# Sesquilinear form

(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)
Jump to: navigation, search

2010 Mathematics Subject Classification: Primary: 15-XX [MSN][ZBL]

A sesquilinear form is a function in two variables on a module (for example, on a vector space) which is linear in one variable and semi-linear in the other. More precisely, a sesquilinear form on a unitary module $E$ over an associative-commutative ring $A$ with an identity, equipped with an automorphism $\def\s{\sigma}a\mapsto a^\s$, is a mapping $q:E\times E\to A$, $(x,y)\mapsto q(x,y)$, linear in $x$ for fixed $y$, and semi-linear in $y$ for fixed $x$ (see Semi-linear mapping). Analogously one defines a sesquilinear mapping $E\times F\to G$, where $E$, $F$, $G$ are $A$-modules. In the case when $a^\s = a$ ($a\in A$), one obtains the notion of a bilinear form (or a bilinear mapping). Another important example of a sesquilinear form is obtained when $V$ is a vector space over the field $\C$ and $a^\s=\bar a$ is complex conjugation. Special cases of sesquilinear forms are Hermitian forms (cf. Hermitian form) (and also skew-Hermitian forms).

Sesquilinear forms can also be considered on modules over a non-commutative ring $A$; in this case it is assumed that $\s$ is an anti-automorphism, that is,

$$(ab)^\s = b^\s a^\s\quad a,b\in A.$$ For sesquilinear forms it is possible to introduce many notions of the theory of bilinear forms, for example the notions of an orthogonal submodule, a left and a right kernel, a non-degenerate form, the matrix of the form in a given basis, the rank of the form, and conjugate homomorphisms.

#### Comments

Let $D$ be a division ring with centre $k$ and $V$ a right vector space over $D$. Let $\s$ be an anti-automorphism of $D$, i.e. $\s$ is an automorphism of the underlying additive group of $D$ and $\s(xy) =\s(y)\s(x)$. A sesquilinear form relative to $\s$ on $V$ is a bi-additive mapping

$$f : V\times V\to D$$ such that

$$f(vx,wy) = \s(x)f(v,w)y.$$ Unless $f=0$, the anti-automorphism $\s$ is obviously uniquely determined by $f$.

Let $\def\e{\epsilon}\e\in k\setminus \{0\}$. A $(\s,\e)$-Hermitian form is a sesquilinear form on $V$ such that moreover

$$f(w,v)=\s (f(v,w))\e$$ One must then have $\e\s(\e)=1$ and $\s^2(x)=\e x\e^{-1}$ for all $x\in D$. The concepts of a Hermitian, anti-Hermitian, symmetric, anti-symmetric, or bilinear form (or matrix) for complex vector spaces (with $\s =$ complex conjugation) arise as the special cases of a $(\s,1)$-Hermitian form, a $(\s,-1)$-Hermitian form, an $({\rm id},1)$-Hermitian form, and an $({\rm id},-1)$ Hermitian form.

Given a subspace $W\subset V$, $W^\perp = \{v\in V : f(v,w)=0 \textrm{ for all } w\in W\}$. A subspace $W$ is totally isotropic if $W\subset W^\perp$. The Witt index of a sesquilinear form is the dimension of a maximal totally-isotropic subspace.