TITLE: How can Hamilton's quaternion equation be true?
QUESTION [8 upvotes]: I'm reading Ken Shoemake's explanation of quaternions in David Eberly's book Game Physics.  In it, he describes the $\mathbf{i}, \mathbf{j},  \mathbf{k}$ components of quaternions to all equal $\sqrt{-1}$.  Then it states Hamilton's quaternion equation:
$\mathbf{i}^2 = \mathbf{j}^2 = \mathbf{k}^2 = \mathbf{ijk} = \mathbf{-1}$
If $\mathbf{i} = \mathbf{j} = \mathbf{k} = \sqrt{-1}$, then it makes sense how $\mathbf{i}^2 = \mathbf{-1}$.  But $\mathbf{ijk}$ should equal $\mathbf{i}^3$, not $\mathbf{i}^2$.  How does $\mathbf{ijk} = \mathbf{-1}$?
The book's notation says that lowercase bold letters denote a vector, so I'm thinking of $\mathbf{i}$, $\mathbf{j}$, and $\mathbf{k}$ as the basis of the quaternion, similar to the basis of a vector, and can be written $(\sqrt{-1}, \sqrt{-1}, \sqrt{-1})$.  Having the result of $\mathbf{ijk}$ as a bold $\mathbf{-1}$ to me implies that it is the vector $(-1, -1, -1)$.  Is this understanding correct?  In this context, what does it mean to square vector $\mathbf{i}$?  If it equals another vector, then the only operation that makes sense is the cross product, but the cross product of a vector and itself is the zero vector.

REPLY [2 votes]: In it, he describes the $\mathbf{i}, \mathbf{j},  \mathbf{k}$ components of quaternions to all equal $\sqrt{-1}$. 

Unless he is constructing quaternions from complex numbers, there is no $\sqrt{-1}$ available to be equated with anything.   
Instead, these are three independent equations   $i^2 = -1, \quad j^2=-1, \quad k^2=-1$ that are used as part of the definition of a multiplication rule on the 4-dimensional vector space generated by a set of four different vectors that are (arbitrarily) assigned the names $1,i,j,k$.   There are many un-interesting multiplication rules such as $x \ast y = x$ for all vectors $x,y$, but Hamilton found a much more interesting one.

Then it states Hamilton's quaternion equation:
$\mathbf{i}^2 = \mathbf{j}^2 = \mathbf{k}^2 = \mathbf{ijk} = \mathbf{-1}$

The equation $ijk=-1$ is a mnenomic device for reproducing the full set of defining equations, but is not itself part of the definition.
There are 4x4=16 products of ordered pairs of the generators, and all of them need to be specified in order to define multiplication.  Here only 3 products of pairs are given, and one is supposed to infer the rest from $ijk=-1$ by multiplying that equation on the left or the right by $i,j$ and $k$ in all possible ways, assuming associativity, and applying the previous three rules.  
The other rules of quaternion multiplication are $1x = x1 = x$ for all $x$; the $i,j,k$ anticommute when distinct pairs are multiplied (so $ij = -ji$ et cetera); and cyclic permutations of $ij = k$.
This multiplication law is linear in each variable, distributive, associative, noncommutative, and (most unusually) every nonzero element has a multiplicative inverse.

But $\mathbf{ijk}$ should equal $\mathbf{i}^3$, not $\mathbf{i}^2$.
  How does $\mathbf{ijk} = \mathbf{-1}$?

For example, $ijk = i(jk) = i(i) = -1$.

The book's notation says that lowercase bold letters denote a vector, so I'm thinking of $\mathbf{i}$, $\mathbf{j}$, and $\mathbf{k}$ as the basis of the quaternion, similar to the basis of a vector,

They are 3 of the 4 basis vectors.  Every quaternion has a unique expression as a1 + bi + cj + dk for some numbers $a,b,c,d$.

Having the result of $\mathbf{ijk}$ as a bold $\mathbf{-1}$ to me implies that it is the vector $(-1, -1, -1)$.

It is the vector (-1)1 where the -1 is a real number and the boldface 1 is one of the basis elements of the quaternions as a vector space.

In this context, what does it mean to square vector $\mathbf{i}$? 

To use Hamilton's multiplication law to multiply that vector by itself.  The result will, by construction, be equal to some other quaternion.  The quaternion that it equals happens to be (-1)1, also known as -1.