5
The Ring of Real Quaternions
The ring of real quaternions was discovered by William Rowan Hamiltion (1805-1865) in 1843. The equations
governing arithmetic in this algebraic structure can still be seen on a plaque on the Broome Bridge of the
Royal Canal. It is reported that Hamiliton was so excited by his insight that he etched the equations in
stone as he walked past the bridge.
In Hamilton’s own words: ”Time is said to have only one dimension, and space to have three dimensions.
The mathematical quaternion partakes of both these elements; in technical language it may be said to be
’time plus space’, or ’space plus time’: and in this sense it has, or at least involves a reference to, four
dimensions.”
In fact Hamilton’s quaternions have many applications other than in physics. They are extesnively used
in computer graphics to describe motion in 3-space, and more recently, they have been used in multiple
antennae communications systems.
In some ways we can think of the quaternions as an extension of the complex numbers.
Definition 5.1. The set of real quaternions, denoted by H, is defined by
H = {q0 + iq1 + jq2 + kq3 : qi ∈ R},
where i2 = j 2 = k 2 = ijk = −1
Note that ij = −ji, ik = −ki, jk = −kj, so multiplication in H is not commutative. For this reason, extra
care has to be taken when performing arbitrary multiplications in H.
We define addition and multiplication in H as follows.
Addition is component-wise, as with addition of the complex numbers. So if a = a0 + a1 i + a2 j + a3 k and
b = b0 + b1 i + b2 j + b3 k are a pair of quaternions then
a + b = (a0 + b0 ) + (a1 + b1 )i + (a2 + b2 )j + (a3 + b3 )k
where addition of the real components ai + bi is the usual addition in R.
For example, if a = 2 − i + 3j + k and b = 3 + 2i − j − k, then
a + b = 5 − i − 2j.
It’s not hard to see that this addition is well defined on H (i.e. given an pair of quaternions a, b, there is
a uniquely determined quaternion a + b), since addition is well defined on R. Note also that
1. addition in H is associative - it is easy to show that (a + b) + c = a + (b + c) for all a, b, c ∈ H, using
the fact that (ai + bi ) + ci = ai + (bi + ci ) for all ai , bi , ci ∈ R,
2. H has an additive identity, namely the real number 0 = 0 + 0i + 0j + 0k,
3. every element of H has an additive inverse - if a = a0 + a1 i + a2 j + a3 k ∈ H then −a = (−a0 ) +
(−a1 )i+(−a2)j +(−a3 )k is another quaternion (as all elements of R have negatives in R) and a−a = 0,
4. since addition is commutative in R, it is also commutative in H, since ai + bi = bi + ai for all
ai , bi ∈ R.
This all shows that the quaternions form an abelian group wrt addition.
Multiplication is a longer calculation. Essentially, when multiplying a pair of quaternions a = a0 +ia1 +ja2 +
ka3 and b = b0 + b1 i + b2 j + b3 k, we take all ordered products uv with u running over the terms of a and v
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running over the terms of b, and use the relations i2 = j 2 = k 2 = ijk = −1 and ij = −ji, ik = −ki, jk = −kj
to get a final answer in the form c = c0 + c1 i + c2 j + c3 k. We compute the product ab as
ab =
+
+
+
(a0 b0 − a1 b1 − a2 b2 − a3 b3 )
(a0 b1 + a1 b0 + a2 b3 − a3 b2 )i
(a0 b2 + a2 b0 + a1 b3 − a3 b1 )j
(a0 b3 + a3 b0 + a1 b2 − a2 b1 )k
For example, if a = 2 − i + 3j + k and b = 3 + 2i − j − k then
ab = (6 + 2 + 3 + 1) + (4 − 3 − 3 + 1)i + (−2 − 1 + 9 + 2)j + (−2 + 1 − 6 + 3)k = 12 − i + 8j − 4k.
Again, multiplication is well-defined in H, since addition and multiplication are well defined in R, so each
pair a, b ∈ H determines a unique product ab ∈ H, which can be observed by looking at the coefficients
a0 bj0 + a1 bj1 + a2 bj2 + a3 bj3 ∈ R. It can also be shown that
1. multiplication in H is associative, which follows from the fact that the assoc and distributive laws
hold in R,
2. H has a multiplicative identity, namely the real number 1 = 1 + 0i + 0j + 0k,
3. the left and right distributive laws hold in H, which follows from the fact that the assoc and
distributive laws hold in R.
It follows that:
Lemma 5.1. The real quaternions form a unital ring wrt addition and multiplication as defined above.
In fact, the ring of quaternions has another property: we can show that every element of H has a multiplicative
inverse.
Definition 5.2. Let q = q0 + q1 i + q2 j + q3 k ∈ H. The norm of q is denoted by N (q) and given by
N (q) = q02 + q12 + q22 + q32
The norm of a quaternion is similar to the modulus of a complex number, and plays a like role in the
compuation of an inverse element of a quaternion. Note that the norm of any quaternion is a non-negative
real number and takes the value zero only on 0.
Definition 5.3. Let q = q0 + q1 i + q2 j + q3 k ∈ H. The conjugate of q is denoted by q̄ and given by
q̄ = q0 − q1 i − q2 j − q3 k.
Let q = q0 + q1 i + q2 j + q3 k ∈ H. We note the following relationship between q and its conjugate.
q q̄
=
+
=
=
(q0 q0 + q1 q1 + q2 q2 + q3 q3 ) + (−q0 q1 + q1 q0 − q2 q3 + q3 q2 )i
(−q0 q2 + q2 q0 − q1 q3 + q3 q1 )j + (−q0 q3 + q3 q0 − q1 q2 + q2 q1 )k
q02 + q12 + q22 + q32
N (q)
Similarly, q̄q = N (q). To summarize, we have that
q q̄ = q̄q = N (q) ∈ R.
So if q 6= 0, then N (q) 6= 0 and hence we get the equation
q
q̄
q̄
=
q = 1,
N (q)
N (q)
which tells us that the inverse of any non-zero quaternion q is given by
q̄
q −1 =
.
N (q)
We have just shown that:
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Theorem 5.1. The ring of real quaternions is a division ring.
(Recall that a division ring is a unital ring in which every element has a multiplicative inverse. It is not
necessarily also a commutative ring. A division ring that is commutative is simply a field.)
Example 5.1. Let q = 1 + 2i + 4j + 8k. Then N (q) = 1 + 4 + 16 + 64 = 85 and q̄ = 1 − 2i − 4j − 8k. It
2
4
8
1
− 85
i − 85
− 85
.
follows that q −1 = 85
Note that q + q̄ = 2 ∈ R (compare this with the complex numbers). This gives the equation
q̄ = 2 − q,
which gives
q q̄ = N (q) = 85 = q(2 − q) = 2q − q 2 ,
so it follows that
q 2 = 2q − N (q) = 2 + 4i + 8j + 16k − 85 = −83 + 4i + 8j + 16k.
We have just found another way to compute q 2 , using the norm and conjugate of q. In fact, this type of
manipulation can be used for any quaternion.
Theorem 5.2. Let q = q0 + q1 i + q2 j + q3 k ∈ H. Then
q 2 − 2q0 q + N (q) = 0.
Proof. Since q + q̄ = 2q0 , we find that q q̄ = N (q) = q(2q0 − q) = 2q0 q − q 2 , and the result follows.
The properties of conjugation in H are similar to those for C.
Theorem 5.3. Let a, b ∈ H. Then
1. a + b = ā + b̄
2. ab = b̄ā
3. N (ab) = N (a)N (b)
Proof. Let a = a0 + ia1 + ja2 + ka3 and b = b0 + b1 i + b2 j + b3 k. Part 1 is straightforward and left as an
exercise. To see that Part 2 holds, note first that
ab
= (a0 b0 − a1 b1 − a2 b2 − a3 b3 ) − (a0 b1 + a1 b0 + a2 b3 − a3 b2 )i
− (a0 b2 + a2 b0 + a1 b3 − a3 b1 )j − (a0 b3 + a3 b0 + a1 b2 − a2 b1 )k.
Now compute ba:
ba
=
+
=
−
=
(b0 a0 − b1 a1 − b2 a2 − b3 a3 ) + (−b0 a1 − b1 a0 + b2 a3 − b3 a2 )i
(−b0 a2 − b2 a0 + b1 a3 − b3 a1 )j + (−b0 a3 − b3 a0 + b1 a2 − b2 a1 )k
(a0 b0 − a1 b1 − a2 b2 − a3 b3 ) − (a0 b1 + a1 b0 + a2 b3 − a3 b2 )i
(a0 b2 + a2 b0 + a1 b3 − a3 b1 )j − (a0 b3 + a3 b0 + a1 b2 − a2 b1 )k
ab
Part 3 has a more elegant proof, if we use Part 2 and the fact that N (q) = q q̄.
N (ab) = (ab)ab
= (ab)(b̄ā)
= a(bb̄)ā
= aN (b)ā
= aāN (b)
= N (a)N (b)
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Example 5.2. Let a = 2 − i + j − k, b = i + j. Then N (ab) = N (a)N (b) = (4 + 1 + 1 + 1)(1 + 1) = 14.
N (ab12 ) = N (a)N (b12 ) = N (a)N (b)12 = (7)(212 ). Suppose we wish to find a5 . We can first use the equation
a2 = 4a − N (a) = 8 − 4i + 4j − 4k − 7 = 1 − 4i + 4j − 4k,
and then use the fact that a4 = (a2 )2 to find that
a4 = 2a2 − N (a2 ) = 2 − 8i + 8j − 8k − 49 = −47 − 8i + 8j − 8k.
Finally, compute
a5 = a4 a = (−47 − 8i + 8j − 8k)(2 − i + j − k) = −118 + 31i − 31j + 31k.
Note that N (a5 ) = N (a)5 = 75 = 16807 = 1182 + 3(312 ), which is consistent with our result.
A quaternion that has the form q1 i + q2 j + q3 k is called pure quaternion. It is particularly convenient to
compute powers of such quaternions as follows. If q is a pure quaternion then
q 2 = −N (q),
so
q 4 = (q 2 )2 = (−N (q))2 = N (q)2 ,
and in general,
q 2k = (−1)k N (q)k .
Note that every even power of a pure quaternion is real, while every odd power of a pure quaternion generates
another pure quaternion.
q 2k+1 = q 2k q = (−1)k N (q)k q.
Example 5.3. Let q = i + 2j − k. Then q 2 = −N (q) = −6.
Suppose we wish to compute q 6 . This is given by q 6 = (q 2 )3 = (−N (q))3 = (−1)3 63 = −63 .
If we wish to compute q 32 , we may observe that q 32 = (q 2 )18 = N (q)18 = 618 .
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