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5 the Ring of Real Quaternions 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 = q + iq + jq + kq : q R , { 0 1 2 3 i ∈ } where i2 = j2 = k2 = 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 + a1i + a2j + a3k and b = b0 + b1i + b2j + b3k 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 (a + b )+ c = a + (b + c ) for all a ,b ,c R, ∈ i i i i i i i i i ∈ 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 + a1i + a2j + a3k H then a = ( a0)+ ( a )i+( a )j +( a )k is another quaternion (as all elements of R have negatives∈ in R)− and a −a = 0, − 1 − 2 − 3 − 4. since addition is commutative in R, it is also commutative in H, since ai + bi = bi + ai for all a ,b R. i i ∈ 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 + b1i + b2j + b3k, we take all ordered products uv with u running over the terms of a and v 15 running over the terms of b, and use the relations i2 = j2 = k2 = ijk = 1 and ij = ji,ik = ki,jk = kj − − − − to get a final answer in the form c = c0 + c1i + c2j + c3k. We compute the product ab as ab = (a0b0 a1b1 a2b2 a3b3) + (a b +− a b +− a b − a b )i 0 1 1 0 2 3 − 3 2 + (a0b2 + a2b0 + a1b3 a3b1)j + (a b + a b + a b − a b )k 0 3 3 0 1 2 − 2 1 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 a b + a∈b + a b + a b R. It can also be∈ shown that 0 j0 1 j1 2 j2 3 j3 ∈ 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 = q + q i + q j + q k H. The norm of q is denoted by N(q) and given by 0 1 2 3 ∈ 2 2 2 2 N(q)= q0 + q1 + q2 + q3 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 = q + q i + q j + q k H. The conjugate of q is denoted by q¯ and given by 0 1 2 3 ∈ q¯ = q q i q j q k. 0 − 1 − 2 − 3 Let q = q + q i + q j + q k H. We note the following relationship between q and its conjugate. 0 1 2 3 ∈ qq¯ = (q q + q q + q q + q q ) + ( q q + q q q q + q q )i 0 0 1 1 2 2 3 3 − 0 1 1 0 − 2 3 3 2 + ( q0q2 + q2q0 q1q3 + q3q1)j + ( q0q3 + q3q0 q1q2 + q2q1)k −2 2 2 −2 − − = q0 + q1 + q2 + q3 = N(q) Similarly,qq ¯ = N(q). To summarize, we have that qq¯ =qq ¯ = N(q) R. ∈ So if q = 0, then N(q) = 0 and hence we get the equation 6 6 q¯ q¯ q = q =1, N(q) N(q) which tells us that the inverse of any non-zero quaternion q is given by q¯ q−1 = . N(q) We have just shown that: 16 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 q¯ = 1 2i 4j 8k. It follows that q−1 = 1 2 i 4 8 . − − − 85 − 85 − 85 − 85 Note that q +¯q =2 R (compare this with the complex numbers). This gives the equation ∈ q¯ =2 q, − which gives qq¯ = N(q)=85= q(2 q)=2q q2, − − so it follows that q2 =2q N(q)=2+4i +8j + 16k 85 = 83+4i +8j + 16k. − − − We have just found another way to compute q2, using the norm and conjugate of q. In fact, this type of manipulation can be used for any quaternion. Theorem 5.2. Let q = q + q i + q j + q k H. Then 0 1 2 3 ∈ q2 2q q + N(q)=0. − 0 Proof. Since q +¯q =2q , we find that qq¯ = N(q)= q(2q q)=2q q q2, and the result follows. 0 0 − 0 − The properties of conjugation in H are similar to those for C. Theorem 5.3. Let a,b H. Then ∈ 1. a + b =¯a + ¯b 2. ab = ¯ba¯ 3. N(ab)= N(a)N(b) Proof. Let a = a0 + ia1 + ja2 + ka3 and b = b0 + b1i + b2j + b3k. Part 1 is straightforward and left as an exercise. To see that Part 2 holds, note first that ab = (a0b0 a1b1 a2b2 a3b3) (a0b1 + a1b0 + a2b3 a3b2)i (a b +− a b +− a b − a b )j− (a b + a b + a b − a b )k. − 0 2 2 0 1 3 − 3 1 − 0 3 3 0 1 2 − 2 1 Now compute ba: ba = (b0a0 b1a1 b2a2 b3a3) + ( b0a1 b1a0 + b2a3 b3a2)i + ( b a− b a−+ b a− b a )j +− ( b −a b a + b −a b a )k − 0 2 − 2 0 1 3 − 3 1 − 0 3 − 3 0 1 2 − 2 1 = (a0b0 a1b1 a2b2 a3b3) (a0b1 + a1b0 + a2b3 a3b2)i (a b +− a b +− a b − a b )j− (a b + a b + a b − a b )k − 0 2 2 0 1 3 − 3 1 − 0 3 3 0 1 2 − 2 1 = ab Part 3 has a more elegant proof, if we use Part 2 and the fact that N(q)= qq¯. N(ab) = (ab)ab = (ab)(¯ba¯) = a(b¯b)¯a = aN(b)¯a = aaN¯ (b) = N(a)N(b) 17 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.
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