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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.

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    $\begingroup$ Your first hint about where things were going wrong is that if $i=j=k$, then Hamilton wasted an awful lot of time using three letters instead of one... $\endgroup$
    – rschwieb
    Dec 11 '15 at 18:53
  • $\begingroup$ The definition Hamilton gave is algebraic; we can't "solve" for $i$, $j$, and $k$ because he means them as placeholder symbols, more or less. A similar concept is that the dual numbers which have some use in computation are the reals plus an element which isn't zero, but whose square is $0$. If this doesn't make sense to you, you might do well just to find explicit formulas for how to add & multiply quaternions $\endgroup$ Dec 11 '15 at 18:58
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    $\begingroup$ He wrote $i^2=j^2=k^2=ijk=-1$, nothing more. You can't conclude what you've added. The solution is as you've seen that $ij=k$, $jk=i$ and $ki=j$. Maybe an interresting question is what minimally we would require more of the rules to infer the complete multiplication table for the units. $\endgroup$
    – skyking
    Dec 11 '15 at 19:14
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It is not true that $i=j=k=\sqrt{-1}$. It is only true that $i^2=j^2=k^2=-1$.

Similarly, it is true that $(-2)^2=2^2=4$, but that does not mean that $-2=2=\sqrt 4$.

In fact, the formulas you have written down are axioms that form the quaternions. You decide that you will look at the division ring (i.e. a ring where you can divide by all numbers but $0$) in which you have $3$ different numbers which all square to $-1$ and which satisfy the equation $ijk=-1$. That's how the quaternions are defined. And no, $ijk$ should not be equal to $i^3$, because in fact, $ij=k$, meaning that $ijk=kk=-1$ which fits with your formulas.

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    $\begingroup$ Not a field (because multiplication is not commutative), but an algebra. $\endgroup$ Dec 11 '15 at 18:48
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    $\begingroup$ @RobertIsrael Thanks. I believe "division ring" is also allowed? In my language, we have a separate word for "noncommutative field", but I don't know about english... $\endgroup$
    – 5xum
    Dec 11 '15 at 18:49
  • $\begingroup$ Division ring is best, although 'field' has been used for noncommutative fields at certain times by some cultures. $\endgroup$
    – rschwieb
    Dec 11 '15 at 18:49
  • $\begingroup$ @5xum: What language is that, and what word would that be, if you don't mind my asking? $\endgroup$
    – Brian Tung
    Dec 11 '15 at 18:55
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    $\begingroup$ @BrianTung In Slovenian, a ring is "kolobar" (it's a rough translation as there is no direct translation of the word ring. I think "obroč" would be a better one), and a division ring is "obseg" (which, strangely enough, would directly translate into "circumference"). A field is then "polje" ("polje" literally means a field even outside mathematics). $\endgroup$
    – 5xum
    Dec 11 '15 at 18:58
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Because $\mathbf{jk} = \mathbf{i}$, so $\mathbf{ijk} = \mathbf{i}\cdot\mathbf{i} = -1$. That's the way they're defined; $\mathbf{i}, \mathbf{j}, \mathbf{k}$ are all three separate quantities, each of whose square is equal to $-1$. Multiplication is associative but not commutative.

We have

$$ \mathbf{ij} = \mathbf{k}, \mathbf{ji} = -\mathbf{k} $$ $$ \mathbf{jk} = \mathbf{i}, \mathbf{kj} = -\mathbf{i} $$ $$ \mathbf{ki} = \mathbf{j}, \mathbf{ik} = -\mathbf{j} $$

These rules can be derived from Hamilton's equation. Provided that multiplication is performed associatively and carefully, adhering to those rules, one avoids the kind of contradiction you describe.

The $-1$ in Hamilton's equation is generally regarded as the ordinary scalar value $-1$, although one can write quaternions in general as $4$-tuples, with $(a, b, c, d)$ representing $a+b\mathbf{i}+c\mathbf{j}+d\mathbf{k}$.

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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.

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