2
$\begingroup$

I am following Theorem 2.3.1 of Maclachlan's and Reid's The Arithmetic of Hyperbolic 3-Manifolds. We define a quaternion algebra $A=\left(\frac{a,b}{F}\right)$ over a field $F$ of characteristic $\neq 2$ by the vector space spanned by $\{1,i,j,k\}$ with $i^2=a,\,j^2=b,$ and $ij=-ji=k$ (of course letting $a,b\in F^\ast$). It is earlier established in this book that $$ \left(\frac{a,b}{F}\right)\cong\left(\frac{ax^2,by^2}{F}\right) $$ for any $x,y\in F^\ast$. Further, it is established that $M_2(F)\cong\left(\frac{1,1}{F}\right)$.

Keep the definition of $A$ from above. What I am attempting to show is that if there exist $x,y\in F^\ast$ such that $ax^2+by^2=1$, then $A\cong M_2(F)$. I suspect this is trivial but I'm a bit stuck. Thanks for your help.

Edit: to add a bit of what I've done, it's clear that we have $$ \frac{1}{a}=\frac{x^2}{1-by^2}. $$ If we can prove the RHS is a square, then we're good.

Edit 2: Equivalently, it could be proven that the existence of those $x,y$ imply that $A$ is not a division algebra, and I actually figured that out. Let $\alpha\in A$ such that $\alpha=1+xi+yj$. Then $\alpha\overline\alpha=1-ax^2-by^2=0$, and clearly $\alpha,\overline\alpha\neq 0$. I'm still interested in how the first assertion might be proven, since it is definitely true.

$\endgroup$
  • $\begingroup$ Is there any assumption that the field is algebraically closed, by the way? I'm not sure it's necessary. I'm just curious. $\endgroup$ – rschwieb Mar 29 '13 at 19:18
  • $\begingroup$ Nope. In fact, if $F$ is algebraically closed, then the only quaternion algebra over $F$ is the matrix algebra $M_2(F)$ for the very reason you're thinking of. $\endgroup$ – Ian Coley Mar 29 '13 at 19:19
  • $\begingroup$ Isomorphism of $F$ algebras? Actually, I get the feeling you mean something a little stronger. Since these are Clifford algebras, I'm expecting a Clifford algebra isomorphism, which has to transfer the bilinear form as well. $\endgroup$ – rschwieb Mar 29 '13 at 19:36
1
$\begingroup$

Using the equation $ax^2+by^2=1$, you can deduce the existence of an orthonormal basis with respect to that bilinear form:

$$ v=\left(\frac{x}{1-by^2},0\right)\\ w=\left(0,\frac{y}{by^2}\right)\\ $$

The transformation of the metric space $(V,B)\to (V,\cdot)$ using the matrix $T=\begin{bmatrix}\frac{1-by^2}{x}&0\\0&\frac{by^2}{y}\end{bmatrix}$ maps $v,w$ onto $(1,0), (0,1)$, and moreover the bilinear forms match:

$$ Tv\cdot Tw=B(v,w) $$

So, they actually match for all pairs of vectors, and the associated Clifford algebras are isomorphic.

$\endgroup$
  • $\begingroup$ Thanks for your answer. I haven't looked too closely into Clifford algebras but it seems a natural extension of what I've been thinking about right now. $\endgroup$ – Ian Coley Mar 29 '13 at 19:53
  • $\begingroup$ @FrankMcGovern No problem.. I may be missing something. Honestly I can't see why how I even used $ax^2+by^2=1$. As far as I can tell, as long as $a$ and $b$ are nonzero, you can use $(x/ax^2,0)$ and $(0,y/by^2)$ no matter what. I'm clearly missing something. In the case of the real numbers, $a=1$ and $b=-1$ is not isomorphic to the case when $a=b=1$. $\endgroup$ – rschwieb Mar 29 '13 at 19:57
  • $\begingroup$ I'll bear that in mind, thanks. $\endgroup$ – Ian Coley Mar 29 '13 at 19:58
  • $\begingroup$ @FrankMcGovern I guess the condition translates into "it's a 2 by 2 matrix ring over $F$, and not an extension of $F$". The first thing that jumped out at me was: "$(x,y)$ is a unit vector", but I couldn't make use of it. $\endgroup$ – rschwieb Mar 29 '13 at 20:01
0
$\begingroup$

See $\S$ 5 of these notes, especially Theorems 92 and 94.

$\endgroup$

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.