1
$\begingroup$

Let $\mathcal A$ be the algebra over the real numbers consisting of matrices of the form $$\begin{pmatrix} z&w\\ - \bar{w}& \bar{z} \end{pmatrix} \ (z, w \in \mathbb C). $$ $\mathcal A$ is in the bijection with the algebra of quaternions via $$a+bi+cj+dk \longleftrightarrow \begin{pmatrix} a+bi& c+di\\ - c+di & a-bi \end{pmatrix}. $$
I saw no other way of verifying the homomorphism condition then by calculation (which I only finished partially until I believed it).

Is there a more natural way of noting that the given bijection is a homomorphism, without the necessity of computing matrix- and quaternion products?

I got a few clues from @martinis answer:

Put $$ I_2 := \begin{pmatrix} 1&0\\0&1 \end{pmatrix} \ ,\ I := \begin{pmatrix} i&0 \\ 0&-i \end{pmatrix},\ J:= \begin{pmatrix} 0&1 \\ -1&0 \end{pmatrix},\ K := \begin{pmatrix} 0&i\\i&0 \end{pmatrix}.$$ Now notice that we have $$a+bi+cj+dk \longleftrightarrow \begin{pmatrix} a+bi& c+di\\ - c+di & a-bi \end{pmatrix} = aI_2 +bI+cJ+dK. $$ Hence the bijection is $\mathbb R$-linear in $(a,b,c,d)$. For the homomorphism condition it now suffices to check whether the above defined matrices satisfy $$I^2=J^2=K^2=IJK=-I_2,$$ because then multiplication rules on both sides agree. It then follows that the bijection is multiplicative. And with that, the isomorphism is established.

For some reason the concept transport of structure pops into my mind.

$\endgroup$
  • 1
    $\begingroup$ I know that this is cheating, but one way is to define $\mathbb{H}$ as this algebra. The nasty multiplication rule is then an immediate consequence. This way, you also don't have to do horrible calculations in order to verify that $\mathbb{H}$ is an algebra. $\endgroup$ – Martin Brandenburg Jul 3 '14 at 16:02
  • 1
    $\begingroup$ @MartinBrandenburg, they are not horrible. You have to check associativity on the basis, so for any three elements $x$, $y$, $z$ you need to show that $x(yz)=(xy)z$. If any of the three is $1$ this is obvious. Since the multiplication is "invariant" under cyclic rotations and somewhat commutative, you have to consider two cases: $(x,y,z)$ is $(i,j,k)$ or $(i,i,j)$. $\endgroup$ – Mariano Suárez-Álvarez Jul 3 '14 at 18:06
  • $\begingroup$ @MarianoSuárez-Alvarez What does associativity has to do with showing that the bijection is a homomorfism? $\endgroup$ – Mussé Redi Jul 3 '14 at 18:10
  • $\begingroup$ @MusséRedi, if you read what I wrote and the comment of Martin to which I was answering, you'll immediately see that I am referring to his third sentence. $\endgroup$ – Mariano Suárez-Álvarez Jul 3 '14 at 18:25
  • $\begingroup$ @MarianoSuárez-Alvarez An algebra is not determined by associativity. I'm missing your point. Could you explain why checking associativity on the basis is helpful? $\endgroup$ – Mussé Redi Jul 3 '14 at 18:39
4
$\begingroup$

Hint: Both sides are $\mathbb R$-linear in $(a,b,c,d)$, so you have to check multiplicativity only on the $\mathbb R$-basis $\{1,i,j,k\}$ of $\mathbb H$, that is one only has to check whether $$ I := \def\p#1#2#3#4{\begin{pmatrix}#1&#2\\#3&#4\end{pmatrix}}\p i00{-i}, J := \p 01{-1}0, K := \p 0ii0 $$ fulfill $I^2 = J^2 = K^2 = IJK = -\mathrm{Id}$.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ ... and the case of $1$ is clear anyway. $\endgroup$ – Martin Brandenburg Jul 3 '14 at 16:03
  • $\begingroup$ Is there an isomorphism between quaternions and $M_n(\mathbb{R})$, nxn square matrices of real values? Or perhaps a 2x2x2 tensor of reals? $\endgroup$ – CogitoErgoCogitoSum Feb 28 '17 at 7:02

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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