Tensor Fields and Natural Operations

Tensors encode multilinear relations among vectors and covectors, and several important operations on them require neither a metric nor a connection. Tensor products, permutation of slots, weighted symmetrization and antisymmetrization, contraction of one vector index with one covector index, and pullback of purely covariant tensors are already natural on a smooth manifold. Raising indices, tracing two covariant indices, lengths, and covariant differentiation are not.

Required background. Calculus and Linear Algebra for Relativity — use multilinear maps, tensor products, dual bases, permutations, and contraction.

Tensors and their coordinate components

At $p\in M$, a tensor of type $(r,s)$ is a multilinear map

$$T:(T_p^*M)^r\times(T_pM)^s\longrightarrow\mathbb R.$$

With a coordinate basis, it has the expansion

$$T=T^{i_1\cdots i_r}{}_{j_1\cdots j_s} \,\partial_{i_1}\otimes\cdots\otimes\partial_{i_r} \otimes dx^{j_1}\otimes\cdots\otimes dx^{j_s}.$$

The components transform with one Jacobian for each upper index and one inverse Jacobian for each lower index. Those transformation laws ensure that evaluating $T$ on geometric arguments gives the same number in every chart. A tensor field is a smooth assignment $p\mapsto T_p$, equivalently one whose components are smooth in every chart. The invariant and component viewpoints are developed in Lee 2013, Chs. 11–12, pp. 272–326.

An upper index corresponds to a vector factor, which accepts a covector argument; a lower index corresponds to a covector factor, which accepts a vector argument. The distinction is already meaningful without a metric. In particular, $T^{i}{}_{j}$ may be contracted to $T^i{}_i$, but $S_{ij}$ cannot be “traced” by summing $S_{ii}$: that expression is not invariant.

For a concrete tensor, let $\alpha=dx$, $\beta=dy$ on the plane, and set $T=\alpha\otimes\beta$. It acts by $T(u,v)=\alpha(u)\beta(v)$, so $T(\partial_x,\partial_y)=1$ but $T(\partial_y,\partial_x)=0$. A tensor product does not automatically symmetrize its factors. The ordered inputs are part of the object being defined.

Natural algebraic operations

If $S$ has type $(r,s)$ and $T$ has type $(r',s')$, then $S\otimes T$ has type $(r+r',s+s')$. Permuting like slots is canonical. For a rank-two covariant tensor, the site’s weighted conventions are

$$T_{(ij)}=\frac12(T_{ij}+T_{ji}), \qquad T_{[ij]}=\frac12(T_{ij}-T_{ji}).$$

For $k$ indices, parentheses or brackets include the factor $1/k!$. Thus every covariant rank-two tensor decomposes uniquely into symmetric and antisymmetric parts: $T_{ij}=T_{(ij)}+T_{[ij]}$.

Contraction pairs one contravariant and one covariant slot using the canonical evaluation $T_p^*M\times T_pM\to\mathbb R$. For example, $A^i{}_{ij}$ is a covector. Contraction commutes with coordinate change because a Jacobian and its inverse cancel. It does not require a volume form or inner product.

Alternating covariant tensors form the exterior algebra. Let $\operatorname{Alt}$ denote antisymmetrization with the factor $1/(p+q)!$. For a $p$-form $\alpha$ and a $q$-form $\beta$, the standard wedge normalization is

$$\alpha\wedge\beta =\frac{(p+q)!}{p!q!}\operatorname{Alt}(\alpha\otimes\beta).$$

In particular, $\alpha\wedge\beta=\alpha\otimes\beta-\beta\otimes\alpha$ for one-forms, with no factor $1/2$. It satisfies

$$\alpha\wedge\beta=(-1)^{pq}\beta\wedge\alpha.$$

This preview remains entirely metric-free; differential forms and their integration are developed in the next chapter.

Tensoriality as a practical test

An expression is tensorial at $p$ if its value depends only on the values of its tensor arguments at $p$ and transforms homogeneously. Partial derivatives of tensor components usually fail this test because differentiating a Jacobian produces an inhomogeneous term. By contrast, algebraic combinations made from tensor products, permutations, contractions, and linear combinations with scalar coefficients remain tensors.

There is a quick diagnostic. If an alleged multilinear object $A(X_1,\ldots,X_k)$ is $C^\infty(M)$-linear in every vector-field argument, then its value at $p$ depends only on $X_1(p),\ldots,X_k(p)$ and defines a tensor. The Lie bracket fails because $[fX,Y]=f[X,Y]-Y(f)X$. A differential one-form succeeds: $df(f_0X)=f_0df(X)$, where $f_0$ is a scalar function and $f$ is the function being differentiated.

As a coordinate check, under $y=y(x)$ a mixed tensor transforms as

$$T'^a{}_b =\frac{\partial y^a}{\partial x^i} \frac{\partial x^j}{\partial y^b}T^i{}_j.$$

Contracting gives $T'^a{}_a=T^i{}_i$. The invariant trace is therefore available for an endomorphism, even without a metric.

Operations that need extra geometry

A nondegenerate metric supplies musical maps $T_pM\leftrightarrow T_p^*M$, so it permits raising and lowering indices and contracting two indices of the same variance. A connection supplies covariant derivatives of tensor fields and comparisons between neighboring fibers. Top-degree forms can be integrated after choosing an orientation; densities can be integrated without one. Conflating these layers obscures which conclusions survive a change of metric, connection, or orientation.

The natural operations here are pointwise. They say nothing about how a tensor varies, whether it is conserved, or how its components are measured by an observer. Those questions belong to later pages on Lie derivatives, connections, frames, and physical interpretation. Tong 2019, §2.3, pp. 69–75, PDF provides a physics-oriented component treatment, while the present scope keeps the underlying operations metric-free.

Symmetry type as invariant information

Symmetry or antisymmetry in a pair of like slots survives every coordinate transformation because it is defined by permuting arguments, not components. A symmetric covariant tensor remains symmetric in all charts, and a two-form remains antisymmetric. By contrast, a statement about a particular component vanishing may depend on the basis. Decomposing a tensor into symmetry types is therefore a coordinate-independent simplification.

Contraction respects this distinction but can annihilate a symmetry type only when the paired slots are defined. For example, if a symmetric contravariant tensor $S^{ij}=S^{ji}$ is supplied together with an antisymmetric covariant tensor $A_{ij}=-A_{ji}$, then the natural pairing vanishes:

$$S^{ij}A_{ij}=S^{ji}A_{ji}=-S^{ij}A_{ij}=0.$$

This conclusion uses two like-slot symmetry statements followed by valid vector–covector contractions. There is no invariant notion of antisymmetry obtained by swapping an upper slot with a lower slot unless extra structure first identifies vectors with covectors.

Exercises

Let $A^i{}_j$ and $B^k{}_\ell$ be $(1,1)$ tensors. Form a scalar from $A\otimes B$ in two inequivalent natural ways, and explain why $A^{ij}B_{ij}$ is not available from two arbitrary $(1,1)$ tensors without extra structure.

Solution

One may contract within each tensor to obtain $(\operatorname{tr}A)(\operatorname{tr}B)=A^i{}_iB^j{}_j$. Alternatively, contract the upper slot of $A$ with the lower slot of $B$ and the upper slot of $B$ with the lower slot of $A$:

$$\operatorname{tr}(AB)=A^i{}_jB^j{}_i.$$

Both use only vector–covector evaluation. Writing $A^{ij}$ would require converting the lower index of $A^i{}_j$ into an upper one, which has no canonical meaning until a nondegenerate bilinear form is chosen.

References

• Lee, John M. Introduction to Smooth Manifolds. 2nd ed., Graduate Texts in Mathematics 218, Springer, 2013, Chs. 11–12, pp. 272–326.
• Tong, David. General Relativity. University of Cambridge lecture notes, 2019, §2.3, pp. 69–75, PDF.