Bilinear, sesquilinear and quadratic forms

Arun Ram
Department of Mathematics and Statistics
University of Melbourne
Parkville, VIC 3010 Australia
aram@unimelb.edu.au

Last updates: 11 June 2011

Bilinear, sesquilinear and quadratic forms

Bilinear forms. Let $A$ and $B$ be rings, $V$ a left $A$-module, $W$ a right $B$-module, and $F$ an $(A,B)$-bimodule.
A bilinear form is a function $⟨,⟩:V\times W\to F$ such that

(a)   If $v_1,v_2\in V$ and $w\in W$ then $⟨v_1+v_2,w⟩=⟨v_1,w⟩+⟨v_2,w⟩$,

(b)   If $v\in V$ and $w_1,w_2\in W$ then $⟨v,w_1+w_2⟩=⟨v,w_1⟩+⟨v,w_2⟩$,

(c)   If $a\in A$, $v\in V$ and $w\in W$ then $⟨av,w⟩=a⟨v,w⟩$,

(d)   If $b\in B$, $v\in V$ and $w\in W$ then $⟨v,wb⟩=⟨v,w⟩b$.

Sesquilinear forms. Let $A$ and $B$ be rings and let $\bar{\phantom{P}}:B\to B$ be an antiautomorphism so that

if   $b_1,b_2\in B$      then    $\overline{b_1{b}_2}=\overline{b_2}\overline{b_1}$.

Let $V$ be a left $A$-module, $W$ a left $B$-module, and $F$ an $(A,B)$-bimodule.
A sesquilinear form is a function $⟨,⟩:V\times W\to F$ such that

(a)   If $v_1,v_2\in V$ and $w\in W$ then $⟨v_1+v_2,w⟩=⟨v_1,w⟩+⟨v_2,w⟩$,

(b)   If $v\in V$ and $w_1,w_2\in W$ then $⟨v,w_1+w_2⟩=⟨v,w_1⟩+⟨v,w_2⟩$,

(c)   If $a\in A$, $v\in V$ and $w\in W$ then $⟨av,w⟩=a⟨v,w⟩$,

(d)   If $b\in B$, $v\in V$ and $w\in W$ then $⟨v,bw⟩=⟨v,w⟩\bar{b}$.

Quadratic forms. Let $A$ be a commutative ring and let $V$ be an $A$-module.
A quadratic form is a function ${\Vert \phantom{x}\Vert }^2:V\to A$ such that

(a)   If $a\in A$ and $v\in V$ then ${\Vert av\Vert }^2=a^2{\Vert v\Vert }^2$,

(b)   The map $⟨,⟩:V\times V\to A$ given by

$⟨v_1,v_2⟩={\Vert v_1+v_2\Vert }^2-{\Vert v_1\Vert }^2-{\Vert v_2\Vert }^2$

is a bilinear form.

Sesquilinear to bilinear. If $B$ is commutative then a bilinear form is a sesquilinear form with $\bar{\phantom{P}}=\mathrm{id}$.
Let $⟨,⟩:V\times W\to F$ be a sesquilinear form. Define a right $B$-module by setting

$W^{\sigma }$   to be   $W$      with     $wb={\sigma }^{-1}\left(b\right)w,\text{for}b\in B,w\in W$.

Then

$⟨,⟩:V\times W^{\sigma }\to F\text{given by}⟨v,w⟩=⟨v,w⟩$

is a bilinear form since $⟨v,wb⟩=⟨v,{\sigma }^{-1}\left(b\right)w⟩=⟨v,w⟩\sigma {\sigma }^{-1}b=⟨v,w⟩b$. This construction converts a sesquilinear form to a bilinear form.

Matrix of a form. Let $⟨,⟩:V\times W\to F$ be a sesquilinear form. Assume that $(e_1,e_2,\dots ,e_n)$ is a basis of $V$ and $(f_1,f_2,\dots ,f_n)$ is a basis of $W$. The matrix of $⟨,⟩$ with respect to the bases $(e_1,e_2,\dots ,e_n)$ and $(f_1,f_2,\dots ,f_n)$ is

$\left(⟨e_i,f_j⟩\right)$,     an element of    $M_{n\times m}\left(F\right)$.

Orthogonals. Let $⟨,⟩:V\times W\to F$ be a sesquilinear form. Let $N$ be a submodule of $W$. The orthogonal to $N$ is

$N^{\perp }=\{v\in V|\text{if}n\in N\text{then}⟨v,n⟩=0\}$.

Rank and nondegeneracy. Let $⟨,⟩:V\times W\to F$ be a bilinear form. Define

$\begin{array}{rccl}s:& V& ⟶& {\mathrm{Hom}}_B(W,F)\\ & v& ⟼& \begin{array}{rccl}⟨v,\cdot ⟩:& W& \to & F\\ & w& \mapsto & ⟨v,w⟩\end{array}\\ \begin{array}{}\end{array}\end{array}\text{and}\begin{array}{rccl}d:& W& ⟶& {\mathrm{Hom}}_A(V,F)\\ & w& ⟼& \begin{array}{rccl}⟨\cdot ,w⟩:& V& \to & F\\ & v& \mapsto & ⟨v,w⟩\end{array}\\ \begin{array}{}\end{array}\end{array}$

satisfy

$s\left(av\right)=as\left(v\right)\text{and}d\left(wb\right)=d\left(w\right)b$.

The form $⟨,⟩:V\times W\to F$ is nondegenerate if both $s$ and $d$ are injective.

If $A=B=F$ is a field then the rank of $⟨,⟩:V\times W\to F$ is

$\mathrm{rk}s=\dim \left(\mathrm{im}s\right)=\dim (V/W^{\perp })=\dim (W/V^{\perp })=\dim \left(\mathrm{im}d\right)=\mathrm{rk}d$.

Adjoints. Let $A=B=F$ be a ring. Let ${⟨,⟩}_1:V_1\times W_1\to F$ and ${⟨,⟩}_2:V_2\times W_2\to F$ be nondegenerate sesquilinear forms. The left adjoint of $S:V_1\to V_2$ is $S^{*}:W_2\to W_1$ given by

${⟨S\left(v_1\right),w_2⟩}_2={⟨v_1,S^{*}\left(w_2\right)⟩}_1$.

The right adjoint of $T:W_1\to W_2$ is $T^{*}:V_2\to V_1$ given by

${⟨v_2,T\left(w_1\right)⟩}_2={⟨T^{*}\left(v_2\right),w_1⟩}_1$.

Discriminants

Let $A=B=F$ be a commutative ring. Let ${⟨,⟩}_1:V_1\times W_1\to F,\dots ,{⟨,⟩}_m:V_m\times W_m\to F$ be sesquilinear forms. The product form is the sesquilinear form

$⟨,⟩:(V_1\otimes \cdots \otimes V_m)\times (W_1\otimes \cdots \otimes W_m)\to F$     given by     $⟨v_1\otimes \cdots \otimes v_m,w_1\otimes \cdots \otimes w_m⟩={⟨v_1,w_1⟩}_1\cdots {⟨v_m,w_m⟩}_m$.

Let $A=B=F$ be a commutative ring. Let $⟨,⟩:V\times W\to F$ be a sesquilinear form. The $m$th exterior product form is the sesquilinear form

$⟨,⟩:{\Lambda }^{m}V\times {\Lambda }^{m}W\to F$     given by     $⟨v_1\wedge \cdots \wedge v_m,w_1\wedge \cdots \wedge w_m⟩=\det \left(⟨v_i,w_j⟩\right)$.

Let $A=B=F$ be a commutative ring. Let $V$ be a free $𝔽$-module of dimension $n$ and let $⟨,⟩:V\times V\to F$ be a sesquilinear form. The discriminant of $⟨,⟩$ is

${\Vert \phantom{P}\Vert }^2:{\Lambda }^{n}V\to F$     given by     ${\Vert v_1\wedge \cdots \wedge v_n\Vert }^2=⟨v_1\wedge \cdots \wedge v_n,v_1\wedge \cdots \wedge v_n⟩=\det \left(⟨v_i,w_j⟩\right)$.

Let $⟨,⟩:V\times V\to F$ be a sesquilinear form. Let $\{e_1,\dots ,e_n\}$ be a basis of $V$. The following are equivalent:

(a)   $s:V\to V^{*}$ is bijective.

(b)   $d:V\to V^{*}$ is bijective.

(c)   ${\Vert e_1\wedge \cdots \wedge e_n\Vert }^2$ is invertible in $F$.

Hermitian, symmetric and skew-symmetric forms

Let $A=B=F$ be a ring and let $\bar{\phantom{P}}:F\to F$ be an involutive antiautomorphism so that

if   $b,b_1,b_2\in B$      then    $\overline{b_1{b}_2}=\overline{b_2}\overline{b_1},\text{and}\overline{\left(\stackrel{‾}{b}\right)}=b$.

Let $\epsilon \in Z\left(F\right)$. An $\epsilon$-Hermitian form is a sesquilinear form $⟨,⟩:V\times V\to F$ such that

if    $v_1,v_2\in V$    then    $⟨v_2,v_1⟩=\epsilon \overline{⟨v_1,v_2⟩}$.

A Hermitian form is a sesquilinear form $⟨,⟩:V\times V\to F$ such that $\epsilon =1$ so that

if $v_1,v_2\in V$ then $⟨v_2,v_1⟩=\overline{⟨v_1,v_2⟩}$.

A symmetric form is a sesquilinear form $⟨,⟩:V\times V\to F$ such that $\bar{\phantom{P}}=\mathrm{id}$ and $\epsilon =1$ so that

if $v_1,v_2\in V$ then $⟨v_2,v_1⟩=⟨v_1,v_2⟩$.

A skew-symmetric form is a sesquilinear form $⟨,⟩:V\times V\to F$ such that $\bar{\phantom{P}}=\mathrm{id}$ and $\epsilon =-1$ so that

if $v_1,v_2\in V$ then $⟨v_2,v_1⟩=-⟨v_1,v_2⟩$.

PUT Gram-Schmidt and principal minors here, following [Bou, Ch. 9 §6 Prop. 1].

Positive Hermitian forms

Let $𝔽$ be a maximal ordered field and let

(a)   $A=𝔽$ and $\bar{\phantom{P}}=\mathrm{id}$,

(b)   $A=𝔽\left[i\right]$ with $i^2=-1$ and $\bar{\phantom{P}}$ the conjugation antiautomorphism,

(c)   $A=𝔽[i,j,k]$ a quaternion algebra over $𝔽$ and $\bar{\phantom{P}}$ the conjugation antiautomorphism,

A positive Hermitian form is a Hermitian form $⟨,⟩:V\times V\to A$ such that

if $v\in V$ then $⟨v,v⟩\in F_{\ge 0}$.

(Note that $⟨v,v⟩\in F$ since $⟨v,v⟩=\overline{⟨v,v⟩}$.)

Assume $A=𝔽$ or $A=𝔽\left[i\right]$, $E$ is finite dimensional and $⟨,⟩:V\times V\to A$ is a positive nondegenerate Hermitian form. Let $k\in {\mathbb{Z}}_{>0}$. Then

(a)  

$⟨,⟩:V^{\otimes k}\times V^{\otimes k}\to A\text{and}⟨,⟩:{\Lambda }^{k}V\times {\Lambda }^{k}W\to A$

are positive nondegenerate.

(b)   If $x,y\in V$ then

$⟨x,y⟩\overline{⟨x,y⟩}\le ⟨x,x⟩⟨y,y⟩$.

(c)   If $A$ is commutative then the principal minors of the matrix of $⟨,⟩$ are positive.

Notes and References

These notes follow Bourbaki [Bou]. In particular the definitions of adjoints follows [Bou, Ch. 9 §1 no. 8], the definitions of product forms follows [Bou, Ch. 9. §1 no. 9], the definition of discriminants follows [Bou, Ch. 9 §2], the definitions of Hermitian forms follows [Bou, Ch. 9 §3] the definitions of positive Hemitian forms follows [Bou, Ch. 9 §7 no. 1]. WARNING: These statements have not been checked carefully, and a careful check, probably a lecture presentation needs to be done to check everything thoroughly. The proof of the Cauchy-Schwartz identity appears in [Bourbaki, Topological Vector Spaces Ch. V §2 Prop. 2] with the same proof as in [Ru]. The proof here comes from [Bou], where it is pointed out that the restriction of a positive form, to the two dimensional space spanned by $x$ and $y$ is positive. A third proof is via Lagrange's identity, found in [Bou, ???], and in the complex case, in [Ah, §1.4 Ex. 5]. In the real case Lagrange's identity and the Schwarz inequality are

$0\le \frac{1}{2}\sum _{i,j=1}^n{(x_i{y}_j-x_j{y}_i)}^2={\Vert x\Vert }^2{\Vert y\Vert }^2-{⟨x,y⟩}^2$

and, in the complex case,

${\left|\sum _{i=1}^n{a}_i{b}_i\right|}^2=\sum _{i=1}^n{\left|a_i\right|}^2\sum _{i=1}^n{\left|b_i\right|}^2-\sum _{1\le i<j\le n}{|a_i\overline{b_j}-a_j\overline{b_i}|}^2$.

SHOULDN'T THERE BE A bar ON THE $b_i$ on the LHS? CHECK THIS. NO, there isn't a bar in Ahlfors, or in any of the identities in [Ah, § 1.5].

References

[Ah] Lars V. Ahlfors, Complex analysis. An introduction to the theory of analytic functions of one complex variable, Third edition, International Series in Pure and Applied Mathematics. McGraw-Hill Book Co., New York, 1978. xi+331 pp. ISBN: 0-07-000657-1, 30-01 MR0510197.

[Bou] N. Bourbaki, Algèbre, Chapitre 9: Formes sesquilinéaires et formes quadratiques, Actualités Sci. Ind. no. 1272 Hermann, Paris, 1959, 211 pp. MR0107661.

[Ru] W. Rudin, Real and complex analysis, Third edition, McGraw-Hill, 1987. MR0924157.

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