Yesterday I wrote about the fact that quaternions, unlike complex numbers, can form conjugates via a series of multiplications and additions. This post will show that you can do something similar with octonions.

If x is an octonion

x = r₀ e₀ + r₁ e₁ + … + r₇ e₇

where all the r‘s are real numbers. The conjugate flips the signs of all the components except the real component:

x* = r₀ e₀ − r₁ e₁ − … − r₇ e₇

The conjugate theorem says

x* = − (x + (e₁ x) e₁ + … + (e₇ x) e₇) / 6

which is analogous to the theorem

q* = − (q + iqi + jqj + kqk) /2

for quaternions.

The internal parentheses are necessary because multiplication in octonions is not associative:

xyz

is ambiguous because it could mean

(xy)z

or

x(yz)

and the two are not necessarily equal.

Update: It is true in general that (xy)z ≠ x(yz) for octonions. However, the subalgebra of octonions generated by any two elements is associative. In particular, (xy)x = x(yz) and so we can write the equation for x* above as simply

x* = − (x + e₁ x e₁ + … + e₇ x e₇) / 6.

Thanks to Martin Erik Horn for pointing this out.

The proof is also analogous to the proof given in the earlier post for quaternions. First work out what the effect is of multiplying on the left and right by one of the imaginary units, then add everything up. You’ll find that the real component is multiplied by -6 and the rest of the components are multiplied by 6.

This post will present a way of multiplying octonions that’s easy to remember.

Please note that there are varying conventions for how to define multiplication for octonions [1].

Octonions

The complex numbers have one imaginary unit i, and the quaternions have three: i, j, and k. The octonions have seven, and so it makes sense to switch over to subscripts rather than venturing further out into the alphabet. Besides, numerical indices will be useful for reasons we’ll see shortly.

Let e₀ = 1 and let e₁ through e₇ be the imaginary units for octonions. Then the e‘s form a basis for ℝ⁸. The octonions are ℝ⁸ with a product that distributes over addition, but the product is not commutative or associative.

Multiplication is determined by how the basis elements multiply, and there are multiple ways of presenting these multiplication rules, the simplest being to write down a multiplication table. Here I’ll present the way I find easiest to remember [2].

Multiplication rules

The squares of each of the imaginary units e₁ through e₇ are equal to −1, as you’d expect based on experience with complex numbers and quaternions. You can derive the rest of the multiplication facts from

e₁ e₂ = e₄

and two simple rules.

The first rule is that the equation above holds when you shift the indices mod 7.

e₁ e₂ = e₄
e₂ e₃ = e₅
e₃ e₄ = e₆
e₄ e₅ = e₇
e₅ e₆ = e₁
e₆ e₇ = e₂
e₇ e₁ = e₃

The second rule is that each of the triplets of units above behave analogous to i, j, and k in quaternions. That is, when you take a triple like (e₂, e₃, e₅), rotations don’t flip the sign but swapping adjacent elements does. That is,

e₂ e₃ = e₅
e₅ e₂ = e₃
e₃ e₅ = e₂

and

e₃ e₂ = −e₅
e₂ e₅ = −e₃
e₅ e₃ = −e₂

So we have seven rotations of the equation

e₁ e₂ = e₄

and six permutations of each rotation, for a total of 42 rules. We already had 7 rules, saying each unit squares to -1, so we have the 49 rules we need to multiply all the non-real units by each other.

More octonion posts

• How close is octonion multiplication to being associative?
• Python code for multiplying octonions and friends

[1] There are multiple conventions for defining octonion multiplication. For example, according to a common definition,

e₁ e₂ = e₃

but according to our definition

e₁ e₂ = e₄

In fact, there are 480 ways to define the multiplication rules for e₁ through e₇. However, all the ways of defining octonion multiplication are isomorphic. You can translate results from one convention to another by renumbering the basis elements. For example, you can convert from the first convention above to our convention by relabeling

(e₁, e₂, e₃, e₄, e₅, e₆, e₇)

as

(e₁, e₂, e₄, – e₇, e₃, e₆, e₅).

I picked the convention used here because it makes the multiplication rules easy to remember: the mod 7 rule doesn’t hold for some other ways of defining multiplication.

[2] There’s a neat way of representing octonion multiplication rules in terms of a Fano plane diagram, but to use it you have to remember how to label the diagram. The presentation in this post is easier to remember, in my opinion. You might remember the rules given here, then use them to fill in a Fano diagram.

Quaternion multiplication is associative but not commutative. An earlier post looked at the average size of the commutator xy − yx as a measure of how far quaternion multiplication is from being commutative.

This post looks at an analogous question for octonions. Octonion multiplication is neither commutative nor associative. So in this post we look at the associator of three octonions (xy)z − x(yz) as a measure of how far octonion multiplication is from being associative.

(A post from yesterday looked at how close octonion multiplication comes to being associative in an algebraic sense, looking at weak forms of associativity. This post looks at how close multiplication comes to being associative in an analytical sense, looking at norm distances rather than algebraic identities.)

We will use a simulation to look at the average norm of the octonion associator over the octonions of unit length, analogous to the earlier post that looked at the commutator of the quaternions. We developed code for octonion multiplication in the previous post and will reuse that code here. We also developed code for generating random unit-length octonions in the same post.

Here’s code to find the average norm of the associator and plot a histogram of its values.

    
    import matplotlib.pyplot as plt

    # omult is octonion multiplication. 
    # See previous post.
    def associator(x, y, z):
        return omult(omult(x, y), z) - 
               omult(x, omult(y, z))

    N = 100000
    s = 0
    h = np.zeros(N)
    for n in range(N):
        x = random_unit_octonion()
        y = random_unit_octonion()
        z = random_unit_octonion()    
        t = norm(associator(x, y, z))
        s += t
        h[n] = t

    print(s/N)
    plt.hist(h, bins=50)
    plt.show()

This gives an average of 1.095 and the histogram below.

Update: Greg Egan calculated the exact mean value for the norm of the associator to be 147456/(42875 π) ≈ 1.0947336. Here are the details.