In this entry, unless otherwise specified, $R$ is a commutative ring with multiplicative identity $1$ and $M=R[X_1,\ldots,X_n]$ is a polynomial ring over $R$ in $n$ indeterminates.

Definition

A homogeneous polynomial of degree 2 in $M$ is called a quadratic form (over $R$ ) in $n$ indeterminates. In general, a quadratic form (without specifying $n$ ) over a ring $R$ is a quadratic form in some polynomial ring over $R$ .

For example, in $\mathbb{Z}[X,Y]$ , $X^2-5XY$ is a quadratic form, while $Y^3+2XY$ and $X^2+Y^2+1$ are not.

In general, a quadratic form $Q$ in $n$ -indeterminates looks like $$Q=a_{11}X_1^2+a_{12}X_1X_2+\cdots+a_{n,n-1}X_nX_{n-1}+a_{nn}X_n^2=\sum_{1\le i,j \le n}a_{ij}X_iX_j$$ where $a_{ij}\in R$ .

Letting $\bX=(X_1,\ldots,X_n)^{\mathrm{T}}$ , and $\bA=\lbrace a_{ij} \rbrace$ the $n\times n$ matrix, then we can rewrite $Q$ as $$Q={\bX}^{\mathrm{T}} \bA \bX.$$

For example, the quadratic form $X^2-5XY$ can be rewritten as $$X^2-5XY=\begin{pmatrix} X & Y \end{pmatrix}\begin{pmatrix} 1 & -2 \\ -3 & 0 \end{pmatrix}\begin{pmatrix} X \\ Y \end{pmatrix}.$$

Now suppose the characteristic of $R$ , $\operatorname{char}(R)\ne 2$ . In fact, suppose that $2$ is invertible in $R$ (its inverse denoted by $\frac{1}{2}$ ). Since $X_iX_j=X_jX_i$ , define $b_{ij}=\frac{1}{2}(a_{ij}+a_{ji})$ . Then $b_{ii}=a_{ii}$ and $b_{ij}=b_{ji}$ . Furthermore, if $\bB=\lbrace b_{ij}\rbrace$ , then $\bB$ is a symmetric matrix and $$Q={\bX}^{\mathrm{T}} \bB \bX.$$

Again, in the example of $X^2-5XY$ , over $\mathbb{Q}$ it can be written as $$X^2-5XY=\begin{pmatrix} X & Y \end{pmatrix}\begin{pmatrix} 1 & -\frac{5}{2} \\ -\frac{5}{2} & 0 \end{pmatrix}\begin{pmatrix} X \\ Y \end{pmatrix}.$$ However, it is not possible to represent $X^2-5XY$ over $\mathbb{Z}$ by a symmetric matrix.

Evaluating a Quadratic Form

It is not hard to see that, given a quadratic form $Q$ in $n$ indeterminates, setting one of its indeterminates to $0$ gives us another quadratic form, in $(n-1)$ indeterminates. This is an informal way of saying the following:

embed $R$ into $N=R[X_1,\ldots,X_{n-1}]$ . Let $\phi:M\to N$ be the (unique) evaluation homomorphism of the embedding, with $\phi(X_i)=X_i$ for $i<n$ and $\phi(X_n)=0$ . Then for any quadratic form $Q\in M$ , $\phi(Q)$ is a quadratic form in $N$ .

In particular, if we take $N=R$ , and $\bs=(s_1,\ldots,s_n)$ with $s_i\in R$ . Then the evaluation homomorphism $\phi$ at $\bs$ for any quadratic form $Q\in M$ is called the evaluation of $Q$ at $\bs$ , and we write it $\phi_{\bs}(Q)$ , or simply $Q(\bs)$ (since $\phi$ is uniquely determined by $\bs$ ). In this way, a quadratic form $Q$ can be realized as a quadratic map, as follows:

Let $Q\in M$ be a qudratic form. Take the direct sum of $n$ copies of $R$ and call this $V$ . Define a map $q:V\to R$ by $q(v)=Q(v)$ . Then $q$ is a quadratic map.

Conversely, if $2$ is invertible in $R$ (so that $\operatorname{char}(R)\ne 2$ is clear), then given a quadratic map $q:M\to R$ , one can find a corresponding quadratic form $Q\in M$ such that $q(v)=Q(v)$ , by setting $$a_{ij}=\frac{1}{2}\big(q(e_i+e_j)-q(e_i)-q(e_j)\big),$$ where $e_i$ and $e_j$ are coordinate vectors whose coordinates are all $0$ except at positions $i$ and $j$ respectively, where the coordinates are $1$ . Then $Q$ defined by ${\bX}^{\mathrm{T}} \bA \bX$ , where $\bA=\lbrace a_{ij}\rbrace$ is the desired quadratic form.

Equivalence of Quadratic Forms

From the above discussion, we shall identify a quadratic form as a quadratic map.

Two quadratic forms $Q_1$ and $Q_2$ are said to be equivalent if there is a non-singular matrix $M$ such that $Q_1(v)=Q_2(Mv)$ , for all $v\in R^n$ . The definition of equivalent quadratic forms is well-defined and it is not hard to see that this equivalence is an equivalence relation.

In fact, if $\bA_1$ and $\bA_2$ are matrices corresponding to (see the definition section) the two equivalent quadratic forms $Q_1$ and $Q_2$ above, then $\bA_1=M^\mathrm{T}\bA_2M$ .

For example, the quadratic form $X^2-Y^2$ is equivalent to $XY$ over any ring $R$ where $2$ is invertible, with $M=\begin{pmatrix} 1 & -1 \\ 1 & 1 \end{pmatrix}$ .

In the case where $R=\mathbb{R}$ is the field of real numbers (or any formally real field), we say that a quadratic form is positive definite, negative definite, or positive semidefinite according to whether its corresponding matrix is positive definite, negative definite, or positive semidefinite. The definiteness of a quadratic form is preserved under the equivalence relation on quadratic forms.

Sums of Quadratic Forms

If $Q_1$ and $Q_2$ are two quadratic forms in $m$ and $n$ indeterminates. We can define a quadratic form $Q$ in $m+n$ indeterminates in terms of $Q_1$ and $Q_2$ , called the sum of $Q_1$ and $Q_2$ , as follows:

write $Q_1={\bX}^{\mathrm{T}} \bA \bX$ and $Q_2={\bY}^{\mathrm{T}} \bB \bY$ , with $\bX=(X_1,\ldots,X_m)^{\mathrm{T}}$ and $\bY=(Y_1,\ldots,Y_n)^{\mathrm{T}}$ . Then $$Q:={\bZ}^{\mathrm{T}} (\bA\oplus \bB) \bZ,$$ where $\bZ=(\bX,\bY)=(X_1,\ldots,X_m,Y_1,\ldots,Y_n)^{\mathrm{T}}$ , and $\bA\oplus \bB$ is the direct sum of matrices $\bA$ and $\bB$ .

Bibliography

1

T. Y. Lam, Introduction to Quadratic Forms over Fields, American Mathematical Society (2004)