The Determinant of a (0,2) Tensor by Raising an Index

The determinant is already settled for $(1,1)$ tensors. By Determinant From Homomorphism, an endomorphism $M$ of an $n$-dimensional space has a determinant

the single-valued Leibniz polynomial; it is multiplicative, $\det(M_1M_2)=\det(M_1)\det(M_2)$, and therefore conjugation-invariant, $\det(P^{-1}MP)=\det(M)$ — an intrinsic scalar.

A $(0,2)$ tensor has no determinant of its own as a scalar: the determinant of its component matrix, $\det(A_{ij})$, is basis-dependent — under a change of basis $A\mapsto P^{\mathsf T}AP$ it picks up a factor $\det(P)^2$, making it a density of weight 2 (the subject of The Determinant of a $(0,2)$ Tensor from Relative Invariance). A metric repairs this by raising an index, exactly as for the trace. This note takes that construction as the definition, reads off existence and every property by piggybacking on the $(1,1)$ determinant, and proves uniqueness from multiplicativity only at the very end.

Setup¶

Let $V$ be $n$-dimensional over $\mathbb C$. A metric is a nondegenerate symmetric $(0,2)$ tensor $g$ with matrix $G=(g_{ij})$ and inverse $g^{ij}=(G^{-1})_{ij}$. In a basis $e_1,\dots,e_n$ we meet three kinds of components:

• an endomorphism — a $(1,1)$ tensor — $M=(M^{i}{}_{j})$;

• a $(0,2)$ tensor $A=(A_{ij})$;

• the metric $G$.

The metric lowers the upper index of $M$, and because $G$ is invertible this is reversible — it raises the lower index of $A$:

Under a change of basis $P\in GL(V)$ the two covariant objects transform by congruence and the endomorphism by similarity,

the last following from the first two. A basis is orthonormal for $g$ when $G=I$ (one exists over $\mathbb C$).

The $(1,1)$ determinant¶

These are the facts we borrow from Determinant From Homomorphism. On endomorphisms the determinant is

• explicit and single-valued — the Leibniz polynomial above, defined for every $M$;

• multiplicative on $GL(V)$, hence conjugation-invariant, $\det(P^{-1}MP)=\det(M)$;

• normalized, $\det(I)=1$;

• universal: every homomorphism $GL(V)\to\mathbb C^{*}$ equals $h\circ\det$ for some homomorphism $h:\mathbb C^{*}\to\mathbb C^{*}$, and $\det$ is the only one taking the value $\prod_i d_i$ on $\operatorname{diag}(d_1,\dots,d_n)$.

Definition¶

The $(0,2)$ tensor $A$ has no intrinsic determinant; the metric supplies one. Raise its index to the endomorphism $M=G^{-1}A$ and define the determinant of $A$ to be that of $M$:

This is the metric-furnished extension of the $(1,1)$ determinant to $(0,2)$ tensors. Everything below is an immediate consequence.

Existence and consequences¶

Existence and component formula. The Leibniz polynomial makes $\det(G^{-1}A)$ an explicit, single-valued polynomial in the entries of $A$, and by multiplicativity

In an orthonormal basis, where $\det(g_{ij})=1$, its value is the component determinant $\det(A_{ij})$. But that component determinant is only a coordinate number: under $A\mapsto P^{\mathsf T}AP$ it scales by $\det(P)^2$, so it is a weight-2 density, not basis-independent, and it agrees with $\det_g(A)$ only because the orthonormal metric has unit determinant. This is exactly why $A$ has no determinant of its own as a scalar — and why an index had to be raised to define one.

It is a scalar. Under (3) the endomorphism transforms by similarity $M\mapsto P^{-1}MP$, so the conjugation invariance of the $(1,1)$ determinant makes $\det_g(A)$ basis-independent — for the full $GL(V)$.

Multiplicative under metric-composition. Composing two $(0,2)$ tensors through the metric, $A_1\,G^{-1}A_2$ is again a $(0,2)$ tensor, and

Homogeneous of degree $n$. $\det_g(cA)=\det(cG^{-1}A)=c^{n}\det_g(A)$ for $c\in\mathbb C$.

The raised index may be either one. Raising the second index of $A$ produces the endomorphism $AG^{-1}$, and $\det(AG^{-1})=\det(G^{-1}A)$, so (4) does not depend on which slot is raised.

Normalization. Taking $A=g$, $\det_g(g)=\det(G^{-1}G)=\det(I)=1$.

That is the whole theory: the definition (4), the formula (5), basis-independence, multiplicativity, the degree, and the normalization, all inherited from the $(1,1)$ determinant without a single new computation.

Uniqueness¶

The construction produces one functional; we close by showing it is the only one of its kind. The hypotheses are the multiplicative analogues of the trace’s linear ones.

Claim. Let $D$ be a function on $(0,2)$ tensors, nonzero on the nondegenerate ones, with

• (M) metric-multiplicativity: $D(A_1G^{-1}A_2)=D(A_1)D(A_2)$ for all nondegenerate $A_1,A_2$;

• (H) homogeneity: $D(cA)=c^{n}D(A)$ for $c\in\mathbb C^{*}$;

• (N) normalization: $D(g)=1$. Then $D=\det_g$ on every nondegenerate $(0,2)$ tensor.

Pull back along the index-raising. For $M\in GL(V)$ set $\tilde D(M):=D(GM)$; then $A=GM$ is nondegenerate, so $\tilde D(M)\in\mathbb C^{*}$. By (M), for $M_1,M_2\in GL(V)$,

so $\tilde D:GL(V)\to\mathbb C^{*}$ is a group homomorphism. By Determinant From Homomorphism it factors,

Now evaluate on the metric ray. By (H) and (N), $D(cg)=c^{n}D(g)=c^{n}$; on the other hand $cg=G(cI)$, so $D(cg)=\tilde D(cI)=h(\det(cI))=h(c^{n})$. Hence

Over $\mathbb C$ every element is an $n$-th power, so $h=\operatorname{id}$, and (8) collapses to $\tilde D=\det$. Therefore, for every nondegenerate $A$,

On a degenerate $A$ both sides vanish — $\det_g(A)=0$ by (5), and the explicit polynomial $\det_g$ extends the identity there by continuity — so $D=\det_g$ outright once $D$ is taken to be the everywhere-defined Leibniz extension.

Density versus scalar: reconciling with the relative-invariance derivation¶

It is worth comparing this result with The Determinant of a $(0,2)$ Tensor from Relative Invariance, which derives a different object — by design. There are two inequivalent notions of “the determinant of a $(0,2)$ tensor,” and the metric is exactly what separates them.

The component determinant is a density, not a scalar. Under (3), $A\mapsto P^{\mathsf T}AP$, so

The bare component determinant is therefore not basis-independent — it rescales by $\det(P)^2$. It is a density of weight 2 (a relative scalar). This is the object the relative-invariance note characterizes, using no metric: the cost of staying metric-free is that the answer is a density, allowed to transform by that factor, rather than a true number.

Raising an index produces a genuine scalar instead. The metric determinant $\det_g(A)=\det(G^{-1}A)$ is invariant under the full $GL(V)$ — weight 0, a true scalar. The two objects differ by exactly one factor of the metric’s own density:

Because $\det(g_{ij})$ is itself a weight-2 density — it too rescales by $\det(P)^2$ — the quotient is weight 0: the two factors of $\det(P)^2$ cancel. Dividing the density of $A$ by the density of $g$ is the standard way a metric turns a density into a number.

Where they agree, and where they part. In an orthonormal basis $\det(g_{ij})=1$, so $\det_g(A)=\det(A_{ij})$: both notes give the same number. In a general basis they differ by the factor $\det(G)$ — the relative-invariance note’s $\det(A_{ij})$ keeps rescaling as a density, while $\det_g(A)$ stays fixed. The two derivations are thus consistent, not competing: one gives the determinant as a density (metric-free), the other as a scalar (metric-bound), equal up to $\det(G)$.

Why this route can only land on the scalar. The construction is built on the $(1,1)$ determinant, which is conjugation-invariant — a genuine scalar — so anything assembled from it is a genuine scalar too. The density $\det(A_{ij})$ is not a scalar, so it cannot be the output here; the method is forced to deliver the weight-0 quotient $\det(A)/\det(G)$.

Why the determinant has two notions while the trace has one. The component determinant survives metric-free, as a density, only because it is a true relative invariant: $\det(P^{\mathsf T}AP)=\det(P)^2\det(A)$ carries the clean character $P\mapsto\det(P)^2$. The trace enjoys no such thing — $\sum_i A_{ii}\mapsto\operatorname{tr}(P^{\mathsf T}AP)$ rescales by no character of $P$, so it is not even a density, only a basis-dependent number with no invariant meaning. Hence there is a “determinant from relative invariance” but no “trace from relative invariance”: the trace has only its metric scalar $g^{ij}A_{ij}$, whereas the determinant has both the metric-free density and the metric scalar.

Remarks¶

Existence piggybacked; uniqueness used multiplicativity. As with the trace, existence, the component formula, and invariance came straight from the $(1,1)$ determinant by raising an index. Uniqueness, however, rested on (M) — multiplicativity — and reduced, through the same index-raising, to the homomorphism theorem of Determinant From Homomorphism. It is the determinant’s counterpart of the trace’s pull-back of cyclicity.

Why isometry-invariance is not enough — the contrast with the trace. For the trace, linearity together with isometry-invariance pinned $\operatorname{tr}_g$ uniquely (a signed-permutation argument). For the determinant the same geometric hypothesis is far too weak: every coefficient of the characteristic polynomial of $M=G^{-1}A$ — the elementary symmetric functions $\operatorname{tr}(M),\ \dots,\ \det(M)$ of its eigenvalues — is an isometry-invariant scalar built from $A$, and $\det_g$ is only the last of them. Linearity selects the first of these (the trace); multiplicativity selects the last (the determinant). This is why the determinant’s uniqueness must invoke its multiplicative structure rather than a symmetry.

Geometric meaning. $M=G^{-1}A$ is the endomorphism of the generalized eigenvalue problem $A\,v=\lambda\,g\,v$, so $\det_g(A)$ is the product of the eigenvalues of $A$ relative to $g$ — the ratio of the volume form determined by $A$ to that determined by $g$. Unlike the trace, which saw only the symmetric part of $A$, the determinant sees all of it: a nonzero antisymmetric $A$ can have $\det_g(A)\neq0$.