.4.2, p. 14], Scharlau [Scha85, Theorem 1.5.3], or O’Meara [O’Me73, Theorem 42:17]. $\square$

Theorem 4.2.22(a) is called Witt cancellation and 4.2.22(b) is called Witt extension.

3 Discriminants, nondegeneracy

For the remainder of this chapter, we suppose that ${{\,\mathrm{char}\,}}F \ne 2$. (We take up the case ${{\,\mathrm{char}\,}}F = 2$ in section 6.3.) Throughout, let $Q:V \rightarrow F$ be a quadratic form with $\dim_F V=n<\infty$ and associated symmetric bilinear form T.

The following result (proven by induction) is a standard application of Gram–Schmidt orthogonalization (Exercise 4.1); working with a quadratic form as a polynomial, this procedure can be thought of as iteratively completing the square.

Lemma 4.3.1

There exists a basis of V such that $Q \simeq \langle a_1,\dots ,a_n \rangle$ with $a_i \in F$.

A form presented with a basis as in Lemma 4.3.1 is called  normalized (or  diagonal). For a diagonal quadratic form Q, the associated Gram matrix [T] is diagonal with entries $2a_1,\dots ,2a_n$.

4.3.2

The determinant $\det ([T])$ of a Gram matrix for Q depends on a choice of basis for V, but by (4.2.3), a change of basis matrix $A \in {{\,\mathrm{GL}}}_n(F)$ operates on [T] by $A^{\textsf {t} }[T]A$, and $\det (A^{\textsf {t} }[T]A)=\det (A)^2\det ([T])$, so we obtain a well-defined element $\det (T) \in F/F^{\times 2}$ independent of the choice of basis.

Definition 4.3.3

The discriminant  of Q is

$$\begin{aligned} {{\,\mathrm{disc}\,}}(Q) :=2^{-n} \det (T) \in F/F^{\times 2}. \end{aligned}$$

The signed discriminant of Q is

$$\begin{aligned} {{\,\mathrm{sgndisc}\,}}(Q) :=(-1)^{n(n-1)/2} {{\,\mathrm{disc}\,}}(Q) \in F/F^{\times 2}. \end{aligned}$$

When it will cause no confusion, we will represent the class of the discriminant in $F/F^{\times 2}$ simply by a representative element in F.

Remark 4.3.4. The extra factor $2^{-n}$ is harmless since ${{\,\mathrm{char}\,}}F \ne 2$, and it allows us to naturally cancel certain factors 2 that appear whether we are in even or odd dimension—it will be essential when we consider the case ${{\,\mathrm{char}\,}}F=2$ (see 6.3.1). The distinction between even and odd dimensional quadratic spaces is not arbitrary: indeed, this distinction is pervasive, even down to the classification of semisimple Lie algebras.

Example 4.3.5

We have ${{\,\mathrm{disc}\,}}(\langle a_1,\dots ,a_n \rangle )=a_1\cdots a_n$ for $a_i \in F$.

Definition 4.3.6

The bilinear form $T:V \times V \rightarrow F$ is nondegenerate if for all $x \in V \smallsetminus \{0\}$, the linear functional $T_x:V \rightarrow F$ defined by $T_x(y)=T(x,y)$ is nonzero, i.e., there exists $y \in V$ such that $T(x,y) \ne 0$. We say that Q (or V) is nondegenerate if the associated bilinear form T is nondegenerate.

4.3.7

The bilinear form T induces a map

$$\begin{aligned} V&\rightarrow {{\,\mathrm{Hom}\,}}(V,F) \\ x&\mapsto (y \mapsto T(x,y)) \end{aligned}$$

and T is nondegenerate if and only if this map is injective (and hence an isomorphism) if and only if $\det (T) \ne 0$. Put another way, Q is nondegenerate if and only if ${{\,\mathrm{disc}\,}}(Q) \ne 0$, and so a diagonal form $\langle a_1,\dots ,a_n \rangle$ is nondegenerate if and only if $a_i\ne 0$ for all i.

Example 4.3.8

Let $B=({a,b} \mid {F})$ be a quaternion algebra. Then by 3.2.9, the quadratic form ${{\,\mathrm{nrd}\,}}:B \rightarrow F$ is normalized with respect to the basis $1,i,j,ij$. Indeed,

$$\begin{aligned} {{\,\mathrm{nrd}\,}}\simeq \langle 1,-a,-b,ab \rangle . \end{aligned}$$

We have ${{\,\mathrm{disc}\,}}({{\,\mathrm{nrd}\,}})=(ab)^2 \ne 0$, so ${{\,\mathrm{nrd}\,}}$ is nondegenerate.

If B is an F-algebra with a standard involution, then the reduced norm defines a quadratic form on B, and we say that the standard involution is nondegenerate if ${{\,\mathrm{nrd}\,}}$ is nondegenerate.

4.3.9

One can often restrict to the case where a quadratic form Q is nondegenerate by splitting off the radical, as follows. We define the radical of Q to be

$$\begin{aligned} {{\,\mathrm{rad}\,}}(Q) :=V^\perp =\{x \in V : T(x,y)=0 \text { for all }y \in V\}. \end{aligned}$$

The radical ${{\,\mathrm{rad}\,}}(Q) \subseteq V$ is a subspace, so completing a basis of ${{\,\mathrm{rad}\,}}(Q)$ to V we can write (noncanonically) $V={{\,\mathrm{rad}\,}}(Q) \boxplus W$, as the direct sum is an orthogonal direct sum by definition of the radical. In this decomposition, $Q|_{{{\,\mathrm{rad}\,}}(Q)}$ is identically zero and $Q|_W$ is nondegenerate.

4 Nondegenerate standard involutions

In this section, we follow Theorem 3.5.1 with a characterization of quaternion algebras beyond division algebras.

Main Theorem 4.4.1

Suppose ${{\,\mathrm{char}\,}}F \ne 2$ and let B be an F-algebra. Then B has a nondegenerate standard involution if and only if one of the following holds:

• (i) $B=F$;

• (ii) $B=K$ is a quadratic F-algebra and either $K \simeq F \times F$ or K is a field; or

• (iii) B is a quaternion algebra over F.

Case (ii) in Main Theorem 4.4.1 is equivalent to requiring that K be a quadratic F-algebra that is reduced (has no nonzero nilpotent elements).

Remark 4.4.2. By Exercise 3.15, there exist F-algebras with standard involution having arbitrary dimension, so it is remarkable that the additional requirement that the standard involution be nondegenerate gives such a tidy result.

Proof of Main Theorem 4.4.1

If $B=F$, then the standard involution is the identity and ${{\,\mathrm{nrd}\,}}$ is nondegenerate. If $\dim_F K=2$, then after completing the square we may write $K \simeq F[x]/(x^2-a)$ and in the basis 1, x we find ${{\,\mathrm{nrd}\,}}\simeq \langle 1,a \rangle$. By Example 4.3.5, ${{\,\mathrm{nrd}\,}}$ is nondegenerate if and only if $a \in F^\times$ if and only if K is a quadratic field extension of F or $K \simeq F \times F$.

Suppose that $\dim_F B > 2$. Let $1,i,j$ be a part of a normalized basis for B with respect to the quadratic form ${{\,\mathrm{nrd}\,}}$. Then $T(1,i)={{\,\mathrm{trd}\,}}(i)=0$, so $i^2=a \in F^\times$, since ${{\,\mathrm{nrd}\,}}$ is nondegenerate. Note in particular that $\overline{i}=-i$. Similarly $j^2=b\in F^\times$, and by (4.2.16) we have ${{\,\mathrm{trd}\,}}(ij)=ij+ji=0$. We have $T(1,ij)={{\,\mathrm{trd}\,}}(ij)=0$, and $T(ij,i)={{\,\mathrm{trd}\,}}(\overline{i}(ij))=-a{{\,\mathrm{trd}\,}}(j)=0$ and similarly $T(ij,j)=0$, hence $ij\in \{1,i,j\}^\perp$. If $ij=0$ then $i(ij)=aj=0$ so $j=0$, a contradiction. Since ${{\,\mathrm{nrd}\,}}$ is nondegenerate, it follows then that the set $1,i,j,ij$ is linearly independent.

Therefore, the subalgebra A of B generated by $i,j$ satisfies $A \simeq ({a,b} \mid {F})$, and if $\dim_F B=4$ we are done. So let $k \in A^\perp$; then ${{\,\mathrm{trd}\,}}(k)=0$ and $k^2=c \in F^\times$. Thus $k \in B^\times$, with $k^{-1}=c^{-1}k$. By 4.2.13 we have $k\alpha =\overline{\alpha }k$ for any $\alpha \in A$ since $\overline{k}=-k$. But then

$$\begin{aligned} k(ij)=(\overline{ij})k=\overline{j}\,\overline{i}k=\overline{j}ki=k(ji). \end{aligned}$$

(4.4.3)

But $k \in B^\times$ so $ij=ji=-ij$, and this is a contradiction. $\square$

Main Theorem 4.4.1 has the following corollaries.

Corollary 4.4.4

Let B be an F-algebra with ${{\,\mathrm{char}\,}}F \ne 2$. Then B is a quaternion algebra if and only if B is noncommutative and has a nondegenerate standard involution.

Proof. Immediate. $\square$

Corollary 4.4.5

Let B have a nondegenerate standard involution, and suppose that $K \subseteq B$ is a commutative F-subalgebra such that the restriction of the standard involution is nondegenerate. Then $\dim_F K \le 2$. Moreover, if $K \ne F$, then the centralizer of $K^\times$ in $B^\times$ is $K^\times$.

Proof. The first statement is immediate; the second follows by considering the algebra generated by the centralizer. $\square$

Remark 4.4.6 Algebras with involutions come from quadratic forms, and the results of this chapter are just one special case of a much more general theory. More precisely, there is a natural bijection between the set of isomorphism classes of finite-dimensional simple F-algebras equipped with an F-linear involution and the set of similarity classes of nondegenerate quadratic forms on finite-dimensional F-vector spaces. More generally, for involutions that act nontrivially on the base field, one looks at Hermitian forms. Consequently, there are three broad types of involutions on central simple algebras, depending on the associated quadratic or Hermitian form: orthogonal, symplectic, and unitary. Accordingly, algebras with involutions can be classified by the invariants of the associated form. This connection is the subject of the tome by Knus–Merkurjev–Rost–Tignol [KMRT98]. In this way the theory of quadratic forms belongs to the theory of algebras with involution, which in turn is a part of the theory of linear algebraic groups, as expounded by Weil [Weil60]: see the survey by Tignol [Tig98] for an overview and further references.

5 Special orthogonal groups

In this section, we revisit the original motivation of Hamilton (Section 2.4) in a more general context, relating quaternions to the orthogonal group of a quadratic form. We retain our running hypothesis that ${{\,\mathrm{char}\,}}F\ne 2$ and $Q:V \rightarrow F$ is a nondegenerate quadratic form with $\dim_F V=n<\infty$.

Definition 4.5.1

An isometry $f \in {{\,\mathrm{O}\,}}(Q)(F)$ is special (or proper) if $\det f = 1$. The special orthogonal group of Q is the group of special isometries of Q:

$$\begin{aligned} {{\,\mathrm{SO}\,}}(Q)(F) :=\{f \in {{\,\mathrm{O}\,}}(Q)(F) : \det (f)=1\}. \end{aligned}$$

The condition “$\det f=1$” is well-defined, independent of the choice of F-basis of V; having chosen a basis of V so that ${{\,\mathrm{O}\,}}(Q)(F) \le {{\,\mathrm{GL}}}_n(F)$, we have ${{\,\mathrm{SO}\,}}(Q)(F) = {{\,\mathrm{O}\,}}(Q)(F) \cap {{\,\mathrm{SL}}}_n(F)$.

4.5.2

Suppose that $V=F^n$ and let $f \in {{\,\mathrm{O}\,}}(Q)$ be a self-isometry of Q, represented in the standard basis by $A \in {{\,\mathrm{GL}}}_n(F)$. Taking determinants in (4.2.10) we conclude that $\det (A)^2=1$ so $\det (A)=\pm 1$. The determinant is surjective (see Exercise 4.15), so we have an exact sequence

$$\begin{aligned} 1 \rightarrow {{\,\mathrm{SO}\,}}(Q)(F) \rightarrow {{\,\mathrm{O}\,}}(Q)(F) \xrightarrow {\det } \{\pm 1\} \rightarrow 1. \end{aligned}$$

If n is odd, then either f or $-f$ is special, so the sequence splits and

$$\begin{aligned} {{\,\mathrm{O}\,}}(Q)(F) \simeq \{\pm 1 \} \times {{\,\mathrm{SO}\,}}(Q)(F). \end{aligned}$$

(4.5.3)

4.5.4

Similarly, if $(f,u) \in {{\,\mathrm{GO}\,}}(Q)(F)$ then from (4.2.11) we get $u^{-n}\det (f)^2=1$. If $n=2m$ is even, then $u^{-m}\det (f)=\pm 1$, and we define the general special orthogonal group (or special similarity group) of Q to be

$$\begin{aligned} {{\,\mathrm{GSO}\,}}(Q)(F) :=\{(f,u) \in {{\,\mathrm{GO}\,}}(Q)(F) : u^{-m}\det (f)=1\} \end{aligned}$$

giving an exact sequence

$$\begin{aligned} 1 \rightarrow {{\,\mathrm{GSO}\,}}(Q)(F) \rightarrow {{\,\mathrm{GO}\,}}(Q)(F) \rightarrow \{\pm 1\} \rightarrow 1. \end{aligned}$$

If n is odd, we have little choice other than to define ${{\,\mathrm{GSO}\,}}(Q)(F) :={{\,\mathrm{GO}\,}}(Q)(F)$.

Example 4.5.5

If $V=\mathbb R ^n$ and Q is the usual Euclidean norm on V, then

$$\begin{aligned} {{\,\mathrm{O}\,}}(Q)(\mathbb R )={{\,\mathrm{O}\,}}(n)=\{A \in {{\,\mathrm{GL}}}_n(\mathbb R ) : AA^{\textsf {t} }=1\} \end{aligned}$$

is the group of linear maps preserving length (but not necessarily orientation), whereas ${{\,\mathrm{SO}\,}}(Q)(\mathbb R )$ is the usual group of rotations of V (preserving orientation). Similarly, ${{\,\mathrm{GSO}\,}}(Q)(\mathbb R )$ consists of orientation-preserving similarities, preserving orientation but allowing a constant scaling.

In particular, if $n=2$ then ${{\,\mathrm{O}\,}}(2) :={{\,\mathrm{O}\,}}(Q)(\mathbb R )$ contains

$$\begin{aligned} {{\,\mathrm{SO}\,}}(2) :={{\,\mathrm{SO}\,}}(Q)(\mathbb R )= \left\{ \begin{pmatrix} \cos \theta &{} \sin \theta \\ -\sin \theta &{} \cos \theta \end{pmatrix} : \theta \in \mathbb R \right\} \simeq \mathbb R /(2\pi \mathbb Z ) \simeq \mathbb {S}^1 \end{aligned}$$

(the circle group) with index 2, with a reflection in any line through the origin representing a nontrivial coset of ${{\,\mathrm{SO}\,}}(2) \le {{\,\mathrm{O}\,}}(2)$.

4.5.6

More generally, we may define reflections in ${{\,\mathrm{O}\,}}(Q)(F)$ as follows. For $x \in V$ anisotropic (so $Q(x) \ne 0$), we define the reflection in x to be

$$\begin{aligned} \tau_x :V&\rightarrow V \\ \tau_x(v)&= v - \frac{T(v,x)}{Q(x)}x. \end{aligned}$$

We have $\tau_x(x)=x-2x=-x$, and

$$\begin{aligned} Q(\tau_x(v))&= Q(v)+Q\left( -\frac{T(v,x)}{Q(x)}x\right) + T\left( v,-\frac{T(v,x)}{Q(x)}x\right) \\&= Q(v) + \frac{T(v,x)^2}{Q(x)^2}Q(x) - \frac{T(v,x)}{Q(x)}T(v,x) = Q(v) \end{aligned}$$

so $\tau_x \in {{\,\mathrm{O}\,}}(Q)(F) \smallsetminus {{\,\mathrm{SO}\,}}(Q)(F)$.

By a classical theorem of Cartan and Dieudonné, the orthogonal group is generated by reflections.

Theorem 4.5.7

(Cartan–Dieudonné). Let (V, Q) be a nondegenerate quadratic space with $\dim_F V=n$. Then every isometry $f \in {{\,\mathrm{O}\,}}(Q)(F)$ is a product of at most n reflections.

Proof. See Lam [Lam2005, §I.7], O’Meara [O’Me73, §43B], or Scharlau [Scha85, Theorem 1.5.4]. The proof is by induction on n, carefully recording the effect of a reflection in an anisotropic vector. $\square$

Since reflections have determinant $-1$, an isometry f is special (and $f \in {{\,\mathrm{SO}\,}}(Q)(F)$) if and only if it is the product of an even number of reflections.

4.5.8

Now let B be a quaternion algebra over F, and recall (3.3.5) that we have defined

$$\begin{aligned} B^0 :=\{v \in B : {{\,\mathrm{trd}\,}}(B)=0\}. \end{aligned}$$

Then there is a (left) action

$$\begin{aligned} \begin{aligned} B^\times \circlearrowright B^0&\rightarrow B^0 \\ \alpha \cdot v&= \alpha v \alpha ^{-1}. \end{aligned} \end{aligned}$$

(4.5.9)

since ${{\,\mathrm{trd}\,}}(\alpha v \alpha ^{-1})={{\,\mathrm{trd}\,}}(v)=0$. Moreover, $B^\times$ acts on V by isometries with respect to the quadratic form $Q={{\,\mathrm{nrd}\,}}|_{B^0} :V \rightarrow F$, since ${{\,\mathrm{nrd}\,}}(\alpha v \alpha ^{-1})={{\,\mathrm{nrd}\,}}(v)$ for all $\alpha \in B$ and $v \in V$.

Proposition 4.5.10

Let B be a quaternion algebra over F. Then the action (4.5.9) induces an exact sequence

$$\begin{aligned} 1 \rightarrow F^\times \rightarrow B^\times \rightarrow {{\,\mathrm{SO}\,}}({{\,\mathrm{nrd}\,}}|_{B^0})(F) \rightarrow 1. \end{aligned}$$

(4.5.11)

If further ${{\,\mathrm{nrd}\,}}(B^\times ) = F^{\times 2}$, then

$$\begin{aligned} 1 \rightarrow \{\pm 1\} \rightarrow B^1 \rightarrow {{\,\mathrm{SO}\,}}({{\,\mathrm{nrd}\,}}|_{B^0})(F) \rightarrow 1, \end{aligned}$$

where $B^1 :=\{\alpha \in B : {{\,\mathrm{nrd}\,}}(\alpha )=1\}$.

Proof. Let $Q={{\,\mathrm{nrd}\,}}|_{B^0}$. We saw in 4.5.8 that the action of $B^\times$ is by isometries, so lands in ${{\,\mathrm{O}\,}}(Q)(F)$. By the Cartan–Dieudonné theorem (Theorem 4.5.7, the weak version of Exercise 4.17 suffices), every isometry is the product of reflections, and by determinants an isometry is special if and only if it is the product of an even number of reflections. A reflection in $x \in V=B^0$ with $Q(x)={{\,\mathrm{nrd}\,}}(x) \ne 0$ is of the form

$$\begin{aligned} \begin{aligned} \tau_x(v)&= v - \frac{T(v,x)}{Q(x)}x = v - \frac{{{\,\mathrm{trd}\,}}(v\overline{x})}{{{\,\mathrm{nrd}\,}}(x)}x \\&= v - (v\overline{x}+x\overline{v})\overline{x}^{-1} = -x\overline{v}\,\overline{x}^{-1} = x\overline{v}x^{-1}, \end{aligned} \end{aligned}$$

(4.5.12)

the final equality from $\overline{x}=-x$ as $x \in B^0$. The product of two such reflections is thus of the form $v \mapsto \alpha v \alpha ^{-1}$ with $\alpha \in B^\times$. Therefore $B^\times$ acts by special isometries, and every special isometry so arises: the map $B^\times \rightarrow {{\,\mathrm{O}\,}}(Q)(F)$ surjects onto ${{\,\mathrm{SO}\,}}(Q)(F)$. The kernel of the action is given by those $\alpha \in B^\times$ with $\alpha v \alpha ^{-1} =v$ for all $v \in B^0$, i.e., $\alpha \in Z(B^\times )=F^\times$.

The second statement follows directly by writing $B^\times = B^1 F^\times$. $\square$

Example 4.5.13

If $B \simeq {{\,\mathrm{M}}}_2(F)$, then ${{\,\mathrm{nrd}\,}}=\det$, so $\det ^0 \simeq \langle 1, -1, -1 \rangle$ and (4.5.11) yields the isomorphism ${{\,\mathrm{PGL}}}_2(F) \simeq {{\,\mathrm{SO}\,}}(\langle 1, -1, -1 \rangle )(F)$.

Example 4.5.14

If $F=\mathbb R$ and $B=\mathbb H$, then $\det (\mathbb H ) = \mathbb R_{>0} = \mathbb R ^{\times 2}$, and the second exact sequence is Hamilton’s (Section 2.4).

To conclude, we pass from three variables to four variables.

4.5.15

In Exercise 2.17, we showed that there is an exact sequence

$$\begin{aligned} 1 \rightarrow \{\pm 1\} \rightarrow \mathbb H ^1 \times \mathbb H ^1 \rightarrow {{\,\mathrm{SO}\,}}(4) \rightarrow 1 \end{aligned}$$

with $\mathbb H ^1 \times \mathbb H ^1$ acting on $\mathbb H \simeq \mathbb R ^4$ by $v \mapsto \alpha v \beta ^{-1}=\alpha v \overline{\beta }$ for $\alpha ,\beta \in \mathbb H ^1$.

More generally, let B be a quaternion algebra over F. Then there is a left action of $B^\times \times B^\times$ on B:

$$\begin{aligned} \begin{aligned} B^\times \times B^\times \circlearrowright B&\rightarrow B \\ (\alpha ,\beta )\cdot v&= \alpha v \beta ^{-1}. \end{aligned} \end{aligned}$$

(4.5.16)

This action is by similarities, since if $a={{\,\mathrm{nrd}\,}}(\alpha )$ and $b={{\,\mathrm{nrd}\,}}(\beta )$, then

$$\begin{aligned} {{\,\mathrm{nrd}\,}}(\alpha v \beta ^{-1})={{\,\mathrm{nrd}\,}}(\alpha ){{\,\mathrm{nrd}\,}}(v){{\,\mathrm{nrd}\,}}(\beta ^{-1})=\frac{a}{b}{{\,\mathrm{nrd}\,}}(v) \end{aligned}$$

for all $v \in V$, with similitude factor $u=a/b$. In particular, if ${{\,\mathrm{nrd}\,}}(\alpha )={{\,\mathrm{nrd}\,}}(\beta )$, then the action is by isometries.

Proposition 4.5.17

With notation as in 4.5.15, the left action (4.5.16) induces exact sequences

$$\begin{aligned} \begin{aligned} 1 \rightarrow F^\times&\rightarrow B^\times \times B^\times \rightarrow {{\,\mathrm{GSO}\,}}({{\,\mathrm{nrd}\,}})(F) \rightarrow 1 \\ a&\mapsto (a,a) \end{aligned} \end{aligned}$$

(4.5.18)

and

$$\begin{aligned} 1 \rightarrow F^\times \rightarrow \{(\alpha ,\beta ) \in B^\times \times B^\times : {{\,\mathrm{nrd}\,}}(\alpha )={{\,\mathrm{nrd}\,}}(\beta )\} \rightarrow {{\,\mathrm{SO}\,}}({{\,\mathrm{nrd}\,}})(F) \rightarrow 1. \end{aligned}$$

If further ${{\,\mathrm{nrd}\,}}(B^\times ) = F^{\times 2}$, then the sequence

$$\begin{aligned} 1 \rightarrow \{\pm 1\} \rightarrow B^1 \times B^1 \rightarrow {{\,\mathrm{SO}\,}}({{\,\mathrm{nrd}\,}})(F) \rightarrow 1 \end{aligned}$$

is exact.

Proof. For the first statement, we first show that the kernel of the action is the diagonally embedded $F^\times$. Suppose that $\alpha v \beta ^{-1} = v$ for all $v \in B$; taking $v=1$ shows $\beta =\alpha$, and then we conclude that $\alpha v = v\alpha$ for all $v \in B$ so $\alpha \in Z(B)=F$.

Next, the map $B^\times \times B^\times \rightarrow {{\,\mathrm{GSO}\,}}({{\,\mathrm{nrd}\,}})(F)$ is surjective. If $f \in {{\,\mathrm{GSO}\,}}({{\,\mathrm{nrd}\,}})(F)$ then ${{\,\mathrm{nrd}\,}}(f(x))=u{{\,\mathrm{nrd}\,}}(x)$ for all $x \in B$, so in particular $u \in {{\,\mathrm{nrd}\,}}(B^\times )$. Every such similitude factor occurs, since the similitude factor of $(\alpha ,1)$ is ${{\,\mathrm{nrd}\,}}(\alpha )$. So it suffices to show that the map

$$\begin{aligned} \{(\alpha ,\beta ) \in B^\times \times B^\times : {{\,\mathrm{nrd}\,}}(\alpha )={{\,\mathrm{nrd}\,}}(\beta )\} \rightarrow {{\,\mathrm{SO}\,}}({{\,\mathrm{nrd}\,}})(F) \end{aligned}$$

is surjective. We again appeal to the Cartan–Dieudonné theorem; by the same computation as in (4.5.12), we calculate that a reflection in $x \in B^\times$ is of the form

$$\begin{aligned} \tau_x(v) = -x\overline{v}\,\overline{x}^{-1}. \end{aligned}$$

The product of two reflections for $x,y \in B^\times$ is thus of the form

$$\begin{aligned} v \mapsto -y\overline{(-x\overline{x}\overline{x}^{-1})} \overline{y}^{-1}=(yx^{-1})v\overline{(yx^{-1})}^{-1} = \alpha v \beta ^{-1} \end{aligned}$$

where $\alpha =yx^{-1}$ and $\beta =\overline{\alpha }$, and in particular the action is by special similarities. We conclude that (4.5.18) and the second sequence are both exact.

The final statement again follows by writing $B^\times = B^1 F^\times$, and seeing the kernel as $F^\times \cap B^1 = \{\pm 1\}$. $\square$

Example 4.5.19

When $B={{\,\mathrm{M}}}_2(F)$, then ${{\,\mathrm{nrd}\,}}(B^\times )=\det ({{\,\mathrm{GL}}}_2(F))=F^\times$, giving the exact sequence

$$\begin{aligned} 1 \rightarrow {{\,\mathrm{GL}}}_1(F) \rightarrow {{\,\mathrm{GL}}}_2(F) \times {{\,\mathrm{GL}}}_2(F) \rightarrow {{\,\mathrm{GSO}\,}}(\det )(F) \rightarrow 1. \end{aligned}$$

6 Exercises

Let F be a field with ${{\,\mathrm{char}\,}}F \ne 2$.

$\triangleright$ 1.:

Give an algorithmic proof that every finite-dimensional quadratic space has a normalized basis (Lemma 4.3.1).

• 2. Let $F=\mathbb R$ and let

  $$ V = \biggl \{ (a_n)_n : a_n \in \mathbb R \text { for all }n \ge 0~\text {and } \sum_{n=0}^{\infty } a_n^2 \text { converges} \biggr \}. $$

  Show that V is an $\mathbb R$-vector space, and the map $Q:V \rightarrow \mathbb R$ by $Q((a_n)_n) = \sum_{n=0}^{\infty } a_n^2$ is a quadratic form, and so V is an example of an infinite-dimensional quadratic space. [This example generalizes to the context of Hilbert spaces.]

• 3. Let B be a quaternion algebra over F. Let $N:B \rightarrow F$ and $\Delta :B \rightarrow F$ be defined by $N(\alpha )={{\,\mathrm{trd}\,}}(\alpha ^2)$ and $\Delta (\alpha )={{\,\mathrm{trd}\,}}(\alpha )^2-4{{\,\mathrm{nrd}\,}}(\alpha )$. Show that $N,\Delta$ are quadratic forms on B, describe their associated bilinear forms, and compute a normalized form (and basis) for each.

• 4. Generalize Exercise 2.15 as follows. Let $Q:V \rightarrow F$ be a quadratic form with $\dim_F V=n$ and let $\gamma \in {{\,\mathrm{O}\,}}(Q)$.

  • (a) If n is odd and $\det \gamma =1$, then $\gamma$ has a nonzero fixed vector (and therefore restricts to the identity on a one-dimensional subspace of V).

  • (b) If n is even and $\det \gamma =-1$, then $\gamma$ has both eigenvalues $-1$ and 1.

• 5. Generalizing part of Exercise 4.3, let B be an F-algebra with a standard involution. Show that the discriminant form

  $$\begin{aligned} \Delta :B&\rightarrow F \\ \Delta (\alpha )&={{\,\mathrm{trd}\,}}(\alpha )^2 - 4{{\,\mathrm{nrd}\,}}(\alpha ) \end{aligned}$$

  is a quadratic form.

• 6. Let $Q :V \rightarrow F$ be a quadratic form with $\dim_F V < \infty$ and associated bilinear form T. The map

  $$\begin{aligned} V&\rightarrow {{\,\mathrm{Hom}}}_F(V,F) \\ x&\mapsto (y \mapsto T(x,y)) \end{aligned}$$

  is F-linear. Show that Q is nondegenerate if and only if this map is an isomorphism.

• 7. Write out the action (4.5.9) explicitly, as follows. Let $B=({a,b} \mid {F})$ and let $\alpha =t+xi+yj+zij$.

  • (a) Show that the matrix of the action $v \mapsto \alpha v \alpha ^{-1}$ in the F-basis $\beta =\{i,j,k\}$ for $B^0$ is $[\alpha ]$ where ${{\,\mathrm{nrd}\,}}(\alpha )[\alpha ]$ is equal to

    $$ \begin{pmatrix} t^2 - ax^2 + by^2 - abz^2 &{} -2a(tz + xy) &{} -2(ty+axz) \\ 2b(tz - xy) &{} t^2 + ax^2 - by^2 - abz^2 &{} 2(tx - byz) \\ 2b(axz-ty) &{} 2a(tx + byz) &{} t^2 + ax^2 + by^2 + abz^2 \end{pmatrix} $$

    and ${{\,\mathrm{nrd}\,}}(\alpha )=t^2-ax^2-by^2+abz^2$.

  • (b) Let $Q={{\,\mathrm{nrd}\,}}|_{B^0}$ and let T be the associated bilinear form. Show that the Gram matrix [T] in the basis $\beta$ is the diagonal matrix with entries $-2a,-2b,2ab$. Then confirm by direct calculation that

    $$ {[}\alpha {]} \in {{\,\mathrm{SO}\,}}(Q)(F) = \{A \in {{\,\mathrm{SL}}}_3(F) : A[T] A^{\textsf {t} }= [T]\}. $$

• 8. In this exercise, we prove the chain lemma. Let $B :=({a,b} \mid {F})$ be a quaternion algebra.

  • (a) Show that if $i'$ is orthogonal to 1, j, then $(i')^2=a' \in F^\times$ and $i',j$ are standard generators for B, so $B \simeq ({a',b} \mid {F})$.

  • (b) Let $B' :=({a',b'} \mid {F})$, and suppose that B is isomorphic to $B'$. Show that there exists $c \in F^\times$ such that

    $$ B = \displaystyle {\biggl (\frac{a,b}{F}\biggr )} \simeq \displaystyle {\biggl (\frac{c,b}{F}\biggr )} \simeq \displaystyle {\biggl (\frac{c,b'}{F}\biggr )} \simeq \displaystyle {\biggl (\frac{a',b'}{F}\biggr )}. $$

    [Hint: let  be the isomorphism, and take an element orthogonal to $1,j,\phi (j')$.]

• 9. In this exercise, we develop some of the notions mentioned in Remark 3.3.8 in the context of quadratic forms.

  Let B be a finite-dimensional F-algebra (not necessarily a quaternion algebra), and let ${{\,\mathrm{Tr}\,}}:B \rightarrow F$ be the left algebra trace (the trace of the endomorphism given by left multiplication).

  • (a) Show that the map $B \rightarrow F$ defined by $x \mapsto {{\,\mathrm{Tr}\,}}(x^2)$ is a quadratic form on B; this form is called the (left) trace form  on B.

  • (b) Compute the trace form of $A \times B$ and $A \otimes_F B$ in terms of the trace form of A and B.

  • (c) Show that if $K \supseteq F$ is a inseparable field extension of finite degree, then the trace form on K (as an F-algebra) is identically zero. On the other hand, show that if K/F is a finite separable field extension (with ${{\,\mathrm{char}\,}}F \ne 2$) then the trace form is nondegenerate.

  • (d) Compute the trace form on $\mathbb Q (\sqrt{5})$ and $\mathbb Q (\alpha )$ where $\alpha =2\cos (2\pi /7)$, so that $\alpha ^3+\alpha ^2-2\alpha -1=0$.

$\triangleright$ 10.:

Let $Q:V \rightarrow F$ and $Q':V' \rightarrow F$ be quadratic forms over F with $\dim_F V=\dim_F V'=n<\infty$, and let $T,T'$ be the associated bilinear forms. Suppose that there is a similarity $Q \sim Q'$ with similitude factor $u \in F^\times$. Show that $\det T'=u^n \det T \in F/F^{\times 2}$.

• 11. Let $Q :V \rightarrow F$ be a nondegenerate quadratic form with $\dim_F V=n< \infty$.

  • (a) A subspace $W \subseteq V$ is totally isotropic if $Q|_W=0$ is identically zero. The Witt index $\nu (Q)$ of Q is the maximal dimension of a totally isotropic subspace. Show that if $\nu (Q)=m$ then $2m \le n$.

  • (b) A Pfister form is a form in $2^m$ variables defined inductively by $\langle \!\langle a \rangle \!\rangle = \langle 1, -a \rangle$ and

    $$\begin{aligned} \langle \!\langle a_1, \dots , a_{m-1}, a_m \rangle \!\rangle = \langle \!\langle a_1,\dots ,a_{m-1} \rangle \!\rangle \boxplus -a_m\langle \!\langle a_1, \dots , a_{m-1} \rangle \!\rangle . \end{aligned}$$

    Show that the reduced norm ${{\,\mathrm{nrd}\,}}$ on $\displaystyle {\biggl (\frac{a,b}{F}\biggr )}$ is the Pfister form $\langle \langle a,b \rangle \rangle$.

  • (c) The hyperbolic plane is the quadratic form ${H :F^2 \rightarrow F}$ with $H(x,y)=xy$. A quadratic form Q is a hyperbolic plane if $Q \simeq H$. A quadratic form Q is totally hyperbolic if $Q \simeq H \boxplus \dots \boxplus H$ where H is a hyperbolic plane. Show that if Q is an isotropic Pfister form, then Q is totally hyperbolic.

  • (d) Suppose that Q is an isotropic Pfister form with $n \ge 4$. Let $W \subset V$ be a subspace of dimension $n-1$. Show that $Q|_W$ is isotropic. [This gives another proof of Main Theorem 5.4.4 (iii) $\Rightarrow$ (iv).]

• 12.

  • (a) Let B be a quaternion algebra over F. Show that the reduced norm is the unique nonzero quadratic form Q on B that is multiplicative, i.e., $Q(\alpha \beta )=Q(\alpha )Q(\beta )$ for all $\alpha ,\beta \in B$.

  • (b) Show that (a) does not necessarily hold more generally, for B an algebra with a standard involution. [Hint: consider upper triangular matrices.]

$\triangleright$ 13.:

In this exercise, we pursue some geometric notions for readers some background in algebraic geometry (at the level of Hartshorne [Har77, Chapter 1]).

Let Q be nonzero quadratic form on V with $\dim_F V=n$. The vanishing locus of $Q(x)=0$ defines a projective variety $X \subseteq \mathbb P (V) \simeq \mathbb P ^n$ of degree 2 called a quadric. Show that the quadratic form Q is nondegenerate if and only if the projective variety X is nonsingular. [For this reason, a nondegenerate quadratic form is also synonymously called nonsingular.]

• 14. In this exercise, we work out from scratch Example 4.5.13: we translate the results on rotations in section 2.4 to $B={{\,\mathrm{M}}}_2(\mathbb R )$, but with respect to a different measure of ‘length’.

  Let

  $$ {{\,\mathrm{M}}}_2(\mathbb R )^0=\{v \in {{\,\mathrm{M}}}_2(\mathbb R ) : {{\,\mathrm{tr}\,}}(v)=0\}=\left\{ \begin{pmatrix} x &{} y \\ z &{} -x \end{pmatrix} : x,y,z \in \mathbb R \right\} . $$

  For $v \in {{\,\mathrm{M}}}_2(\mathbb R )^0$, we have $\det (v)=-x^2-yz$. Show that the group

  $$ {{\,\mathrm{M}}}_2(\mathbb R )^1={{\,\mathrm{SL}}}_2(\mathbb R )=\{\alpha \in {{\,\mathrm{M}}}_2(\mathbb R ) : \det (\alpha )=1\} $$

  acts linearly on ${{\,\mathrm{M}}}_2(\mathbb R )^0$ by conjugation (the adjoint representation) preserving the determinant, giving rise to an exact sequence

  $$ 1 \rightarrow \{\pm 1\} \rightarrow {{\,\mathrm{SL}}}_2(\mathbb R ) \rightarrow {{\,\mathrm{SO}\,}}(\det ) \rightarrow 1. $$

$\triangleright$ 15.:

Let $Q:V \rightarrow F$ be a quadratic form with V finite-dimensional over F. Show that ${{\,\mathrm{SO}\,}}(Q) \le {{\,\mathrm{O}\,}}(Q)$ is a (normal) subgroup of index 2. What can you say about ${{\,\mathrm{GSO}\,}}(Q) \le {{\,\mathrm{GO}\,}}(Q)$?

$\triangleright$ 16.:

In this exercise, we prove Theorem 4.2.22. Let $Q :V \rightarrow F$ be a quadratic form with $\dim_F V <\infty$ and let T be its associated bilinear form.

(a):

Let $v \in V$ be anisotropic. Define the reflection along v by

$$\begin{aligned} \tau_v&:V \rightarrow V \\ \tau_v(x)&= x - \frac{T(v,x)}{Q(v)}v. \end{aligned}$$

Observe that $\tau_v$ is F-linear, and then show that $\tau_v \in {{\,\mathrm{O}\,}}(V)$ with $\det \tau_v=-1$. [Hint: extend v to a basis of the orthogonal complement of V.] Why is $\tau_v$ called a reflection?

(b):

If $x,y \in V$ are anisotropic with $Q(x)=Q(y)$, show that there exists $f \in {{\,\mathrm{O}\,}}(V)$ such that $f(x)=y$. [Hint: reflect along either $v=x+y$ or $v=x-y$ as at least one is anisotropic, in the former case postcomposing with reflection along x.]

(c):

Let $Q' :V' \rightarrow F$ be another quadratic form, and let be an isometry. For $W \subseteq V$, show that $f(W^\perp ) = f(W)^\perp$.

(d):

Prove Theorem 4.2.22(a). [Hint: reduce to the case where $\dim_F W_1=\dim_F W_1'=1$; apply parts (b) and (c).]

(e):

Prove Theorem 4.2.22(b). [Hint: compare the isometry $V \simeq V'$ with the isometry g.]

$\triangleright$ 17.:

Prove the following weakened version of the Cartan–Dieudonné theorem (Theorem 4.5.7): Let (V, Q) be a nondegenerate quadratic space with $\dim_F V=n$. Show that every isometry $f \in {{\,\mathrm{O}\,}}(Q)(F)$ is a product of at most $2n-1$ reflections. [Hint: in the proof of Exercise 4.16(b), note that f can be taken to be a product of at most 2 reflections, and finish by induction.]