2
$\begingroup$

I have to prove that the ring of real quaternions, is isomorphic to

$$ \begin{pmatrix} z & w\\ -\bar{w} & \bar{z}\\ \end{pmatrix} $$ Where $z$ and $w$ are complex numbers, and $\bar{z}$, $\bar{w}$ their conjugates.

My question is, if I define $1=\begin{pmatrix} 1 & 0\\ 0 & 1\\ \end{pmatrix}$ , $i= \begin{pmatrix} i & 0\\ 0 & -i\\ \end{pmatrix}$ , $j= \begin{pmatrix} 0 & 1\\ -1 & 0\\ \end{pmatrix} $ and $k= \begin{pmatrix} 0 & i\\ i & 0\\ \end{pmatrix} $

As my isomorphism, and I notice that $i^2=j^2=k^2=ijk=-1$ and that also happens in the matrix. Is that enough to prove the isomorphism? Or do I have to show something else?

$\endgroup$
  • 1
    $\begingroup$ Yes, basically that's it. But please rather use an assignment notation $\mapsto$ rather than equality, as an isomorphism is a function which has inverse. $\endgroup$ – Berci May 31 '17 at 0:07
  • $\begingroup$ Ooo yes you are right. My mistake. Thanks :) $\endgroup$ – J.Rodriguez May 31 '17 at 0:52
0
$\begingroup$

You have to prove that the the quaternions and the matrices of this form are isomorphic as skew fields. Tis means that the addition of two quaternions correspond to the addition of the corresponding matrices and the same for the product and vice-versa, and that all required properties are satisfied ( this is easy). Also you have to prove that the inverse of a quaternion correspond to the inverse of a matrix, this is a consequence of the fact that , if the quaternion $z$ correspond to the matrix $Z$, we have $\det Z=|z|^2$. To complete the proof, we have also to consider that the quaterions are a ring with an involution ( the conjugation) so we have to prove also tha we can define an involution in the ring of matrices and that the conjugate of a quaternion is the conjugate of the corresponding matrix.

$\endgroup$
  • 1
    $\begingroup$ I don't know why you say you need to prove the inverse property separately. It suffices, and I think is more straightforward, to prove there is a bijective ring homomorphism. $\endgroup$ – Kimball Jun 2 '17 at 5:01

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.