Two quadratic spaces $(V_{1},Q_{1})$ and $(V_{2},Q_{2})$ are said to be isomorphic   if there exists an isomorphic linear transformation $T:V_{1}\to V_{2}$ such that for any $v\in V_{1}$, $Q_{1}(v)=Q_{2}(Tv)$. Since $T$ is easily seen to be an isometry between $V_{1}$ and $V_{2}$ (over the symmetric bilinear forms  induced by $Q_{1}$ and $Q_{2}$ respectively), we also say that $(V_{1},Q_{1})$ and $(V_{2},Q_{2})$ are isometric.

Example of a Qudratic Space. The Generalized Quaternion Algebra.

Let $F$ be a field and $a,b\in\dot{F}:=F-\{0\}$. Let $H$ be the algebra over $F$ generated by $i,j$ with the following defining relations:

1. 1.

$i^{2}=a$,

2. 2.

$j^{2}=b$, and

3. 3.

$ij=-ji$.

Then $\{1,i,j,k\}$, where $k:=ij$, forms a basis for the vector space $H$ over $F$. For a direct proof, first note $(ij)^{2}=(ij)(ij)=i(ji)j=i(-ij)j=-ab\neq 0$, so that $k\in\dot{F}$. It’s also not hard to show that $k$ anti-commutes with both $i,j$: $ik=-ki$ and $jk=-kj$. Now, suppose $0=r+si+tj+uk$. Multiplying both sides of the equation on the right by $i$ gives $0=ri+sa+tji+uki$. Multiplying both sides on the left by $i$ gives $0=ri+sa+tij+uik$. Adding the two results and reduce, we have $0=ri+sa$. Multiplying this again by $i$ gives us $0=ra+sai$, or $0=r+si$. Similarly, one shows that $0=r+tj$, so that $si=tj$. This leads to two equations, $sa=tij$ and $sa=tji$, if one multiplies it on the left and right by $i$. Adding the results then dividing by 2 gives $sa=0$. Since $a\neq 0$, $s=0$. Therefore, $0=r+si=r$. Same argument shows that $t=u=0$ as well.

Next, for any element $\alpha=r+si+tj+uk\in H$, define its conjugate   $\overline{\alpha}$ by $r-si-tj-uk$. Note that $\alpha=\overline{\alpha}$ iff $\alpha\in F$. Also, it’s not hard to see that

• $\overline{\overline{\alpha}}=\alpha$,

• $\overline{\alpha+\beta}=\overline{\alpha}+\overline{\beta}$,

• $\overline{\alpha\beta}=\overline{\beta}\overline{\alpha}$,

We next define the norm $N$ on $H$ by $N(\alpha)=\alpha\overline{\alpha}$. Since $\overline{N(\alpha)}=\overline{\alpha\overline{\alpha}}=\overline{\overline{% \alpha}}\overline{\alpha}=\alpha\overline{\alpha}=N(\alpha)$, $N(\alpha)\in F$. It’s easy to see that $N(r\alpha)=r^{2}N(\alpha)$ for any $r\in F$.

Finally, if we define the trace $T$ on $H$ by $T(\alpha)=\alpha+\overline{\alpha}$, we have that $N(\alpha+\beta)-N(\alpha)-N(\beta)=T(\alpha\overline{\beta})$ is bilinear (linear each in $\alpha$ and $\beta$).

Therefore, $N$ defines a quadratic form on $H$ ($N$ is commonly called a norm form), and $H$ is thus a quadratic space over $F$. $H$ is denoted by

 $\Big{(}\frac{a,b}{F}\Big{)}.$

It can be shown that $H$ is a central simple algebra over $F$. Since $H$ is four dimensional over $F$, it is a quaternion algebra  . It is a direct generalization  of the quaternions $\mathbb{H}$ over the reals

 $\Big{(}\frac{-1,-1}{\mathbb{R}}\Big{)}.$

In fact, every quaternion algebra (over a field $F$) is of the form $\displaystyle{\Big{(}\frac{a,b}{F}\Big{)}}$ for some $a,b\in F$.