Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

So, I understand complex number multiplication, and how it represents $2D$ rotations.

What I don't understand is, how you add two more imaginary numbers $j$ and $k$, and get $3D$ rotations. I believe Hamilton "made up rules" on how to add these extra imaginary numbers, but how did he construct it in a way that worked?

I also don't understand the purpose of adding two imaginary numbers instead of just $1$ more for $3D$ rotations.

In this paragraph, I will explain my attempt to visualize / construct the quaternion, you don't have to read this, it's probably utterly wrong. I tried to imagine a $3D$ space with the axis vectors $1$ (the real number line, side to side), $i$ (up), and $j$ (depth). Next, I imagined a vector pointing out from the origin with a real component and a $j$ component. Next, I imagined multiplying this vector by $i$ to rotate it $90$ degrees up, so I would get a resultant vector with an '$i$' component and an '$j$' component. The problem is, multiplying (real + $j$) with ($i$) algebraically gives you an '$i$' component and an '$ij$' component, which does not make sense to me. This is where I got stuck.

share|cite|improve this question
If I remember right, Hamilton spent a decade trying to do it with just one more instead of two more, and then just realized in a flash of inspiration that two were necessary so that there was "enough room". I guess you're kind of asking for a back-explanation of that flash of inspiration? – rschwieb Mar 14 '13 at 0:34
@rschwieb yes, that would be helpful. I want to know how he constructed it though – Dan Webster Mar 14 '13 at 0:37
If I remember right, Hamilton struggled for a decade trying to use "just one more" before he realized it wasn't possible, and then he found that four worked. – rschwieb Mar 14 '13 at 1:16
up vote 2 down vote accepted

So in the complex picture, rotations and dilations of the plane are acheived just by multiplying with a complex number.

I see here you are transferring that idea, but as you said, you get stuck if your three axes are $1,i,j$, because $ij=k$, which is a fourth axis perpendicular to the first two. If you're willing to imagine a four-dimensional space, then you have a similar picture to the complex numbers: multiplication by $i$ would permute the set $\{\pm 1, \pm i,\pm j,\pm k\}$ just as multiplying by $i$ permutes $\{\pm 1, \pm i\}$.

There is a more useful way to view quaternions as rotations in 3-space, though. Imagine $i,j,k$ acting as orthogonal unit vectors, as in physics. Linear combinations of these model every point in that 3-space as a "pure" quaternion $v$ with no real part. Now given another nonzero quaternion $q$, conjugating $v$ to become $qvq^{-1}$ produces a tranformation of the 3 space! When $q$ is a unit-length quaternion, this transformation is actually a rotation.

Try it out with $q=i$ and see what happens to the $i,j,k$ axes! If you feel like skipping $j$ and $k$, you can try $\sqrt{2}/2 +i\sqrt{2}/2$ (whose inverse is, naturally, $\sqrt{2}/2 -i\sqrt{2}/2$.)

This is not useful in the complex numbers because $aba^{-1}=b$ by commutativity, so no change occurs. However, the noncommutativity of the quaternions allows interesting stuff to happen :)

One thing to keep in mind about this conjugation action modeling rotations is that many rotations are produced by a pair of unit-length quaternions (and not just one unit-length quaternion). All that means is that there is a slight redundancy in the way unit-length quaternions model rotations.

EDIT: I just wanted to add a little more trying to justify why "just one more" doesn't seem to work. In analogy with the complex numbers, you want to get $1,i,j$ such that the produce of any two lands in $\{\pm1, \pm i \pm j\}$. As you noted, we have to figure out where $ij$ goes. But if $ij=\pm i$, that implies that $-j=\pm -1$, and then $j$ is not orthogonal to 1. Similarly, $ij\neq \pm j$. The only thing left is if $ij=\pm 1$, but then $-j=\pm i$, and $j$ is not perpendicular to $i$.

share|cite|improve this answer

Here's one way to think about it. Let's restrict ourselves to unit quaternions (they are of the form $a+bi+cj+dk$ such that $a^2+b^2+c^2+d^2 = 1).$ Unit quaternions aren't supposed to give you elements of $\mathbb{R}^3,$ they're supposed to give you orthogonal transformations. So you shouldn't just use $i,j,$ and $k$ to label the axes of $\mathbb{R}^3,$ you should be using them to label certain kinds of rigid motions (although we will end up labeling the axes $i,j,$ and $k$ later to see how these rigid motions work).

So what do rigid motions of $\mathbb{R}^3,$ look like? Well they are all given by fixing an axis of rotation and an angle of rotation about that axis. Unit quaternions are three dimensional, so there better be 3 independent parameters here. Choosing the axis of rotation amounts to choosing a unit vector in $\mathbb{R}^3,$ so a point on the 2 sphere. Choosing an angle is an additional 1 parameter, so we should have a 3 dimensional group of orthogonal transformations.

To see what linear transformation a quaternion $q=a+bi+cj+dk$ gives, label the axes $i,j$ and $k.$ This will be our basis for $\mathbb{R}^3.$ The transformation will be given by conjugating by the quaternion, so $i$ goes to $qiq^{-1}.$ For example, what does the linear transformation given by $q=i$ do? It sends $i$ to $(i)(i)(-i) = i$ and thus fixes the first axis. It sends $j$ to $-j$ and $k$ to $-k,$ so rotates by 180 degrees in the $j,k$ plane. This also gives us a matrix representation for the transformation $i,$ it is $$\begin{pmatrix} 1 & 0& 0 \\ 0 &-1& 0 \\ 0& 0&-1 \\ \end{pmatrix}$$ when we use the ordered basis $i,j,k.$ Note that $i$ and $-i$ actually induce the same linear transformation, so the unit quaternions are not exactly the special orthogonal transformations, but map 2-1 onto them. Geometrically, the unit quaternions look like $S^3$ and $SO(3)$ is homeomorphic to $RP^3,$ and this map gives a double cover.

share|cite|improve this answer

I like to think of the quaternions as $\mathbb{H},$ the set of complex matrices of the form $\left( \begin{array}{clcr} z & w\\ -\bar{w} &\overline{z} \end{array} \right)$, where $z,w$ run through all the complex numbers. Note that any such complex matrix has strictly positive real determinant, except when $z = w =0.$ Hence this is an example of a division ring (a ring in which every non-zero element has an inverse). Thinking of quaternions this way, $i$ corresponds to the case $z = i, w = 0$, $j$ corresponds to the case $z=0, w =i$ and $k$ corresponds to the case $z=0, w = 1.$ Note that such a matrix has trace zero exactly when $z$ is pure imaginary (as a complex number). In fact, such a matrix gives rise to a rotation of $3$-dimensional real space, because we can identify the matrices of trace $0$ with the $\mathbb{R}$-linearcombinations of $i,j,k$ in the above correspondence. And for any such matrix $M,$ and matrix $T \in \mathbb{H}$ of trace zero, $M^{-1}TM$ also has trace zero. This allows us to think of $M$ as a linear transformation on $\mathbb{R}^{3},$ and as such it is length preserving and has determinant $1$ (this takes a little checking which I omit). To recover $M$ from the rotation of $\mathbb{R}^{3}$ which it induces, we also have to include ${\rm det}(M) = |z|^{2} + |w|^{2},$ for notice that replacing $T$ by a non-zero real positive multiple does not change its effect on the matrix $T.$ This gives a way of describing the quaternions as ordered pairs, the first component being a non-negative real number, and the second being a rotationof $\mathbb{R}^{3},$ but it is easier to think of the multiplication as given by the usual matrix multiplication of complex $2 \times 2$ matrices.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.