Generalised Quaternion Algebras

Q: Why is there no intermediate case between division algebra and M_2(F)?

Because the algebra is simple of dimension 4. Wedderburn–Artin writes it as M_r(E) with 4 = r^2_F E, so r is 1 or 2. If r=1 it is a division algebra; if r=2 then _F E = 1, forcing E=F. The same argument works for any cyclic algebra of prime degree.

Q: How do I actually find a zero divisor when the algebra splits?

Solve the conic au^2+bv^2=w^2 for a nontrivial point, then build a quaternion whose norm vanishes. For example over Q with a=-1, b=2, the point (1,1,1) corresponds to q = 1+i+j with N(q) = 1 - (-1)(1) - 2(1) = 0; then q and q are nonzero with q q = 0.

Q: What changes in characteristic 2?

The relation ij = -ji becomes ij = ji, which would make the algebra commutative, so the presentation is useless. The correct object is [a,b) with i^2+i=a, j^2=b and jij^-1 = i+1 — an Artin–Schreier extension in place of a Kummer one. It is still central simple of dimension 4, and the splitting criterion is again a norm condition.

Q: Why do rotations use unit quaternions rather than 3 × 3 matrices?

The unit quaternions form a group isomorphic to SU(2), a double cover of SO(3), and composition is 4 multiplications' worth of arithmetic with no trigonometric evaluation and no coordinate singularities. The conjugation action v qv q^-1 preserves the pure quaternions and the norm form, which is the rotation.

Q: How does a quaternion algebra relate to a quadratic form?

Two ways, and they agree. The norm form 1,-a,-b,ab is a 2-fold Pfister form, and the algebra is a division algebra exactly when that form is anisotropic. The restriction to pure quaternions, -a,-b,ab, determines the algebra up to isomorphism; that is the classical correspondence between quaternion algebras and ternary quadratic forms.

Executive Summary

Hamilton's quaternions generalise verbatim. Fix a field $F$ with $char F \neq 2$ and two scalars $a, b \in F^{\times}$ . The algebra with $F$ -basis $1, i, j, k$ and relations $i^{2} = a$ , $j^{2} = b$ , $i j = - j i = k$ is a central simple $F$ -algebra of dimension $4$ . It is exactly the degree- $2$ case of the cyclic algebra construction.

Because $4 = 2^{2}$ leaves Wedderburn–Artin only two options, a quaternion algebra is either a division algebra or ** $M_{2} (F)$ ** — there is no intermediate case. Which one it is is decided by a quadratic form: the reduced norm $N (q) = q \bar{q}$ . The algebra is a division algebra precisely when $N$ has no nontrivial zero, equivalently when the conic $a u^{2} + b v^{2} = w^{2}$ has no nontrivial $F$ -point.

$4$

\dim_{F}

$2$ Degree

$2$ Possible structures

$a u^{2} + b v^{2} = w^{2}$ Splits iff solvable

Overview

In §14 the cyclic algebra $(K / F, σ, a)$ is built from a cyclic extension of degree $s$ . Taking $s = 2$ gives the smallest interesting case: $K = F (\sqrt{a})$ is a quadratic extension, $σ$ is the nontrivial automorphism, and the adjoined symbol $x$ satisfies $x^{2} = b$ and $x c = σ (c) x$ . Renaming $\sqrt{a} = i$ and $x = j$ turns the cyclic relations into Hamilton's.

(\frac{a, b}{F}) = F \cdot 1 \oplus F i \oplus F j \oplus F k, i^{2} = a, j^{2} = b, i j = - j i = k .

(Q)

The presentation. Consequences: $k^{2} = - a b$ , $i k = - k i = a j$ , $j k = - k j = - b i$ .

Two features distinguish this case from higher degree. First, the algebra carries a canonical anti-automorphism of order $2$ — the standard involution — which does not exist in general degree. Second, the resulting norm is a quadratic form in $4$ variables, so the entire theory becomes the theory of one small quadratic form. Every splitting question turns into a question about representing values by a conic.

Hamilton's $ℍ = (\frac{- 1, - 1}{ℝ})$ is the case treated on Division Rings and the Real Quaternions; the higher-degree generalisation is Cyclic Algebras; and the twisted series ring of Twisted Laurent Series Division Rings with $σ$ of order $2$ produces a quaternion algebra over a Laurent series field.

Learning Objectives

State the presentation $(Q)$ and derive $k^{2} = - a b$ and the anticommutation rules.
Identify $(\frac{a, b}{F})$ with the cyclic algebra $(F (\sqrt{a}) / F, σ, b)$ when $a \notin (F^{\times})^{2}$ .
Compute the standard involution and the reduced norm $N (q) = w^{2} - a x^{2} - b y^{2} + a b z^{2}$ .
Prove that $q$ is invertible if and only if $N (q) \neq 0$ .
Prove the dichotomy: division algebra or $M_{2} (F)$ , never anything else.
Decide splitting over $ℝ$ and over $ℚ$ in explicit cases.

Definitions

Definition(Q)Generalised quaternion algebra

Let $F$ be a field with $char F \neq 2$ and let $a, b \in F^{\times}$ . The quaternion algebra $(\frac{a, b}{F})$ is the associative $F$ -algebra with identity, free of rank $4$ on $1, i, j, k$ , with multiplication determined by

i^{2} = a, j^{2} = b, i j = k = - j i .

The scalars $a$ and $b$ matter only up to nonzero squares: $(\frac{a, b}{F}) ≅ (\frac{a c^{2}, b d^{2}}{F})$ for all $c, d \in F^{\times}$ .

DefinitionConjugation, trace and norm

For $q = w + x i + y j + z k$ set $\bar{q} = w - x i - y j - z k$ . Then $q \mapsto \bar{q}$ is an $F$ -linear anti-automorphism with $\bar{\bar{q}} = q$ , and

tr (q) = q + \bar{q} = 2 w \in F, N (q) = q \bar{q} = \bar{q} q = w^{2} - a x^{2} - b y^{2} + a b z^{2} \in F .

$N$ is the reduced norm; it is multiplicative, $N (q q') = N (q) N (q')$ , and every $q$ satisfies $q^{2} - tr (q) q + N (q) = 0$ .

$(\frac{a, b}{F})$: The quaternion symbol; also written $(a, b)_{F}$ .
Pure part: $q^{-} = x i + y j + z k$ , the $tr$ -zero component. $q$ is pure iff $\bar{q} = - q$ iff $q^{2} \in F$ with $q \notin F$ or $q = 0$ .
$⟨ 1, - a, - b, a b ⟩$: The diagonal quadratic form of $N$ in the basis $1, i, j, k$ — the norm form of the algebra.
$⟨ - a, - b, a b ⟩$: The restriction of $- N$ to pure quaternions, the pure norm form.
Anisotropic: A form $ϕ$ with $ϕ (v) = 0$ only for $v = 0$ .
Split: Isomorphic to $M_{2} (F)$ . Equivalently the norm form is isotropic.

Characteristic 2 needs a different presentation, with i squared plus i equal to a; the results below are stated only for characteristic not 2.

Core Concepts

The multiplication table

Everything follows from $(Q)$ by moving generators past each other. From $k = i j$ : $k^{2} = i j i j = - i^{2} j^{2} = - a b$ ; $i k = i (i j) = a j$ while $k i = (i j) i = - i^{2} j = - a j$ ; similarly $j k = - b i$ and $k j = b i$ .

Products of the basis elements (row times column)
$\cdot$	$i$	$j$	$k$
$i$	$a$	$k$	$a j$
$j$	$- k$	$b$	$- b i$
$k$	$- a j$	$b i$	$- a b$

A quaternion algebra is a cyclic algebra

Suppose $a \notin (F^{\times})^{2}$ . Then $K = F (i) = F (\sqrt{a})$ is a quadratic — hence cyclic Galois, as $char F \neq 2$ — extension of $F$ , with $σ (i) = - i$ generating $Gal (K / F)$ . The relation $j i = - i j$ says exactly $j c = σ (c) j$ for $c \in K$ , and $j^{2} = b \in F^{\times}$ . Hence

(\frac{a, b}{F}) = K \oplus K j ≅ (K / F, σ, b) (a \notin (F^{\times})^{2}) .

(C2)

So the whole structure theory of cyclic algebras applies with $s = 2$ , which is prime.

If instead $a = c^{2}$ is a square, the algebra splits outright: mapping $i \mapsto c (\begin{matrix} 1 & 0 \\ 0 & - 1 \end{matrix})$ and $j \mapsto (\begin{matrix} 0 & b \\ 1 & 0 \end{matrix})$ respects all relations and is an isomorphism onto $M_{2} (F)$ by dimension count and simplicity.

Why the norm decides everything

The identity $q \bar{q} = N (q)$ makes invertibility a scalar condition: if $N (q) \neq 0$ then $q^{- 1} = N (q)^{- 1} \bar{q}$ ; if $N (q) = 0$ with $q \neq 0$ then $q \bar{q} = 0$ and $\bar{q} \neq 0$ , so $q$ is a zero divisor. Quaternion algebras are therefore the one family where "is it a division algebra?" is literally a question about a quadratic form.

N

anisotropicno zero divisorsdivision algebranot

≅ M_{2} (F)

Key Results

Proposition(Q.1)Central simple of dimension four

Let $char F \neq 2$ and $a, b \in F^{\times}$ , and set $Q = (\frac{a, b}{F})$ . Then $Q$ is a central simple $F$ -algebra with $\dim_{F} Q = 4$ and $Z (Q) = F$ .

Proof

If $a \in (F^{\times})^{2}$ then $Q ≅ M_{2} (F)$ by the explicit matrices above, and $M_{2} (F)$ is central simple of dimension $4$ . If $a \notin (F^{\times})^{2}$ then by $(C 2)$ , $Q ≅ (K / F, σ, b)$ with $K = F (\sqrt{a})$ cyclic of degree $2$ , and the structure theorem for cyclic algebras gives simplicity, $Z (Q) = F$ and $\dim_{F} Q = 2^{2} = 4$ .

Proposition(Q.2)The reduced norm

With $Q$ as above and $q = w + x i + y j + z k$ : $N (q) = q \bar{q} = \bar{q} q = w^{2} - a x^{2} - b y^{2} + a b z^{2}$ lies in $F$ , $N$ is multiplicative, and

q \in U (Q) ⟺ N (q) \neq 0, in which case q^{- 1} = N (q)^{- 1} \bar{q} .

Proof

Expand $q \bar{q}$ using the multiplication table. The cross terms cancel in pairs — for instance the $i$ -terms are $- w x i + x w i = 0$ , and the $k$ -coefficient contributions from $x i \cdot (- y j)$ and $y j \cdot (- x i)$ are $- x y k$ and $+ x y k$ — leaving $w^{2} - a x^{2} - b y^{2} - z^{2} k^{2} = w^{2} - a x^{2} - b y^{2} + a b z^{2}$ , since $k^{2} = - a b$ . The same computation gives $\bar{q} q = N (q)$ .

Multiplicativity follows from $\bar{q q'} = \bar{q}' \bar{q}$ : indeed $N (q q') = q q' \bar{q'} \bar{q} = q N (q') \bar{q} = N (q') q \bar{q} = N (q) N (q')$ , using that $N (q') \in F$ is central.

If $N (q) \neq 0$ then $q \cdot N (q)^{- 1} \bar{q} = N (q)^{- 1} \bar{q} \cdot q = 1$ , so $q$ is a unit. If $N (q) = 0$ and $q \neq 0$ then $\bar{q} \neq 0$ (conjugation is bijective) and $q \bar{q} = 0$ , so $q$ is a left zero divisor and cannot be a unit.

Theorem(Q.3)The dichotomy and the splitting criterion

Let $char F \neq 2$ , $a, b \in F^{\times}$ , $Q = (\frac{a, b}{F})$ . The following are equivalent:

$Q$ is not a division algebra;
$Q ≅ M_{2} (F)$ ;
the norm form $⟨ 1, - a, - b, a b ⟩$ is isotropic over $F$ ;
the conic $a u^{2} + b v^{2} = w^{2}$ has a solution in $F$ other than $(0, 0, 0)$ ;
$a \in (F^{\times})^{2}$ , or $a \notin (F^{\times})^{2}$ and $b \in N_{F (\sqrt{a}) / F} (F (\sqrt{a})^{\times})$ .

In particular $Q$ is either a division algebra or isomorphic to $M_{2} (F)$ ; no other structure occurs.

Proof

**(1) $\Leftrightarrow$ (2).** By $(Q . 1)$ , $Q$ is simple of dimension $4$ , so Wedderburn–Artin gives $Q ≅ M_{r} (E)$ with $E$ a division $F$ -algebra and $4 = r^{2} \dim_{F} E$ . Hence $r \in {1, 2}$ . If $r = 1$ , $Q = E$ is a division algebra; if $r = 2$ then $\dim_{F} E = 1$ , so $E = F$ and $Q ≅ M_{2} (F)$ . These are the only two possibilities, which is the dichotomy and also the equivalence of (1) and (2).

**(1) $\Leftrightarrow$ (3).** By $(Q . 2)$ , $Q$ fails to be a division algebra exactly when some $q \neq 0$ has $N (q) = 0$ , which is exactly isotropy of $⟨ 1, - a, - b, a b ⟩$ .

**(5) $\Leftrightarrow$ (2).** If $a$ is a square, the explicit matrices give $Q ≅ M_{2} (F)$ . If $a$ is not a square, $(C 2)$ identifies $Q$ with the cyclic algebra $(K / F, σ, b)$ , $K = F (\sqrt{a})$ , and the splitting criterion for cyclic algebras says $Q ≅ M_{2} (F)$ iff $b \in N_{K / F} (K^{\times})$ .

**(5) $\Leftrightarrow$ (4).** Assume $a \notin (F^{\times})^{2}$ , so $N_{K / F} (u + v \sqrt{a}) = u^{2} - a v^{2}$ . If $b = u^{2} - a v^{2}$ then $(u, v, 1) \mapsto$ the point $a v^{2} + b \cdot 1^{2} = u^{2}$ solves the conic with $w = u$ . Conversely let $a u^{2} + b v^{2} = w^{2}$ with $(u, v, w) \neq 0$ . If $v = 0$ then $w^{2} = a u^{2}$ ; $u \neq 0$ would make $a = (w / u)^{2}$ a square, excluded, so $u = w = 0$ , contradicting nontriviality. Hence $v \neq 0$ and $b = (w / v)^{2} - a (u / v)^{2} = N_{K / F} (w / v + (u / v) \sqrt{a}) \in N_{K / F} (K^{\times})$ , the value being $b \neq 0$ . If instead $a$ is a square, $(u, v, w) = (1, 0, \sqrt{a})$ solves the conic, so (4) holds automatically and (5) holds by its first clause.

Corollary(Q.4)Symbol relations

For $a, b, c \in F^{\times}$ and $char F \neq 2$ :

$(\frac{a, b}{F}) ≅ (\frac{b, a}{F})$ — interchange $i$ and $j$ .
$(\frac{a, b}{F}) ≅ (\frac{a, - a b}{F})$ — replace the pair $(i, j)$ by $(i, k)$ , using $i^{2} = a$ , $k^{2} = - a b$ , $i k = - k i$ .
$(\frac{a, - a}{F})$ and $(\frac{a, 1 - a}{F})$ (for $a \neq 1$ ) are split, since $- a = N (\sqrt{a})$ and $1 - a = N (1 + \sqrt{a})$ .
$(\frac{a, b c^{2}}{F}) ≅ (\frac{a, b}{F})$ — rescale $j$ by $c$ .

Corollary(Q.5)Quaternion algebras over the reals

Over $F = ℝ$ every $a \in ℝ^{\times}$ is $\pm$ a square, so up to isomorphism there are exactly two quaternion algebras: $(\frac{- 1, - 1}{ℝ}) = ℍ$ , a division algebra because $w^{2} + x^{2} + y^{2} + z^{2}$ is anisotropic, and $M_{2} (ℝ)$ , which is $(\frac{a, b}{ℝ})$ whenever $a > 0$ or $b > 0$ .

Proof Techniques and Method

How these proofs work, and which move to reuse.

Move 1

Turn invertibility into a scalar

An involution with $q \bar{q} \in F$ converts a noncommutative question into a quadratic-form question. This works only in degree $2$ ; in higher degree the reduced norm is a degree- $s$ form and there is no involution to produce it so cheaply.

Move 2

Count with Wedderburn–Artin

$\dim_{F} Q = p^{2}$ with $p$ prime leaves only $r = 1$ and $r = p$ . The dichotomy is a dimension count, not a computation — the same argument that makes prime-degree cyclic algebras easy.

Move 3

Change generators to change the symbol

Any pair of anticommuting elements with squares in $F^{\times}$ generates the whole algebra. Choosing $(i, k)$ instead of $(i, j)$ proves $(\frac{a, b}{F}) ≅ (\frac{a, - a b}{F})$ in one line.

Move 3 generalises: if $Q$ is a quaternion division algebra and $u$ is any pure quaternion with $u^{2} = a' \neq 0$ , then $u$ and any pure $v$ anticommuting with it give $Q ≅ (\frac{a', b'}{F})$ with $b' = v^{2}$ . So the symbol is far from unique, and this is precisely why the classification must be by Brauer class rather than by the pair $(a, b)$ .

A caution about Move 1: the cancellation in $q \bar{q}$ uses $char F \neq 2$ nowhere directly, but the definition of the pure part as a complement to $F \cdot 1$ does — in characteristic $2$ , $tr (q) = 2 w = 0$ identically and the whole apparatus must be rebuilt.

Worked Example

Three quaternion algebras over $ℚ$

Take $F = ℚ$ and $a = - 1$ , so $K = ℚ (i)$ and $N_{K / ℚ} (u + v i) = u^{2} + v^{2}$ . By $(Q . 3)$ , $(\frac{- 1, b}{ℚ})$ splits exactly when $b$ is a sum of two rational squares.

$b = - 1$ : a division algebra

The norm form is $w^{2} + x^{2} + y^{2} + z^{2}$ , which over $ℚ$ vanishes only at the origin. So $(\frac{- 1, - 1}{ℚ})$ is a $4$ -dimensional division algebra over $ℚ$ — the rational Hamilton quaternions.

$b = 2$ : split

$2 = 1^{2} + 1^{2} = N_{K / ℚ} (1 + i)$ , so $2$ is a norm and $(\frac{- 1, 2}{ℚ}) ≅ M_{2} (ℚ)$ . Explicitly the conic $- u^{2} + 2 v^{2} = w^{2}$ has the point $(u, v, w) = (1, 1, 1)$ , and the corresponding zero divisor in the algebra is $q = 1 + i + j$ with $N (q) = 1 + 1 - 2 = 0$ ; indeed $q \bar{q} = 0$ with $\bar{q} = 1 - i - j \neq 0$ .

$b = 3$ : a division algebra

Claim: $3$ is not a sum of two rational squares. Suppose $3 = (p / m)^{2} + (q / m)^{2}$ with $p, q, m \in ℤ$ , $m > 0$ minimal, so $3 m^{2} = p^{2} + q^{2}$ . Since $- 1$ is not a square modulo $3$ , a prime $\equiv 3 (\mod 4)$ — in particular $3$ — divides $p^{2} + q^{2}$ only if it divides both $p$ and $q$ . Then $9 ∣ p^{2} + q^{2} = 3 m^{2}$ , so $3 ∣ m^{2}$ and $3 ∣ m$ , and dividing through by $3$ contradicts minimality of $m$ . Hence $3 \notin N_{K / ℚ} (K^{\times})$ and $(\frac{- 1, 3}{ℚ})$ is a division algebra, with anisotropic norm form $w^{2} + x^{2} - 3 y^{2} - 3 z^{2}$ .

A quaternion algebra from a twisted series ring

Let $F = ℝ ((t))$ and $K = ℂ ((t))$ , with $σ$ acting as complex conjugation on coefficients. Then $ℂ ((x; σ))$ with $x^{2} = t$ is $(\frac{- 1, t}{F})$ . It is a division algebra: a norm $N_{K / F} (f) = f \bar{f}$ has even $t$ -order, while $t$ has order $1$ , so $t$ is not a norm.

Process and Workflow

Normalise the symbolReplace

a

and

b

by squarefree representatives modulo

(F^{\times})^{2}

; use

(\frac{a, b}{F}) ≅ (\frac{b, a}{F}) ≅ (\frac{a, - a b}{F})

to pick the most convenient pair.

Check for an easy splitIf

a

b

is a square, or

b = - a

, or

a + b = 1

, the algebra splits immediately.

Set up the conicOtherwise test whether

a u^{2} + b v^{2} = w^{2}

has a nontrivial

F

-point.

Localise if $F$ is a number fieldBy Hasse–Minkowski the conic has a rational point iff it has a point over every completion; only finitely many places can fail.

Read off the structurePoint exists:

Q ≅ M_{2} (F)

. No point:

Q

is a division algebra with

q^{- 1} = N (q)^{- 1} \bar{q}

What do you actually need from the algebra?

Just division or notTest isotropy of

⟨ 1, - a, - b, a b ⟩

, or equivalently solve the conic.

An explicit zero divisorFind a nontrivial conic point

(u, v, w)

with

a u^{2} + b v^{2} = w^{2}

; then

q = w + u i + v j

has

N (q) = w^{2} - a u^{2} - b v^{2} = 0

, so

q \bar{q} = 0

with both factors nonzero.

An explicit matrix representationWrite

b = N_{K / F} (c)

, replace

j

c^{- 1} j

to reduce to

b = 1

, then use the left-regular action on

K

Its Brauer classThe class has order dividing

2

; over a number field it is determined by the finite even set of ramified places.

Comparison and Classification

Quaternion algebras over familiar base fields
Field $F$	Quaternion algebras up to isomorphism	Reason
$ℂ$ , or any algebraically closed field	only $M_{2} (F)$	every element is a square, so $a \in (F^{\times})^{2}$
$ℝ$	$M_{2} (ℝ)$ and $ℍ$	$a, b$ matter only by sign; $⟨ 1, 1, 1, 1 ⟩$ is anisotropic
$𝔽_{q}$ , $q$ odd	only $M_{2} (𝔽_{q})$	Wedderburn's little theorem; also every ternary form over a finite field is isotropic
$ℚ_{p}$	$M_{2} (ℚ_{p})$ and one division algebra	the Brauer group of a $p$ -adic field has a unique element of order $2$
$ℚ$	infinitely many	one for each finite even set of places, by Hasse–Brauer–Noether
$ℝ ((t))$	$M_{2}$ and $(\frac{- 1, t}{ℝ ((t))})$ among others	the residue field $ℝ$ already supports $ℍ$ , and $t$ contributes ramification

Which properties survive as the degree grows
	Quaternion, $s = 2$	Cyclic, $s$ prime	Cyclic, $s$ composite
Central simple of dimension $s^{2}$	yes	yes	yes
Only two possible structures	yes	yes	no
Standard involution of order $2$	yes	no	no
Division tested by a quadratic form	yes	no	no
Division $\Leftrightarrow$ scalar is not a norm	yes	yes	no
Brauer class of order dividing $s$	yes	yes	yes

Which properties survive as the degree grows

Relationship Map

$(\frac{a, b}{F})$ — how it connects
- specialises from
  - cyclic algebras $(K / F, σ, a)$ with $s = 2$
  - crossed products with group $ℤ / 2$
- specialises to
  - $ℍ = (\frac{- 1, - 1}{ℝ})$
  - $M_{2} (F) = (\frac{1, b}{F})$
  - $(\frac{- 1, t}{ℝ ((t))})$ from the twisted series ring
- is classified by
  - the norm form $⟨ 1, - a, - b, a b ⟩$ up to isometry
  - its class in the $2$ -torsion of $Br (F)$
  - the Hilbert symbol $(a, b)_{v}$ at each place, for $F$ a number field
- supports
  - the Niven–Jacobson analysis of roots of polynomials in a quaternion division ring
  - arithmetic of orders and quaternionic modular forms

The link to Niven–Jacobson Theorem on Quaternion Roots is worth flagging: in $ℍ$ the equation $q^{2} = - 1$ has a two-sphere of solutions, so polynomial equations over a quaternion division algebra behave nothing like their commutative counterparts. That phenomenon is visible already from the norm form: $q$ pure with $N (q) = 1$ gives $q^{2} = - 1$ .

Applications and Industry Use

Applications here means where this structure is used — inside mathematics and in the engineering and computing disciplines that consume it.

Robotics and graphics

Rotations without gimbal lock

Unit quaternions in $ℍ$ double-cover $SO (3)$ ; the norm-one condition and the conjugation action $v \mapsto q v {\bar{q}}^{- 1}$ come straight from the standard involution on this page. Attitude control and skeletal animation use this representation.

Number theory

Orders, modular forms and lattices

Maximal orders in definite rational quaternion algebras give class numbers, Brandt matrices and modular forms; the $ℍ$ -order of Hurwitz quaternions produces the $D_{4}$ lattice.

Cryptography

Isogeny-based schemes

The endomorphism ring of a supersingular elliptic curve is a maximal order in a quaternion algebra ramified at $p$ and infinity; the Deuring correspondence turns isogeny problems into quaternion arithmetic.

Signal processing

Quaternionic filters

Colour image and polarised signal processing use quaternion-valued transforms, where noncommutativity forces separate left and right filter conventions.

Quadratic form theory

Witt groups and invariants

Quaternion algebras are the geometric content of the Hasse–Witt invariant; splitting behaviour of the symbol is the same data as isotropy of a ternary form.

Symbolic computation

Library primitives

Magma, Sage and Pari all implement quaternion algebras over number fields with splitting tests, maximal orders and explicit matrix representations when split.

The rotation application uses only $ℍ$ , but the general symbol is what allows the same computations over other fields — for instance over $ℚ$ for exact arithmetic, and over finite fields in cryptographic settings.

Standards and Notation

Standards here covers notation, symbol and markup standards, and reference implementations, rather than material or design codes.

Symbol notation

(\frac{a, b}{F})

, also

(a, b)_{F}

and

(a, b)

when

F

is fixed

Basis convention

1, i, j, k

with

k = i j

. Some sources set

k = j i

, flipping the sign of

k

Hilbert symbol

(a, b)_{v} = + 1

if the algebra splits at

v

- 1

if not

Hamilton

ℍ = (\frac{- 1, - 1}{ℝ})

; ISO 80000-2 reserves

ℍ

for the real quaternions

Characteristic $2$ Written

[a, b)

with

i^{2} + i = a

j^{2} = b

j i j^{- 1} = i + 1

Magma / SageQuaternionAlgebra(F,a,b), IsMatrixRing, MaximalOrder

Pari/GPalginit with a quaternion symbol; algsplit for an explicit splitting

MarkupPresentation MathML per ISO/IEC 40314; symbols per ISO 80000-2

Computational Notes

Computational notes cover algorithms, cost and library behaviour rather than manufacturing process.

Multiplication costs $16$ base-field multiplications naively, $8$ with a Karatsuba-style scheme. Inversion is one norm evaluation plus a conjugation plus $4$ divisions.
**Splitting over $ℚ$ ** reduces to Legendre's theorem on the conic $a u^{2} + b v^{2} = w^{2}$ : the algebra splits iff the Hilbert symbol $(a, b)_{v} = + 1$ at every place, and only places dividing $2 a b \infty$ need checking. Cost is dominated by factoring $a$ and $b$ .
Finding a zero divisor in a split algebra is equivalent to finding a rational point on a conic — solvable in polynomial time given factorisations, by Simon's or Cremona–Rusin methods.
Maximal orders in a quaternion algebra over a number field are computed by Voight's algorithms, again requiring factorisation of the discriminant.
Over finite fields nothing is left to decide: by Wedderburn's little theorem there are no finite noncommutative division rings, so every quaternion algebra over $𝔽_{q}$ is $M_{2} (𝔽_{q})$ .

Failure Modes and Common Mistakes

Do not assume $q \bar{q} = | q |^{2} \geq 0$ ; positivity is special to $ℍ$ over $ℝ$ and fails as soon as $a$ or $b$ is positive.
Do not treat $tr$ and $N$ as the matrix trace and determinant of the regular representation: the reduced trace and reduced norm are the square roots of those, ${tr}_{reg} = 2 tr$ and $\det_{reg} = N^{2}$ .
Do not expect the polynomial $x^{2} + 1$ to have at most $2$ roots — in $ℍ$ it has infinitely many.
Do not conclude from $Q_{1} \otimes Q_{2} ≅ M_{4} (F)$ that $Q_{1} ≅ Q_{2}^{op}$ is a different algebra: for quaternions $Q^{op} ≅ Q$ via the standard involution.

Quick Reference

Presentation

i^{2} = a

j^{2} = b

i j = k = - j i

char F \neq 2

Derived

k^{2} = - a b

i k = a j

j k = - b i

Conjugate

\bar{q} = w - x i - y j - z k

;

\bar{q q'} = \bar{q}' \bar{q}

Reduced norm

N (q) = w^{2} - a x^{2} - b y^{2} + a b z^{2}

Inverse

q^{- 1} = N (q)^{- 1} \bar{q}

when

N (q) \neq 0

Dichotomydivision algebra or

M_{2} (F)

, never else

Splits iff

a u^{2} + b v^{2} = w^{2}

has a nontrivial

F

-point

Cyclic form

(\frac{a, b}{F}) ≅ (F (\sqrt{a}) / F, σ, b)

for

a

a non-square

Statements and their hypotheses
Statement	Hypotheses	Reference
$Q$ central simple, $\dim_{F} Q = 4$	$char F \neq 2$ , $a, b \in F^{\times}$	(Q.1)
$q$ a unit $\Leftrightarrow N (q) \neq 0$	same	(Q.2)
division algebra or $M_{2} (F)$	same	(Q.3)
splits $\Leftrightarrow b \in N_{F (\sqrt{a}) / F}$	same, and $a \notin (F^{\times})^{2}$	(Q.3), (14.7)–(14.8)
$(\frac{a, b}{F}) ≅ (\frac{a, - a b}{F})$	same	(Q.4)
exactly two algebras over $ℝ$	$F = ℝ$	(Q.5)

Frequently Asked Questions

Why is there no intermediate case between division algebra and

M_{2} (F)

Because the algebra is simple of dimension $4$ . Wedderburn–Artin writes it as $M_{r} (E)$ with $4 = r^{2} \dim_{F} E$ , so $r$ is $1$ or $2$ . If $r = 1$ it is a division algebra; if $r = 2$ then $\dim_{F} E = 1$ , forcing $E = F$ . The same argument works for any cyclic algebra of prime degree.

How do I actually find a zero divisor when the algebra splits?

Solve the conic $a u^{2} + b v^{2} = w^{2}$ for a nontrivial point, then build a quaternion whose norm vanishes. For example over $ℚ$ with $a = - 1$ , $b = 2$ , the point $(1, 1, 1)$ corresponds to $q = 1 + i + j$ with $N (q) = 1 - (- 1) (1) - 2 (1) = 0$ ; then $q$ and $\bar{q}$ are nonzero with $q \bar{q} = 0$ .

Is the pair

(a, b)

recoverable from the algebra?

No. The symbol is highly non-unique: swapping the entries, replacing $b$ by $- a b$ , and scaling either entry by a square all give isomorphic algebras. What is an invariant is the isometry class of the norm form, equivalently the class in the Brauer group.

What changes in characteristic

2

The relation $i j = - j i$ becomes $i j = j i$ , which would make the algebra commutative, so the presentation is useless. The correct object is $[a, b)$ with $i^{2} + i = a$ , $j^{2} = b$ and $j i j^{- 1} = i + 1$ — an Artin–Schreier extension in place of a Kummer one. It is still central simple of dimension $4$ , and the splitting criterion is again a norm condition.

Why do rotations use unit quaternions rather than

3 \times 3

matrices?

The unit quaternions form a group isomorphic to $SU (2)$ , a double cover of $SO (3)$ , and composition is $4$ multiplications' worth of arithmetic with no trigonometric evaluation and no coordinate singularities. The conjugation action $v \mapsto q v {\bar{q}}^{- 1}$ preserves the pure quaternions and the norm form, which is the rotation.

How does a quaternion algebra relate to a quadratic form?

Two ways, and they agree. The norm form $⟨ 1, - a, - b, a b ⟩$ is a $2$ -fold Pfister form, and the algebra is a division algebra exactly when that form is anisotropic. The restriction to pure quaternions, $⟨ - a, - b, a b ⟩$ , determines the algebra up to isomorphism; that is the classical correspondence between quaternion algebras and ternary quadratic forms.

References

T. Y. Lam, A First Course in Noncommutative Rings, Graduate Texts in Mathematics 131, Springer-Verlag, 1991, §14, pp. 231–237; the case $s = 2$ of (14.5)–(14.9), with $ℍ$ identified as a cyclic algebra on p. 230.
T. Y. Lam, Introduction to Quadratic Forms over Fields, Graduate Studies in Mathematics 67, American Mathematical Society, 2005, Chapter III (quaternion algebras and Pfister forms).
J. Voight, Quaternion Algebras, Graduate Texts in Mathematics 288, Springer, 2021.
R. S. Pierce, Associative Algebras, Graduate Texts in Mathematics 88, Springer-Verlag, 1982, Chapter 1 and Chapter 15.
N. Jacobson, Basic Algebra I, 2nd edition, W. H. Freeman, 1985, §7.4 (quaternion algebras and the Frobenius theorem).
J.-P. Serre, A Course in Arithmetic, Graduate Texts in Mathematics 7, Springer-Verlag, 1973, Chapters III–IV (Hilbert symbols and the Hasse–Minkowski theorem).

AI Suggested Questions

Prove that $(\frac{a, b}{F}) \otimes_{F} (\frac{a, b}{F}) ≅ M_{4} (F)$ and deduce the order of the Brauer class.
Work out the characteristic- $2$ theory of $[a, b)$ and its splitting criterion in detail.
Given a quaternion algebra over $ℚ$ by a symbol, describe an algorithm computing its ramified places.
Explain the correspondence between quaternion algebras and ternary quadratic forms up to similarity.
How does the Deuring correspondence turn supersingular isogeny problems into quaternion order problems?
Compare the reduced norm on a quaternion algebra with the reduced norm of a cyclic algebra of degree $3$ .
Show that in a quaternion division algebra every element satisfies a quadratic equation over the centre, and describe the maximal subfields.

01Executive Summary

02Overview

03Learning Objectives

04Definitions

05Core Concepts

The multiplication table

A quaternion algebra is a cyclic algebra

Why the norm decides everything

06Key Results

07Proof Techniques and Method

Turn invertibility into a scalar

Count with Wedderburn–Artin

Change generators to change the symbol

08Worked Example

Three quaternion algebras over ℚ

b=−1: a division algebra

b=2: split

b=3: a division algebra

A quaternion algebra from a twisted series ring

09Process and Workflow

10Comparison and Classification

11Relationship Map

12Applications and Industry Use

Rotations without gimbal lock

Orders, modular forms and lattices

Isogeny-based schemes

Quaternionic filters

Witt groups and invariants

Library primitives

13Standards and Notation

14Computational Notes

15Failure Modes and Common Mistakes

16Quick Reference

17Frequently Asked Questions

18Related KEVOS Topics

19References

20AI Suggested Questions

Continue learning

Executive Summary

Overview

Learning Objectives

Definitions

Core Concepts

Key Results

Proof Techniques and Method

Worked Example

Three quaternion algebras over $ℚ$

$b = - 1$ : a division algebra

$b = 2$ : split

$b = 3$ : a division algebra

Process and Workflow

Comparison and Classification

Relationship Map

Applications and Industry Use

Standards and Notation

Computational Notes

Failure Modes and Common Mistakes

Quick Reference

Frequently Asked Questions

Related KEVOS Topics

References

AI Suggested Questions