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.

  • $\begingroup$ 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

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.

  • $\begingroup$ 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

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


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