Three easy ways to distinguish quadratic forms: Orthogonal Basis, Discriminant and number of Representations.

 When talking about representations of quadratic forms, we need to specify its domain. 

Example. Question: can the form \(q(x) = x^2 \) represents \(-1\)? 

Answer. It depends, over \(\mathbb{C}\) yes, but not over \( \mathbb{R}\). 

This motivates us to change our viewpoint from studying quadratic forms as just some polynomial expressions to studying \(R\)-modules equipped with a quadratic form.  We call such modules quadratic \(R\)-modules. Two such modules \(M, N\) are called isometric if there exist a bijection \( \varphi: M \to N\) that preserves the quadratic form, i.e. \( q_M(x, y) = q_n(\varphi(x), \varphi(y) \) for all \(x, y \in M \). 

Below we discuss three elementary ways to check if two quadratic modules are isometric. 

1. Orthogonal Basis 

A useful tool in studying quadratic \(R\)-modules is its orthogonal basis. Every quadratic form is induced from a unique bilinear form B(x,y). Using B(x, y), we can define the notion of orthogonality in an obvious way. The question is: Does every quadratic \(R\)-module have an orthogonal basis?

Answer: No. Remember in Linear Algebra, we produce orthogonal basis via Gram-Schmidt, which involves division by norms of vectors. So to be able to reproduce Gram-Schmidt, we need to be able to divide by elements of \(R\)- i.e. we want \(R\) to be a field. Indeed we have the following theorem.

Theorem. If V is a quadratic \(R\)-module over a field \(R = F\) (in which case we call it a quadratic \(F\)-space), then V has an orthogonal basis \(\{e_1, \ldots, e_n\}\). If \(B(e_i, e_i) = \alpha_i \), then we write \(V \cong \langle \alpha_1, \ldots, \alpha_n \rangle \) in \( \{e_1, \ldots, e_n\}\). 

Proof. The proof essentially use  Gram-Schmidt with one caveat: some non-zero element \(v\)  of \(V\) might have zero "norm", \(q(v) = 0\). (We call such a \(v\) isotropic.)

Now scaling an orthonormal basis gives an orthonormal basis. As \(B(c e_i, c e_i) = c^2 B (e_i, e_i)\) we see that if  \(V \cong \langle \alpha_1, \ldots, \alpha_n \rangle \) in \( \{e_1, \ldots, e_n\}\) then  \(V \cong \langle \alpha_1/c_1^2, \ldots, \alpha_n/c_n^2 \rangle \) in \( \{e_1/c_1, \ldots, e_n/c_n\}\). In other words, the tuples \( \alpha_i \) are only determined up to a square in \(F^*\). 

Example. Are the spaces \( \langle 1, 2\rangle, \langle 2, 2\rangle\) and \( \langle 2, 3\rangle\) isometric over \( \mathbb{R}\)?

Answer: Yes, because by the above discussion, they are all isometric to \( \langle 1, 1\rangle\). For example if \(V \cong \langle 2, 3\rangle \) in \(\{e_1, e_2\}\) then \(V \cong \langle 1, 1\rangle \) in \( \{e_1/\sqrt{2}, e_2/\sqrt{3}\}\). 

More generally, observe that over \(\mathbb{R}\) if \( \alpha_i / \beta_i \geq 0 \) and is thus a square in \( \mathbb{R}*\) then \( \langle \alpha_1, \ldots, \alpha_n \rangle \cong \langle \beta_1, \ldots, \beta_n\rangle \). In particular, every quadratic space over \(\mathbb{R}\) is of the form \( \langle \alpha_1, \ldots, \alpha_m, \beta_1, \ldots, \beta_n\rangle \) for \( \alpha_i \geq 0, \beta_i < 0\). Thus every quadratic space over \( \mathbb{R}\) is isometric to one of form \(\langle 1, \ldots, 1, - 1, \ldots -1 \rangle\). 

Even more generally, we have the following.

Lemma. Let \(F\) be a field. Suppose a full set of coset representatives of \(F^*/(F^*)^2\) is \(\{ \epsilon_1= 1, \ldots, \epsilon_n\}\). Then  every quadratic space over \(F\) is isometric to one of form \[ \langle \epsilon_1 =1, \epsilon_1, \ldots, \epsilon_1, \epsilon_2, \ldots, \epsilon_2, \ldots, \epsilon_n, \ldots, \epsilon_n\rangle. \]

Example. Since every element of \( \mathbb{C}^*\) is a square, every quadratic space over \( \mathbb{C}\) is isometric to one of form \( \langle 1, 1, \ldots, 1 \rangle \).

Example. Let \(F = \mathbb{F}_q\) be a finite field. Since the kernel of the square map on \( F^*\) is \( \{\pm 1\} \), we must have its image, the group of squares \( (F^*)^2\), must be of side \(|F^*|/2\), i.e. it is a subgroup of index \(2\). Thus every quadratic space over a finite field \(F\) is isometric to one of form \(\langle 1, 1, \ldots, 1, \epsilon, \ldots, \epsilon\rangle\) for some nonsquare \(\epsilon \in F^*\).   

 

2. Discriminant

To each quadratic \(R\)-module \( (M, q, B) \) with a fixed basis \( \{v_1, \ldots, v_n\} \) we can associate a Gram matrix \(A = \left(B(v_i, v_j) \right) \). We write \(M \cong A\) in \( \{ v_1, \ldots, v_n \} \). If we change basis then the new matrix \(B\) will be congruent to \(A\), i.e. \(B = T^t A T\) for some invertible matrix \(T\). Thus \(\det B = (\det T)^2 \det A \), so the determinant of the Gram matrix of \(M\) is uniquely determined up to a square in \(R^*\). We define the discriminant of \(M\), denoted \(dM\) to be the image of \(\det A\) in \(R^*/(R^*)^2\).

How does the discriminant help with classification? Well, for one thing, if \(dM \neq dN\), then \(M \not \cong N\). 

Example. The \(\mathbb{Q}\)-modules \(  M \cong \langle 1, 1 \rangle\) and \(N \cong \langle 2, 3 \rangle \) are not isometric over \( \mathbb{Q}\) since  \( dM = 1 \not \equiv 6 = dN \mod (\mathbb{Q}^*)^*\). 

Note that above we showed \(M \cong N\) over \(\mathbb{R}\) and indeed their discriminant differ by a square in \( \mathbb{R}^*\): \(dN = (\sqrt{6})^2 dM \equiv dM\) in \( (\mathbb{R}/\mathbb{R}^*)^2.\)

3. Number of Representations

Example. Are the following \( \mathbb{Z} \)-modules isometric?

\[ M \cong \begin{pmatrix}2 & 1 & 0 \\ 1 & 2 & 1 \\ 0 & 1 & 5 \end{pmatrix}; \]

and

\[ N \cong \begin{pmatrix}2 & 1 & 0 \\ 1 & 3 & 1 \\ 0 & 1 & 3 \end{pmatrix}; \]

Answer. Since \(dM = dN = 13 \) we cannot use discriminant to distinguish them. However we observe that \(M\) represents \(2\) exactly 6 times while \(N\) represents \(2\) only twice. So they cannot be isometric.

Here we have used the following idea.

If the quadratic \(R\)-modules \(M, N\) are isometric, then for every \(r \in R\), we have a bijection \[ \{ m \in M \mid q_M(m) = r\} \leftrightarrow \{ n \in N \mid q_N(n) = r\}. \]