quadratic space
A quadratic space (over a field) is a vector space^{} $V$ equipped with a quadratic form^{} $Q$ on $V$. It is denoted by $(V,Q)$. The dimension^{} of the quadratic space is the dimension of the underlying vector space. Any vector space admitting a bilinear form^{} has an induced quadratic form and thus is a quadratic space.
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.
A quadratic space equipped with a regular quadratic form^{} is called a regular quadratic space.
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.
${i}^{2}=a$,

2.
${j}^{2}=b$, and

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 ${(ij)}^{2}=(ij)(ij)=i(ji)j=i(ij)j=ab\ne 0$, so that $k\in \dot{F}$. It’s also not hard to show that $k$ anticommutes 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 $a\ne 0$, $s=0$. 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 $rsitjuk$. 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
$$\left(\frac{a,b}{F}\right).$$ 
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
$$\left(\frac{1,1}{\mathbb{R}}\right).$$ 
In fact, every quaternion algebra (over a field $F$) is of the form $\left({\displaystyle \frac{a,b}{F}}\right)$ for some $a,b\in F$.
Title  quadratic space 
Canonical name  QuadraticSpace 
Date of creation  20130322 15:05:55 
Last modified on  20130322 15:05:55 
Owner  CWoo (3771) 
Last modified by  CWoo (3771) 
Numerical id  14 
Author  CWoo (3771) 
Entry type  Definition 
Classification  msc 15A63 
Classification  msc 11E88 
Synonym  nondegenerate quadratic space 
Related topic  QuadraticForm 
Related topic  QuaternionAlgebra 
Defines  norm form 
Defines  isomorphic quadratic spaces 
Defines  isometric quadratic spaces 
Defines  generalized quaternion algebra 
Defines  regular quadratic space 