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I'm currently trying to understand the following part of a script (translated from German to English). It is the first part where quaternions get introduced, so I don't know anything about them except what is written there:

Quaternions are an expansion of the concept of complex numbers on structures with four (instead of two) components. A quaterion $h$ can be written as a vector or in the form of $h = h_0 + ih_1 + j h_2 + > kh_3$, where $i$, $j$ and $k$ are related to the $i$ in complex numbers. Accordingly $h_0$ is often called real part and $h_1, h_2, > h_3$ are called imaginary part of a quaternion.

For $i$, $j$ and $k$ the following rules are applied:

$$i^2 = j^2 = k^2 = -1$$ $$ijk=-1$$

From these rules follows:

$\begin{align} ij &= k\\ ji &= -k\\ jk &= i\\ kj &= -i\\ ki &= j\\ ik &= -j \end{align}$

I don't understand why these rules follow.

My thoughts

$ij = k$ and $jk = i$

$$\begin{align} ijk &= -1\\ \Leftrightarrow ijk \cdot (k^3) &= (-1) \cdot (k^3)\\ \Leftrightarrow ij \cdot (-1) \cdot (-1) &= (-1) \cdot (-1) \cdot k\\ \Leftrightarrow ij &= k \end{align}$$

You get $jk = i$ with the same idea.

$-j = ik$ and $-k = ji$

$$\begin{align} ij &= k\\ \Leftrightarrow iij &= ik\\ \Leftrightarrow -j &= ik \end{align}$$

you get $-k = ji$ with the same idea.

Question

Why is $ki=j$ and $kj = -i$?

First I thought I could proof it like this:

$\begin{align} -k &= ji\\ \Leftrightarrow (-1) \cdot ki &= jii\\ \Leftrightarrow (-1) \cdot ki &= j \cdot (-1)\\ \Leftrightarrow ki &= (-1) \cdot j \cdot (-1)\\ \stackrel{*}{\Leftrightarrow} ki &= j \end{align}$

But for the last transformation * I would need commutativity. Obviously, $ij \neq ji$ so how can I know that $(-1) \cdot j = j \cdot (-1)$?

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    $\begingroup$ $i, j, k$ do not commute, but the real numbers do around them. So $(-1)\cdot j=-j=j\cdot(-1)$ is not an issue. $\endgroup$ – abiessu Sep 8 '13 at 10:19
  • $\begingroup$ It is interesting, from a modern perspective, $ij=k$, $jk=i$ and $ki=j$ all are motivated from corresponding cross-products of the unit-vectors ${\bf i}=\langle 1,0,0 \rangle$, ${\bf j}=\langle 0,1,0 \rangle$ and ${\bf k}=\langle 0,0,1 \rangle$. The cross-product is not associative, yet, this quaternionic multiplication is associative. Curious. $\endgroup$ – James S. Cook Sep 8 '13 at 11:57
  • $\begingroup$ @JamesS.Cook It's wrong to say $Q_8$ (or $\mathbb H$) comes from the cross-product. That might be the perspective of people whose experiences are based in application. Hamilton's perspective is the modern one: In order to define cross-product of vectors (a nonassociative algebra), you need a vector space, which requires a field, which requires a ring, which requires a group. $Q_8$ comes from trying to develop a group that can be turned into a field larger than $\mathbb C$. Hamilton proved this cannot be done, but the quaternions form a noncommutative division ring, the next best thing. $\endgroup$ – Travis Bemrose Nov 15 '15 at 2:06
  • $\begingroup$ @TravisBemrose I merely indicate the source of the notation for the usual vector algebra in modern texts. I agree that it is a dangerous game to play precisely for the reason my original comment indicated and for the reasons you outline; associativity in quaternions is not transferred to the cross-product. That said, I don't think it's at all crazy to suppose the quaternions are coming from the cross product in a certain sense. Perhaps if we look for a product which includes the cross and dot products in an associative fashion this leads us to quaternions... $\endgroup$ – James S. Cook Nov 15 '15 at 19:17
  • $\begingroup$ but, this of course was not the historical development as the cross-product came after the popularization of the $i,j,k$ notation. In any event, if you want a product on vectors which somehow includes rotations, this will pick up on the cross-product since cross-products are tied to rotations. $\endgroup$ – James S. Cook Nov 15 '15 at 19:19
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it's very trivial. $$(-1)j=(j^2)j=j(j^2)=j(-1).$$ Because $.$ operator in $Q_8$ is associative.

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$$(-1)\cdot x=x\cdot (-1), x\in Q_8$$ because $$-1\in Z(Q_8)=\{+1,-1\}$$

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  • $\begingroup$ I'd rather say $-1 \in Z(Q_8)$, because $\forall g \in Q_8: (-1) \cdot g = g \cdot (-1)$ which is the definiton of $Z(Q)$. But when I doubt that $(-1) \cdot k = k \cdot (-1)$ and you argue that $-1 \in Z(Q_8)$, your argument becomes circular (except you provide a reason why $-1 \in Z(Q_8)$) $\endgroup$ – Martin Thoma Sep 8 '13 at 10:57
  • $\begingroup$ @moose: If we want to see the problem in this way, yes you're right but I assumed that you already didn't know the centre of the group. (-: $\endgroup$ – mrs Sep 8 '13 at 11:03

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