Cotangent Bundle
A tangent vector acts on a smooth function by directional differentiation. The differential of the function, in turn, is a linear functional on tangent vectors: a covector. Cotangent spaces collect these functionals, and the cotangent bundle lets them vary smoothly over the manifold.
We begin with duality in linear algebra, then construct the bundle and describe covector fields in local coordinates. Pullback will be the natural operation for transporting them along a smooth map.
As in linear algebra, let \(V\) be a finite-dimensional real vector space. A covector on \(V\) is a real-valued linear function on \(V\). The space of all covectors on \(V\) is a real vector space, under pointwise scalar multiplication and addition. We call this space the dual space of \(V\), denoted \(V^*\).
Lemma 1. Let \(V\) be a finite-dimensional real vector space. Let \(\{e_1,\cdots,e_n\}\) be a basis for \(V\). Let \(E^1,\cdots,E^n \in V^*\) be the covectors defined by \(E^i(e_j) = \delta_{ij}\), where \(\delta_{ij}\) is the Kronecker delta symbol. Then, \(\{E^1,\cdots,E^n\}\) is a basis for \(V^*\). This is called the dual basis to \(\{e_1,\cdots,e_n\}\). As such, \(\dim V^* = \dim V\).
Proof. It follows from assuming \(\{e_1,\cdots,e_n\}\) is a basis for \(V\) that, for all \(x \in V\),
for some functions \(e_1^*,\cdots,e_n^* : V \to \mathbb R\). Clearly, \(e_k^* (e_i) = \delta_{ik}\) by inspection, and \(e_k^*\) is linear, so \(e_k^* \in V^*\) for all \(k\). It follows that, for any \(f \in V^*\) and by linearity of \(f\),
so these covectors span the dual. They are linearly independent because evaluating a zero linear combination at each basis vector gives every coefficient equal to zero. Thus \(\{e_1^*, \cdots, e_n^*\}\) forms a basis for \(V^*\). ◻
Note that in general, any vector \(v = v^i e_i \in V\) can be written as
Any arbitrary covector \(\omega \in V^*\) can be written in terms of the dual basis as
Therefore, it follows that \(\omega\) acts on \(v \in V\) via the formula \(\omega(v) = \omega_i v^i\).
Between vector spaces \(V\) and \(W\), suppose \(A : V \to W\) is a linear map. We define the dual map \(A^* : W^* \to V^*\) via the formula
It remains to verify that \(A^*\omega\) is a linear functional on \(V\). The verification is routine. Note that \(A^* \omega\) is a composition of linear maps \(\omega \circ A\), therefore it is linear as a map from \(V \to \mathbb R\). Take any \(\omega_1,\omega_2 \in W^*\), \(a, b \in \mathbb R\), and \(v \in V\); then,
and
as desired.
Lemma 2. The dual map satisfies the properties
\((A \circ B)^* = B^* \circ A^*\).
\((\operatorname{id}_V)^*: V^* \to V^*\) is the identity map of \(V^*\).
Proof of Lemma 2. For composable maps \(B:U\to V\), \(A:V\to W\), a covector \(\omega\in W^*\) and \(u\in U\),
Evaluation at every \(u\) proves the first identity. Likewise \(((\operatorname{id}_V)^*\eta)(v)=\eta(v)\) proves the second. \(\square\)
There is also the notion of a second dual space, defined simply as \(V^{**} = (V^*)^*\). There is a natural, basis-independent map \(\xi(v) : V^* \to \mathbb R\), defined as
If \(V\) is a vector space, then \(\xi(v)(\omega)\) depends linearly on \(\omega\). To show this, let \(\omega_1, \omega_2 \in V^*\). Then,
The same evaluation formula gives homogeneity in the covector, so \(\xi(v)\in V^{**}\). Linearity in the vector follows directly: for real \(a,b\),
Thus \(\xi:V\to V^{**}\) is linear, independently of any chosen basis.
Lemma 3. For any finite-dimensional vector space \(V\), the map \(\xi : V \to V^{**}\) is an isomorphism.
Proof. It suffices to verify that \(\xi\) is injective, because \(\dim V = \dim V^{**}\). Take a nonzero element \(v \in V\). Extend \(v\) to a basis for \(V\), i.e. \(\{e_1,\cdots,e_n\}\). Let \(\{E^1,\cdots,E^n\}\) be the dual basis for \(V^*\). Then, \(\xi(v)(E^1)=E^1(v)=E^1(e_1)=1\), i.e. \(\xi(v)\neq 0\). ◻
Remark 4. There is a canonical identification of \(V\) with its second dual, when \(V\) is finite-dimensional. There is no such canonical identification between \(V\) and \(V^*\).
Apply the dual-space construction separately to each tangent space. Coordinate changes determine how the resulting covectors fit together.
Definition 5. Let \(M\) be a differentiable manifold. The cotangent space at \(p \in M\) is defined as
Elements of \(T_p^* M\) are called tangent covectors at \(p \in M\).
Given smooth local coordinates on an open subset \(U \subset M\), denoted \(\{x^i\}\), at each \(p \in M\), \(\{\pd{}{{x^i}}\big|_p\}\) gives rise to a dual basis for \(T_p^* M\). Call this dual basis \(\{d^{\: i}\}\). As such, any covector \(\omega \in T_p^* M\) has the unique representation as
We can now speak of transitioning between different coordinate systems whose domains contain \(p\). Suppose \(\{{\widetilde x}^j\}\) denote another coordinate system. Then, straightforward calculations show that
This means for \(\omega \in T^*_p M\), we have the representations \(\omega = \omega_i d^{\: i}\big|_p = \widetilde \omega_j \widetilde d^j\big|_p\), and
By way of nomenclature, on an open subset \(U \subset M\), given any smooth local coordinates \(\{x^i\}\), we have the basis dual to \(\{\pd{}{{x^i}}\}\), denoted \(\{d^{\: i}\}\). This defines maps \(d^1,\cdots,d^n : U \to T^* M\). These are called coordinate covector fields. The following result is standard.
Proposition 6. Let \(M\) be a differentiable manifold. With its standard projection map and the natural vector space structure on each fibre, the cotangent bundle \(T^* M\) has a unique topology and smooth structure making it into a differentiable vector bundle of rank \(n\) over \(M\), for which all coordinate covector fields are smooth local sections.
Proof. We will prove this in two parts: the existence of such a topology and smooth structure, and its uniqueness. Let \(\pi : T^* M \to M\) denote the standard projection map given by \(\pi(\omega) = p\) for any \(\omega \in T_p^* M\).
Existence: Let \(\mathscr{A} = \{(U_\alpha, x_\alpha)\}\) be a smooth atlas for \(M\), where \(x_\alpha(p) = (x^1(p), \cdots, x^n(p))\). For each chart \((U_\alpha, x_\alpha)\), the coordinate covector fields \(\{d^{\: 1}, \cdots, d^{\: n}\}\) form a basis for \(T_p^* M\) at every \(p \in U_\alpha\). We define a bijection \(\varphi_\alpha : \pi^{-1}(U_\alpha) \to x_\alpha(U_\alpha) \times \mathbb R^n \subset \mathbb R^{2n}\) by mapping each covector \(\omega = \omega_i d^{\: i}\big|_p\) to its natural coordinates:
To verify that this collection of maps forms a smooth atlas for \(T^* M\), we must examine the transition maps between overlapping charts. Consider two charts \((U, \{x^i\})\) and \((V, \{\widetilde x^j\})\) on \(M\) with non-empty intersection, and let \(\varphi_U\) and \(\varphi_V\) be their respective natural coordinate charts for \(T^* M\). The transition map \(\varphi_V \circ \varphi_U^{-1}\) maps the coordinates \((x^1, \cdots, x^n, \omega_1, \cdots, \omega_n)\) to \((\widetilde x^1, \cdots, \widetilde x^n, \widetilde \omega_1, \cdots, \widetilde \omega_n)\). As derived previously, the components of a covector transform as \(\omega_i = \dpd{{\widetilde x}^j}{{x^i}} (p) \widetilde \omega_j\). Multiplying by the inverse Jacobian matrix yields the transformation for \(\widetilde \omega_j\):
Since \(M\) is a differentiable manifold, the coordinate transition map \(x \mapsto \widetilde x\) is a smooth diffeomorphism, meaning its partial derivatives \(\dpd{x^i}{{\widetilde x}^j}\) are smooth functions. Thus, the transition map for \(T^* M\) is given by
This mapping is smooth because its first \(n\) components are smooth functions of \(x\), and its last \(n\) components are linear in \(\omega\) with coefficients that are smoothly varying functions of \(x\).
By the Vector Bundle Construction Lemma (or Chart Lemma, Lemma 4), the existence of these bijective maps whose transition functions are smooth and fibre-wise linear guarantees that we can endow \(T^* M\) with a unique topology and smooth structure. In this structure, \(\pi\) is smooth, \(\pi^{-1}(U_\alpha)\) are open sets, and the maps \(\varphi_\alpha\) are smooth diffeomorphisms. This makes \(T^* M\) a differentiable vector bundle of rank \(n\) over \(M\).
Furthermore, under this constructed smooth structure, the coordinate covector field \(d^{\: j}\) on \(U_\alpha\) maps to the natural coordinates \((x^1(p), \cdots, x^n(p), 0, \cdots, 1, \cdots, 0)\), where the \(1\) is strictly in the \((n+j)\)-th position. The base coordinate functions are smooth and the fibre components are constant, so the entire coordinate representation is smooth. Therefore, the coordinate covector fields are explicitly smooth local sections.
Uniqueness: Suppose \(T^* M\) is endowed with any topology and smooth structure making it a rank \(n\) differentiable vector bundle over \(M\) for which all coordinate covector fields \(\{d^{\: i}\}\) are smooth local sections. Let \((U, \{x^i\})\) be any coordinate chart on \(M\). By hypothesis, \(\{d^{\: 1}, \cdots, d^{\: n}\}\) are smooth local sections over \(U\). Since they form a basis for \(T_p^* M\) at each \(p \in U\), they form a smooth local frame for the vector bundle over \(U\). By the standard properties of vector bundles, any smooth local frame uniquely defines a smooth local trivialization \(\Psi : U \times \mathbb R^n \to \pi^{-1}(U)\) given by
The inverse of this diffeomorphism, mapping into \(\mathbb R^{2n}\) via the chart on \(U\), is precisely the natural coordinate chart \(\varphi_U\) we constructed. Because the topology and smooth structure of a manifold are completely and uniquely determined by its maximal atlas of smooth compatible charts, and any such vector bundle structure must admit these exact natural charts as smooth diffeomorphisms, the alternative structure has the same charts as the constructed one, so the two structures agree. ◻
As is the case with tangent bundles, on any open subset \(U \subset M\) and with any smooth local coordinates \(\{x^i\}\), we can write down a smooth coordinate chart for \(T^* M\), as
We call \((x^i, \omega_i)\) the natural coordinates for \(T^*M\) associated with \(\{x^i\}\). Terminology-wise, a (local or global) section of \(T^* M\) is called a covector field, or a (differential) 1-form. Following standard ideas, there are several ways to check for the smoothness of a covector field.
Proposition 7 (Smoothness criteria for covector fields). Let \(M\) be a differentiable manifold, and \(\omega : M \to T^* M\) be a covector field. The following statements are equivalent:
\(\omega\) is smooth;
In every smooth coordinate chart, the component functions of \(\omega\) are smooth;
Each point of \(M\) is in some coordinate chart in which \(\omega\) has smooth component functions;
For every smooth vector field \(X \in \mathscr X(M)\), the function \(\omega(X)\) is smooth on \(M\);
For every open subset \(U \subset M\), and every smooth vector field \(X\) on \(U\), the function \(\omega(X) : U \to \mathbb R\) is smooth on \(U\).
Proof. We will prove the equivalence of these statements by establishing the cyclic chain of implications (a) \(\implies\) (b) \(\implies\) (c) \(\implies\) (d) \(\implies\) (e) \(\implies\) (a).
(a) \(\implies\) (b): Suppose \(\omega: M \to T^* M\) is a smooth covector field (a smooth section). Let \((U, \{x^i\})\) be any smooth coordinate chart on \(M\). The natural coordinates on \(T^* M\) associated with \(U\) are given by the chart \(\varphi : \pi^{-1}(U) \to \mathbb{R}^{2n}\), mapping \(\eta = \eta_i d^{\:i}\big|_p \mapsto (x^1(p), \dots, x^n(p), \eta_1, \dots, \eta_n)\). The local representation of the section \(\omega\) mapped in these charts is \(\varphi \circ \omega : U \to \mathbb{R}^{2n}\), given by \(p \mapsto (x^1(p), \dots, x^n(p), \omega_1(p), \dots, \omega_n(p))\). Because \(\omega\) is a smooth map between manifolds, its coordinate representation must be smooth. Therefore, its individual component functions \(\omega_i : U \to \mathbb{R}\) are necessarily smooth.
(b) \(\implies\) (c): This implication is immediate. Because \(M\) is a differentiable manifold, it is covered by an atlas of smooth coordinate charts. Thus, each point \(p \in M\) is contained in at least one such coordinate chart, and by (b), the component functions of \(\omega\) in this chart are smooth.
(c) \(\implies\) (d): Suppose (c) holds, and let \(X \in \mathscr X(M)\) be a globally defined smooth vector field on \(M\). To show that the function \(\omega(X): M \to \mathbb R\) is smooth, it suffices to establish that it is smooth in an open neighborhood of every point \(p \in M\). By (c), there exists a smooth coordinate chart \((U, \{x^i\})\) containing \(p\) in which \(\omega\) can be written locally as \(\omega = \omega_i d^{\: i}\), where each component \(\omega_i\) is a smooth function on \(U\). In this same chart, the vector field \(X\) has the local representation \(X = X^j \dpd{}{{x^j}}\), where the components \(X^j\) are also smooth functions on \(U\). Evaluating \(\omega(X)\) on \(U\) yields
Since \(\omega_i\) and \(X^i\) are smooth functions on \(U\), their finite sum of products \(\omega_i X^i\) is smooth on \(U\). Thus, \(\omega(X)\) is smooth in a neighborhood of every point \(p \in M\), meaning it is smooth globally on all of \(M\).
(d) \(\implies\) (e): Suppose (d) holds. Let \(U \subset M\) be an arbitrary open subset, and let \(X\) be a smooth vector field defined strictly on \(U\). We want to show that \(\omega(X)\) is smooth on \(U\). Let \(p \in U\). We can choose a smooth bump function \(\rho: M \to \mathbb R\) such that \(\rho \equiv 1\) on some open neighborhood \(V\) of \(p\) such that \(\overline{V} \subset U\), and with support contained in \(\operatorname{supp}(\rho) \subset U\). We then define a global vector field \(\widetilde X \in \mathscr{X}(M)\) by
Since \(\widetilde X\) is a globally smooth vector field on \(M\), assumption (d) guarantees that \(\omega(\widetilde X)\) is a globally smooth function on \(M\). On the neighborhood \(V\) of \(p\), we have \(\rho \equiv 1\), so \(\widetilde X = X\), which implies that \(\omega(X) = \omega(\widetilde X)\) exactly on \(V\). Therefore, \(\omega(X)\) is smooth at \(p\). Since \(p \in U\) was chosen arbitrarily, \(\omega(X)\) is smooth everywhere on \(U\).
(e) \(\implies\) (a): Suppose (e) holds. To show that \(\omega : M \to T^* M\) is smooth globally, it suffices to show that its local representation in any natural coordinate chart \((x^i, \omega_i)\) associated with a smooth coordinate chart \((U, \{x^i\})\) on \(M\) is smooth. This holds if and only if the component functions \(\omega_i : U \to \mathbb R\) are smooth. The coordinate vector fields \(X_j = \dpd{}{{x^j}}\) are naturally smooth vector fields defined on the open subset \(U\). By (e), the function \(\omega(X_j)\) must be smooth on \(U\). However, by the definition of the dual basis, \(\omega(X_j) = \omega\left(\dpd{}{{x^j}}\right) = \omega_j\). Thus, the component functions \(\omega_j\) are smooth on \(U\). Consequently, the local representation of \(\omega\) is smooth on every chart mapping \(M\), which formally establishes that \(\omega\) is a smooth map. ◻
As before, we can build frames using an ordered tuple of covector fields. The definitions are analogous to before: a local coframe for \(M\) over an open subset \(U \subset M\) is an ordered \(n\)-tuple of covector fields, \(\{E^1,\cdots,E^n\}\) defined on \(U\), such that \(\{E^i\big|_p\}\) form a basis for \(T_p^* M\) at each \(p \in U\). If \(U=M\), then this is called a global coframe. The most straightforward example of coframes are the collection of coordinate covector fields; this is called the coordinate coframe. The desired smoothness property of the coordinate coframe follows from the fact that its component functions in the given chart are constants.
If we have a local frame for \(TM\), say, \(\{e_1,\cdots,e_n\}\), then there is a uniquely determined local coframe, \(\{E^1,\cdots,E^n\}\), that is the dual basis to \(\{e_1|_p,\cdots,e_n|_p\}\) for each \(p \in U\). The equivalent formulation of this statement is \(E^i(e_j) = \delta_{ij}\). This coframe is called the coframe dual to \(\{e_1,\cdots,e_n\}\). There is an analogous definition for if we start with a local coframe; the uniquely determined local frame is called the frame dual to \(\{E^1,\cdots,E^n\}\). This uniqueness in the determination of the local (co)frames gives rise to the following statement.
Lemma 8. Let \(M\) be a differentiable manifold. Let \(\{e_1,\cdots,e_n\}\) be a local frame over an open subset \(U \subset M\), and let \(\{E^1,\cdots,E^n\}\) denote its dual coframe. Then, \(\{e_i\}\) is smooth if and only if \(\{E^i\}\) is smooth.
Proof. To prove this statement, it suffices to show that for each \(p \in U\), the local frame \(\{e_i\}\) is smooth in a neighbourhood of \(p\), if and only if \(\{E^i\}\) is. This can be done with the help of smooth local coordinate charts. Let \((V,\{x^i\})\) denote a smooth local coordinate chart around \(p \in U\), chosen such that \(p \in V \subset U\). Then, in \(V\), we can write \(e_i = a_i^k \pd{}{{x^k}}\), and \(E^j = b_\ell^j d^\ell\). Here, \((a_i^k)\) and \((b_\ell^j)\) are matrices of real-valued functions defined on \(V\). Note that the vector fields \(\{e_i\}\) and the covector fields \(\{E^i\}\) are smooth on \(V\), if and only if their respective component functions are smooth. Clearly, the fact that \(E^i(e_j) = \delta_{ij}\) implies the matrices \((a_i^k)\) and \((b_\ell^j)\) are inverses of each other. Because matrix inversion is a smooth map from \(\operatorname{GL}_n (\mathbb R)\) to \(\operatorname{GL}_n (\mathbb R)\), it follows that one of these matrix-valued functions is smooth if and only if the other one is smooth. ◻
Proposition 9. Let \(M\) be a differentiable manifold, and \(\omega\) be a covector field on \(M\). Let \(\{E^1,\cdots,E^n\}\) denote a smooth coframe on an open subset \(U \subset M\). Then, \(\omega\) is smooth on \(U\) if and only if its component functions with respect to \(\{E^1,\cdots, E^n\}\) are smooth.
Proof of Proposition 9. Let \(e_j\) be the smooth frame dual to the coframe, using Lemma 8. If \(\omega\) is smooth, its coefficients are \(\omega_j=\omega(e_j)\) and are smooth by Proposition 7. Conversely, if each \(\omega_j\) is smooth, then \(\omega=\sum_j\omega_jE^j\) is a finite sum of smooth sections multiplied by smooth functions, hence is smooth. \(\square\)
The nomenclature above comes from the same observations as before: because, given any local coframe over an open subset of \(M\), any covector field \(\omega\) can be written as \(\omega = \omega_i E^i\), it follows that we call \(\omega_1,\cdots,\omega_n : U \to \mathbb R\) the component functions of \(\omega\) with respect to the coframe \(\{E^1,\cdots,E^n\}\).
Remark 10. The space of all smooth covector fields on a differentiable manifold \(M\) is denoted using \(\mathscr X^* (M)\). Note that if \(f \in \mathscr C^\infty (M)\) and \(\omega \in \mathscr X^* (M)\), we have the covector field \(f\omega\) defined by the formula
Note that \(\mathscr X^* (M)\) is the space of smooth sections of a vector bundle, therefore it is a module over \(\mathscr C^\infty (M)\).
The differential of a function supplies the basic example of a covector field. Its smoothness can be checked by evaluating it on the coordinate vector fields.
Lemma 11. The differential of a smooth function is a smooth covector field.
Proof. The verification is routine: for each \(p \in M\), \((\dif f)_p (v)\) depends linearly on \(v\), hence it is a covector at \(p\). Smoothness is a consequence of the fact that \(\dif f(X) = Xf\) and by an application of Proposition 7. ◻
We can directly calculate the coordinate representation of \(\dif f\) for some smooth function \(f\). With ongoing notations, i.e. \(\{x^i\}\) as smooth local coordinates on an open subset \(U \subset M\) and \(\{d^{\: i}\}\) as the corresponding coordinate coframe on \(U\), we have
Then, we have
We can examine a special case, i.e. if \(f\) is a coordinate function \(x^k : U \to \mathbb R\), then
This is why we can interchangeably use \(\dif x^j\) and the coordinate covector field \(d^j\). This gives us the relation
This is why, in the main text, we use \(\dif x^i\) to mean the coordinate covector field.
Two consequences relate the differential to constancy and to derivatives along curves.
Proposition 12. Let \(M\) be a differentiable manifold, and \(f : M \to \mathbb R\) be a smooth function.
\(\dif f = 0\) if and only if \(f\) is constant on each component of \(M\).
Let \(J \subset \mathbb R\) be an interval, and \(\gamma : J \to M\) be a differentiable curve. Then, \((f \circ \gamma)'(t) = (\dif f)_{\gamma(t)} (\gamma'(t))\).
Proof of Proposition 12. If \(df=0\), the coordinate derivatives of \(f\) vanish on every coordinate ball. The fundamental theorem of calculus along line segments in that ball makes \(f\) locally constant. On a connected component, a locally constant function is constant: a nonempty level set and its complement are both open, so connectedness forces the complement to be empty. Conversely, components of a manifold are open, and constancy on each makes \(df=0\) locally. For a differentiable curve on an interval \(J\), the ordinary coordinate chain rule gives
Duality also explains how covectors transform under smooth maps. Composing a covector with the differential pulls it back to the source tangent space. Let \(F: M \to N\) denote a smooth map between differentiable manifolds. Let \(p \in M\). The differential \((\dif F)_p : T_p M \to T_{F(p)} N\) induces a dual linear map
The map \((\dif F)_p^*\) is called the pullback map by \(F\) at \(p \in M\). The ongoing discussions show that the pullback map is characterized by the formula
Remark 13. Covector fields pull back under every smooth map. A vector field on the source does not in general determine a vector field on the target: its differential images must agree at all points of a fibre and define a smooth target field. For example, under \(F(x,y)=x\), the field \(y\partial_x\) has different differential images at different points of the same fibre. A diffeomorphism always allows pushforward; additional cases require the corresponding relatedness condition.
This follows from definitions. Given a smooth map \(F: M \to N\) and a covector field \(\omega\) on \(N\), we can define a covector field \(F^* \omega\) on \(M\), which we call the pullback of \(\omega\) by \(F\), by the formula
Its action on a tangent vector \(v \in T_p M\) is
There is no ambiguity here about what point to pull back from: \(F^* \omega\) at \(p\) is the pullback of \(\omega\) at \(F(p)\)!
Lemma 14. Let \(F: M \to N\) denote a smooth map between differentiable manifolds. Let \(u\) be a continuous, real-valued function on \(N\), and \(\omega\) be a covector field on \(N\). Then,
Additionally, if \(u\) is smooth, then
Proof. These facts follow from definitions and direct calculations. For the first identity, note that
For the second identity, let \(v \in T_p M\); then,
as desired. ◻
Lemma 15. Let \(F: M \to N\) denote a smooth map between differentiable manifolds, and \(\omega\) be a continuous covector field on \(N\). Then, \(F^* \omega\) is a continuous covector field on \(M\). If, additionally, \(\omega\) is smooth, then so is \(F^* \omega\).
Proof. Take any \(p \in M\). Choose smooth local coordinates \(\{z^j\}\) in a neighbourhood \(V\) of \(F(p) \in N\). Let \(U = F^{-1}(V)\). Clearly, \(U\) is a neighbourhood of \(p\). Writing \(\omega\) in coordinates, we have \(\omega = \omega_j \dif z^j\), where \(\omega_j\) are continuous functions on \(V\). Applying Lemma 14 twice, to the restriction of \(F\) to \(U\), we have
The coefficient functions \(\omega_j\circ F\) are continuous, and the one-forms \(\dif(z^j\circ F)\) are smooth. Their finite sum is therefore continuous. If \(\omega\) is smooth, the coefficients are smooth as well, so the pullback is smooth. There is no converse for an arbitrary map: if \(F\) is constant, its differential vanishes and \(F^*\omega=0\) for every \(\omega\). ◻
It is customary to write the pullback of a covector field via components of the function \(F\) in coordinates, i.e.
where \(F^j\) denotes the \(j\)th component function of \(F\) in the chosen coordinates.
Remark 16. We can think of changing coordinates as a pullback under the identity map. For demonstration, take \(H\) to be the right-half plane, i.e. \(H = \{(x,y) : x>0\}\), and say we use \((r,\theta)\) as the domain, and \((x,y)\) as the codomain. Consider the identity map on \(H\). Then, the pullback formula tells us that we can calculate the polar coordinate expression for a covector field by straightforward substitution and expansion. For instance, the coordinate change map is evaluated as
These calculations follow from what we have shown above in the general case.
Reading: Carmo (1992), Riemannian geometry.