Symmetric Tensors
Among covariant tensors, those whose value is unaffected by reordering their arguments form a
distinguished class. An inner product has this property, and so does any object that measures a
quantity attached symmetrically to a collection of vectors. Isolating this property algebraically is
the first step toward the tensor that carries the geometry of lengths and angles on a manifold.
Definition: Symmetric Tensor
Let \(V\) be a finite-dimensional real vector space. A
covariant \(k\)-tensor
\(\alpha\) on \(V\) is symmetric if its value is unchanged whenever any two of
its arguments are interchanged: for all vectors and every pair of indices \(i < j\),
\[
\alpha(v_1, \ldots, v_i, \ldots, v_j, \ldots, v_k)
= \alpha(v_1, \ldots, v_j, \ldots, v_i, \ldots, v_k) .
\]
The symmetric covariant \(k\)-tensors form a linear subspace of the space of all covariant
\(k\)-tensors, denoted \(\Sigma^k(V^*)\).
Interchanging a single adjacent pair is enough to control every reordering, because an arbitrary
permutation of the arguments is a composition of transpositions. To make this precise, the group of
all permutations acts on tensors. Let \(S_k\) denote the
symmetric group
on \(k\) elements, the group of permutations of \(\{1, \ldots, k\}\). Given a covariant \(k\)-tensor
\(\alpha\) and a permutation \(\sigma \in S_k\), define a new covariant \(k\)-tensor
\({}^{\sigma}\alpha\) by reindexing the arguments,
\[
{}^{\sigma}\alpha(v_1, \ldots, v_k) = \alpha\big(v_{\sigma(1)}, \ldots, v_{\sigma(k)}\big) .
\]
Placing \(\sigma\) as a left superscript records the order of composition correctly. Writing out
both sides on arguments \((v_1, \ldots, v_k)\),
\[
{}^{\tau}({}^{\sigma}\alpha)(v_1, \ldots, v_k)
= {}^{\sigma}\alpha\big(v_{\tau(1)}, \ldots, v_{\tau(k)}\big)
= \alpha\big(v_{\tau(\sigma(1))}, \ldots, v_{\tau(\sigma(k))}\big)
= {}^{\tau\sigma}\alpha(v_1, \ldots, v_k) ,
\]
where the middle equality applies the definition of \({}^{\sigma}(\,\cdot\,)\) to the relabelled
arguments \(w_i = v_{\tau(i)}\). Thus \({}^{\tau}({}^{\sigma}\alpha) = {}^{\tau\sigma}\alpha\), where
\(\tau\sigma\) is the permutation \(i \mapsto \tau(\sigma(i))\), so the assignment
\(\sigma \mapsto {}^{\sigma}\alpha\) is a genuine left action of \(S_k\). A tensor is symmetric
exactly when \({}^{\sigma}\alpha = \alpha\) for every \(\sigma \in S_k\), since the transpositions
generate the whole group.
Averaging a tensor over the action of \(S_k\) produces a symmetric tensor and fixes those that are
already symmetric. This averaging is a projection onto the symmetric subspace.
Theorem: Symmetrization
The symmetrization of a covariant \(k\)-tensor \(\alpha\) on \(V\) is the tensor
\[
\operatorname{Sym}\alpha = \frac{1}{k!} \sum_{\sigma \in S_k} {}^{\sigma}\alpha ,
\]
so that on vectors
\[
(\operatorname{Sym}\alpha)(v_1, \ldots, v_k)
= \frac{1}{k!} \sum_{\sigma \in S_k}
\alpha\big(v_{\sigma(1)}, \ldots, v_{\sigma(k)}\big) .
\]
Then \(\operatorname{Sym}\alpha\) is symmetric, and \(\operatorname{Sym}\alpha = \alpha\) if and
only if \(\alpha\) is symmetric. Consequently \(\operatorname{Sym}\) is a projection of the space
of covariant \(k\)-tensors onto the subspace \(\Sigma^k(V^*)\).
Proof:
To see that \(\operatorname{Sym}\alpha\) is symmetric, apply an arbitrary permutation
\(\tau \in S_k\) to its arguments:
\[
{}^{\tau}(\operatorname{Sym}\alpha)
= \frac{1}{k!} \sum_{\sigma \in S_k} {}^{\tau}({}^{\sigma}\alpha)
= \frac{1}{k!} \sum_{\sigma \in S_k} {}^{\tau\sigma}\alpha
= \frac{1}{k!} \sum_{\eta \in S_k} {}^{\eta}\alpha
= \operatorname{Sym}\alpha ,
\]
where the substitution \(\eta = \tau\sigma\) is a bijection of \(S_k\) onto itself, so the sum
runs over the same set of permutations. Since \({}^{\tau}(\operatorname{Sym}\alpha)\) equals
\(\operatorname{Sym}\alpha\) for every \(\tau\), the tensor \(\operatorname{Sym}\alpha\) is
symmetric. If \(\alpha\) is itself symmetric, then \({}^{\sigma}\alpha = \alpha\) for every
\(\sigma\), so the sum has \(k!\) equal terms and \(\operatorname{Sym}\alpha = \alpha\).
Conversely, if \(\operatorname{Sym}\alpha = \alpha\), then \(\alpha\) is symmetric because
\(\operatorname{Sym}\alpha\) is. Applying \(\operatorname{Sym}\) once lands in
\(\Sigma^k(V^*)\), and applying it again changes nothing, so \(\operatorname{Sym}\) is a
projection onto that subspace.
The Symmetric Product
The tensor product of two symmetric tensors need not be symmetric: interchanging an argument from the
first factor with one from the second generally changes the value. Composing the tensor product with
the symmetrization projection repairs this and yields a product that stays within the symmetric
tensors.
Definition: Symmetric Product
Let \(\alpha \in \Sigma^k(V^*)\) and \(\beta \in \Sigma^l(V^*)\) be symmetric tensors. Their
symmetric product is the symmetric \((k+l)\)-tensor obtained by symmetrizing
their tensor product, written by juxtaposition with no intervening symbol,
\[
\alpha\beta = \operatorname{Sym}(\alpha \otimes \beta) .
\]
Writing out the symmetrization, its value on vectors \(v_1, \ldots, v_{k+l}\) is
\[
\alpha\beta(v_1, \ldots, v_{k+l})
= \frac{1}{(k+l)!} \sum_{\sigma \in S_{k+l}}
\alpha\big(v_{\sigma(1)}, \ldots, v_{\sigma(k)}\big)\,
\beta\big(v_{\sigma(k+1)}, \ldots, v_{\sigma(k+l)}\big) .
\]
The symmetric product behaves like the multiplication of polynomials: it is commutative, distributes
over addition, and is compatible with scalars. The proof of these properties, together with a closed
form in the most important special case, completes the algebraic picture.
Theorem: Properties of the Symmetric Product
Let \(\alpha, \beta, \gamma\) be symmetric covariant tensors on a finite-dimensional real vector
space \(V\), and let \(a, b \in \mathbb{R}\).
(a) The symmetric product is symmetric and bilinear:
\[
\alpha\beta = \beta\alpha, \qquad
(a\alpha + b\beta)\gamma = a\,\alpha\gamma + b\,\beta\gamma
= \gamma(a\alpha + b\beta) .
\]
(b) If \(\alpha\) and \(\beta\) are covectors, then
\[
\alpha\beta = \tfrac{1}{2}\big(\alpha \otimes \beta + \beta \otimes \alpha\big) .
\]
Proof:
For part (a), commutativity follows from reindexing the symmetrizing sum. The map that splits the
first \(k\) arguments from the last \(l\) for \(\alpha \otimes \beta\) is converted into the
splitting for \(\beta \otimes \alpha\) by composing each \(\sigma\) with the permutation that
swaps the first \(k\) positions with the last \(l\); this composition is a bijection of
\(S_{k+l}\), so the two symmetrized sums agree and \(\alpha\beta = \beta\alpha\). Bilinearity is
inherited from the tensor product: \(\otimes\) is bilinear and \(\operatorname{Sym}\) is linear,
so their composite is bilinear in each factor, which gives the displayed identities. (The second
equality uses commutativity already established.)
For part (b), take \(k = l = 1\), so \(S_{k+l} = S_2 = \{e, \tau\}\) with \(\tau\) the
transposition. The defining sum has \((k+l)! = 2\) terms,
\[
\alpha\beta(v_1, v_2)
= \tfrac{1}{2}\big(\alpha(v_1)\beta(v_2) + \alpha(v_2)\beta(v_1)\big)
= \tfrac{1}{2}\big((\alpha \otimes \beta)(v_1, v_2)
+ (\beta \otimes \alpha)(v_1, v_2)\big) ,
\]
the identity term contributing \(\alpha(v_1)\beta(v_2) = (\alpha \otimes \beta)(v_1, v_2)\) and
the transposition contributing \(\alpha(v_2)\beta(v_1) = (\beta \otimes \alpha)(v_1, v_2)\). Since
this holds for all \(v_1, v_2\), the tensors agree.
The normalization adopted here defines the symmetric product as \(\operatorname{Sym}\) applied to the
tensor product, with the factor \(1/(k+l)!\) built into \(\operatorname{Sym}\); some sources instead
insert a binomial factor \(\binom{k+l}{k}\), under which the rank-two identity reads
\(\alpha\beta = \alpha \otimes \beta + \beta \otimes \alpha\) without the factor \(\tfrac{1}{2}\).
The convention chosen here is the one for which a symmetric \(2\)-tensor with components
\(\beta_{ij}\) satisfies \(\beta = \beta_{ij}\,\varepsilon^i\varepsilon^j\) and
\(\beta(v, v) = \beta_{ij} v^i v^j\), the form in which the symmetric product will be used to write a
metric.
The rank-two case in part (b) is that form. A symmetric covariant \(2\)-tensor built from coordinate
differentials is a symmetric product of two covectors, and the symmetrized combination
\(\tfrac{1}{2}(\alpha \otimes \beta + \beta \otimes \alpha)\) is precisely what is needed so that the
off-diagonal coordinate components of such a tensor pair correctly. When this construction is
performed fiberwise over a manifold, the symmetric product of coordinate differentials produces the
standard coordinate expression for a field of inner products.
Tensor Bundles and Tensor Fields
Everything established so far is pointwise linear algebra on a single vector space. On a smooth
manifold the relevant vector space is the tangent space, which varies from point to point. Performing
the tensor constructions in each tangent space and assembling the results into a single smooth object
produces the tensor bundles, of which the
tangent bundle
and the
cotangent bundle
are the simplest cases.
Tensor bundles
Let \(M\) be a smooth manifold with or without boundary. The bundle of covariant
\(k\)-tensors on \(M\) is the disjoint union of the covariant tensor spaces over all points,
\[
T^k T^* M = \coprod_{p \in M} T^k\big(T_p^* M\big) ,
\]
where \(T^k(T_p^* M)\) is the space of covariant \(k\)-tensors on the tangent space \(T_p M\); the
bracket-free symbol \(T^k T^* M\) denotes the bundle assembled from these fibers.
Replacing covariant tensors by contravariant or mixed tensors gives the bundle of
contravariant \(k\)-tensors \(T^k TM = \coprod_p T^k(T_p M)\) and the bundle of mixed
\((k,l)\)-tensors \(T^{(k,l)} TM = \coprod_p T^{(k,l)}(T_p M)\). Each is called a
tensor bundle over \(M\). The low-rank cases recover familiar bundles: there are
natural identifications
\[
T^{(0,0)} TM = M \times \mathbb{R}, \qquad
T^{(0,1)} TM = T^1 T^* M = T^* M, \qquad
T^{(1,0)} TM = T^1 TM = TM ,
\]
and more generally \(T^{(0,k)} TM = T^k T^* M\) and \(T^{(k,0)} TM = T^k TM\), so the tangent and
cotangent bundles are the rank-one tensor bundles.
Theorem: Tensor Bundles are Smooth Vector Bundles
For a smooth manifold \(M\) of dimension \(n\), each tensor bundle \(T^k T^* M\), \(T^k TM\), and
\(T^{(k,l)} TM\) has a natural structure as a smooth
vector bundle
over \(M\), of rank \(n^k\), \(n^k\), and \(n^{k+l}\) respectively.
Proof:
Treat \(T^k T^* M\); the other cases are identical with the roles of \(T_p M\) and \(T_p^* M\)
adjusted. Each fiber \(T^k(T_p^* M)\) is a real vector space of dimension \(n^k\), since the
tensor spaces over an \(n\)-dimensional space have that dimension. To produce local
trivializations, fix a smooth coordinate chart on an open set \(U \subseteq M\), giving at each
\(p \in U\) the coordinate basis \((\partial/\partial x^i|_p)\) of \(T_p M\) and the dual basis of
coordinate covectors of \(T_p^* M\). The tensor products of \(k\) coordinate covectors form a
basis of \(T^k(T_p^* M)\) at each point, depending smoothly on \(p\), and assigning to each
tensor its \(n^k\) components in this basis defines a bijection from \(T^k T^* M|_U\) to
\(U \times \mathbb{R}^{n^k}\) that is linear on fibers. On an overlap of two such charts the
transition map is built from tensor products of the cotangent transition functions, hence is
smooth and invertible. The hypotheses of the
vector bundle chart lemma
are therefore met, so the local trivializations assemble \(T^k T^* M\)
into a smooth vector bundle of rank \(n^k\). The mixed bundle uses \(k\) tangent and \(l\)
cotangent factors, giving rank \(n^{k+l}\).
Tensor fields
A
section
of a tensor bundle assigns to each point of \(M\) a tensor on the tangent space at that point, of the
appropriate variance. Such a section is a tensor field.
Definition: Tensor Field
Let \(M\) be a smooth manifold with or without boundary. A tensor field on \(M\)
is a section of a tensor bundle: a covariant, contravariant, or mixed tensor field is a section
of \(T^k T^* M\), \(T^k TM\), or \(T^{(k,l)} TM\) respectively. A smooth tensor
field is one that is smooth as a section of a smooth vector bundle. The space of smooth
covariant \(k\)-tensor fields is denoted \(\mathcal{T}^k(M)\); it is a real vector space and a
module over the ring of smooth functions on \(M\).
The rank-one and rank-zero cases reduce to objects already studied. Under the identifications above,
a contravariant \(1\)-tensor field is a vector field, a covariant \(1\)-tensor field is a covector
field, and a \(0\)-tensor field is a continuous real-valued function. The covariant tensor fields
receive most of the attention because they admit a pullback under arbitrary smooth maps, as the final
section shows.
Smoothness criteria
A section of a tensor bundle is smooth precisely when its components in coordinates are smooth, in
direct generalization of the criterion for vector fields. Expressing a covariant \(k\)-tensor field
in coordinates uses the
component representation
of a tensor at each point, with the coordinate differentials playing the role of the dual basis.
Proposition: Coordinate Representation of a Tensor Field
In any smooth local coordinates \((x^i)\) on an open set \(U \subseteq M\), a covariant
\(k\)-tensor field \(A\) is written using the
coordinate tensor basis
as
\[
A = A_{i_1 \cdots i_k}\, dx^{i_1} \otimes \cdots \otimes dx^{i_k} ,
\]
with the summation convention in force and the component functions given pointwise by
\[
A_{i_1 \cdots i_k} = A\left(
\frac{\partial}{\partial x^{i_1}}, \ldots, \frac{\partial}{\partial x^{i_k}}
\right) .
\]
The contravariant and mixed cases are written the same way with coordinate vector fields
\(\partial/\partial x^i\) in the contravariant slots and coordinate differentials \(dx^i\) in the
covariant slots.
Theorem: Smoothness Criteria for Tensor Fields
Let \(M\) be a smooth manifold with or without boundary, and let \(A\) be a section of the
covariant tensor bundle \(T^k T^* M\). The following are equivalent.
(a) \(A\) is smooth.
(b) In every smooth coordinate chart, the component functions of \(A\) are smooth.
(c) Each point of \(M\) lies in some coordinate chart in which \(A\) has smooth component
functions.
Moreover, if \(A\) is a smooth covariant \(k\)-tensor field and \(X_1, \ldots, X_k\) are smooth
vector fields, then the function
\[
A(X_1, \ldots, X_k) : M \to \mathbb{R}, \qquad
p \mapsto A_p\big(X_1|_p, \ldots, X_k|_p\big) ,
\]
is smooth.
Proof:
The coordinate tensor products \(dx^{i_1} \otimes \cdots \otimes dx^{i_k}\) form a smooth local
frame for \(T^k T^* M\) over the chart domain, by the construction of the bundle structure. With
respect to this frame the components of \(A\) are exactly the functions \(A_{i_1 \cdots i_k}\) of
the coordinate representation. The general criterion for a section of a smooth vector bundle to
be smooth, namely that its component functions relative to a smooth local frame be smooth,
applied to this frame, gives the equivalence of (a), (b), and (c); the only point to check beyond
the general statement is that smoothness in one chart of an atlas propagates, which holds because
the transition functions between coordinate tensor frames are smooth. For the last assertion, in
a chart write \(X_j = X_j^{\,i}\, \partial/\partial x^i\) with smooth components; then
multilinearity gives
\(A(X_1, \ldots, X_k) = A_{i_1 \cdots i_k}\, X_1^{\,i_1} \cdots X_k^{\,i_k}\), a sum of products
of smooth functions, hence smooth on the chart domain, and therefore smooth on \(M\).
The tensor characterization lemma
Pairing a smooth covariant \(k\)-tensor field with \(k\) smooth vector fields produces a smooth
function, and the resulting operation on vector fields is linear over the ring of smooth functions,
not merely over the reals. This stronger linearity characterizes exactly those operations that arise
from tensor fields, and provides a coordinate-free test for tensoriality.
Lemma: Tensor Characterization
Let \(M\) be a smooth manifold with or without boundary, and let
\[
\mathcal{A} : \mathfrak{X}(M) \times \cdots \times \mathfrak{X}(M) \to C^\infty(M)
\]
be a map of \(k\) smooth vector field arguments to smooth functions. Then \(\mathcal{A}\) is
induced by a smooth covariant \(k\)-tensor field \(A\), through
\(\mathcal{A}(X_1, \ldots, X_k) = A(X_1, \ldots, X_k)\), if and only if \(\mathcal{A}\) is
multilinear over the ring of smooth functions: for all smooth functions \(f, f'\) and all smooth
vector fields,
\[
\mathcal{A}(X_1, \ldots, f X_i + f' X_i', \ldots, X_k)
= f\, \mathcal{A}(X_1, \ldots, X_i, \ldots, X_k)
+ f'\, \mathcal{A}(X_1, \ldots, X_i', \ldots, X_k) .
\]
When it exists, the tensor field \(A\) is unique.
Proof:
If \(\mathcal{A}\) comes from a smooth tensor field \(A\), then at each point the value depends
multilinearly on the argument vectors, and a scalar function multiplying a vector field factors
out pointwise, so \(\mathcal{A}\) is multilinear over smooth functions. The substance is the
converse. Assume \(\mathcal{A}\) is multilinear over smooth functions. The key step is that the
value \(\mathcal{A}(X_1, \ldots, X_k)\) at a point \(p\) depends only on the values
\(X_1|_p, \ldots, X_k|_p\), not on the fields away from \(p\).
First, the value at \(p\) is unchanged if one argument is altered outside a neighborhood of
\(p\). Suppose \(X_i\) vanishes on a neighborhood \(U\) of \(p\). Choose a smooth bump function
\(\psi\) supported in \(U\) with \(\psi(p) = 1\). Then \(\psi X_i\) is identically zero, so by
multilinearity \(0 = \mathcal{A}(\ldots, \psi X_i, \ldots) = \psi\, \mathcal{A}(\ldots, X_i,
\ldots)\); evaluating at \(p\) and using \(\psi(p) = 1\) shows
\(\mathcal{A}(\ldots, X_i, \ldots)(p) = 0\). By linearity the value at \(p\) depends only on the
\(X_i\) near \(p\).
Next, the value depends only on the \(X_i\) at \(p\). Suppose \(X_i|_p = 0\). In a coordinate
chart centered at \(p\) write \(X_i = X_i^{\,j}\, \partial/\partial x^j\) with component functions
\(X_i^{\,j}\) vanishing at \(p\). Fix a closed coordinate ball \(\overline{B}\) around \(p\)
contained in the chart domain; the coordinate vector fields \(\partial/\partial x^j\) and the
components \(X_i^{\,j}\), restricted to \(\overline{B}\), are smooth data on a closed set, so the
extension lemma for vector fields
(and its analogue for functions) produces global smooth fields \(E_j\) and functions
\(f_i^{\,j}\) on \(M\) agreeing with \(\partial/\partial x^j\) and \(X_i^{\,j}\) throughout
\(\overline{B}\), hence on a neighborhood of \(p\); there \(f_i^{\,j} E_j = X_i\).
Multilinearity over smooth functions gives, near \(p\),
\[
\mathcal{A}(\ldots, X_i, \ldots)
= \mathcal{A}(\ldots, f_i^{\,j} E_j, \ldots)
= f_i^{\,j}\, \mathcal{A}(\ldots, E_j, \ldots) ,
\]
and evaluating at \(p\), where every \(f_i^{\,j}(p) = X_i^{\,j}(p) = 0\), yields zero. By
linearity the value at \(p\) depends only on \(X_i|_p\).
Consequently, for tangent vectors \(v_1, \ldots, v_k \in T_p M\), choose smooth vector fields
\(V_1, \ldots, V_k\) with \(V_j|_p = v_j\) — such fields exist, since any tangent vector is
the value at \(p\) of a smooth field, for instance the one obtained by extending a constant
coordinate field with a bump function — and set
\[
A_p(v_1, \ldots, v_k) = \mathcal{A}(V_1, \ldots, V_k)(p) .
\]
This is well defined: by the pointwise dependence just established, the right-hand side is
unchanged if any \(V_j\) is replaced by another smooth field with the same value at \(p\), so it
depends only on \(v_1, \ldots, v_k\). The assignment is multilinear on \(T_p M\), so it defines a
covariant \(k\)-tensor at each point and hence a rough tensor field \(A\). Its component functions
in any chart are
\(\mathcal{A}(\partial/\partial x^{i_1}, \ldots, \partial/\partial x^{i_k})\), which are smooth,
so \(A\) is smooth by the smoothness criteria. Uniqueness is immediate, since the values of \(A\)
on coordinate vector fields are determined by \(\mathcal{A}\).
The Pullback of Tensor Fields
A smooth map carries covectors backward by composition with its differential, and the same mechanism
extends to covariant tensors of every rank. This is the operation that transports a field of inner
products from one manifold to another, and it is the reason the covariant tensors are the natural
setting for the geometry to come: unlike vectors, covariant tensors require no inverse and no
diffeomorphism hypothesis to be pulled back.
Pointwise pullback
Let \(F : M \to N\) be a smooth map, with
differential
\(dF_p : T_p M \to T_{F(p)} N\) at each point. Given a covariant \(k\)-tensor \(\alpha\) on
\(T_{F(p)} N\), the
pointwise pullback
of a covector extends to a covariant \(k\)-tensor \(dF_p^*(\alpha)\) on \(T_p M\) by precomposing
each argument with the differential,
\[
dF_p^*(\alpha)(v_1, \ldots, v_k) = \alpha\big(dF_p(v_1), \ldots, dF_p(v_k)\big),
\qquad v_1, \ldots, v_k \in T_p M .
\]
Because the differential needs no inverse, this is defined for any smooth map, and it reduces to the
pullback of a covector when \(k = 1\).
Definition: Pullback of a Tensor Field
Let \(F : M \to N\) be smooth and \(A\) a covariant \(k\)-tensor field on \(N\). The
pullback \(F^* A\) is the covariant \(k\)-tensor field on \(M\) whose value at
each point is the pointwise pullback of the value of \(A\) at the image point,
\[
(F^* A)_p = dF_p^*\big(A_{F(p)}\big) .
\]
Equivalently, on vectors \(v_1, \ldots, v_k \in T_p M\),
\[
(F^* A)_p(v_1, \ldots, v_k)
= A_{F(p)}\big(dF_p(v_1), \ldots, dF_p(v_k)\big) .
\]
Properties
The pullback respects the entire algebraic structure of covariant tensor fields, is functorial in the
map, and preserves smoothness.
Theorem: Properties of Tensor Pullbacks
Let \(F : M \to N\) and \(G : N \to P\) be smooth maps, let \(A\) and \(B\) be covariant tensor
fields on \(N\), and let \(f\) be a real-valued function on \(N\).
(a) \(F^*(f B) = (f \circ F)\, F^* B\).
(b) \(F^*(A \otimes B) = F^* A \otimes F^* B\).
(c) \(F^*(A + B) = F^* A + F^* B\).
(d) \(F^* B\) is a continuous tensor field, and is smooth if \(B\) is smooth.
(e) \((G \circ F)^* B = F^*(G^* B)\).
(f) \((\operatorname{Id}_N)^* B = B\).
Proof:
Parts (a), (b), (c) are pointwise identities of the linear-algebraic pullback \(dF_p^*\), checked
on arbitrary vectors. For (b), with \(A\) of rank \(k\) and \(B\) of rank \(l\), evaluating both
sides on \(v_1, \ldots, v_{k+l} \in T_p M\) gives
\(A_{F(p)}(dF_p v_1, \ldots, dF_p v_k)\, B_{F(p)}(dF_p v_{k+1}, \ldots, dF_p v_{k+l})\) either
way, since the tensor product splits the arguments and \(dF_p\) is applied to each. Part (a) is
the rank-zero case of (b), reading the function \(f\) as a \(0\)-tensor field and noting that its
pullback is the composition \(f \circ F\); part (c) is the linearity of \(dF_p^*\). For (e), the
chain rule \(d(G \circ F)_p = dG_{F(p)} \circ dF_p\) gives, on vectors,
\[
(G \circ F)^* B_p(v_1, \ldots, v_k)
= B_{G(F(p))}\big(d(G \circ F)_p v_1, \ldots\big)
= B_{G(F(p))}\big(dG_{F(p)} dF_p v_1, \ldots\big)
= \big(F^*(G^* B)\big)_p(v_1, \ldots, v_k) ,
\]
and (f) holds because the differential of the identity is the identity of each tangent space.
For (d), smoothness is local, so it suffices to check it in coordinates; the coordinate formula
established next exhibits the components of \(F^* B\) as smooth functions whenever those of \(B\)
are, which gives the claim.
Computation in coordinates
In coordinates the pullback is computed by the same substitution rule used for covector fields:
replace each coordinate function of \(N\) by its expression in terms of \(F\), and replace each
coordinate differential by the differential of that expression. The
coordinate formula
for the pullback of a covector field is the rank-one instance of the following.
Proposition: Coordinate Formula for Tensor Pullbacks
Let \(F : M \to N\) be smooth and let \(B\) be a covariant \(k\)-tensor field on \(N\). If
\((y^i)\) are smooth coordinates on an open subset of \(N\) and
\(B = B_{i_1 \cdots i_k}\, dy^{i_1} \otimes \cdots \otimes dy^{i_k}\) there, then on the preimage
\[
F^* B = \big(B_{i_1 \cdots i_k} \circ F\big)\,
d\big(y^{i_1} \circ F\big) \otimes \cdots \otimes d\big(y^{i_k} \circ F\big) .
\]
In practice one substitutes the component functions of \(F\) for the coordinates \(y^i\)
throughout the expression for \(B\) and expands each \(d(y^i \circ F)\) in the coordinates of
\(M\).
Proof:
The pullback is linear and multiplicative over tensor products by the properties just proved, so
it suffices to pull back each factor. A function pulls back by composition, giving the
coefficient \(B_{i_1 \cdots i_k} \circ F\) by part (a). Each coordinate differential \(dy^i\),
being the differential of the coordinate function \(y^i\), pulls back to the differential of the
composite, \(F^*(dy^i) = d(y^i \circ F)\), since pullback commutes with the differential of a
function. Assembling the factors with the tensor-product compatibility of part (b) yields the
stated expression. The coefficients \(B_{i_1 \cdots i_k} \circ F\) are continuous, and smooth
when \(B\) is smooth, while the differentials \(d(y^i \circ F)\) are smooth; a sum of products of
such fields is continuous in general and smooth when the coefficients are, which settles the
regularity claim (d) left open above.
The pullback closes the algebraic development of covariant tensors and prepares their geometric use.
A field of symmetric, positive-definite covariant \(2\)-tensors transported by this operation retains
its symmetry automatically, since precomposition with \(dF_p\) treats the two arguments alike. Its
positive-definiteness, however, is not automatic: from
\((F^* g)_p(v, v) = g_{F(p)}(dF_p v,\, dF_p v)\) and the positivity of \(g\), the pulled-back tensor
satisfies \((F^* g)_p(v, v) > 0\) for \(v \neq 0\) exactly when \(dF_p v \neq 0\) for \(v \neq 0\),
that is, exactly when \(dF_p\) is injective at every point — the condition that \(F\) be an
immersion. At points where the differential has a kernel the pullback is only positive
semidefinite. Thus an immersion transports a field of inner products to a genuine field of inner
products on the source, the mechanism by which an ambient metric induces one on a submanifold, the
construction at the foundation of the geometry of curved spaces.