# why is multiplication of unit quaternion not commutative?

I have two unit quaternions q1(rotation theta1 around axis u1) and q2(rotation theta2 around axis u2)  Now I should get the same result whether I multiply q1*q2 or q2*q1 as I am multiplying two exponentials. So why is multiplication of two unit quaternions non commutative? is there something wrong with adding the arguments of the exponentials when dealing with quaternions. The multiplication above yeilds same expression whether q1*q2 or q2*q1 is performed.

I am from engineering background so a simple explanation would be easier for me to comprehend.

• Because it is defined to be. $ij=k$ but $ji=-k$, and $i,j$ are unit quaternions. Jun 24, 2019 at 10:30
• i understand that ij=k and ji =-k by definition. But I cant understand why the above expression of q1*q2 or q2*q1 is same irrespective of the order of multiplication. i have revised the question. kindly have a look. Jun 24, 2019 at 11:11
• The only finite-dimensional commutative and associative division algebras over $\Bbb R$ are $\Bbb R$ and $\Bbb C$ Jun 24, 2019 at 11:12
• @user27665 No! $i=\exp(\frac\pi2i)$ and $j=\exp(\frac\pi2j)$ are two exponentials and they do not commute. Jun 24, 2019 at 11:15
• How do you define the exponential of a quaternion ??
– user65203
Jun 24, 2019 at 11:42

When you write something like $$q_1 = \exp(\theta_1/2 + u_1)$$ you are extending the exponential function from the domain $$\mathbb R$$ or $$\mathbb C$$ (where you are familiar with it, and where the rule for adding the exponents is valid) to a new domain, $$\mathbb H.$$

You cannot just assume that all the rules you learned that are valid in $$\mathbb R$$ and $$\mathbb C$$ will also be valid in $$\mathbb H.$$ You have to prove that they are valid if you want to use them.

In this case you will not be able to prove that this rule is valid, because in fact it is not valid!

We know that the unit quaternions $$i$$ and $$j$$ represent rotations, and therefore they can be expressed as exponentials using their rotation axes and angles, but $$ij \neq ji$$ whereas the old rule for adding exponents says the multiplication should commute. The rule contradicts the plain facts of quaternions, so it must not be a correct rule to apply to quaternions.

To put it another way: put two six-sided dice on the table in front of you, oriented the same way. Rotate one a quarter turn around the $$x$$-axis and then a quarter turn around the $$y$$-axis. Now perform the same rotations on the other die, but do the $$y$$-axis rotation before the $$x$$-axis rotation. Did both dice end up in the same orientation?

Rotations do not generally commute under composition (where you do one rotation and then the other). They commute only in some special cases. Therefore any theory that assigns each rotation to some "number" (say, a quaternion) and then claims that the composition of two rotations is the product of those two numbers, and claims that multiplication of those two numbers commutes, is telling you that the two dice must end up in the same orientation no matter which order you performed the operations, and therefore that theory is wrong.

As it turns out there is a valid theory that maps every rotation to a unit quaternion, and composition of rotations corresponds to multiplication of those quaternions. This works only because the multiplication of the quaternions is not commutative. For example, the two quarter-turn rotations of the dice in the example above are the quaternions $$1/\sqrt2 + i/\sqrt2$$ and $$1/\sqrt2 + j/\sqrt2.$$ Multiplying one way we get $$\left(\frac1{\sqrt2} + \frac 1{\sqrt2}i\right) \left(\frac1{\sqrt2} + \frac 1{\sqrt2}j\right) = \frac12 + \frac12 i + \frac12 j + \frac12 k$$ whereas in the other order we get $$\left(\frac1{\sqrt2} + \frac 1{\sqrt2}j\right) \left(\frac1{\sqrt2} + \frac 1{\sqrt2}i\right) = \frac12 + \frac12 i + \frac12 j - \frac12 k.$$ These work out to $$120$$-degree rotations around the axes $$\frac1{\sqrt3}(i+j+k)$$ and $$\frac1{\sqrt3}(i+j-k),$$ which is the same result you get if you do the two rotations on an actual object in the real world. So the theory matches the reality, which is a good property for a theory to have.

The exponential $$\exp(q)$$ is defined as the sum of the usual power series, which is absolutely convergent. It follows that $$\exp(q+r) = \exp(q)\exp(r)$$ can be guaranteed only when $$q$$ and $$r$$ commute, because you need commutativity to manipulate the series.