General Tensors and the Tensor Product

Every quantitative structure in general relativity — the metric that measures proper time, the curvature that bends light, the stress–energy that curves space — is one kind of object wearing different numbers of indices. That object is the tensor, and it is built by a single construction repeated as often as you like. Hand me the tangent space $T_xM$ at a point, established in aq-gr-tangent-vectors-tangent-space, and its dual the cotangent space $T_x^*M$ from aq-gr-cotangent-space-one-forms, and I can manufacture the entire zoo: a tensor of type $(p,q)$ is nothing more than a machine with $p$ slots that swallow covectors and $q$ slots that swallow vectors, spitting out a real number, linear in each slot on its own. Two facts make this humble definition the load-bearing wall of the whole subject. First, multilinearity means the machine is pinned down completely by what it does to the basis vectors and basis one-forms — and those answers are exactly the components $T^{\mu_1\cdots\mu_p}{}_{\nu_1\cdots\nu_q}$ you see stacked with indices in every equation. The abstract object and the index array are two views of one thing. Second, the operation that welds small machines into big ones — the tensor product $\otimes$ — is associative and bilinear but pointedly not commutative, because slot one is not slot two. That single asymmetry is why index order carries physical meaning and why symmetrization exists at all. Get this construction right and the metric, the Riemann tensor, contraction, and general covariance are all downstream corollaries of one pointwise piece of linear algebra.

Key takeaways

A type- $(p,q)$ tensor at a point is a multilinear map that eats $p$ covectors and $q$ vectors and returns a real number — equivalently an element of $T_xM^{\otimes p}\otimes T_x^*M^{\otimes q}$ .
Multilinearity forces the map to be fixed by its values on basis arguments; those $n^{p+q}$ values are the components $T^{\mu_1\cdots}{}_{\nu_1\cdots}=T(\mathrm{d}x^{\mu_1},\dots,\partial_{\nu_1},\dots)$ .
The tensor product $\otimes$ glues a $(p,q)$ tensor to an $(r,s)$ tensor by multiplying their outputs on separate slots, giving a $(p{+}r,\,q{+}s)$ tensor; it is bilinear and associative but not commutative.
The space of $(p,q)$ tensors is itself a vector space of dimension $n^{p+q}$ ; in $n=4$ spacetime a $(1,3)$ tensor like Riemann has $4^4=256$ components before symmetry cuts it to $20$ .
Tensors carry physical law because a genuine tensor that vanishes in one chart vanishes in all — the precise content of general covariance.
Key idea: a type- $(p,q)$ tensor IS a multilinear map $\underbrace{T_x^*M\times\cdots}_{p}\times\underbrace{T_xM\times\cdots}_{q}\to\mathbb{R}$ ; multilinearity means it is fixed by its values on basis arguments, and those values are the components $T^{\mu_1\cdots\mu_p}{}_{\nu_1\cdots\nu_q}$ .

Where this lives today

The tensor-product construction is not a historical formality; it is the daily working language of gravitational physics. On the right-hand side of Einstein's equation $G_{\mu\nu}=8\pi G\,T_{\mu\nu}/c^4$ sits the stress–energy tensor, a symmetric type- $(0,2)$ object whose ten independent components in four dimensions encode energy density, momentum flux, and pressure all at once — and it is because both sides are genuine tensors that the field equation holds in every coordinate system, the operational meaning of background independence. When LIGO and Virgo matched the waveform of GW150914 in 2015 against templates, those templates were solutions of tensor equations, and the numerical-relativity codes that generate them (the BSSN and moving-puncture formulations) are engineered to evolve tensors and tensor densities correctly while the coordinate grid sloshes through the merging horizons; a code that silently treated a non-tensor as a tensor would emit coordinate-dependent garbage dressed up as gravitational waves. The same $\otimes$ that concatenates slots here is, remarkably, the exact operation that builds the state space of a composite quantum system: the Hilbert space of two particles is the tensor product of the individual spaces, and the non-factorizable elements of that product are precisely the entangled states. This shared algebra is why the "tensor network" methods now central to condensed-matter and quantum-information research, and the multilinear machinery of general relativity, are recognisably the same mathematics — a crystallographer's word for a stretched solid now spanning quantum entanglement and the curvature of the cosmos.

The physical picture

Picture a tensor as a machine with labelled slots. A type- $(p,q)$ machine has $p$ slots on top that only accept covectors and $q$ slots below that only accept vectors. Feed the right object into every slot and a single real number falls out, and the machine is linear in each slot separately — double the input in one slot and the output doubles, holding the others fixed. Nothing else is required of it. There is no metric, no connection, no notion of length hiding inside; the machine is pure multilinear bookkeeping, assembled fresh at each point of the manifold from the tangent and cotangent spaces already sitting there. The components are simply the machine's answers on the "unit test objects." Feed in the basis one-forms $\mathrm{d}x^\mu$ where the covector slots go and the basis vectors $\partial_\nu$ where the vector slots go, and the numbers that come out are $T^{\mu_1\cdots}{}_{\nu_1\cdots}$ . Because the machine is multilinear, these numbers are everything: any other input is a weighted sum of basis inputs, and linearity reassembles the full answer from the component table. This is why an equation full of indices and a coordinate-free geometric statement are the same claim written in two alphabets.

Figure

A type (1,2) tensor drawn as a machine with one covector slot on top and two vector slots below. Feeding the basis one-form and two basis vectors reads off a single component; in n = 4 spacetime the machine holds 4³ = 64 such numbers

The tensor product is the act of bolting two machines together side by side so the combined machine owns all the slots of both. Its non-commutativity is not a technicality but the whole reason index order matters: putting object $A$ in the first slot and $B$ in the second is a genuinely different arrangement from the reverse, so $v\otimes w$ and $w\otimes v$ are different tensors even though they are built from the same two vectors. The payoff of the entire framework is coordinate independence at the level of the object: an equation between tensors, $A=B$ , becomes $A^{\cdots}{}_{\cdots}=B^{\cdots}{}_{\cdots}$ in every chart at once, so you may prove a law in whatever frame is easiest and trust it everywhere. The second figure shows how slots concatenate under $\otimes$ and how fast the component count grows with rank.

Figure

The tensor product concatenates slots: an S of type (1,1) glued to a T of type (0,1) makes an S ⊗ T of type (1,2), and dimensions multiply. On n = 4 spacetime the Riemann machine of type (1,3) carries 4⁴ = 256 components, which its symmetries reduce to 20 independent numbers

The core equations

The starting point is the definition itself: a type- $(p,q)$ tensor at the point $x$ is a real-valued function of $p$ covector arguments and $q$ vector arguments, linear in each argument separately. Everything downstream is an unpacking of this one line.

T:\;\underbrace{T_x^*M\times\cdots\times T_x^*M}_{p}\times\underbrace{T_xM\times\cdots\times T_xM}_{q}\longrightarrow\mathbb{R} \qquad (1)

where: $T_xM$ — tangent space at $x$ , the vectors; $T_x^*M$ — cotangent space at $x$ , the covectors or one-forms; $p$ — contravariant rank, the number of covector slots and hence upper indices; $q$ — covariant rank, the number of vector slots and hence lower indices. Physical reading: a tensor is a linear-in-everything answer machine, defined pointwise, with no metric required. Because the tensor is multilinear, its whole content lives in what it returns on the basis arguments. Those returns are its components, and they rely on the one structural fact linking the two bases: the coordinate one-forms are dual to the coordinate vectors.

\mathrm{d}x^\mu(\partial_\nu)=\delta^\mu{}_\nu,\qquad T^{\mu_1\cdots\mu_p}{}_{\nu_1\cdots\nu_q}\equiv T\!\left(\mathrm{d}x^{\mu_1},\dots,\mathrm{d}x^{\mu_p},\partial_{\nu_1},\dots,\partial_{\nu_q}\right) \qquad (2)

where: $\partial_\mu=\partial/\partial x^\mu$ — coordinate basis vector of $T_xM$ ; $\mathrm{d}x^\nu$ — coordinate basis one-form dual to it; $\delta^\mu{}_\nu$ — Kronecker delta, $1$ if $\mu=\nu$ else $0$ . Physical reading: probe the machine with the standard basis and record its answers — those $n^{p+q}$ numbers are the tensor in coordinates. Running the definition backwards, the tensor is reconstructed from its components by re-attaching the basis objects with tensor products. This expansion is the coordinate face of the geometric object.

T=T^{\mu_1\cdots\mu_p}{}_{\nu_1\cdots\nu_q}\;\partial_{\mu_1}\otimes\cdots\otimes\partial_{\mu_p}\otimes\mathrm{d}x^{\nu_1}\otimes\cdots\otimes\mathrm{d}x^{\nu_q} \qquad (3)

where: the product $\partial_{\mu_1}\otimes\cdots\otimes\mathrm{d}x^{\nu_q}$ — a basis tensor of type $(p,q)$ ; Einstein summation over each repeated index pair is implied. Physical reading: the components are the coefficients of a tensor in the basis $\{\partial_{\mu}\otimes\cdots\otimes\mathrm{d}x^{\nu}\}$ , exactly as $v^\mu$ are the coefficients of a vector in $\{\partial_\mu\}$ . The simplest non-trivial tensor product welds a vector to a one-form. It produces a rank-two machine whose value on a covector and a vector is just the product of the two separate pairings — the archetype from which the general rule is read off.

(v\otimes\omega)(\alpha,u)=\langle\alpha,v\rangle\,\langle\omega,u\rangle=\alpha(v)\,\omega(u) \qquad (4)

where: $v$ — a vector of type $(1,0)$ , answering the covector slot; $\omega$ — a one-form of type $(0,1)$ , answering the vector slot; $\alpha,u$ — the covector and vector arguments supplied to the new tensor. Physical reading: each factor answers its own argument and the answers multiply — slots do not talk to each other. The general tensor product obeys the same law: split the arguments between the two factors, let each factor evaluate its share, and multiply. In components this is simply the outer product of the two component arrays.

(S\otimes T)^{\mu_1\cdots\mu_p\,\rho_1\cdots\rho_r}{}_{\nu_1\cdots\nu_q\,\sigma_1\cdots\sigma_s}=S^{\mu_1\cdots\mu_p}{}_{\nu_1\cdots\nu_q}\;T^{\rho_1\cdots\rho_r}{}_{\sigma_1\cdots\sigma_s} \qquad (5)

where: $S$ — a type- $(p,q)$ tensor; $T$ — a type- $(r,s)$ tensor; $S\otimes T$ — the resulting type- $(p{+}r,\,q{+}s)$ tensor; $\rho,\sigma$ — indices belonging to the second factor. Physical reading: $\otimes$ is bilinear and associative but not commutative, since swapping $S$ and $T$ reassigns which object owns which slot. Counting basis tensors gives the dimension of the whole space. Each of the $p+q$ index positions ranges over $n$ values independently, so the component arrays fill an $n^{p+q}$ -dimensional vector space.

\dim\!\big(T^{(p,q)}_xM\big)=n^{\,p+q},\qquad n=\dim M \qquad (6)

where: $n$ — dimension of the manifold, $4$ for spacetime; $T^{(p,q)}_xM$ — the vector space of all type- $(p,q)$ tensors at $x$ . Physical reading: tensors of a fixed type add and scale like ordinary vectors, so linear algebra applies to them wholesale. Among all types, $(1,1)$ is special: one up and one down index make it a linear map from vectors to vectors, and exactly one such tensor has the same components in every chart.

v^\mu=T^\mu{}_\nu\,u^\nu \;\;\Longleftrightarrow\;\; T:T_xM\to T_xM,\qquad \delta^\mu{}_\nu=\text{the unique invariant }(1,1) \qquad (7)

where: $T^\mu{}_\nu$ — components of the endomorphism; $u^\nu$ — input vector; $\delta^\mu{}_\nu$ — Kronecker delta, whose two Jacobian factors cancel so it reads identically in all coordinates. Physical reading: a $(1,1)$ tensor is a coordinate-free matrix, and the identity map is the only matrix every observer agrees on.

Derivation AWhy the component array completely determines the tensor

The goal is to prove the claim on which the entire index formalism rests: that a multilinear map is fixed with no remaining freedom by the finite table of numbers it returns on basis arguments. If this failed, two distinct tensors could share every component and the index notation would be ambiguous. We show instead that multilinearity forces any input to be reconstructed from basis inputs, so the $n^{p+q}$ components pin the map exactly. We argue the representative $(1,1)$ case, and the pattern extends verbatim to any $(p,q)$ .

Step 1. Fix the dual bases and define the components by probing, $T^\mu{}_\nu\equiv T(\mathrm{d}x^\mu,\partial_\nu)$ , using $\mathrm{d}x^\mu(\partial_\nu)=\delta^\mu{}_\nu$ . Why: the components are not extra data bolted onto the tensor — they are literally its outputs on the standard test objects, so they are guaranteed to exist and be finite.

Step 2. Expand arbitrary arguments in the bases, $\alpha=\alpha_\mu\,\mathrm{d}x^\mu$ and $u=u^\nu\,\partial_\nu$ , where $\alpha_\mu=\alpha(\partial_\mu)$ and $u^\nu=\mathrm{d}x^\nu(u)$ . Why: every covector and vector is a coordinate combination of basis elements, so no input lies outside the reach of the basis — this is what finite dimension buys us.

Step 3. Push the expansion through the map using linearity in each slot, $T(\alpha,u)=\alpha_\mu\,u^\nu\,T(\mathrm{d}x^\mu,\partial_\nu)=T^\mu{}_\nu\,\alpha_\mu\,u^\nu$ . Why: multilinearity lets the scalar coefficients $\alpha_\mu,u^\nu$ come outside, leaving only the basis evaluations — so the answer on any input is a fixed contraction of the component table.

Step 4. Conclude uniqueness. If two tensors $T$ and $T'$ have $T^\mu{}_\nu=T'^\mu{}_\nu$ for all $\mu,\nu$ , then $T(\alpha,u)=T'(\alpha,u)$ for every $\alpha,u$ , hence $T=T'$ as maps. Why: equal component tables give equal outputs everywhere, so the map is determined — the array and the object are one, and the same argument with more slots settles the general $(p,q)$ case.

Go deeper · Derivation B · 1 minThe tensor product builds every higher tensor and fixes the dimension count

Derivation BThe tensor product builds every higher tensor and fixes the dimension count

The goal is to show that the products of basis vectors and basis one-forms are not merely a set of tensors but a genuine basis of the whole type- $(p,q)$ space, which simultaneously proves the dimension formula $n^{p+q}$ and justifies writing any tensor as a sum of tensor-product terms. This is the abstract engine behind the index expansion of equation (3) and behind the universal property that characterizes $\otimes$ as the space where multilinear maps become linear.

Step 1. Form the candidate basis, the $n^{p+q}$ tensors $\partial_{\mu_1}\otimes\cdots\otimes\partial_{\mu_p}\otimes\mathrm{d}x^{\nu_1}\otimes\cdots\otimes\mathrm{d}x^{\nu_q}$ , one for each choice of the $p+q$ indices. Why: these are the outputs of $\otimes$ on basis inputs, and by construction each is a legitimate type- $(p,q)$ multilinear map, so the candidate set lives in the right space.

Step 2. Show they span. Given any $T$ , define numbers $T^{\mu\cdots}{}_{\nu\cdots}=T(\mathrm{d}x^\mu,\dots,\partial_\nu,\dots)$ and check that $T$ and the sum $T^{\mu\cdots}{}_{\nu\cdots}\,\partial_\mu\otimes\cdots\otimes\mathrm{d}x^\nu$ agree on all basis arguments; by the previous derivation they are therefore equal. Why: agreement on basis inputs is enough for equality, so every tensor is a coordinate combination of the candidate set — the set spans.

Step 3. Show they are independent. If a combination $c^{\mu\cdots}{}_{\nu\cdots}\,\partial_\mu\otimes\cdots\otimes\mathrm{d}x^\nu$ is the zero tensor, evaluate it on $(\mathrm{d}x^{\alpha},\dots,\partial_{\beta},\dots)$ to read off $c^{\alpha\cdots}{}_{\beta\cdots}=0$ . Why: probing with basis arguments isolates one coefficient at a time via the deltas, so the only vanishing combination is the trivial one — the set is linearly independent.

Step 4. Count. A spanning, independent set is a basis, and it has one member per assignment of $p+q$ indices each ranging over $n$ values, so $\dim T^{(p,q)}_xM=n^{p+q}$ . Why: with $n=4$ this yields $4^2=16$ for the metric, $4^4=256$ for Riemann, and $4^5=1024$ for a $(2,3)$ tensor — the numbers used throughout this article, now derived rather than asserted.

Go deeper · Derivation C · 1 minComponents carry one Jacobian per index

Derivation CComponents carry one Jacobian per index

The goal is to derive how the component array must change between two charts so that the geometric object $T$ stays fixed, recovering the transformation law that most textbooks promote to the very definition of a tensor. Demanding invariance of the object and substituting the chain-rule transformations of the basis factors will attach exactly one Jacobian to each contravariant index and one inverse Jacobian to each covariant index — the tensor product of the vector and one-form rules. The full machinery is owned by aq-gr-tensor-transformation-law; here we extract the law directly from the basis expansion.

Step 1. Write the same object in the chart $x$ , $T=T^{\mu_1\cdots}{}_{\nu_1\cdots}\,\partial_{\mu_1}\otimes\cdots\otimes\mathrm{d}x^{\nu_1}\otimes\cdots$ . Why: the tensor is basis-independent, so a coordinate change alters only the split into components times basis — track how each basis factor moves and the component rule follows.

Step 2. Substitute the basis transformations $\partial_\mu=\dfrac{\partial x^{\mu'}}{\partial x^\mu}\,\partial_{\mu'}$ for each vector factor and $\mathrm{d}x^\nu=\dfrac{\partial x^\nu}{\partial x^{\nu'}}\,\mathrm{d}x^{\nu'}$ for each one-form factor. Why: a basis vector is a directional derivative and drags the Jacobian $\partial x'/\partial x$ ; a basis one-form is a coordinate gradient and drags the inverse — the two opposite rules are the origin of upper versus lower.

Step 3. Collect the coefficient of the new basis tensor $\partial_{\mu'_1}\otimes\cdots\otimes\mathrm{d}x^{\nu'_1}\otimes\cdots$ , giving $T^{\mu'_1\cdots\mu'_p}{}_{\nu'_1\cdots\nu'_q}=\dfrac{\partial x^{\mu'_1}}{\partial x^{\mu_1}}\cdots\dfrac{\partial x^{\mu'_p}}{\partial x^{\mu_p}}\dfrac{\partial x^{\nu_1}}{\partial x^{\nu'_1}}\cdots\dfrac{\partial x^{\nu_q}}{\partial x^{\nu'_q}}\,T^{\mu_1\cdots}{}_{\nu_1\cdots}$ . Why: whatever multiplies the new basis is by definition the new component array, so the law reads off — one Jacobian per upper index, one inverse Jacobian per lower, matching Carroll Eq. (1.51) and Wald (2.3.6).

Step 4. Check the corollary that makes tensors carry physics: the law is linear and homogeneous in the old components, so if every $T^{\mu\cdots}{}_{\nu\cdots}=0$ in one chart, every $T^{\mu'\cdots}{}_{\nu'\cdots}=0$ in all charts. Why: a genuine tensor equation, once true, is true for every observer — this homogeneity is exactly the operational content of general covariance.

Limiting cases

Every familiar geometric object is a low-rank tensor; dialling $(p,q)$ selects the number and kind of slots.

Case	Result	Meaning
Type $(0,0)$ — scalars	$f:M\to\mathbb{R}$	No slots: a plain number or field, invariant under any change of basis.
Type $(1,0)$ — vectors	$v=v^\mu\,\partial_\mu$	One covector slot; the tangent vectors, acting on one one-form.
Type $(0,1)$ — one-forms	$\omega=\omega_\mu\,\mathrm{d}x^\mu$	One vector slot; the cotangent vectors, acting on one vector.
Type $(0,2)$ — bilinear forms	$g=g_{\mu\nu}\,\mathrm{d}x^\mu\otimes\mathrm{d}x^\nu$	Two vector slots; the symmetric instance is the metric, the antisymmetric a 2-form.
Type $(1,1)$ — linear maps	$T=T^\mu{}_\nu\,\partial_\mu\otimes\mathrm{d}x^\nu$	An endomorphism of $T_xM$ ; the identity $\delta^\mu{}_\nu$ is the unique chart-independent one.
Type $(1,3)$ — Riemann	$R^\rho{}_{\sigma\mu\nu}$	Three vector slots and one covector slot; $4^4=256$ components, $20$ independent in $n=4$ .

Common trapA tensor is the same thing as its array of components

It is not, and conflating the two is the single most common source of confusion in the subject. A tensor is a coordinate-free geometric object — a multilinear machine — that exists before any chart is chosen; its component array is only the machine's readout in one particular basis, and that array changes the instant you relabel the coordinates while the object itself never budges. Two consequences follow. First, you may never assume $\otimes$ commutes: $v\otimes w\neq w\otimes v$ in general, because the first assigns $v$ to slot one and $w$ to slot two while the second swaps them, and their component arrays $v^\mu w^\nu$ and $w^\mu v^\nu$ are transposes, not equals. Only after explicit symmetrization may indices be freely permuted. Second, carrying indices does not make an array a tensor — the components must transform by the law of the third derivation under every admissible coordinate change, which the Christoffel symbols and the bare partial derivative $\partial_\mu V^\nu$ notoriously fail. Always test the transformation before trusting the name.

Worked examples

1 · Counting the components of curvature (fully numeric)

On $n=4$ spacetime, how many components does a general type- $(2,3)$ tensor carry with no symmetry imposed, and how does the naive count for the Riemann tensor compare with its true number of independent components?

Step 1. Apply the dimension formula $n^{p+q}$ . For the $(2,3)$ tensor, $p+q=5$ , so the count is $4^{5}=1024$ . Every one of the five index slots ranges independently over the four spacetime directions, and $4\times4\times4\times4\times4=\mathbf{1024}$ .

Step 2. Now the Riemann tensor $R^{\rho}{}_{\sigma\mu\nu}$ is type $(1,3)$ , so its naive count is $4^{1+3}=4^{4}=\mathbf{256}$ . This is what the raw tensor-product basis of the second derivation delivers before any of curvature's symmetries are used.

Step 3. The symmetries — antisymmetry in the last pair, in the first pair after lowering, pair-exchange symmetry, and the first Bianchi identity — collapse this dramatically. The standard result is $\tfrac{1}{12}n^{2}(n^{2}-1)$ , which for $n=4$ gives $\tfrac{1}{12}\cdot16\cdot15=\tfrac{240}{12}=\mathbf{20}$ . So the tensor-product construction hands us $256$ slots, of which only $20$ are physically independent in four dimensions — the rest are fixed by symmetry. The tensor-product count is the ceiling; symmetry does the carving.

2 · The tensor product of two vectors, and why order matters (fully numeric)

Work on a 2-dimensional tangent space with basis $\{\partial_1,\partial_2\}$ . Take $v=2\,\partial_1+3\,\partial_2$ and $w=\partial_1-\partial_2$ . Compute the components of the $(2,0)$ tensor $v\otimes w$ , compare with $w\otimes v$ , and verify the result by feeding in basis one-forms.

Step 1. By equation (5) the components are the outer product $(v\otimes w)^{\mu\nu}=v^\mu w^\nu$ , with $v^\mu=(2,3)$ and $w^\nu=(1,-1)$ . This gives $(v\otimes w)^{11}=2\cdot1=2$ , $(v\otimes w)^{12}=2\cdot(-1)=-2$ , $(v\otimes w)^{21}=3\cdot1=3$ , $(v\otimes w)^{22}=3\cdot(-1)=-3$ , i.e. the matrix $\begin{pmatrix}2&-2\\3&-3\end{pmatrix}$ .

Step 2. Now reverse the factors: $(w\otimes v)^{\mu\nu}=w^\mu v^\nu$ gives $\begin{pmatrix}1\cdot2 & 1\cdot3\\ -1\cdot2 & -1\cdot3\end{pmatrix}=\begin{pmatrix}2&3\\-2&-3\end{pmatrix}$ . Comparing entry by entry, $(v\otimes w)^{12}=-2$ but $(w\otimes v)^{12}=3$ : the two are transposes of each other and are not equal, the concrete meaning of $v\otimes w\neq w\otimes v$ .

Step 3. Verify against the defining action $(v\otimes w)(\alpha,\beta)=\alpha(v)\,\beta(w)$ . Feed the basis one-forms $\alpha=\mathrm{d}x^1$ and $\beta=\mathrm{d}x^2$ : then $\alpha(v)=v^1=2$ and $\beta(w)=w^2=-1$ , so the value is $2\cdot(-1)=\mathbf{-2}$ , which matches the component $(v\otimes w)^{12}=-2$ read off in Step 1. The machine and its array agree exactly, as the first derivation guaranteed.

3 · A (1,1) tensor as a linear map and its invariant trace (fully numeric)

On the same 2-dimensional space let a type- $(1,1)$ tensor have components $T^\mu{}_\nu=\begin{pmatrix}2&1\\0&3\end{pmatrix}$ . Act with it on the vector $u=\partial_1+4\,\partial_2$ , and compute its trace as an invariant scalar. Contrast with the identity tensor.

Step 1. By equation (7) the image has components $v^\mu=T^\mu{}_\nu\,u^\nu$ with $u^\nu=(1,4)$ . The first component is $v^1=T^1{}_1u^1+T^1{}_2u^2=2\cdot1+1\cdot4=\mathbf{6}$ ; the second is $v^2=T^2{}_1u^1+T^2{}_2u^2=0\cdot1+3\cdot4=\mathbf{12}$ . So $T$ sends $u\mapsto 6\,\partial_1+12\,\partial_2$ , acting as an honest endomorphism of the tangent space.

Step 2. Contract the upper index against the lower to form the trace $T^\mu{}_\mu=T^1{}_1+T^2{}_2=2+3=\mathbf{5}$ . Because this pairs one contravariant slot against one covariant slot, its single Jacobian and inverse Jacobian cancel — the trace is a genuine scalar, the same number in every chart, unlike any individual component.

Step 3. Finally the identity tensor $\delta^\mu{}_\nu=\begin{pmatrix}1&0\\0&1\end{pmatrix}$ maps every $u$ to itself and has trace $\delta^\mu{}_\mu=n=\mathbf{2}$ here. It is the unique $(1,1)$ tensor whose components are identical in all coordinate systems, because its Jacobian and inverse-Jacobian factors cancel identically — the invariant heart of the $(1,1)$ space.

Physics of the portrait

The word "tensor" reached physics through crystals, not gravity. Long before spacetime curved, someone needed a name for a quantity that points differently in different directions inside a stressed solid — and that name, coined in the elasticity laboratories of Göttingen, is the one we now stamp on the curvature of the universe.

Origins · 1 minWoldemar Voigt and how this was discovered

Woldemar Voigt

1850–1919 · the physicist who gave the tensor its name, in the crystal elasticity of Göttingen

Woldemar Voigt was a German physicist who spent most of his career at Göttingen studying the physics of crystals — their elasticity, their optical and thermal behaviour, and the way a single crystal responds differently along different axes. It was precisely this direction-dependence that demanded a new vocabulary. In his 1898 treatise Die fundamentalen physikalischen Eigenschaften der Krystalle, Voigt introduced the word Tensor for a physical quantity that, unlike a scalar or a simple vector, carries a whole array of directional data — the stress inside a strained crystal being the prototype. He took the name from the Latin tendere, to stretch, the same root as tension: a tensor was originally the mathematical embodiment of the stretching and shearing stresses a solid can sustain. The irony of history is that Voigt's name outgrew Voigt's subject. The multilinear algebra behind the object had already been built by Grassmann in 1844 and turned into a calculus by Ricci-Curbastro and Levi-Civita by 1900, and when Einstein needed a language in which the laws of gravitation would read the same to every observer, he adopted that calculus wholesale — and with it Voigt's word. Today the stress tensor of a crystal and the Riemann curvature tensor of spacetime are recognised as the same species of object, both type- $(p,q)$ multilinear machines, and both go by the name a crystallographer chose to describe how a solid resists being stretched.

History

1844

— Grassmann publishes Die lineale Ausdehnungslehre, founding the multilinear algebra behind the tensor product

1854

— Riemann's habilitation lecture founds the intrinsic geometry of curved manifolds

1898

— Voigt adopts the word tensor for direction-dependent physical quantities in crystal physics

1900

— Ricci-Curbastro and Levi-Civita publish the absolute differential calculus, systematizing tensor methods

1916

— Einstein's review of general relativity writes the laws of gravitation as equations between tensors

Going further

Texts. Misner, Thorne & Wheeler, Gravitation (1973), Ch. 3 (tensors as machines with slots) · Wald, General Relativity (1984), Ch. 2 and Appendix A (the abstract tensor-product construction) · Carroll, Spacetime and Geometry (2004), Ch. 1.4–1.5 · Schutz, A First Course in General Relativity (2009), Ch. 3 · Hartle, Gravity (2003), Ch. 20. The key numerical checks are: the dimension formula $n^{p+q}$ yields $4^5=1024$ , $4^4=256$ , and $4^2=16$ ; the Riemann independent-component count is $\tfrac{1}{12}n^2(n^2-1)=20$ at $n=4$ ; the duality relation is $\mathrm{d}x^\mu(\partial_\nu)=\delta^\mu{}_{\nu}$ ; the worked outer products $v\otimes w$ and $w\otimes v$ are transposes of each other, and the trace of the example $(1,1)$ tensor is $T^\mu{}_{\mu}=5$ . All of these are exact mathematical identities — no empirical inputs appear — and the word tensor in this sense follows Voigt (1898).

Builds on: Tangent Vectors and the Tangent Space · Cotangent Space and One-Forms · Smooth Manifolds and Coordinate Charts · Atlases, Transition Functions and Differentiable Structure. Leads to: The Tensor Transformation Law · Tensor Algebra: Contraction, Symmetrization and Antisymmetrization · The Metric Tensor and the Line Element · Raising and Lowering Indices · Tensor Densities and Weights.

Ask the tutor: ask me to build any tensor you like from vectors and one-forms, evaluate it slot by slot on explicit inputs, and watch which contractions come out as chart-independent invariants.

What is the difference between a tensor and its components?

A tensor is a coordinate-free geometric object — a multilinear machine — that exists before you pick any chart. Its components are the finite table of numbers the machine returns when you feed it the basis one-forms and basis vectors of a chosen coordinate system. Change the chart and every component can change, yet the machine itself is untouched, the same way an arrow on a page is one arrow whether you measure it in inches or centimetres. The two are tied together by the fact that the components determine the tensor completely (multilinearity), so nothing is lost by working with indices — but you must remember that a component is a chart-dependent readout, not the object.

Why does the order of indices matter if the tensor product is "just multiplication"?

Because the multiplication happens on labelled slots, and the labels are part of the object. When you form $v\otimes w$ , the first factor is assigned to slot one and the second to slot two, so its components are $v^\mu w^\nu$ ; forming $w\otimes v$ assigns them the other way, giving $w^\mu v^\nu$ , which is the transpose. Numerically these differ: with $v=(2,3)$ and $w=(1,-1)$ the $(1,2)$ entry is $-2$ in one and $3$ in the other. The scalars multiply commutatively, but the arrangement into slots does not, and that arrangement is exactly what an index position records. Only after symmetrizing — averaging over slot orderings — may you permute indices freely.

Do I need a metric to define tensors or the tensor product?

No, and this is worth stating sharply because it is easy to assume otherwise. The tensor-product construction is purely linear-algebraic: it needs only the tangent space $T_xM$ , its dual $T_x^*M$ , and the ability to form multilinear maps, all of which exist at every point of a bare smooth manifold with no extra structure. A metric enters only later, and for a different job — it lets you raise and lower indices, identifying $T_xM$ with $T_x^*M$ and turning, say, a $(0,2)$ tensor into a $(2,0)$ or $(1,1)$ one. Without a metric those types remain genuinely distinct, but tensors and $\otimes$ are already fully defined.

How is a (1,1) tensor a matrix, and why isn't every matrix a tensor?

A type- $(1,1)$ tensor has exactly one upper and one lower index, so its components $T^\mu{}_\nu$ form a square array, and its action $v^\mu=T^\mu{}_\nu u^\nu$ is matrix-times-vector — it is a linear map from the tangent space to itself. But not every square array of numbers you write down is a tensor. To qualify, the array must transform correctly under a change of chart: one Jacobian on the upper index, one inverse Jacobian on the lower. A generic table of numbers, or the Christoffel symbols $\Gamma^\lambda{}_{\mu\nu}$ , or the bare partial derivative $\partial_\mu V^\nu$ , will pick up extra inhomogeneous terms and fail the test. Wearing indices is necessary but not sufficient; obeying the transformation law is what makes a matrix a tensor.

The universal property is stated abstractly — what does it actually buy me?

The universal property says the tensor product is the unique space that turns every multilinear map into an ordinary linear map: any bilinear $B(u,v)$ factors as $B(u,v)=\tilde B(u\otimes v)$ for a single linear $\tilde B$ . Concretely, this is the guarantee that "multilinear algebra is just linear algebra on a bigger space." It is why the products $\{e_i\otimes f_j\}$ form a genuine basis, why the dimension multiplies to $n^{p+q}$ , and why you may manipulate tensors with the full toolkit of vector spaces — add them, expand them in a basis, solve linear systems in their components. Without it, the leap from "linear in each slot separately" to "an element of one vector space with one basis" would be an assertion rather than a theorem.

Why is the Kronecker delta the only invariant (1,1) tensor?

Under a change of chart a $(1,1)$ tensor picks up one Jacobian $\partial x'^\mu/\partial x^\alpha$ on its upper index and one inverse Jacobian $\partial x^\beta/\partial x'^\nu$ on its lower index. For the delta, $\delta'^\mu{}_\nu=\frac{\partial x'^\mu}{\partial x^\alpha}\frac{\partial x^\beta}{\partial x'^\nu}\delta^\alpha{}_\beta=\frac{\partial x'^\mu}{\partial x^\alpha}\frac{\partial x^\alpha}{\partial x'^\nu}=\delta^\mu{}_\nu$ by the chain rule — the two factors collapse to the identity, so its components are literally unchanged in every chart. No other $(1,1)$ array has this property: a general $T^\mu{}_\nu$ gets genuinely reshuffled by the Jacobians. The delta is invariant precisely because it is the identity map, and the identity is the one linear transformation every observer agrees on.

Is the tensor product the same as the outer product I learned in linear algebra?

In components, yes — that is the whole point. The outer product of two arrays, $(v\otimes w)^{\mu\nu}=v^\mu w^\nu$ , is exactly the component form of the tensor product of the underlying geometric objects, and the Kronecker product of matrices is the same operation stacked up. What the geometric language adds is the meaning: the outer product isn't just a bookkeeping array but a genuine multilinear machine with a definite type $(p{+}r,\,q{+}s)$ , living in a vector space with a basis and a transformation law. So you already know how to compute $\otimes$ if you can take outer products; the tensor formalism tells you what the result is, how it transforms, and why its slots must not be reordered casually.