Classification of Quardratic Forms over \(\mathbb Q\)
First Attempts and General Approachs
Foreword
References relied on heavily:
J.P. Serre “A Course in Arithmetic” [1]
Shiva Chidambaram, MIT18.782 Introduction to Arithmetic Geometry (Spring 2023) Lecture Notes
Arushi Gupta, Participant Papers of The University of Chicago Mathematics REU 2018, The p-adic Integers, Analytically and Algebraically
May serve as a guidance of the first part of [1]
Assume familarity of quadratic residues and basic knowledge of \(p\)-adic numbers
Skip most of the proofs
Apology in advance for potential mistakes
Notations
We denote by \(K\) an arbitrary field. All fields are assumed to be of characteristic \(\neq 2\).
\(\nu_p : \mathbb Q_p \to \mathbb Z\) being the \(p\)-adic valuation.
\(\left(\dfrac a p \right)\) being the Legendre symbol. \(a\) is understanded as \(p^{-\nu_p(a)} a \bmod p\) if \(a \in \mathbb Q_p\). Define this similarly in \(\mathbb F_q\).
Let \(f \oplus g\) denote the direct sum of two quadratic forms \(f\) and \(g\).
Example: Quadratic Forms over \(\mathbb R\) and \(\mathbb F_q\)
Review: Quadratic Forms over \(\mathbb R\)
A quadratic form \(f: V \to K\) may be identified by a symmetric matrix \(A \in M_n(K)\) by \(f(v) = v^\mathrm{T}A v\).
Their equivalence is defined by congruence: \(A \sim B \iff A = Q^\mathrm{T}B Q\).
Real symmetric matrices may be diagonalized orthogonally.
Scale each eigenvalue by multiplying a square. Only their sign matters.
- the rank \(n\), an invariant
- the signature \((r,s) := (\# \text{positive eigenvalues}, \# \text{negative eigenvalues})\).
Same rank and signature implies the equivalence.
Sylvester’s law of inertia: signature is also an invariant.
Some Refinement
On an abitrary field \(K\):
All symmetric matrix is equivalent to a diagonal one.
- Pick a non-isotropic vector \(v\) (exists when the form is nonzero), its orthogonal complement is a hyperplane and does not include \(v\). Change basis and do the induction.
The rank is always a invariant. We may (and we shall always) reduce to classify the non-degenerate quadratic forms of rank \(n\).
The squares \((K^\times)^2\) give us the ability to scale. Knowledge of distribution of diagonal elements in \(K^\times / (K^\times)^2\) suffices to show the equivalence1.
- \(\mathbb C^\times / (\mathbb C^\times)^2 \cong \{1\}\), suffices to classify by the rank.
- \(\mathbb R^\times / (\mathbb R^\times)^2 \cong \{1,-1\}\), signature is also needed.
- \(\mathbb F_q^\times / (\mathbb F_q^\times)^2 \cong \{1,a\}\), where \(a \in \mathbb F_q\) is a quadratic nonresidue.
- For \(\mathbb Q_p\) and \(\mathbb Q\)?
1 Though working in the refined structure \(\{0\} \cup K^\times / (K^\times)^2\) is probably a better idea if one wish to deal with the degenerate case in a uniform manner.
Another Example: Quadratic Forms over \(\mathbb F_q\)
We classify the non-degenerate quadratic forms of rank \(n\).
Refined signature: counting nonzero quadratic residues and nonresidues. It may serve as a sufficient criterion of the equivalence.
But it’s not an invariant. \(a X^2 + a Y^2 \sim X^2 + Y^2\) over \(\mathbb F_q\).
Do a change of basis \(X=sU+tV\) and \(Y=tU-sV\). If we require \(a U^2 + a V^2 = X^2 + Y^2\), then \(s^2+t^2=a\).
It always has a nonzero solution in \(\mathbb F_q\): \(s^2\) and \(a - t^2\) have both \((q+1)/2\) possible values, thus must reach a common value.
The discriminant \(d := \left( \dfrac{\det(A)}{q} \right) \in \mathbb F_q^\times / (\mathbb F_q^\times)^2\) is an invariant and reveals the parity of the signature. It classifies the non-degenerate quadratic forms over \(\mathbb F_q\).
Insight: Existence of nonzero solutions of the equation \(a X^2 + b Y^2 = Z^2\) in \(K\) seems to be of great importantance.
Quadratic Spaces
Quadratic Spaces
The structure of a quadratic space, i.e. vector space equipped with a symmetric bilinear form, is much more subtle than its positive-definite counterpart over \(\mathbb R\) or \(\mathbb C\). For example, for a non-degenerate quadratic space \(V\) and a subspace \(U\) of \(V\) ([1] p. 28, chap. 4, sec. 1.2):
\(U \cap U^\perp = \operatorname{rad}(U)\), \(\dim U + \dim U^\perp = \dim V\), \((U^\perp)^\perp = U\)
\(U \oplus U^\perp = V\) iff \(U + U^\perp = V\) iff \(\operatorname{rad}(U) = 0\)
It’s much harder to show that an orthogonal basis of \(U\) expands to an orthogonal basis of \(V\).
Structure of Quadratic Spaces
We mention some results here without details.
The Common Represented Element Method
Another Invariant: The Range
On an abitrary field \(K\), we say that a quadratic form \(f\) represents \(a \in K\) if there exists a nonzero \(v \in V\) such that \(f(v) = a\).
The range of \(f\), \(\operatorname{Im}f\), is an invariant.
It may be viewed in \(\{0\} \cup K^\times / (K^{\times})^2\).
Is it complete?
Insights from the Range
- Insight from line 3: To understand the range, it suffices to examine when a quadratic form represents \(0\).
- Insight from line 2 (the common represented element method): Say \(f_1\), \(f_2\) are nozero and represent a common \(a \in K^\times\). Reducing \(Z \mapsto aZ^2\), if only \(g_1\) and \(g_2\) also share a common represented element…
Insights from the Range
Sadly, range is not always a complete invariant.
- Otherwise all indefinite quadratic forms over \(\mathbb R\) are equivalent, absurd.
But we shall show that when \(K = \mathbb Q_p\) and moreover \(K = \mathbb Q\), it plays a subtle role in the classification of quadratic forms. This requires a more precise characterization of the range.
In fact, if only there are some simple invariants that can fully determines the range…
Global and Local Equivalence
Global and Local Equivalence
Fact: Field extensions preserve the equivalence of quadratic forms.
- Example: Equivalence classes are finer over \(\mathbb R\) than those over \(\mathbb C\).
\(\mathbb Q \hookrightarrow \mathbb R\), thus the rank and the signature are invariants. But we need more information to classify.
Other field extension of \(\mathbb Q\)? \(\mathbb Q \hookrightarrow \mathbb Q_p\)
- To gain more invariants for \(\mathbb Q\) (especially those related to the range), let’s classify quadratic forms over \(\mathbb Q_p\) first.
Quadratic Forms over \(\mathbb Q_p\) and \(\mathbb Q\)
Structure of \(\mathbb Q_p^\times / (\mathbb Q_p^\times)^2\)
Structure of \(\mathbb Q_p^\times\)
\(\mathbb Q_p^\times \cong \mathbb Z \times \mathbb Z_p^\times\) by collecting common powers of \(p\)
\(\mathbb Z_p^\times \cong \mathbb F_p^\times \times (1 + p \mathbb Z_p)\) by \(a \mapsto a \bmod p\)
- It splits by the explicit construction of a primitive root of order \(p\), via Hensel’s lemma / Teichmüller lift \(\lim_{n \to \infty} g^{p^n}\), where \(g\) is a primitive root of \(\mathbb F_p^\times\).
Structure of \(1 + p \mathbb Z_p\) and the log / exp map
For \(p \neq 2,\, \alpha \geq 1\) or \(p = 2,\, \alpha \geq 2\): \[ \begin{aligned} 1 + p^\alpha \mathbb Z_p &\cong (p^\alpha \mathbb Z_p, +) \cong (\mathbb Z_p, +) \\ 1 + p^\alpha a &\mapsto \log(1+p^\alpha a) \end{aligned} \]
For \(p=2,\,\alpha = 1\),
\[ 1 + 2 \mathbb Z_2 \cong \mathbb Z / 2 \mathbb Z \times (1 + 4 \mathbb Z_2) \]
by \(1 + 2a \mapsto a \bmod 2\)
It splits by the explicit construction of a primitive root of order \(2\): \((-1,-1,\dots) = \sum_{n=0}^{+\infty} 2^n\).
\((1 + 4 \mathbb Z_2) \cong (4 \mathbb Z_2, +) \cong (\mathbb Z_2, +)\) by the log map
Thus \[ 1 + 2 \mathbb Z_2 \cong \mathbb Z / 2 \mathbb Z \times (\mathbb Z_2, +) \]
Quadratic residues of \(\mathbb Q_p\)
For \(p \neq 2\):
\(\mathbb Q_p^\times \cong \mathbb Z \times \mathbb F_p^\times \times (\mathbb Z_p, +)\)
\(2\) is a unit in \(\mathbb Z_p\). Thus \(a \in (\mathbb Q_p^\times)^2\) iff \(\nu_p(a) \bmod 2 = 0\) and \(a \bmod p \in \mathbb F_p^\times\) is a quadratic residue.
\(\mathbb Q_p^\times / (\mathbb Q_p^\times)^2 \cong \mathbb Z / 2\mathbb Z \times \mathbb Z / 2\mathbb Z\), generated by \(p\) and \(a\), where \(a \bmod p\) is a quadratic nonresidue．
For \(p = 2\):
\(\mathbb Q_2^\times \cong \mathbb Z \times \mathbb Z / 2 \mathbb Z \times (\mathbb Z_2, +)\)
Quadratic residues of \((\mathbb Z_2, +)\) are \((2 \mathbb Z_2, +)\), which pulls back to \(1 + 8 \mathbb Z_2\).
\(a \in (\mathbb Q_2^\times)^2\) iff \(\nu_2(a) \bmod 2 = 0\) and \(a \bmod 8 \equiv 1\).
\(\mathbb Q_2^\times / (\mathbb Q_2^\times)^2 \cong \mathbb Z / 2\mathbb Z \times \mathbb Z / 2\mathbb Z \times \mathbb Z / 2\mathbb Z\), generated by \(2\), \(3\) and \(5\).
The Hilbert Symbol
The Hilbert Symbol
The Hilbert symbol over \(\mathbb Q_p\) is defined as: \[ \langle a, b \rangle := \begin{cases} 1 & \text{if } a X^2 + b Y^2 = Z^2 \text{ has a nonzero solution in } \mathbb Q_p \\ -1 & \text{otherwise} \end{cases} \] The symbol may be also viewed in \(\{0\} \cup \mathbb Q_p^\times / (\mathbb Q_p^\times)^2\) or even more simply in \(\mathbb Q_p^\times / (\mathbb Q_p^\times)^2\) when working with non-degenerate forms.2
2 Lots of the resources, even [1], switch between these three views without enough warning. Sadly we shall also commit this ususal mild sin (and have already done to other innocent invariants such as the discriminant…)
Properties of the Hilbert Symbol
\(\langle a, -a \rangle = 1\)
\(\langle a, b \rangle = \langle b, a \rangle\) (symmetric)
If \(\langle a_2,b \rangle = 1\), then \(\langle a_1 a_2, b \rangle = \langle a_1,b \rangle\)
- In fact, \(\langle a_1 a_2, b \rangle = \langle a_1, b \rangle \langle a_2, b \rangle\) (multiplicatively bilinear)
\(\langle a, b \rangle = 1\) for all \(b\) iff \(a \in \mathbb Q_p^2\) (nondegenerate)
the Hilbert symbol is a non-degenerate symmetric bilinear form of the \(\mathbb F_2\)-vector space \(\mathbb Q_p^\times / (\mathbb Q_p^\times)^2\)
This is a non-trivial result and is said to be, to some extent, a generalization of the law of quadratic reciprocity in local class field theory.
To show above over \(\mathbb Q_p\), we develop an explicit formula for the Hilbert symbol.
The Explicit Formula of the Hilbert Symbol
| \(0\) | \(1\) | \(a\) | \(p\) | |
|---|---|---|---|---|
| \(0\) | \(1\) | \(1\) | \(1\) | \(1\) |
| \(1\) | \(1\) | \(1\) | \(1\) | |
| \(a\) | \(1\) | \(-1\) | ||
| \(p\) | \((-1)^{\frac{p-1}{2}}\) |
Invariants that Determines the Range
The Hasse Invariant
Recall that we have reduced to work with non-degenerate diagonalized quadratic forms of rank \(n\). Recall that the discriminant \[ d(f) = a_1 a_2 \dots a_n \in \mathbb Q_p^\times / (\mathbb Q_p^\times)^2 \] is an invariant.
Define the Hasse invariant \(\varepsilon(f) := \prod_{1 \leq i < j \leq n} \langle a_i, a_j \rangle\)
It is an invariant: \[ \varepsilon(f) = \prod_{1 \leq i < j \leq n} \langle a_i, a_j \rangle = \varepsilon(f_1) \prod_{2 \leq j \leq n} \langle a_1, a_j \rangle = \varepsilon(f_1) \cdot \langle a_1, a_1 d(f) \rangle \] Thus \(\varepsilon\) is preserved under contiguous change of orthogonal bases (fixes one of the vector of the basis)
- For \(n \geq 3\), orthorgonal bases are transitive under contiguous change ([1] p. 30, sec. 4.1.4, theorem 2)
\(d\) and \(\varepsilon\) Determine the Range
Recall that \(f\) represents \(a \in \mathbb Q_p^\times / (\mathbb Q_p^\times)^2\) iff \(f \oplus (Z \mapsto -a Z^2)\) represents \(0\), thus above fully characterizes the range.
Classification
Classification of Quadratic Forms over \(\mathbb Q_p\)
\(f,g\) have same \(d\) and \(\varepsilon\), thus have the same range. Say they both represent \(a \in \mathbb Q_p^\times / (\mathbb Q_p^\times)^2\).
Then \(f \sim f_1 \oplus (Z \mapsto a Z^2)\), where \(f_1\) is of rank \(n-1\).
\(d\) and \(\varepsilon\) of \(f_1\) can be determined:
- \(d(f_1) = a d(f)\)
- \(\varepsilon(f_1) = \varepsilon(f) \cdot (a, a d(f))\) (shown when discussing the invariance of \(\varepsilon\))
The same for \(g\). Thus \(f_1, g_1\) shares the same \(d\) and \(\varepsilon\) (thus also their range). QED by induction.
Classification of Quadratic Forms over \(\mathbb Q\)
Say \(f,g\) are equivalent over each local field (\(\mathbb Q_p\) and \(\mathbb R\)), thus they share the same range locally.
By Hasse-Minkowski theorem, they also share the same range globally over \(\mathbb Q\).
Then \(f \sim f_1 \oplus (Z \mapsto a Z^2)\) globally, where \(f_1\) is of rank \(n-1\). The same for \(g\).
\(f_1 \sim g_1\) locally by Witt’s cancellation theorem. QED by induction.
Problem Remains
Proof of the Hasse-Minkowski theorem
- essentially needs some understanding of the global property of the Hilbert symbol, which we have not discussed (cf. [1])
Refine the theory for degenerate quadratic forms (relatively easy)
Enumerate all the equivalence classes of quadratic forms over \(\mathbb Q_p\) and \(\mathbb Q\) (cf. [1])
To what extent can we use the common represented element method to classify quadratic forms over other fields?
For which fields, the range of a quadratic form is a complete invariant? (At least \(\mathbb R\) fails. \(\mathbb Q\), \(\mathbb Q_p\)?)
What can we say about \(\mathbb Q^\times / (\mathbb Q^\times)^2\)?
Classification of quadratic forms over commutative rings (e.g. \(\mathbb Z\), \(\mathbb Z / m \mathbb Z\))