Tensor Contraction and the Trace

Contraction is the arithmetic primitive of every large tensor computation being run right now. In numerical general relativity, codes that model colliding black holes evaluate $R_{\mu\nu}$ and $R$ as literal contractions of a discretized Riemann tensor at billions of grid points, and the LIGO-Virgo-KAGRA waveform templates that matched the 2015 gravitational-wave detection came from exactly this pipeline. In quantum many-body physics and quantum chemistry, tensor-network methods — matrix product states, PEPS, tensor trains — represent a wavefunction as a graph whose every edge is a contracted index, and the feasibility of a simulation is decided entirely by the order in which those contractions are performed. The same operation is the einsum call at the heart of modern machine-learning frameworks, where the attention mechanism of a transformer is a chain of contractions over sequence and feature indices. And in statistical and quantum field theory the trace is inescapable: a partition function is $Z=\mathrm{tr}\,e^{-\beta H}$, a thermal average is a normalized trace, and the cyclicity $\mathrm{tr}(AB)=\mathrm{tr}(BA)$ is the algebraic root of the Kubo-Martin-Schwinger condition defining thermal equilibrium. One index-pairing rule, proven invariant by a single Kronecker delta, runs from a first tensor-algebra lecture to the frontier of computation.

Read an upper index as a plug — an outgoing vector slot — and a lower index as a socket, an incoming covector slot. A tensor is a little machine with some plugs and some sockets sticking out; the tensor product parks two machines side by side and keeps all their legs, while contraction takes one plug and feeds it back into one of the machine's own sockets and sums over the connection. What comes out no longer mentions that particular leg: it has been wired shut. The trace is the extreme case — a $(1,1)$ tensor has exactly one plug and one socket, and joining them leaves a machine with no external legs at all, which is precisely what a scalar is.

The wiring picture also explains why the variance must be opposite. Rotate the basis: the plug is re-expressed with $\Lambda$ and the socket with $\Lambda^{-1}$, and only when a plug plugs into a socket do the two sit on the same summed index and multiply to $\Lambda^{-1}\Lambda=\delta$, cancelling exactly. Wire a plug to another plug and you get $\Lambda\Lambda$ with no inverse in sight — the number changes when you turn your head, so it was never a property of the tensor. This is the geometric content of "you need a metric to dot two vectors": two vectors live in the same space with no built-in pairing, whereas a vector and a covector pair for free through $\delta^{i}{}_{j}$. The single sharpest illustration that the trace is a real invariant, not an artifact of the diagonal you happen to see, is the rotation matrix. A rotation about any axis by angle $\theta$ has eigenvalues $1,e^{i\theta},e^{-i\theta}$, so its trace is $1+2\cos\theta$ — regardless of the axis. Turn the same physical rotation into a basis where its matrix has an all-zero diagonal and the trace is still $1+2\cos\theta$; the off-diagonal entries reshuffle but the contraction does not budge.

Contraction is defined on components: choose one upper slot and one lower slot, set their labels equal, and sum over that repeated label. The map lowers both the contravariant and covariant degree by one.

where: $T$ — a type-$(p,q)$ tensor with $p$ upper and $q$ lower indices; $a$ — the repeated (summed) index, one chosen upper slot equated to one chosen lower slot; $C$ — the contraction map, whose output is type $(p-1,q-1)$. Reading: pick a plug and a socket, wire them together, sum; the degree drops by two. The trace is the case $p=q=1$, the total contraction of an operator. There is only one pair to contract, so the trace is unique and produces a pure number.

where: $A^{i}{}_{j}$ — components of a linear operator on an $n$-dimensional space $V$; $n$ — the dimension of $V$. Reading: the trace sums the diagonal, but because it is a contraction it is basis-free — the diagonal you sum is not. Two identities make the trace usable. Cyclicity follows from relabelling dummy indices, and the trace of the identity counts dimensions.

where: $A,B$ — operators (square matrices) on $V$; $\mathbb{1}_{n}$ — the $n\times n$ identity, with $\mathrm{tr}\,\mathbb{1}_{n}=\delta^{i}{}_{i}=n$. Reading: you may cycle factors inside a trace but not arbitrarily permute them, and $\mathrm{tr}(ABC)=\mathrm{tr}(BCA)=\mathrm{tr}(CAB)$. The engine underneath every contraction is the dual pairing, whose components are the Kronecker delta — the identity on indices, whose own full contraction is the dimension.

where: $e_{j}$ — a basis vector of $V$; $e^{i}$ — the dual basis covector of $V^{*}$; $\langle\cdot,\cdot\rangle$ — the canonical, metric-free pairing. Reading: $\delta^{i}{}_{j}$ is the only $(1,1)$ tensor with the same components in every basis, which is why it is the kernel that survives a contraction. When no covector is available, a metric supplies one: contracting a vector with $g_{ij}$ lowers its index, and $g_{ij}g^{jk}$ returns the delta. This is where inner products and index gymnastics enter as contractions.

where: $g_{ij}$ — the metric (type $(0,2)$); $g^{jk}$ — its inverse (type $(2,0)$); $u^{i},v^{j},V^{j}$ — vector components; $V_{i}$ — the covariant version obtained by contracting with $g$. Reading: raising and lowering indices is nothing but contraction with $g^{ij}$ or $g_{ij}$, which is why a metric is the missing ingredient for same-variance contraction. The payoff on a manifold is curvature. Contracting the Riemann tensor once gives the Ricci tensor; contracting it again with the inverse metric gives the scalar curvature — the two invariants from which Einstein built gravity.

where: $R^{\lambda}{}_{\mu\lambda\nu}$ — the Riemann curvature tensor with its first upper index contracted against its third (lower) index; $R_{\mu\nu}$ — the Ricci tensor; $R$ — the Ricci (scalar) curvature. Reading: the whole chain from curvature to Einstein's equations is a sequence of contractions. Finally, contraction commutes with the covariant derivative — the property that makes contracted tensor fields, and hence conservation laws, well defined on curved space.

where: $\nabla_{k}$ — the covariant derivative; $T^{j}{}_{i}$ — a $(1,1)$ tensor field. Reading: the connection acts on an upper index with $+\Gamma$ and on a lower index with $-\Gamma$, so on a contracted pair the two Christoffel terms cancel and the derivative passes straight through the contraction.

Contraction is not one operation among many — it is the operation that most familiar products turn out to be. Reading a bilinear expression as "which index is summed against which" identifies it instantly:

1 · The trace of a rotation is a genuine invariant (fully worked)

A proper rotation $R\in SO(3)$ turns space about some axis by angle $\theta$. Show that $\mathrm{tr}\,R=1+2\cos\theta$ independent of the axis, and check it on two rotations whose matrices look nothing alike.

Step 1. Fix the eigenvalues. A real rotation preserves lengths (unit-modulus eigenvalues), fixes its axis (eigenvalue $1$), and being real has conjugate pairs $e^{\pm i\theta}$ — so the spectrum is $\{1,e^{i\theta},e^{-i\theta}\}$ for every axis. By Derivation 3, $\mathrm{tr}\,R=1+e^{i\theta}+e^{-i\theta}=\mathbf{1+2\cos\theta}$, with no reference to the axis.

Step 2. Test it on a rotation with a friendly diagonal. Rotating by $\theta=60^{\circ}$ about the $z$-axis gives $R_{z}$ with diagonal $(\cos60^{\circ},\cos60^{\circ},1)=(0.5,0.5,1)$, so $\mathrm{tr}\,R_{z}=0.5+0.5+1=\mathbf{2}$. The formula predicts $1+2\cos60^{\circ}=1+2(0.5)=2$. They agree.

Step 3. Test it on a rotation with a hostile diagonal. The cyclic permutation $x\to y\to z\to x$ is the matrix with rows $(0,0,1),(1,0,0),(0,1,0)$ — a rotation by $120^{\circ}$ about the body diagonal $(1,1,1)$, with $\det=+1$, so it lies in $SO(3)$. Its diagonal is all zeros, so $\mathrm{tr}=0+0+0=\mathbf{0}$, while the formula gives $1+2\cos120^{\circ}=1+2(-0.5)=0$. They agree again, even though not a single diagonal entry survived.

Step 4. The moral: the two matrices share no entries and one has an empty diagonal, yet $R^{i}{}_{i}$ returns the invariant $1+2\cos\theta$ both times, because it equals the basis-blind eigenvalue sum. The trace is a property of the rotation, not of the array you wrote to represent it.

2 · Curvature by double contraction: the scalar curvature of a sphere (fully worked)

A 2-sphere of radius $a$ has metric $g_{\theta\theta}=a^{2}$, $g_{\phi\phi}=a^{2}\sin^{2}\theta$ (diagonal). Its single independent Riemann component corresponds to Gaussian curvature $K=1/a^{2}$, so the Ricci tensor is $R_{\mu\nu}=K\,g_{\mu\nu}$. Contract twice to find the scalar curvature.

Step 1. Build the Ricci tensor from the maximally symmetric relation $R_{\mu\nu}=K\,g_{\mu\nu}=\tfrac{1}{a^{2}}g_{\mu\nu}$. Component-wise, $R_{\theta\theta}=\tfrac{1}{a^{2}}\cdot a^{2}=\mathbf{1}$ and $R_{\phi\phi}=\tfrac{1}{a^{2}}\cdot a^{2}\sin^{2}\theta=\mathbf{\sin^{2}\theta}$. This Ricci tensor is itself the first contraction $R_{\mu\nu}=R^{\lambda}{}_{\mu\lambda\nu}$ of the Riemann tensor.

Step 2. Assemble the inverse metric needed for the second contraction: because $g$ is diagonal, $g^{\theta\theta}=1/a^{2}$ and $g^{\phi\phi}=1/(a^{2}\sin^{2}\theta)$.

Step 3. Contract Ricci with the inverse metric, $R=g^{\mu\nu}R_{\mu\nu}=g^{\theta\theta}R_{\theta\theta}+g^{\phi\phi}R_{\phi\phi}$. Substituting, $R=\dfrac{1}{a^{2}}\cdot 1+\dfrac{1}{a^{2}\sin^{2}\theta}\cdot\sin^{2}\theta=\dfrac{1}{a^{2}}+\dfrac{1}{a^{2}}=\mathbf{\dfrac{2}{a^{2}}}$.

Step 4. Sanity-check. The result is a pure number times $a^{-2}$, as a curvature must be, positive and constant as it should be for a round sphere; the unit sphere $a=1$ gives $R=\mathbf{2}$, and it reproduces $R=2K$ in two dimensions. Contracting Riemann twice has collapsed the entire curvature of the sphere into the single invariant $2/a^{2}$ — the smaller the sphere, the larger its curvature.

3 · Cyclicity and the Frobenius norm (compact check)

For $A=\left(\begin{smallmatrix}1&2\\3&4\end{smallmatrix}\right)$, $B=\left(\begin{smallmatrix}0&1\\1&0\end{smallmatrix}\right)$: verify cyclicity and read the Frobenius norm as a contraction. Here $AB=\left(\begin{smallmatrix}2&1\\4&3\end{smallmatrix}\right)$ and $BA=\left(\begin{smallmatrix}3&4\\1&2\end{smallmatrix}\right)$ share no entries, yet $\mathrm{tr}(AB)=2+3=\mathbf{5}=3+2=\mathrm{tr}(BA)$ — cyclicity confirmed. And the full double contraction $\mathrm{tr}(A^{\mathsf T}A)=\sum_{i,j}A_{ij}A_{ij}=1+4+9+16=\mathbf{30}$ is the squared length of $A$ read as a vector, the same pattern that makes $g_{ij}u^{i}v^{j}$ an inner product.

Contraction and the trace look like linear-algebra bookkeeping until you follow them onto a curved manifold, where they become the machinery that turns raw curvature into the invariants of gravity. The man who built that machinery — the absolute differential calculus, in which contraction is a named, coordinate-free operation — is the mathematician whose contraction of the Riemann tensor still carries his name.

Texts. Stone & Goldbart, Mathematics for Physics (2009), Ch. 10 (Vectors and Tensors) · Arfken, Weber & Harris, Mathematical Methods for Physicists, 7th ed., Ch. 4 (Tensors) · Hassani, Mathematical Physics, 2nd ed., Chs. 25–26 · Nakahara, Geometry, Topology and Physics, 2nd ed., Ch. 5 · Misner, Thorne & Wheeler, Gravitation, Ch. 3 · Byron & Fuller, Mathematics of Classical and Quantum Physics, Ch. 2. The key numerical results in this article are: the $\Lambda^{-1}\Lambda=\delta$ cancellation underlying each invariance proof; the cyclicity check $\mathrm{tr}(AB)=\mathrm{tr}(BA)=5$ for the given matrices; the rotation traces $1+2\cos60^{\circ}=2$ and $1+2\cos120^{\circ}=0$; and the sphere scalar curvature $R=g^{\theta\theta}R_{\theta\theta}+g^{\phi\phi}R_{\phi\phi}=2/a^{2}$.

Builds on: Tensor Algebra and the Tensor Product · Covariant and Contravariant Tensors · The Metric Tensor. Leads to: Raising and Lowering Indices · The Covariant Derivative · The Riemann Curvature Tensor · The Hodge Star and the Codifferential.

Ask the tutor: Ask me to contract any tensor expression you meet — a matrix product, a divergence, or the Riemann-to-Ricci step — and I will show you which index pairs with which and why the result is invariant.