# associativity of quaternion multiplication

I understand that quaternion multiplication is non-commutative, but what association does it have.

putting into context when we have the statement

when given general numbers, and algebra such as: $A*B*C$ we generally are taught left to right $(A*B)*C$ even though it does not directly matter, but all of the examples that I have found for multiplying quaternions is only the case of multiplying $q_1*q_2$, and no examples of 3, or more.

so because quaternion multiplication is non-commutative, and probably has a implicit associative (might be using this word our of context) property. what is it?

when given $q_1 * q_2 * q_3$ do I treat this as $q_1 * (q_2 * q_3)$, or as $(q_1 * q_2) * q_3$, or does it actually matter as long as the absolute order is maintained?

• Yes, it is associative. There is a fairly simple way to write quaternions as 4 by 4 matrices with real number entries, where the ordinary real number $1$ corresponds to the identity matrix, and each of the symbols $i,j,k$ corresponds to a matrix with all entries equal to one of $0,1,-1.$ The closure property follows from the fact that the product of two of these matrices results in one with the same "pattern" of entries, then associativity follows from associativity for matrix multiplication. en.wikipedia.org/wiki/Quaternion#Matrix_representations – Will Jagy May 1 '12 at 22:00
• Associativity means $(A*B)*C = A*(B*C)$; note the order is always the same. Quaternion multiplication is associative; not every operation is (look up the "octonions" to see that the "next generalization" gives you a non-commutative, non-associative mutliplication. – Arturo Magidin May 1 '12 at 22:01
• And if you continue your very interesting line of questioning ("what kind of association property does it have?") the answer would be that they are alternative, which is a weaker form of associativity. This is:$(xx)y=x(xy)$ and $y(xx)=(yx)x$ for all $x,y$ in the octonions. – rschwieb May 2 '12 at 1:40

## 2 Answers

It doesn't matter - quaternions are associative.

• ...as long as order is maintained... – GEdgar May 1 '12 at 21:59
• As long as order is maintained. – Chris Taylor May 1 '12 at 21:59
• Saying an operation is "associative as long as order is maintained" seems slightly weird to me, in roughly the same way as when I hear talk about verifying the "closure property" of a binary operation. Each one makes me want to repeat the definitions slowly and clearly: associativity is, by definition, the property of a binary operation on a set $X$ that $(x*y)*z = x*(y*z)$ for all $x,y,z \in X$. If you change the order of the elements -- or change the identity in any way -- it's not associativity anymore! – Pete L. Clark May 4 '12 at 0:19
• @Pete I agree. I think that GEdgar was trying to avoid potential confusion on the OP's part by emphasising that it doesn't matter in which order you multiply the quaternion pairs, but the order of the quaternions themselves does matter. – Chris Taylor May 4 '12 at 7:50

Let $K$ be a field, and let $A$ be a not-necessarily-associative-or-unital algebra over $K$. For $x,y,z \in A$, we define the associator

$[x,y,z] = (xy)z - x(yz)$.

The associator defines a $K$-linear map $\varphi: A^3 \rightarrow A$. So the following are equivalent:

(i) $A$ is associative.
(ii) The map $\varphi$ is identically zero.
(iii) For any $K$-basis $\{e_i\}_{i \in I}$ of $A$, all associators $[e_i,e_j,e_k]$ of not necessarily distinct triples of basis elements are zero.

Thus, if $A$ has finite dimension $d$ over $K$, showing its associativity is a purely finite calculation involving checking all $d^3$ associators involving elements of a given basis are zero. This calculation is performed for an arbitrary quaternion algebra over a field $K$ (of characteristic not $2$) in $\S 1.6$ of these notes. This is by no means the best way of showing associativity of quaternion algebras -- e.g. it would be better to embed them, possibly after base extension, into a matrix algebra -- but I find it comforting that one can verify associativity of finite-dimensional algebras "while sleeping", so to speak, if one cares to.

• By the way, I suspect this answer may be somewhat over the head of the OP; anyway, s/he has already accepted another answer. But I hope that other passers by may find this answer to be helpful. – Pete L. Clark May 4 '12 at 0:13
• I never took a any course on proving mathematical properties, so I will have to take your word for it on this, and according to the up-votes I am presuming that it is accurate. – gardian06 May 4 '12 at 6:47