Determinant of Tensors: A Comprehensive Guide

From Classical Matrices to Higher-Order Generalizations and Intrinsic Definitions for Covariant Bilinear Forms¶

Table of Contents¶

1. Introduction

2. The Determinant for Matrices (Rank-2 Endomorphisms)

3. Higher-Order Tensors and the Hyperdeterminant

4. Rank-2 Covariant Tensors: Bilinear Forms of Type (0,2)

5. Basis-Independent Constructions for (0,2) Tensors

6. Weight-2 Densities and Transformation Laws

7. Axiomatic Characterization: Uniquely Deriving the Determinant

8. Special Cases and Geometric Applications

9. Conclusion and Further Directions

10. References and Further Reading

1. Introduction¶

The concept of the determinant is one of the most fundamental in linear algebra. For an $n \times n$ matrix $A$, $\det(A)$ is a scalar that encodes critical information: it tells us whether $A$ is invertible, measures the signed volume scaling factor of the linear transformation represented by $A$, and appears in countless formulas (Cramer’s rule, characteristic polynomial, change of variables in integration, etc.).

When we move from matrices to tensors (multi-linear objects of higher rank or different variance types), the situation becomes more subtle. A matrix can be viewed as a rank-2 tensor of type (1,1) (an endomorphism) or (0,2) (a bilinear form). For higher-order tensors (order $d > 2$), or even for pure covariant rank-2 tensors without additional structure, there is no universally agreed “determinant” that preserves all the familiar properties.

This document provides a complete, self-contained explanation, starting from the classical case and building up rigorously to intrinsic (basis-free) definitions, with special emphasis on:

• The hyperdeterminant for higher-order tensors.

• Why a pure (0,2) covariant tensor does not have a canonical scalar determinant.

• How the natural object lives in a weight-2 density bundle.

• An axiomatic characterization that uniquely determines the determinant (up to scale) assuming only natural properties, including the weight-2 transformation law.

All constructions are presented step-by-step, with explicit formulas, geometric interpretations, and proofs of key properties where feasible.

2. The Determinant for Matrices (Rank-2 Endomorphisms)¶

2.1 Classical Definitions (with Basis)¶

Let $V$ be an $n$-dimensional vector space over a field $K$ (usually $\mathbb{R}$ or $\mathbb{C}$) and let $f: V \to V$ be a linear endomorphism. Fix a basis $\{e_1, \dots, e_n\}$ of $V$. The matrix of $f$ with respect to this basis is the $n \times n$ matrix $A = (a_{ij})$ where

The determinant of $A$ (and thus of $f$) can be defined in several equivalent ways:

Leibniz formula (permutation expansion):

where $S_n$ is the symmetric group of all permutations of $\{1,\dots,n\}$ and $\operatorname{sgn}(\sigma) = +1$ if $\sigma$ is even, -1 if odd.

Levi-Civita symbol expression:

where $\epsilon_{1 2 \dots n} = +1$ and $\epsilon$ is totally antisymmetric.

Cofactor expansion (recursive definition along a row or column) and many other equivalent characterizations exist.

These definitions depend on the choice of basis, but the resulting scalar $\det(A)$ is independent of the basis in the following sense: if we change basis with an invertible matrix $P$, the new matrix is $A' = P^{-1} A P$, and

Thus $\det(f)$ is intrinsically associated to the linear map $f$.

2.2 Geometric Meaning¶

If the columns of $A$ are vectors $v_1, \dots, v_n \in V$, then $|\det(A)|$ equals the signed $n$-dimensional volume of the parallelepiped they span. In particular:

• $\det(A) = 0$ if and only if the vectors are linearly dependent (the parallelepiped has zero volume).

• $\det(A)$ measures the local volume distortion factor of the linear transformation $f$.

2.3 Basis-Free Definition via Exterior Algebra (The Clean Intrinsic Way)¶

This is the most elegant and basis-independent definition. It uses only the universal property of the exterior algebra.

Step 1: The determinant line.
Let $L = \bigwedge^n V$ be the top exterior power of $V$. This is a 1-dimensional vector space (the determinant line). Any nonzero element of $L$ can be thought of as a “volume element” on $V$.

Step 2: Induced map on the determinant line.
Any linear map $f: V \to V$ induces a unique linear map on the exterior algebra

defined on decomposable elements by

and extended by linearity. Because $\bigwedge^n V$ is 1-dimensional, any linear endomorphism of a 1-dimensional space is multiplication by a scalar $\lambda \in K$:

Step 3: Definition of the determinant.
We define $\det(f) := \lambda$. This scalar is independent of any basis and satisfies all the classical properties (multiplicativity $\det(f \circ g) = \det(f)\det(g)$, etc.).

Why this works without a basis:

• The construction uses only the universal property of the exterior algebra (alternating multilinear maps factor uniquely through $\bigwedge$).

• The 1-dimensionality of $L$ (“homogeneity”) guarantees that the induced map is multiplication by a scalar.

• No coordinates or choice of volume form are required because domain and codomain are the same line $L$.

This is the definition that generalizes most naturally and will serve as our model when we discuss (0,2) tensors.

2.4 Key Properties (Summary)¶

• Multiplicativity: $\det(f \circ g) = \det(f) \det(g)$.

• Invertibility: $f$ is invertible $\iff \det(f) \neq 0$.

• Eigenvalues: $\det(f) = \prod \lambda_i$ (product of eigenvalues, counted with multiplicity).

• Characteristic polynomial: $\det(\lambda I - f) = \lambda^n - \operatorname{tr}(f)\lambda^{n-1} + \cdots + (-1)^n \det(f)$.

• Volume interpretation: As described above.

3. Higher-Order Tensors and the Hyperdeterminant¶

3.1 The Problem with Higher Order¶

A tensor of order $d > 2$ (a $d$-linear map, or multidimensional array) does not have a direct analogue of the determinant that preserves all the nice properties (multiplicativity, simple geometric meaning, easy computation). Naïve attempts (e.g., summing over multi-permutations with signs) generally fail to be invariant or to detect the correct notion of “singularity.”

3.2 The Hyperdeterminant – The Standard Generalization¶

The hyperdeterminant (introduced by Arthur Cayley in the 1840s and systematically developed by Gelfand–Kapranov–Zelevinsky in the 1990s) is the correct algebraic generalization.

Definition (geometric/algebraic):
Let $A$ be a tensor in $V_1^* \otimes \cdots \otimes V_r^*$ (an $r$-linear form $f$). The hyperdeterminant $\operatorname{Det}(A)$ is the homogeneous polynomial (with integer coefficients) in the components of $A$ that vanishes if and only if there exist nonzero vectors $x^{(1)} \in V_1, \dots, x^{(r)} \in V_r$ such that all partial derivatives of the multilinear form $f$ vanish at $(x^{(1)},\dots,x^{(r)})$. In other words, it is the equation of the discriminant hypersurface of the multilinear map.

It exists (as a non-trivial polynomial) only when the format $(k_1+1, \dots, k_r+1)$ (where $\dim V_i = k_i + 1$) satisfies the convexity condition:

For $r=2$ this reduces exactly to square matrices.

3.3 Explicit Example: Cayley’s Hyperdeterminant for $2 \times 2 \times 2$ Tensors¶

Consider a tensor $A = (a_{ijk})$ with indices $i,j,k \in \{0,1\}$ (format $2\times2\times2$).

The hyperdeterminant is the following quartic polynomial: $$

$$

It vanishes if and only if the following system of six equations has a nontrivial solution $(x_0,x_1,y_0,y_1,z_0,z_1)$: $$

$$

These equations are precisely the conditions that all partial derivatives of the trilinear form vanish.

There are also compact expressions using (generalized) Levi-Civita symbols, but the expanded polynomial above is fully explicit.

3.4 Properties of the Hyperdeterminant¶

• It is a relative invariant under the action of $\mathrm{SL}(k_1+1) \times \cdots \times \mathrm{SL}(k_r+1)$.

• For boundary formats it has a geometric interpretation as a discriminant.

• Computation is hard for large formats (related to tensor rank problems).

• It reduces to the ordinary determinant when $r=2$ and the format is square.

Higher-order hyperdeterminants (e.g., for $2\times2\times2\times2$) exist but become extremely complicated.

4. Rank-2 Covariant Tensors: Bilinear Forms of Type (0,2)¶

4.1 Definition¶

A rank-2 covariant tensor (type (0,2)) on $V$ is a bilinear map

(and similarly in the second argument). Equivalently, $T \in V^* \otimes V^*$.

4.2 Matrix Representation and the Usual Determinant¶

Choose any basis $\{e_1,\dots,e_n\}$. Define the Gram matrix (or component matrix) $M$ by

Then one can form the ordinary scalar $\det(M)$. This number depends on the basis.

4.3 Transformation Law – Why It Is a Density, Not a Scalar¶

Let $\{e'_i\}$ be a new basis related to the old one by

($P$ is the change-of-basis matrix). The new components are

Therefore

Conclusion: Under basis change, the “determinant” transforms by the factor $[\det(P)]^2$. It is therefore a tensor density of weight 2, not a true scalar invariant. This is the fundamental reason why a pure (0,2) tensor does not possess a canonical basis-independent scalar determinant in the same way an endomorphism does.

5. Basis-Independent Constructions for (0,2) Tensors¶

5.1 The Induced Linear Map $\phi_T$¶

Any (0,2) tensor $T$ defines a linear map

$T$ is non-degenerate if and only if $\phi_T$ is an isomorphism.

5.2 Induced Map on Determinant Lines¶

Apply the exterior functor:

Since $\bigwedge^n V^* \cong (\bigwedge^n V)^*$, we obtain a canonical linear map between dual lines:

where $L = \bigwedge^n V$.

This map $L \to L^*$ is the intrinsic object associated to $T$. It can be viewed as:

• A quadratic form on the determinant line $L$, or

• An element of $L^* \otimes L^*$.

5.3 Why No Canonical Scalar Exists¶

In the endomorphism case, $\bigwedge^n f : L \to L$ (same line), so the map is multiplication by a scalar $\det(f) \in K$.

Here the map goes from $L$ to its dual $L^*$. A map $L \to L^*$ is not multiplication by an element of $K$; it encodes more data (a density). To extract a number one must choose an additional structure (a volume form $\mathrm{vol} \in L^*$) to normalize and pair.

Thus the “determinant” of a pure (0,2) tensor lives naturally in the weight-2 density bundle, not in the scalars.

6. Weight-2 Densities and Transformation Laws¶

A density of weight $w$ on $V$ is an object that, under basis change with matrix $P$, transforms by the factor $[\det(P)]^w$.

From Section 4.3 we see that the ordinary matrix determinant of the Gram matrix of a (0,2) tensor transforms with weight $w=2$.

The map $\bigwedge^n \phi_T : L \to L^*$ transforms with exactly this weight because:

• $L$ itself transforms with weight -1 (or +1, depending on convention) under $\mathrm{GL}(V)$,

• The dual $L^*$ transforms with the opposite weight,

• The induced map therefore acquires weight 2 overall.

This confirms that the exterior-algebra construction automatically produces an object with the correct transformation law.

In Riemannian geometry, when $T = g$ is a metric, one uses $g$ itself to define a canonical volume form $\mathrm{vol}_g$. Then $\sqrt{|\det(g_{ij})|}$ appears as the coefficient that makes $\mathrm{vol}_g$ have the proper normalization with respect to $g$-orthonormal frames. The raw $\det(g_{ij})$ is recovered squared in the density.

7. Axiomatic Characterization: Uniquely Deriving the Determinant¶

We now show that, assuming only that the object transforms as a weight-2 density, a small set of natural axioms uniquely determines the determinant (up to a constant factor).

7.1 The Five Natural Axioms¶

Let $d$ be a map that assigns to each bilinear form $T \in \mathcal{B}(V)$ a weight-2 density $d(T)$. We require:

1. Homogeneity of degree $n$
   

   $$d(\lambda T) = \lambda^n \, d(T).$$

   (19)

2. Detects degeneracy
   $d(T) = 0$ if and only if $\phi_T : V \to V^*$ is not an isomorphism.

3. Correct transformation law (weight 2)
   

   $$d(g \cdot T) = [\det(g)]^2 \, d(T) \quad \text{for all } g \in \mathrm{GL}(V).$$

   (20)

4. Multiplicativity for direct sums
   If $V = V_1 \oplus V_2$ and $T = T_1 \oplus T_2$, then

   $$d(T) = d(T_1) \cdot d(T_2)$$

   (21)

   (densities multiply compatibly).

5. Polynomial (or algebraic) character
   In any basis, $d(T)$ is given by a homogeneous polynomial of degree $n$ in the components of $T$. For symmetric forms this can be weakened all the way to mere continuity — in fact it can be dropped entirely; see Section 7.5.

These axioms are all intrinsic and motivated by the classical properties of the determinant, adapted to the density setting.

7.2 Uniqueness: Complete Derivation¶

We now carry out the full argument that the five axioms force $d(T)$ to equal, in components, the ordinary determinant of the Gram matrix.

Theorem. Let $\dim V = n$ and fix a basis $\{e_1,\dots,e_n\}$, so that a bilinear form $T$ is represented by its Gram matrix $M=(T_{ij})$ with $T_{ij}=T(e_i,e_j)$. The only map $d$ satisfying Axioms 1–5 is, in components,

for a single constant $c\in K$, and the normalization of Section 7.3 forces $c=1$, so that

We work over an algebraically closed field $K$ (for instance $K=\mathbb{C}$); the field $K=\mathbb{R}$ is recovered at the very end by complexification.

Step 0 — Reduce to a single polynomial in the components¶

A weight-2 density on an $n$-dimensional space has a single top component. Relative to the fixed basis, $d(T)$ is therefore encoded by one scalar

By Axiom 5, $D:\mathrm{Mat}_n(K)\to K$ is a homogeneous polynomial of degree $n$ in the $n^2$ entries $x_{ij}:=T_{ij}$. The entire problem reduces to identifying this one polynomial. The remaining axioms translate into the following statements about $D$:

• (A1) Homogeneity. $D(\lambda M)=\lambda^n D(M)$.

• (A2) Degeneracy. $D(M)=0 \iff \det M = 0$, because $\phi_T$ is an isomorphism iff $M$ is invertible.

• (A3) Weight-2 law. $D(P^{\mathsf T} M P) = (\det P)^2\,D(M)$ for all $P\in GL_n(K)$ (this is exactly the computation of Section 4.3).

• (A4) Block multiplicativity. $D\!\begin{pmatrix}M_1&0\\0&M_2\end{pmatrix} = D_{k}(M_1)\,D_{n-k}(M_2)$, where $D_m$ denotes the degree-$m$ function in dimension $m$.

Step 1 — The determinant is an irreducible polynomial¶

This is the structural fact that makes the degeneracy axiom so powerful.

Lemma. For every $n\ge 1$, $\det_n=\det(x_{ij})$ is irreducible in the polynomial ring $K[x_{ij} : 1\le i,j\le n]$.

Proof. Induct on $n$. For $n=1$, $\det_1=x_{11}$ is irreducible. Let $n\ge 2$ and expand along the first column:

where $A_{i1}$ is the minor obtained by deleting row $i$ and column 1. Observe:

1. $A_{11}$ is the determinant of the block on rows and columns $\{2,\dots,n\}$, hence irreducible by the inductive hypothesis;

2. no minor $A_{i1}$ contains any column-1 variable, and $A_{11}$ contains no row-1 variable.

Because the variables $x_{i1}$ of column 1 occur in $\det_n$ only through the explicit factors above, $\det_n$ is linear in $x_{11}$, with coefficient $A_{11}$.

Suppose $\det_n=PQ$ with $P,Q$ non-constant. Linearity in $x_{11}$ forces $x_{11}$ to appear in exactly one factor, say $P$; then $Q$ is free of $x_{11}$. Comparing the coefficients of $x_{11}$ gives $A_{11}=(\partial_{x_{11}}P)\,Q$, so $Q\mid A_{11}$. Since $A_{11}$ is irreducible, either $Q$ is a nonzero constant — contradicting non-constancy — or

We rule out the latter. As $A_{11}$ is free of $x_{11}$, the divisibility persists after setting $x_{11}=0$:

Now $\det_n$ is linear in $x_{21}$ with coefficient $-A_{21}$ (every column-1 variable appears only once), and $A_{11}$ is free of $x_{21}$; hence $A_{11}\mid A_{21}$. Both polynomials have degree $n-1$ and $A_{11}$ is irreducible, so $A_{21}=c'A_{11}$ for a constant $c'$. This is impossible: the minor $A_{21}$ (delete row 2, column 1) involves the row-1 variable $x_{12}$, whereas $A_{11}$ does not. The contradiction shows $Q$ must be constant, so $\det_n$ is irreducible. $\qquad\square$

Step 2 — The determinant divides $D$¶

By Axiom A2 the polynomial $D$ vanishes at every point of the determinantal hypersurface $\{M : \det M = 0\}$. Since $\det$ is irreducible (Step 1), it is in particular squarefree, so the principal ideal $(\det)$ is prime and therefore radical. Hilbert’s Nullstellensatz then gives

As $D$ lies in this vanishing ideal, $D\in(\det)$; that is, $\det\mid D$, so there is a polynomial $E$ with

Step 3 — A degree count pins down $D=c\det$¶

Both $\det$ and (by Axiom 5 / A1) $D$ are homogeneous of degree exactly $n$. Hence the quotient $E=D/\det$ is homogeneous of degree 0, i.e. a constant:

This already establishes uniqueness up to a single scalar, using only Axioms 2 and 5 together with the irreducibility of $\det$. (Note that the weight law A3 has not yet been used — in agreement with the fact that the weight alone does not characterize the determinant; the degeneracy and polynomiality axioms are what do the work.)

Step 4 — Identifying the constant intrinsically¶

Axioms 1 and 4 compute $c$ without any arbitrary basis-dependent choice. Apply A4 repeatedly to a diagonal Gram matrix $M=\operatorname{diag}(\lambda_1,\dots,\lambda_n)$, arising from an orthogonal direct sum $T=T_1\oplus\cdots\oplus T_n$ of one-dimensional forms with $\lambda_i=T_i(e_i,e_i)$:

In dimension one, A1 gives $D_1(\lambda)=\lambda\,D_1(1)$. Writing $\kappa:=D_1(1)$,

Comparing with Step 3 on diagonal matrices yields

The overall constant is thus the $n$-th power of the single one-dimensional normalization constant $D_1(1)$ — precisely the residual freedom removed in Section 7.3.

Step 5 — Consistency with the weight-2 law (and a Nullstellensatz-free route)¶

The solution $D=c\det$ automatically satisfies A3, since

So A3 imposes no further constraint: it is exactly the transformation law $\det$ already obeys, and its real content is that the answer is a genuine weight-2 density, not a bare number.

Conversely, A3 by itself gives a quick elementary derivation on the dense set of nondegenerate symmetric forms, bypassing the Nullstellensatz. Over $K=\overline{K}$ with $\operatorname{char}K\ne 2$, every nondegenerate symmetric $M$ is congruent to the identity, $M=P^{\mathsf T}P$, so

because $\det M=(\det P)^2$. Since the nondegenerate matrices are Zariski-dense and $D-c\det$ is polynomial, $D=c\det$ everywhere on symmetric forms. (For non-symmetric bilinear forms the symmetric and antisymmetric parts decouple under congruence and cannot be simultaneously diagonalized — which is exactly why the general statement genuinely needs the irreducibility argument of Steps 1–3.)

Step 6 — The real field¶

Over $K=\mathbb{R}$ the identity $D=c\det$ follows by complexification. By Axiom 5, $D\in\mathbb{R}[x_{ij}]$; extend it to $D_{\mathbb{C}}\in\mathbb{C}[x_{ij}]$. Steps 1–3 apply over $\mathbb{C}$ and give $D_{\mathbb{C}}=c\det$ with $c\in\mathbb{C}$; evaluating at a single real matrix (e.g. $D(I)$) shows $c\in\mathbb{R}$. Equivalently, an identity of polynomials that holds on the Zariski-dense set of real points holds identically.

Conclusion and the role of each axiom¶

Putting the steps together, every solution of Axioms 1–5 is

and imposing the normalization $d(T_0)=1$ of Section 7.3 (e.g. taking $T_0$ to be the form with Gram matrix $I$) forces $c=1$:

The five axioms play sharply distinct roles:

• Axiom 5 (polynomial of degree $n$) places $D$ inside $K[x_{ij}]$ and bounds its degree.

• Axiom 2 (degeneracy) forces $D$ to vanish on the irreducible hypersurface $\{\det=0\}$; with Axiom 5 this already yields $D=c\det$ (Steps 1–3). These two axioms alone give uniqueness up to scale.

• Axioms 1 and 4 (homogeneity and direct sums) pin the constant intrinsically as the $n$-th power of the one-dimensional value (Step 4).

• Axiom 3 (weight-2 law) is automatically satisfied; it supplies the alternative congruence derivation and, crucially, certifies that the output is the correct density rather than a mere scalar. By itself it is not enough to single out the determinant — uniqueness rests on Axioms 2 and 5.

7.3 Normalization – Removing the Ambiguity¶

Choose any fixed non-degenerate reference bilinear form $T_0$ that can be characterized intrinsically (for example, the standard dot product on $K^n$ when a preferred basis or inner-product structure is available, or the form induced by an endomorphism via a fixed volume form). Require

This fixes the constant completely. The resulting function is then the unique object satisfying all the axioms.

7.4 The Exterior-Algebra Construction Satisfies All Axioms¶

The map $\bigwedge^n \phi_T : L \to L^*$ (or the associated quadratic form on $L$) satisfies:

• Homogeneity of degree $n$ (because each factor of $\phi_T$ contributes degree 1).

• Vanishes exactly when $\phi_T$ is singular (the induced map on top exterior power is zero precisely then).

• Transforms with weight exactly 2 (as shown in Section 6).

• Is multiplicative for direct sums (exterior algebra respects direct sums).

• Is polynomial of degree $n$.

When a volume form is chosen to normalize, it yields a scalar density satisfying the normalization condition on reference forms. Therefore it is the unique object characterized by the axioms.

7.5 Continuity in Place of Polynomiality¶

Axiom 5 was used in Section 7.2 in an essential, but rather heavy, way: it placed $D$ inside the polynomial ring so that the Nullstellensatz could be applied. It is natural to ask whether the much weaker hypothesis of continuity suffices. The answer is a clean yes for symmetric forms — which is exactly the geometrically relevant case, since metrics and quadratic forms are symmetric — and there the regularity hypothesis can in fact be removed altogether. We also explain the one genuine pitfall over $\mathbb{R}$ that makes the precise statement matter.

Throughout, $D(M)$ denotes the component function of Section 7.2 (Step 0), and we read the axioms as holding for forms on spaces of every finite dimension (Axiom 4 already forces us to consider subspaces), so in particular homogeneity (Axiom 1) holds in each dimension.

The symmetric case: Axioms 1–4 already suffice (no regularity at all)¶

Theorem. Let $T$ range over symmetric bilinear forms. Then Axioms 1–4 alone force

and the normalization of Section 7.3 gives $\kappa=1$. No polynomiality and no continuity are needed.

Proof.

Step 1 (dimension one). A one-dimensional form is a single number $\lambda$, and Axiom 1 (in dimension one) reads $D_1(\lambda)=D_1(\lambda\cdot 1)=\lambda\,D_1(1)=\kappa\lambda$. Thus $D_1$ is linear — there is no room for, say, $|\lambda|$.

Step 2 (diagonal forms). Iterating the block-multiplicativity Axiom 4 over the orthogonal splitting into one-dimensional pieces,

Step 3 (all symmetric forms). By the real spectral theorem (or Sylvester’s law of inertia) every symmetric $M$ is congruent to a diagonal matrix, $M=P^{\mathsf T}\Lambda P$ with $\Lambda$ diagonal. If $M$ is nondegenerate, Axiom 3 gives

If $M$ is degenerate, Axiom 2 gives $D(M)=0=\kappa^{\,n}\det M$. Hence $D=\kappa^{\,n}\det$ on all symmetric forms. $\qquad\square$

The point is that congruence is transitive enough on symmetric forms — every symmetric matrix has a diagonal representative — so Axiom 3 alone propagates the value from the diagonal forms (where Axioms 1 and 4 already determine it) to every symmetric form, exactly, with no limiting argument. This is why no regularity is needed here. If one nevertheless prefers to state a fifth axiom, continuity is more than enough and is the “much better” hypothesis: it is far weaker than polynomiality and is automatically satisfied by the exterior-algebra construction of Section 7.4.

Why “continuity in dimension $n$ alone” is not enough¶

It is essential that homogeneity be imposed in every dimension (so that the one-dimensional piece is forced to be linear). If one only assumes the axioms in the fixed dimension $n$ and merely asks for continuity, the determinant is not singled out. The clean counterexample is, for even $n$,

Indeed $|\det|$ is continuous and satisfies, in dimension $n$ even,

so it obeys Axioms 1–4 within dimension $n$ and the degeneracy Axiom 2, yet it is not a constant multiple of $\det$ (it never takes negative values). What rescues uniqueness is precisely the cross-dimensional reading above: $|\det|$ corresponds to the one-dimensional rule $D_1(\lambda)=|\lambda|$, which violates Axiom 1 in dimension one ($D_1(-1)=1\neq -1=(-1)\cdot D_1(1)$). Equivalently, $|\det|$ is exactly the sign-twisted impostor $D=\operatorname{sign}(\det)\cdot\det$, and over $\mathbb{R}$ the nondegenerate locus is disconnected (the connected components are the signature classes $(p,q)$), so a hypothesis that does not link the components — bare continuity in one dimension — cannot exclude a different constant on each component. The polynomial Axiom 5 excluded these impostors automatically, because $|\det|$ and its relatives are not polynomials; the cross-dimensional homogeneity above excludes them just as effectively, and much more cheaply.

The general (non-symmetric) case: continuity is provably insufficient¶

For a general (0,2) tensor the situation changes completely, and the honest answer is negative: continuity does not single out the determinant. The reduction of Step 3 is unavailable, because congruence $M\mapsto P^{\mathsf T}MP$ preserves the symmetric and antisymmetric parts separately and therefore cannot diagonalize a non-symmetric Gram matrix. More to the point, congruence has a moduli of orbits on the nondegenerate locus: the cosquare (or “asymmetry operator”)

transforms by conjugation, $S(P^{\mathsf T}MP)=P^{-1}S(M)P$, so the entire conjugacy class of $S$ — in particular its eigenvalues, which occur in reciprocal pairs $\{\mu,\mu^{-1}\}$ — is a continuous congruence invariant. The orbit space is thus positive-dimensional, and a continuous, congruence-invariant function need not be constant on it. This is enough to manufacture impostors.

Counterexample. Fix any bounded continuous $\beta:\mathbb{R}\to\mathbb{R}$ with $\beta(2)=0$ (for instance $\beta(w)=\frac{(w-2)^2}{1+(w-2)^2}$), and define, in every dimension,

on nondegenerate $M$, extended by $D(M)=0$ where $\det M=0$. Then $D$ satisfies all of Axioms 1–4, is continuous, and meets the normalization $D(I)=1$, yet $D\neq c\det$:

• Homogeneity (A1). $S(\lambda M)=S(M)$ for every scalar $\lambda\neq0$, so the exponential factor is scale-invariant and $D(\lambda M)=\lambda^n D(M)$.

• Degeneracy (A2). $|\beta|\le 1$ gives $|\exp(\operatorname{tr}\beta(\cdot))|\le e^{n}$, a bounded nonzero factor; hence $D=0\iff\det=0$, and $D=\det\cdot(\text{bounded})\to0$ at the degenerate locus, so the extension by 0 is continuous.

• Weight-2 law (A3). $S\mapsto P^{-1}SP$ is a conjugation, so $\operatorname{tr}\,\beta(S+S^{-1})$ is unchanged and $D(P^{\mathsf T}MP)=(\det P)^2 D(M)$.

• Block multiplicativity (A4). $S(M_1\oplus M_2)=S_1\oplus S_2$, so the trace adds and the exponential factorizes: $D(M_1\oplus M_2)=D(M_1)\,D(M_2)$. (In dimension one $S=1$, $\beta(2)=0$, so $D_1(\lambda)=\lambda$ — the linear, $\kappa=1$ normalization.)

• It is genuinely different. On symmetric forms $S=I$, $S+S^{-1}=2I$, $\beta(2)=0$, so $D=\det$ there — but on a non-symmetric form the factor is $\neq1$. For example $M=\left(\begin{smallmatrix}1&2\\0&1\end{smallmatrix}\right)$ has $S$ with the double eigenvalue -1, giving $w=\mu+\mu^{-1}=-2$, $\operatorname{tr}\beta(S+S^{-1})=2\beta(-2)=\tfrac{32}{17}$, hence $D(M)=e^{32/17}\det M\neq\det M$.

Varying $\beta$ produces an infinite-dimensional family of distinct continuous solutions. The polynomial Axiom 5 excludes every one of them at a stroke, because none is a polynomial (indeed none is even algebraic). Equivalently: the determinant is the unique solution that is algebraic in the components; the non-uniqueness lives entirely in the transcendental, cosquare-dependent directions that polynomiality forbids but continuity permits.

Moral. The symmetric case is special precisely because congruence orbits exhaust the symmetric forms (the spectral theorem), so there are no continuous invariants beyond $\det$ to exploit. For an arbitrary (0,2) tensor those invariants exist, so one must keep a hypothesis that kills them. The cheap, natural choice is the polynomial/algebraic Axiom 5 of Section 7.2. (Continuity could only be salvaged by enlarging the symmetry group to two-sided equivalence $M\mapsto AMB$, under which rank is the sole invariant — but that action is not a symmetry of a genuine (0,2) tensor, since a single vector space offers only the diagonal congruence $A=P^{\mathsf T},B=P$.) In short:

• Symmetric forms (metrics, quadratic forms): Axioms 1–4 already give $d(T)=\det(T_{ij})$ after normalization; a regularity axiom is optional, and if included, continuity is the natural and sufficient choice.

• Arbitrary bilinear forms: continuity is not sufficient (the counterexample above); keep the polynomial Axiom 5 (Section 7.2), or replace it by the structural axiom of Section 7.6.

7.6 The Natural Non-Regularity Alternative: the Two-Slot Weight Law¶

There is a single extra assumption that removes the need for polynomiality (or continuity) and works for arbitrary (0,2) tensors. It is obtained by noticing exactly where the impostors of Section 7.5 slip through.

The weight-2 Axiom 3 only constrains $d$ under the diagonal congruence $T\mapsto T(P\,\cdot\,,P\,\cdot\,)$, in which the same map $P$ acts on both arguments. But a (0,2) tensor is an element of $V^*\otimes V^*$, whose two covariant slots are a priori independent: for any $A,B\in GL(V)$ the tensor

is again a perfectly good (0,2) tensor. The impostors exploit precisely the directions with $A\neq B$: their cosquare $S=M^{-\mathsf T}M$ measures the mismatch between the two slots, and is invariant only under the diagonal $A=B$. Demanding the correct behavior under independent slot transformations destroys them.

Axiom 3′ (separate / bi-weight $(1,1)$ law). For all $A,B\in GL(V)$,

This says $d(T)$ is a density of weight 1 in each covariant slot separately, rather than merely of total weight 2 — manifestly the honest functorial requirement for an object built from the two tensor factors. It contains Axiom 3 as the special case $A=B$ (and Axiom 1 as $A=B=\lambda^{1/2}\,\mathrm{Id}$).

Theorem. Axiom 3′ together with the degeneracy Axiom 2 and the normalization $d(T_0)=1$ (with $T_0$ the form whose Gram matrix is $I$) forces, in any basis,

for every (0,2) tensor, with no polynomiality, no continuity, and indeed without Axioms 1 or 4.

Proof. Let $M$ be the Gram matrix of $T$. If $M$ is nondegenerate, factor it as $M=A^{\mathsf T}\,I\,B$ with $A=I$, $B=M$ (any factorization $M=A^{\mathsf T}B$ works). Then $T=(T_0)_{A,B}$ and Axiom 3′ gives

If $M$ is degenerate, Axiom 2 gives $d(T)=0=\det(T_{ij})$. $\qquad\square$

The proof is a one-liner because two-sided equivalence $M\mapsto A^{\mathsf T}MB$ acts transitively on the nondegenerate matrices (rank is its only invariant), so the single value $d(T_0)$ propagates to every nondegenerate form exactly — no limiting argument, no irreducibility, no Nullstellensatz. The cosquare invariants that powered the Section 7.5 counterexamples are not invariant under Axiom 3′, so the family of impostors collapses to the single function $\det$. Numerically, the impostor $D(M)=\det(M)\exp\!\big(\operatorname{tr}\beta(S+S^{-1})\big)$ satisfies the diagonal law $D(P^{\mathsf T}MP)=(\det P)^2D(M)$ but fails $D(A^{\mathsf T}MB)=\det(A)\det(B)D(M)$ whenever $A\neq B$, exactly as Axiom 3′ requires.

This is the same statement as the functorial definition. Axiom 3′ is nothing but the coordinate form of the exterior-algebra construction of Sections 5 and 7.4: applying $\bigwedge^n$ to $\phi_{T_{A,B}}=B^{*}\circ\phi_T\circ A$ gives

so the induced map $\bigwedge^n\phi_T:\bigwedge^n V\to\bigwedge^n V^*$ automatically obeys Axiom 3′. In other words, defining the determinant functorially as $d(T)=\bigwedge^n\phi_T$ builds Axiom 3′ in for free and needs no regularity hypothesis whatsoever — the cleanest resolution of all, and the one to prefer over any classification by polynomial or continuity axioms.

8. Special Cases and Geometric Applications¶

8.1 Symmetric Bilinear Forms / Quadratic Forms¶

When $T$ is symmetric, the associated quadratic form $Q(v) = T(v,v)$ has a well-studied discriminant, which is $\det(M)$ considered as an element of the quotient group $K^\times / (K^\times)^2$. This is a true invariant (independent of basis) and classifies quadratic forms up to isomorphism over many fields (together with the Hasse invariant, signature, etc.).