The Trace of a (0,2) Tensor from a Metric

Starting from a covariant rank-2 tensor $T$ together with a metric $g$, and a few invariance assumptions, we derive — with no skipped steps — that the only linear scalar one can attach to $T$ is the metric trace $g^{ij}T_{ij}$, which in any orthonormal basis is the ordinary trace $\sum_i T_{ii}$ of the component matrix.

This is the additive, metric-dependent companion of The Determinant of a $(0,2)$ Tensor from Relative Invariance. The contrast is the whole point. The determinant of a $(0,2)$ tensor needs no metric: it is a relative scalar (a density of weight 2) under the full $GL$ congruence group. A $(0,2)$ tensor has no intrinsic trace — under congruence $M\mapsto A^{\mathsf T}MA$ the coordinate sum $\sum_i M_{ii}$ is not preserved — and a metric is exactly the extra datum that repairs this, by letting one index be raised.

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 matrix

A metric is a nondegenerate symmetric $(0,2)$ tensor $g$ with components $G_{ij}=g(e_i,e_j)$; its inverse is the $(2,0)$ tensor with components $g^{ij}:=(G^{-1})_{ij}$, so that $\sum_j g^{ij}G_{jk}=\delta^i_k$.

For $A\in GL(V)$ (the matrix of a change of basis), both covariant tensors transform by congruence, and the inverse metric contravariantly:

A basis is orthonormal for $g$ when $G=I$, i.e. $g(e_i,e_j)=\delta_{ij}$. Over $\mathbb C$ every nondegenerate symmetric form admits an orthonormal basis (diagonalize, then rescale each diagonal entry to 1 using a complex square root). The changes of basis that preserve $g$ form the isometry group

in an orthonormal basis $O(g)=O(n,\mathbb C)=\{A:A^{\mathsf T}A=I\}$, and there congruence by an isometry coincides with conjugation, $A^{\mathsf T}MA=A^{-1}MA$.

We seek a scalar $D(M)\in\mathbb C$ built from the components of $T$, with the metric fixed as background data.

Assumptions¶

• (L) Linearity. $D(M+M')=D(M)+D(M')$ and $D(cM)=c\,D(M)$ for all $M,M'$ and $c\in\mathbb C$. The trace is additive and homogeneous of degree 1 — the additive analogue of the determinant’s degree-$n$ homogeneity.

• (I) Isometry invariance. There is a character $\kappa:O(g)\to\mathbb C^{*}$ with

  $$D\big(A^{\mathsf T}MA\big)=\kappa(A)\,D(M)\qquad(\forall M,\ \forall A\in O(g)),$$

  (4)

  i.e. $D$ is a relative scalar under isometries — the analogue of the determinant’s relative invariance, now restricted to the subgroup that the metric singles out.

• (N) Normalization. $D(G)=n$ (the metric’s own trace; in an orthonormal basis $D(I)=n$).

• $D\not\equiv 0$.

Step 1 — the character is trivial¶

Set $M=G$ in (I). Since $A^{\mathsf T}GA=G$ for every isometry,

By (N), $D(G)=n\neq0$, so $\kappa(A)=1$ for all $A\in O(g)$:

So $D$ is an absolute invariant under isometries — a scalar of weight 0, in contrast with the determinant, whose weight-2 density survives only as a relative scalar. The normalization is what trivializes the character; note $O(n,\mathbb C)$ does carry the nontrivial character $A\mapsto\det A=\pm1$, but $D(G)\neq0$ excludes it.

Step 2 — work in an orthonormal basis¶

Choosing an orthonormal basis makes $G=I$, the isometry group $O(n,\mathbb C)$, and (6) the conjugation invariance

Everything below is computed in this basis; the basis-free form is restored in Step 4.

Step 3 — signed permutations force $D=\lambda\operatorname{tr}$¶

By linearity (L), $D$ is determined by its values on the matrix units $E_{ij}$:

Two families of isometries pin the coefficients $c_{ij}$. Both are signed permutation matrices, which satisfy $A^{\mathsf T}A=I$ and so lie in $O(n,\mathbb C)$.

Diagonal sign matrices kill the off-diagonal coefficients. For $\varepsilon\in\{\pm1\}^n$ let $S=\operatorname{diag}(\varepsilon_1,\dots,\varepsilon_n)$, so $S^{-1}=S$ and $(S^{-1}MS)_{ij}=\varepsilon_i\varepsilon_j M_{ij}$. Invariance (7) gives $\sum_{ij}c_{ij}\varepsilon_i\varepsilon_j M_{ij}=\sum_{ij}c_{ij}M_{ij}$ for all $M$, hence $c_{ij}(\varepsilon_i\varepsilon_j-1)=0$. For $i\neq j$, take $\varepsilon_i=-1$ and the rest +1; then $\varepsilon_i\varepsilon_j=-1$ and

Permutations equate the diagonal coefficients. For a transposition $\tau=(i\;j)$ the permutation matrix $P_\tau\in O(n,\mathbb C)$ acts by $(P_\tau^{-1}MP_\tau)_{kl}=M_{\tau(k)\tau(l)}$, so (7) forces $c_{\tau(k)\tau(l)}=c_{kl}$; on the diagonal this is $c_{ii}=c_{jj}$. Thus all diagonal coefficients share one value

Combining (9) and (10),

Step 4 — normalization and the covariant formula¶

In the orthonormal basis $G=I$, so (N) reads $D(I)=\lambda n=n$, giving $\lambda=1$ and

To see the same scalar in an arbitrary basis, contract $T$ with the inverse metric:

which collapses to $\sum_i M_{ii}$ exactly when $G=I$. The next section verifies that (13) is itself a genuine scalar satisfying (L), (I), (N), so by the uniqueness just proved it is the answer.

Existence: the metric trace is a scalar¶

Equation (13) defines one explicit functional, which we check directly.

• Linear in $T$, since each $M_{ij}$ is.

• Invariant under every change of basis (not merely isometries): under (2),

  $$\operatorname{tr}\!\big(G^{-1}M\big)\mapsto\operatorname{tr}\!\big(A^{-1}G^{-1}A^{-\mathsf T}\,A^{\mathsf T}MA\big)=\operatorname{tr}\!\big(A^{-1}G^{-1}MA\big)=\operatorname{tr}\!\big(G^{-1}M\big),$$

  (14)

  by the cyclic property of the trace. In particular (I) holds with $\kappa\equiv1$.

• Normalized: $\operatorname{tr}_g(g)=\sum_{ij}g^{ij}G_{ij}=\sum_i\delta^i_i=n$, so (N) holds.

Quick route: raise an index¶

The metric trace need not be rederived from scratch. It collapses, in two lines, to the trace of a genuine endomorphism — to which Trace From Linearity and Cyclicity applies verbatim, since that endomorphism is an ordinary matrix.

A $(0,2)$ tensor carries no trace of its own; the metric supplies one by raising an index. Lowering the upper index of an endomorphism $S^{k}{}_{j}$ with $g$ produces a $(0,2)$ tensor,

and because $G$ is invertible this is a linear bijection between endomorphisms and $(0,2)$ tensors. Inverting it raises the index back:

Now $S^{k}{}_{j}$ is an ordinary matrix, so Trace From Linearity and Cyclicity hands us its trace at once, $\operatorname{tr}(S)=S^{i}{}_{i}$. Since the trace is intrinsic only on the $(1,1)$ side, we define the trace of the $(0,2)$ tensor $T$ to be that of its raised form — the metric-furnished extension of the $(1,1)$ trace. Substituting (16),

which reproduces (13) — and $\sum_i M_{ii}$ when $G=I$ — with no axiomatic machinery at all. The defining $:=$ is deliberate: (17) is a definition, not a consequence of (L), (I), (N). It is the constructive counterpart of the axiomatic characterization, and the two name the same scalar by the existence-and-uniqueness theorem of Steps 1–4.

The raised index may be either one. Raising the second index of $T$ instead gives the endomorphism $MG^{-1}$, whose trace $\operatorname{tr}(MG^{-1})=\operatorname{tr}(G^{-1}M)$ by cyclicity; so $\operatorname{tr}_g(T)$ does not depend on which slot is raised. (As $g$ is symmetric it sees only the symmetric part of $T$ — see the Remarks.)

Invariance is automatic. Under a change of basis (2), $M\mapsto A^{\mathsf T}MA$ and $G^{-1}\mapsto A^{-1}G^{-1}A^{-\mathsf T}$, so the endomorphism transforms by similarity,

and the conjugation invariance of the trace — Step 1 of Trace From Linearity and Cyclicity, i.e. cyclicity — makes $\operatorname{tr}_g(T)=g^{ij}M_{ij}$ a scalar under the full $GL(V)$, not merely the isometries.

Why uniqueness still needs Step 3. The shortcut yields the formula and its invariance, but not uniqueness — and the obstruction is a commutant. Every linear functional of the endomorphism $S$ has the form $\tilde D(S)=\operatorname{tr}(CS)$ for a unique matrix $C$, and invariance under conjugation by a group $H$,

holds exactly when $C$ commutes with every $A\in H$. So the linear $H$-invariants, up to scale, form the commutant of $H$, and uniqueness means that commutant is only the scalars $\{\lambda I\}$. For $H=GL(V)$ this is the content of Trace From Linearity and Cyclicity: the standard representation is irreducible, so the commutant is scalar. But with $G$ fixed the bijection (16) turns isometry invariance of $T$ only into invariance of $S$ under $H=O(g)$, a smaller group with an a priori larger commutant — so the full-$GL$ theorem does not apply. That the commutant of $O(g)$ collapses to the scalars anyway is the separate fact proved elementarily in Step 3: diagonal sign matrices force $C$ diagonal, permutations force it constant. Index-raising buys existence and invariance; Step 3 buys uniqueness.

Conclusion¶

In every orthonormal basis the linear, isometry-invariant, normalized scalar of a $(0,2)$ tensor is the trace of its component matrix, and in an arbitrary basis it is the metric contraction:

Linearity (L), isometry invariance (I), and the scale (N) force the trace of a $(0,2)$ tensor — taken against the metric — to be the ordinary component trace in any orthonormal frame. $\qquad\blacksquare$

Remarks¶

Only the symmetric part is seen. Since $g^{ij}$ is symmetric, $g^{ij}M_{ij}=g^{ij}M_{(ij)}$ depends only on the symmetric part of $T$; equivalently, an antisymmetric $M_{ij}=-M_{ji}$ has $M_{ii}=0$, so $\sum_i M_{ii}=0$. The metric trace is blind to the 2-form part of $T$.

Why a metric is unavoidable. Without the metric the only invariance available on a $(0,2)$ tensor is the full congruence group, under which no nonzero linear scalar is invariant: writing $D(M)=\operatorname{tr}(C^{\mathsf T}M)$ and taking the scaling $A=cI$ gives $D(A^{\mathsf T}MA)=c^2D(M)$, so $D(A^{\mathsf T}MA)=D(M)$ for all $A\in GL(V)$ already forces $c^2D(M)=D(M)$ for every $c$, hence $D\equiv0$. The determinant escapes this because it is relative, scaling by $\det(A)^2$ rather than staying fixed; the trace cannot, so it must borrow a metric to lower the symmetry to the isometry subgroup $O(g)$, on which $\sum_i M_{ii}$ is invariant. The metric-free home of the trace is instead the mixed $(1,1)$ tensor of Trace From Linearity and Cyclicity, recovered here as $\operatorname{tr}(G^{-1}M)$.

Two routes to uniqueness. The signed-permutation argument of Step 3 is needed only because the assumed symmetry is the geometric one — isometry invariance (I), which the metric cuts down to $O(g)$, strictly weaker than full $GL$. Trading (I) for a stronger algebraic hypothesis removes it. Pulling cyclicity back through the index-raising (16), with $\tilde D(S):=D(GS)$ the cyclicity $\tilde D(S_1S_2)=\tilde D(S_2S_1)$ becomes

“metric-cyclicity” — the trace of a metric-product is symmetric in its two factors. Assuming (L) together with (21) in place of (I) makes $\tilde D$ a linear cyclic functional on ordinary matrices, so Trace From Linearity and Cyclicity delivers $\tilde D=\lambda\operatorname{tr}$ at full $GL(V)$ at once, giving $D(M)=\lambda\,g^{ij}M_{ij}$ with no signed-permutation step. It is a genuine trade-off: the weaker geometric (I) needs Step 3, the stronger algebraic (21) inherits uniqueness directly. Either way uniqueness rests on an assumption — only the constructive definition (17) needs none.

Parallel with the determinant. The determinant note reduces its per-slot relative invariance to two homomorphisms $GL(n,\mathbb C)\to\mathbb C^{*}$ and identifies them with $\det$ via Determinant From Homomorphism. Here the relative invariance reduces, after the character is trivialized in Step 1, to a single linear functional invariant under $O(n,\mathbb C)$-conjugation, identified with $\operatorname{tr}$ via the signed-permutation argument of Step 3 — the orthogonal-group shadow of Trace From Linearity and Cyclicity.