The Determinant of a (0,2) Tensor from Relative Invariance

Starting from relative-invariance (covariance) equations for a scalar attached to a covariant rank-2 tensor, we derive — with no skipped steps — that the scalar must be the ordinary component determinant $\det(M_{ij})$ in any basis.

Setup¶

Let $V$ be an $n$-dimensional vector space over $\mathbb{C}$ and $T$ a $(0,2)$ tensor, i.e. a bilinear form $T:V\times V\to\mathbb{C}$. Fix a basis $e_1,\dots,e_n$ and form the component (Gram) matrix

We seek a scalar $D(M)\in\mathbb{C}$ built from these components. For $A,B\in GL(V)$ (matrices in the chosen basis; repeated indices summed), transforming each argument of $T$ separately gives

Assumptions¶

$D$ is a relative scalar in each slot: a change of basis in one argument rescales $D$ by a factor depending only on that change. That is, there exist $\alpha,\beta:GL(V)\to\mathbb{C}$ with

and $D\not\equiv 0$. We also fix the degree and scale:

Step 1: the two-slot law (R$'$)¶

Apply (R) in the two slots successively (the operations commute, $A^{\mathsf T}(MB)=(A^{\mathsf T}M)B$): by the second equation of (R), $D(MB)=\beta(B)D(M)$, then by the first, $D\big(A^{\mathsf T}(MB)\big)=\alpha(A)D(MB)$. So the two-slot operation rescales $D$ by a factor depending only on $A$ and $B$, which we name $\chi(A,B)$:

Alternative starting point. One may take (R$'$) itself as the assumption, in place of (R): a single two-slot relative-invariance equation, with $\chi:GL(V)\times GL(V)\to\mathbb{C}$ an unspecified function and $D\not\equiv0$. It is equivalent to (R) — setting $B=I$ and then $A=I$ in (R$'$) returns the two equations of (R). Everything below uses only (R$'$), (H), and (N).

Step 2: $\chi$ splits into two characters¶

Working from (R$'$) alone, set $B=I$, then $A=I$:

and define $\alpha(A):=\chi(A,I)$, $\beta(B):=\chi(I,B)$ (the two displayed equations are exactly (R)). Also $\chi(I,I)=1$: take $A=B=I$, $M=I$ in (R$'$) and use (N).

Factorization. $D(A^{\mathsf T}MB)=D\big(A^{\mathsf T}(MB)\big)=\alpha(A)\,D(MB)=\alpha(A)\beta(B)\,D(M)$; comparing with (R$'$) at $M=I$,

Multiplicativity. Since $(A_1A_2)^{\mathsf T}M=A_2^{\mathsf T}(A_1^{\mathsf T}M)$,

at $M=I$, $\alpha(A_1A_2)=\alpha(A_1)\alpha(A_2)$, and likewise $\beta(B_1B_2)=\beta(B_1)\beta(B_2)$. Each is nonzero: $\alpha(A)\alpha(A^{-1})=\alpha(I)=\chi(I,I)=1$. Thus

Step 3: $\chi(A,B)=\det(A)\det(B)$¶

A homomorphism $GL(n,\mathbb{C})\to\mathbb{C}^{*}$ factors through the determinant: by the factorization theorem of Determinant From Homomorphism,

for homomorphisms $g_1,g_2:\mathbb{C}^{*}\to\mathbb{C}^{*}$, where $\det$ is the explicit Leibniz polynomial. We pin $g_1,g_2$ with (H). Since $(\lambda I)^{\mathsf T}M=\lambda M$, the first equation of Step 2 gives $D(\lambda M)=\alpha(\lambda I)\,D(M)$; comparing with (H) at $M=I$,

But $\alpha(\lambda I)=g_1(\det(\lambda I))=g_1(\lambda^{n})$, so $g_1(\lambda^{n})=\lambda^{n}$ for every $\lambda\in\mathbb{C}^{*}$. Every $t\in\mathbb{C}^{*}$ is an $n$-th power, hence $g_1=\operatorname{id}$ and $\alpha=\det$. The same argument with $M(\lambda I)=\lambda M$ gives $\beta=\det$. Therefore

Step 4: $D(M)=\det(M)$¶

Invertible $M$. Apply the second equation of Step 2 with base point $I$ and multiplier $M$, using $\beta=\det$ and (N):

Singular $M$. If $\det M=0$ there is a vector $u\neq0$ with $Mu=0$. Pick a covector $w^{\mathsf T}$ with $w^{\mathsf T}u=c$ where $c\neq 0,-1$, and set $B=I+u\,w^{\mathsf T}$, which is invertible with $\det B=1+w^{\mathsf T}u=1+c\neq1$. Since $Mu=0$,

so the second equation of Step 2 gives $D(M)=D(MB)=\beta(B)D(M)=\det(B)\,D(M)=(1+c)\,D(M)$, forcing $D(M)=0=\det(M)$.

Hence in every basis

The two-slot law (R$'$) — equivalently the per-slot equations (R) — together with the degree (H) and scale (N), forces the determinant of a $(0,2)$ tensor to be the ordinary component determinant. $\qquad\blacksquare$