5
$\begingroup$

Why there is no $5$-dimensional analogous of Quaternions? Why the following definition is not well-defined? $$i^2=j^2=k^2=\ell^2=ijk=jk\ell=k\ell i=-1,\quad ijkl=1.$$

$\endgroup$
9
$\begingroup$

If you are looking for a real vector space $V$ with basis $\{1,i,j,k,\ell\}$ and an associative product such that $V$ becomes an $\Bbb{R}$-algebra, then we run into the following difficulty.

The rule that $i^2=-1$ means that $V$ also has a structure as a vector space over the field $\Bbb{C}=\Bbb{R}(i)$.

But a vector space over $\Bbb{C}$ necessarily has an even dimension as a vector space over $\Bbb{R}$.


The above is probably not the shortest route to a contradiction (see the comment by verret). But it also rules out many modifications to the suggested relations defining the product.

$\endgroup$
  • $\begingroup$ In the same vein, if $\Bbb{H}$ is the usual division algebra of quaternions then any associative algebra $A$ containing $\Bbb{H}$ as a subalgebra is also a free left (or right) $\Bbb{H}$-module. Therefore $\dim_{\Bbb{R}}A$ is necessarily divisible by four. $\endgroup$ – Jyrki Lahtonen Nov 5 '17 at 8:10
  • $\begingroup$ I think we can actually relax associativity a bit, but I'm not sure about the best way of phrasing that :-) $\endgroup$ – Jyrki Lahtonen Nov 5 '17 at 8:24
  • 1
    $\begingroup$ Slightly more relaxed: if $A$ is a real alternative algebra with $n$ anticommuting square roots of negative one, then $A$ is a representation of the Clifford algebra ${\rm Cliff}(n)$. Since $\mathrm{Cliff}(n)\cong M_k(\Bbb K)$ for some $k$, the dimensions of its reps are all multiples of $km$ where $m=1,2,2,4,4,8$ and $\Bbb K=\Bbb R,\Bbb R^2,\Bbb C,\Bbb C^2,\Bbb H,\Bbb H^2$. Note $k$ and $\Bbb K$ can be determined from $n$ using the "Clifford clock." $\endgroup$ – anon Nov 5 '17 at 21:52

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.