You are here
Homequadratic space
Primary tabs
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 $(ij)^{2}=(ij)(ij)=i(ji)j=i(ij)j=ab\neq 0$, 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 $a\neq 0$, $s=0$. 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 $rsitjuk$. 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
$\Big(\frac{a,b}{F}\Big).$ 
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 $a,b\in F$.
Mathematics Subject Classification
15A63 no label found11E88 no label found Forums
 Planetary Bugs
 HS/Secondary
 University/Tertiary
 Graduate/Advanced
 Industry/Practice
 Research Topics
 LaTeX help
 Math Comptetitions
 Math History
 Math Humor
 PlanetMath Comments
 PlanetMath System Updates and News
 PlanetMath help
 PlanetMath.ORG
 Strategic Communications Development
 The Math Pub
 Testing messages (ignore)
 Other useful stuff
 Corrections