Why should quaternions exist?

Question

Why do quaternions exist? I want to believe they exist, but all I can think of are reasons they should not exist.

These are my reasons.

1. The quaternions are defined by the following equation:

$$i^2 = j^2 = k^2 = ijk = -1$$

There are four equalities and three variables. There should be no solution.

2. Somehow there's a workaround for dilemma $1$. They exist. So how do you define the quantity $ijkijk$? Is it interpreted as $(ijk)(ijk)$, or $-1 \times -1$, or $1$? Or is it interpreted as $(i^2)(j^2)(k^2)$, or $-1 \times -1 \times -1$, or $-1$? $ijkijk$ can't be two values at once.

3. I read that quaternions are not multiplication commutative. Why multiplication commutative? Why not multiplication associative or some other property?

4. Okay, so there is a reason for number 3. Why can't I define a new set of numbers where the multiplicative identity doesn't hold? Like: $1x = x + 1?$

I've been pondering these questions for a couple of days now, so I would really appreciate an answer.

Answer 1

1. They are not exactly variables, they are symbols. And we are just defining what happens when we apply a binary operator to them. You can always have more equations than variables, that doesn't mean you don't have a solution (even in your highschool algebra), it just means its possible there might not be a solution.

2. You are assuming the quaternions are commutative to make your argument here. They are not. You cannot rearrange terms. ij is different than ji.

3. Commutative means we can rearrange the terms. For example x * y = y * x. The quaternions are in fact associative which means (x * y) * z = x * (y * z). This is why it is not ambiguous to write x * y * z since the grouping doesn't matter. The quaternions are indeed associative. This is why $ijkijk$ is not ambiguous.

4. You can define whatever kind of structure you want. The quaternions are a Group which have the identity property. However there are other algebraic structures, such as Magmas, which impose less restrictions

If you are really interested in detailed answers, check out some abstract algebra books from your local university library.

Answer 2

Presumably, you accept that quadruples of real numbers exist. Like, $(3,2.4,-\pi,6)$ is a thing that exists. It's just an ordered collection of four numbers.

One way to think about the quaternions is just that they are another way of writing quadruples. So, instead of writing $(3,2.4,-\pi,6)$, we can write $3+2.4i-\pi j+6k$, and think of it as a notational difference.

Then, we define what $+$ and $\times$ means for these entities. We can define $+$ to mean anything we want, here; let's go with the following definition: $$(a,b,c,d)+(A,B,C,D):=(a+A,b+B,c+C,d+D)$$ (where $:=$ means "is defined to be equal to"). Similarly, we can define multiplication in a similar way; I won't write it out in the general case, but we can define it to be such that, in the new notation, $ijk=-1$, etc.

Then, the rest of the work is just to go through a bunch of properties, one by one, and see if they work.

• Is addition commutative? Yes.

• Does the distributive property hold? Yes.

• Does every nonzero quaternion have a multiplicative inverse? Yes. (Hard to show but doable)

• Are multiplication and addition associative? Yes.

• Does multiplication commute? No.

• etc.

Of course, what I haven't touched on is why they're defined this way. Or why so many of the properties above seem to coincidentally hold. But I hope that I've convinced you that these numbers exist, since you can think of $a+bi+cj+dk$ to be another way of writing the ordered quadruple $(a,b,c,d)$.

Answer 3

Your logic for #1 is faulty. Just because you have four equations for three variables, there is nothing that concludes that there is no solution. For example, there exists a solution for the system of equations:

$$x^2=1\\ y^2=1\\ z^2=1\\ xyz=1$$

even though there are four equations for three variables.

For #2, you forgot that quaternions do not commute, thus, the expression $ijkijk$ is not equal to $iijjkk$. The expression can, however, be interpreted as$(ijk)(ijk)$ or $i(j(k(i(jk))))$ or any other combination of parentheses. No matter how you place the parentheses, the result will be $1$.

For #3: The quaternions are associative but not commutative. That is the property of the multiplication we defined.

For #4: Sure, you can define such a set. Why not? The thing is that unlike quaternions, the structure you will create will not have as nice a structure as quaternions, which are the only $4$ dimensional vector space over $\mathbb R$ on which multiplication is defined in such a way that we have a division ring.

Answer 4

It is easy to check multiplication is not commutative from the axioms you mentioned. For instance:

From $ijk=-1$ and $k^2=-1$, one deduces $$(ijk)k=-k=(ij)k^2=-ij,\enspace\text{whence}\enspace ij=k.$$ Similarly, we have $\;jk=i,\enspace ki=j$. We deduce: $$ ji=-jk^2i=-(jk)(ki)=-ij.$$

Answer 5

The quaternions are an example of a finite-dimensional associative real division algebra. Frobenius proved that there are only three: the reals, the complex numbers, and the quaternions (the octonions are nonassociative). So if you are inclined to believe division algebras exist, then the quaternions probably do.

On the other hand, it is a consequence of Bott's demonstration of a conjecture by Hirzebruch on the Pontrjagin classes of vector bundles over n-spheres that the sphere $S^{n-1}$ is parallelizable only for $n-1$ equal to 1, 3, or 7. So if you believe 3-spheres exist and that you can parallelize them (put an everywhere full-rank smooth vector field on them), then you should believe that $\mathbf R^4$ possesses a bilinear product operation without zero divisors, and hence that the quaternions exist. But you don't have to.