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The quaternionic group $\mathcal{Q}$ consists of the elements $1$, $-1$, $i$, $-i$,$j$,$-j$,$k$,$-k$ that satisfy the multiplication rules

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

$$ ij=-ji=k$$

$$jk=-kj=i$$

$$ki=-ik=j$$

The quaternionic numbers $$a+ib+cj+dk$$ form a division dividion algebra.

In Group Theory in a Nutshell on p61 A.Zee writes that those two structures are completely unrelated, but I almost cant swallow this.

Are the quaternionic group and the quaternionic numbers really completely unrelated?

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    $\begingroup$ Do physicists really say "quaternionic numbers" rather than "quaternions"? Barf! $\endgroup$ – Rob Arthan Mar 7 '17 at 23:08
  • $\begingroup$ @RobArthan at least A.Zee does ;-) $\endgroup$ – Dilaton Mar 7 '17 at 23:10
  • $\begingroup$ Oh dear, pour souls, bless! $\ddot{\frown}$. $\endgroup$ – Rob Arthan Mar 7 '17 at 23:12
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    $\begingroup$ "Unrelated" here must refer to how he's used to using these things. There is an obvious algebraic relationship since $\mathbb H$ is a quotient of the group algebra $\mathbb R[Q_8]$. $\endgroup$ – rschwieb Mar 8 '17 at 17:06
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It's silly to say that they are completely unrelated. If you take just those quaternions for which precisely one of $a,b,c,d$ is non-zero, and the one that is non-zero is either 1 or $-1$, you obtain the quaternionic group. On the other hand, if you use the elements of the quaternionic group as the basis of an 8-dimensional vector space over $\mathbb R$, and then add relations to make $-x$ the negation of $x$ for each element of the group (creating a 4-dimensional vector space), and define a multiplication on that vector space by using the multiplication rules of the quaternionic group, you get precisely the algebra of the quaternions.

Presumably, your book just wanted to emphasize that you should not confuse the two structures, as one is a group under multiplication, while one is an ($\mathbb R$-, or $\mathbb C$-)algebra.

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  • $\begingroup$ +1 The answer I wanted to give is already described in the first paragraph here, but to rephrase it in my words: $\mathbb R[Q_8]/\langle 1_\mathbb R\cdot 1+1_\mathbb R\cdot(-1)\rangle\cong \mathbb H$. $\endgroup$ – rschwieb Mar 8 '17 at 17:03
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    $\begingroup$ Mees de Vries, @rschwieb: more directly, $Q_8$ is a subgroup of $\Bbb H^\times$... $\endgroup$ – arctic tern Mar 9 '17 at 14:58
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    $\begingroup$ @arctictern Also a good observation. I only mentioned the one above because it is compelling in its relationship to the whole of $\mathbb H$. Our two observations seem to form upper and lower bounds of the relationship between the two :) $\mathbb R[Q_8]\twoheadrightarrow \mathbb H$ and $Q_8\rightarrowtail\mathbb H^\times$. $\endgroup$ – rschwieb Mar 9 '17 at 15:19
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The quaternions $\mathbb{H}$ are a four-dimensional algebra spanned by basis elements $1,\mathbf{i},\mathbf{j},\mathbf{k}$, where $1$ is the multiplicative identity, $\mathbf{i},\mathbf{j},\mathbf{k}$ are three different square roots of $-1$, and $\mathbf{i},\mathbf{j},\mathbf{k}$ satisfy a "cyclic" relation (multiply any two of them in cyclic order, and you get the third, otherwise if you multiply them in the "wrong" order you get the negative of the third). This yields a multiplication table

$$ \begin{array}{l|cccc} & 1 & \mathbf{i} & \mathbf{j} & \mathbf{k} \\ \hline 1 & 1 & \mathbf{i} & \mathbf{j} & \mathbf{k} \\ \mathbf{i} & \mathbf{i} & -1 & \mathbf{k} & -\mathbf{j} \\ \mathbf{j} & \mathbf{j} & -\mathbf{k} & -1 & \mathbf{i} \\ \mathbf{k} & \mathbf{k} & \mathbf{j} & -\mathbf{i} & -1 \end{array} $$

This is just the multiplication table for $\{1,i,j,k\}$ inside $Q_8$! So it turns out, $Q_8$ is literally a group of quaternions (so, a subgroup of $\mathbb{H}^\times$), the same way $\{1,-1\}$ is a group of real numbers and the same way that $\{1,i,-1,-i\}$ is a group of complex numbers. Moreoever, $Z(Q_8)=Q_8\cap Z(\mathbb{H})$ since $Z(\mathbb{H})=\mathbb{R}$, and any transversal for $Q_8/Z(Q_8)$ (so, $\{\pm1,\pm i,\pm j,\pm k\}$ for four choices of signs) will be an orthonormal basis for $\mathbb{H}$ with respect to its natural inner product.

(Hamilton wrote down the relations $i^2=j^2=k^2=ijk=-1$ on the bridge. Exercise: Can you derive the multiplication table from these relations?)

(The reason I use boldface is that it is often helpful to think of a quaternion as a formal combination of a scalar and a vector, $q=r+\mathbf{u}$. Then the product of two purely imaginary quaternions is seen to be $\mathbf{uv}=-\mathbf{u}\cdot\mathbf{v}+\mathbf{u}\times\mathbf{v}$ - notice how the imaginary elements in the above multiplication table match that for the cross product! - and then we can multiply two arbitrary quaternions $(r_1+\mathbf{u}_1)(r_2+\mathbf{u}_2)$ with the distributive property.)

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