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 $k$ 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 $a\ne 0$, $s=0$. 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 $r-si-tj-uk$. 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

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

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 $ℍ$ over the reals

$$\left(\frac{-1,-1}{ℝ}\right).$$

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	2013-03-22 15:05:55
Last modified on	2013-03-22 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	non-degenerate 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