11: The Quaternions

The quaternions were invented by Sir William Rowan Hamilton about 1850. Hamilton was perhaps the first to note that complex numbers could be thought of as a way to multiply points in the plane. He then had the idea of trying to find a way to multiply points in \(\mathbb{R}^3\) so that the field axioms would be satisfied. He was unable to do this, but he finally found a way to define multiplication on \(\mathbb{R}^4\) so that the multiplication together with ordinary vector addition of elements of \(\mathbb{R}^4\) would satisfy all the field axioms except for commutativity of multiplication. He called these new objects quaternions. They turned out, like complex numbers, to have many applications in engineering and physics. This “number system" is denoted by \(\mathbb{H}\) for Hamilton since \(\mathbb{Q}\) is already taken to denote the rational numbers.

Definition 11.1:

The ring of quaternions is the ring \((\mathbb{H},+,\cdot)\) where \[\mathbb{H} = \mathbb{R}^4 = \{ (a,b,c,d) \ | \ a,b,c,d \in \mathbb{R}\}\] and where \(+\) and \(\cdot\) are defined by the rules: \[\begin{aligned} (x,y,z,w) + (a,b,c,d) &=& (x+a,y+b,z+c,w+d) \\ (x,y,z,w) \cdot (a,b,c,d) &=& (xa - yb -zc-wd, \\ & &xb+ya+zd-wc,\\ & &xc-yd+za+wb,\\ & &xd+yc-zb+wa)\end{aligned}\] where \(x,y,z,w,a,b,c,d \in \mathbb{R}\). The addition and multiplication inside the 4-tuples on the right represent addition and multiplication in \(\mathbb{R}\).

Stated this way the rules for multiplication are hard to remember. There is a simpler way to describe them: Let \[\begin{aligned} 1&=&(1,0,0,0) \\ i&=&(0,1,0,0)\\ j&=&(0,0,1,0)\\ k&=&(0,0,0,1)\end{aligned}\] Note that here we are being a little lazy in letting 1 stand for both the vector \((1,0,0,0)\) and the real number 1. The set \(\{1,i,j,k\}\) is what is called in linear algebra a basis for \(\mathbb{R}^4\). This means that if we define for \(a \in \mathbb{R}\) and \((x,y,z,w) \in \mathbb{R}^4\) the scalar by vector product \[a(x,y,z,w)=(ax,ay,az,aw),\] the quaternion \(q=(x,y,z,w)\) may be written uniquely in the form \[q=x1+yi+zj+wk.\] Now if we abbreviate \(x=x1\), the quaternion takes the form \[q= x+yi+zj+wk.\] Addition now becomes \[(x+yi+zj+wk) + (a+bi+cj+dk) = (x+a) +(y+b)i+(z+c)j+(w+d)k.\] Products of the basis elements \(1,i,j,k\) are defined as follows: \[1q=q1=q \mbox{ for all } q \in \mathbb{H},\] \[i^2=j^2=k^2=-1,\] \[ij=-ji=k,\] \[jk=-kj=i,\] \[ki=-ik=j.\] Using these rules, the distributive law, and the fact that if \(q_1\) and \(q_2\) are any quaternions and \(a \in \mathbb{R}\) then \[a(q_1q_2) = (aq_1)q_2 = q_1(aq_2),\] one easily calculates the product of two quaternions \(q_1 =x+yi+zj+wk\) and \(q_2=a+bi+cj+dk\).

Problem 11.1 Use the above rules to calculate the product \(q_1q_2\) of the quaternions \(q_1=1+i+2j+3k\) and \(q_2=1-i-2j-3k\). Write the product in standard form \(a+bi+cj+dk\), where \(a,b,c,d \in \mathbb{R}\).

Problem 11.2 Show that \((1,0,0,0)\) acts as an identity for \(\mathbb{H}\) and that \(\mathbb{H}\) is not a commutative ring.

Problem 11.3 Show that the quaternion \(q =x+yi+zj+wk\) has an inverse given by \(q^* =c(x-yi-zj-wk)\) where \(c = 1/(x^2+y^2+z^2+w^2)\) provided that \(q \ne 0\). Here \(0 = (0,0,0,0)\).

Problem 11.4 Show that there are infinitely many quaternions \(q\) satisfying \(q^2=-1\). Hint: consider quaternions of the form \(q=xi+yj+zk\).

Problem 11.5 Show that the 8 element set \[Q = \{ 1, -1, i, -i, j, -j, k,-k \}\] under quaternion multiplication is a group. This is one of the five non-isomorphic groups of order 8. It is called, naturally enough, the quaternion group.

Definition 11.2

A ring which satisfies all the field axioms except possibly for commutativity of multiplication is called a division ring.

Note that a division ring may be defined as a ring whose non-zero elements form a group under multiplication. All fields are division rings. A commutative ring which is a division ring is a field.

Theorem \(\PageIndex{1}\)

\(\mathbb{H}\) is a division ring.

Proof. From linear algebra we already know that vector addition on \(\mathbb{R}^4\) is an abelian group. From the above problems we know that \(\mathbb{H}\) has an identity and every non-zero element has an inverse. It remains only to prove associativity for multiplication and the two distributive laws. The proofs of these properties are straightforward and we leave them for the interested reader.

The ring of quaternions is one of the rare examples of a non-commutative division ring. The following theorem shows why Hamilton had difficulty finding a division ring whose underlying set is \(\mathbb{R}^3\). \(\blacksquare\)

Theorem \(\PageIndex{2}\) (Frobenius)

Let \(D\) be a division ring which is algebraic over \(\mathbb{R}\). Then \(D\) is isomorphic to \(\mathbb{R}\), \(\mathbb{C}\), or \(\mathbb{H}\). \(\blacksquare\)

See Chapter 7 of to see what it means to be algebraic over \(\mathbb{R}\) and how to prove this theorem. This result implies that there is no “nice" way of defining multiplication on \(\mathbb{R}^n\) so that it becomes a division ring unless \(n \in \{1,2,4\}\). There are many interesting and useful ways to make \(\mathbb{R}^n\) into a ring which is not a division ring for other values of \(n\). However, we do not have time to go into these matters.

Problem 11.6 Define \[\mathcal{H} = \left \{ \left ( \begin{array}{cr} z & -\overline{w} \\ w&\overline{z} \end{array} \right ) \ | \ z,w \in \mathbb{C} \right \}.\]

1. Prove that \(\mathcal{H}\) is a subring of the ring \(M_2(\mathbb{C})\).
2. Prove that \(\mathcal{H}\) is a division ring. Hint: it suffices to show that the each non-zero matrix in \(\mathcal{H}\) has an inverse that is also in \(\mathcal{H}\).
3. Define the matrices \[\mathbf{1} = \left(\begin{array}{cr} 1 & 0 \\ 0&1 \end{array} \right ), I=\left(\begin{array}{cr} i & 0 \\ 0&-i \end{array} \right ), J=\left(\begin{array}{cr} 0 & i \\i&0 \end{array} \right ), K=\left(\begin{array}{cr} 0 & -1 \\ 1&0\end{array} \right )\]
   1. Show that every element of \(\mathcal{H}\) can be written in the form: \[a\mathbf{1} + bI + cJ + dK\] where \(a,b,c,d \in \mathbb{R}\).
   2. Show that \[I^2 = J^2 = K^2 = -\mathbf{1},\] \[IJ = K, JI = -K,\] \[JK = I, KJ = -I,\] \[KI = J, IK = -J\]

Remark

You need not verify it, but it follows from this that \(\mathcal{H} \cong \mathbb{H}\).