Tangent Bundles

The tangent bundle \(TM\) "bundles" together all of the tangent spaces \(T_pM\) for every point \(p \in M\) of a smooth manifold \(M\). As a set, it is just the disjoint union of all of the tangent bundles:

\[TM = \bigsqcup_{p \in M}T_pM.\]

We may take any representative set that satisfies the universal property of this disjoint union (coproduct) as our working definition (since all such sets are isomorphic). The usual candidate is the set of all ordered pairs \((p,v)\) with \(p \in M\) and \(v \in T_pM\).

We can define a projection map \(\pi : TM \rightarrow M\) as

\[\pi(p,v) = p.\]

We can equip the set \(TM\) with a topology and a smooth structure that turns it into a smooth manifold.

The Chart Lemma

First, we will prove a technical lemma. We have the set \(TM\), but we want to equip it with the structure of a manifold. Our goal is to build a topology and smooth structure given a set \(M\) and a countable collection of subsets \(U_i \subseteq M\) that cover \(M\) and maps \(\varphi_i : U_i \rightarrow \mathbb{R}^n\) so that the collection \(\{(U_i,\varphi_i)\}\) will serve as an atlas for the induced manifold.

Such a collection of potential charts must satisfy certain conditions. First of all, the image \(\varphi_i(U_i) \subseteq \mathbb{R}^n\) must be an open set, and the map \(\varphi_i\) must be a bijection onto its image. Next, for each chart to be smooth, the transition map \(\varphi_j \circ \varphi_i^{-1} : \varphi_i(U_i \cap U_j) \rightarrow \varphi_j(U_i \cap U_j)\) must be smooth whenever \(U_i \cap U_j \neq \emptyset\). In particular, this means that the domain \(\varphi_i(U_i \cap U_j)\) and codomain \(\varphi_j(U_i \cap U_j)\) must be open sets in \(\mathbb{R}^n\). As we will show, the countability of the cover will ensure that the induced topology is second-countable. Finally, to ensure that the resultant topology is Hausdorff, we need to require that each pair of distinct points \(p,q \in M\) is either contained in a single domain \(U_i\) (in which case, as we will show, we can rely on the fact that \(\mathbb{R}^n\) is Hausdorff to show that the induced topology is Hausdorff) or that \(p \in U_i\) and \(q \in U_j\) for some disjoint pair of domains \(U_i\) and \(U_j\).

Since we require each map \(\varphi_i : U_i \rightarrow \mathbb{R}^n\) to be a homemorphism onto its image, we can guarantee this if we specify the preimages \(\varphi^{-1}(V)\) of the open sets\(V \subseteq \mathbb{R}^n\) as the basis of the induced topology. This implies that we only consider open sets \(V\) such that \(V \subseteq \varphi(U_i)\) (those in the subspace topology). Since, by construction, the preimage of each open set is an open set, each map \(\varphi_i\) is continuous. Since \(\varphi_i(\varphi_i^{-1}(V)) = V\), each map \(\varphi_i\) is an open map. Since we required \(\varphi_i\) to be bijective, it is a bijective, continuous, open map, and hence a homeomorphism.

The Basis of the Induced Topology

We need to verify that this collection of sets is a basis, however. This means that we need to verify two things:

• The sets \(\varphi_i^{-1}(V)\) cover \(M\)
• For each open set \(V,W \subseteq \mathbb{R}^n\) and every point \(p \in \varphi_i^{-1}(V )\cap \varphi_j^{-1}(W)\), there exists some open set \(C \subseteq \mathbb{R}^n\) such that \(p \in \varphi_i^{-1}(C) \subseteq \varphi_i^{-1}(V )\cap \varphi_j^{-1}(W)\).

Since the sets \(U_i\) are, by definition, the preimages of open sets and cover \(M\), the first condition is satisfied.

For the second condition, it is sufficient to demonstrate that the set \(\varphi_i^{-1}(V) \cap \varphi_j^{-1}(W)\) is already an element of the basis. First, note that, for any subsets \(A,B\) of the domain of a function \(f\), we have that \(f(A \cap B) = f(A) \cap f(B)\). Also, note that, since \(V \subseteq \varphi_i(U_i)\) and \(W \subseteq \varphi_j(U_j)\), it follows that \(\varphi_i^{-1}(V) \subseteq U_i\) and \(\varphi_j^{-1}(W) \subseteq U_j\) and so \(\varphi_i^{-1}(V) = \varphi_i^{-1}(V) \cap U_i\) and \(\varphi_j^{-1}(W) = \varphi_j^{-1}(W) \cap U_j\). Using these facts, we determine that

\begin{align}\varphi_i^{-1}(V) \cap \varphi_j^{-1}(W) &= (\varphi_i^{-1}(V) \cap U_i) \cap (\varphi_j^{-1}(W) \cap U_j) \\&= \varphi_i^{-1}(V) \cap \varphi_j^{-1}(W) \cap (U_i \cap U_j) \\&= \varphi_i^{-1}(V) \cap \varphi_j^{-1}(W) \cap \varphi_j^{-1}(\varphi_j(U_i \cap U_j)) \\&= \varphi_i^{-1}(V) \cap \varphi_j^{-1}(W \cap \varphi_j(U_i \cap U_j)) \\&= \varphi_i^{-1}(V) \cap \varphi_i^{-1}(\varphi_i(\varphi_j^{-1}(W \cap \varphi_j(U_i \cap U_j)))) \\&= \varphi_i^{-1}(V \cap \varphi_i(\varphi_j^{-1}(W \cap \varphi_j(U_i \cap U_j)))) \\&= \varphi_i^{-1}(V \cap (\varphi_i \circ \varphi_j^{-1})(W \cap \varphi_j(U_i \cap U_j))) \\&= \varphi_i^{-1}(V \cap (\varphi_j \circ \varphi_i^{-1})^{-1}(W \cap \varphi_j(U_i \cap U_j))).\end{align}

Next, note that \(\varphi_j(U_i \cap U_j)\) is open in \(\mathbb{R}^n\) by assumption, and since \(W\) is open in \(\mathbb{R}^n\), it follows that \(W \cap \varphi_j(U_i \cap U_j)\) is open in the subspace topology on \(\varphi_j(U_i \cap U_j)\). Then, since the transition map \(\varphi_j \circ \varphi_i^{-1}\) is continuous by assumption, the set \((\varphi_j \circ \varphi_i^{-1})^{-1}(W \cap \varphi_j(U_i \cap U_j))\) is open in the subspace topology on \(\varphi_i(U_i \cap U_j)\). Then, the set \(V \cap (\varphi_j \circ \varphi_i^{-1})^{-1}(W \cap \varphi_j(U_i \cap U_j))\) is open in \(\varphi(U_i)\), and so \(\varphi_i^{-1}(V \cap (\varphi_j \circ \varphi_i^{-1})^{-1}(W \cap \varphi_j(U_i \cap U_j)))\) is an element of the basis by definition.

Induced Manifold

We need to ensure that the induced topology is that of a topological manifold, i.e. it is locally Euclidean, Hausdorff, and second-countable.

Locally Euclidean

Next, we confirm that \(M\), with the induced topology, is locally Euclidean. This is essentially by construction, since each map \(\varphi_i\) is a homeomorphism onto its image.

Hausdorff

Next, we confirm that \(M\) with the induced topology is Hausdorff. For any two points \(p,q \in M\), if \(p,q \in U_i\) for some \(U_i\), then \(\varphi(U_i)\) is open in \(\mathbb{R}^n\) by assumption, and thus we can find disjoint open balls \(B(\varphi(p))\) and \(B(\varphi(q))\) containing \(\varphi(p)\) and \(\varphi(q)\), respectively. Then, the inverse images \(\varphi_i^{-1}(B(\varphi(p)))\) and \(\varphi_i^{-1}B(\varphi(q)))\) are disjoint (since \(\varphi\) is bijective) open sets in the induced topology on \(M\) which contain \(p\) and \(q\), respectively. Otherwise, if there are disjoint sets \(U_i\) and \(U_j\) containing \(p\) and \(q\), respectively, then, since these sets are open, each point is contained in disjoint open sets. Thus, \(M\) is Hausdorff.

Second Countable

Next, we confirm that \(M\) with the induced topology is second-countable. We will use a technical lemma for this purpose. Suppose \(X\) is a topological space with a countable open cover \(\{U_i\}\) such that each \(U_i\) has a countable basis \(\mathcal{B}_i\) under the subspace topology. Since each element of such a basis is open in \(U_i\) and \(U_i\) is open in \(X\), it is, by definition, the intersection of \(U_i\) with an open set in \(X\) and is therefore open in \(X\). The countable union of countable sets is again a countable set, so we take \(\cup_i \mathcal{B}_i\) as a countable basis for \(X\). Consider any open set \(U \subseteq X\) and point \(p \in U\). Since the sets \(\{U_i\}\) cover \(X\), there exists some \(U_i\) such that \(p \in U_i\). Now, \(U_i \cap U\) is open in \(X\) and is also open in \(U_i\) and \(p \in U_i \cap U\). Since \(\mathcal{B}_i\) is a basis for \(U_i\), there exists some \(B \in \mathcal{B}_i\) such that \(B \subseteq U_i \cap U \subseteq U\) and \(p \in B\). Thus, since \(U\) was arbitrary and \(p \in B \subseteq U\), \(\cup_i \mathcal{B}_i\) is a basis for \(X\) as well. Applying this lemma, we conclude that the induced topology on \(M\) is second-countable, since, by assumption, the cover of \(M\) is countable, and each subspace \(U_i\) is second-countable since \(\mathbb{R}^n\) is second-countable.

Thus, we finally conclude that \(M\) with the induced topology is a locally Euclidean, Hausdorff, second-countable topological space, which is the standard definition of a topological manifold.

Generalization of the Atlas

In fact, this same proof goes through with almost no modification if we drop the requirement that the collection of charts be countable and instead merely require that the cover contains a countable subcover.

Uniqueness of the Induced Topology

We want to confirm that the induced topology is unique. That is, we want to confirm that, given any candidate topology such that each \((U_i,\varphi_i)\) is a chart (\(U_i\) is open in the topology and \(\varphi_i\) is a homeomorphism), this candidate topology coincides with the induced topology described above. This is essentially by construction, since each map \(\varphi_i : U_i \rightarrow \mathbb{R}^n\) is a homeomorphism. Thus, any open set in the induced topology, which is of the form \(\varphi_i^{-1}(V)\) for some open set \(V \subseteq \mathbb{R}^n\), must necessarily be open in any topology such that the map \(\varphi_i : U_i \rightarrow \mathbb{R}^n\) is a homeomorphism. Likewise, for any open set \(U\) in any other candidate topology, since \(\{U_i\}\) is a cover, for each point \(p \in U\) there is some \(U_{i_p}\) such that \(p \in U_{i_p}\). Then \(U \cap U_{i_p}\) is open in the candidate topology, and since \(\varphi_{i_p}\) is a homeomorphism, \(\varphi(U \cap U_{i_p})\) is open in \(\mathbb{R}^n\). Then, \(U \cap U_{i_p} = \varphi_{i_p}^{-1}(\varphi_{i_p}(U \cap U_{i_p}))\) is open in the induced topology. Since \(U = \cup_{p \in U}U \cap U_{i_p}\), \(U\) is open in the induced topology. Thus, the two topologies coincide.

Next, we want to show that, whenever \(\{(U_i,\varphi_i)\} \subseteq \{(U_j,\varphi_j)\}\), the induced topologies coincide. The argument is similar to the previous argument. Any open set in the former topology is in the latter topology, since its respective covering is a subcovering of the latter. We only need to consider the charts \((U_j, \varphi_j)\) that are not in the former topology. For any open set \(U\) in the latter topology, since \(\{U_i\}\) is a covering, for each point \(p \in U\), there exists some \(U_{i_p}\) such that \(p \in U_{i_p}\). Then \(p \in U_{i_p} \cap U\) and \(U_{i_p} \cap U\) is open in the latter topology, which means that \(U_{i_p} \cap U\) is the inverse image under \(\varphi_{i_p}\) for some open set in \(\mathbb{R}^n\). This means that it is also open in the former topology. Since \(U = \cup_{p \in U}U_{i_p} \cap U\), this means that \(U\) is also open in the former topology. Thus, the two topologies coincide. Thus, we can use any suitable atlas which is an extension of a countable atlas.

The collection of charts is smooth by construction. Each atlas determines a unique maximal atlas (the smooth structure).

The Induced Topology on \(TM\)

Now that we have a method for constructing a smooth atlas, we can apply it to the tangent bundle. We construct an atlas for \(TM\) using the atlas of an \(n\)-dimensional manifold \(M\). For any chart \((U,\varphi)\) on \(M\), \(\pi^{-1}(U) \subseteq TM\) is the set of all tangent vectors for \(M\) at all points of \(U\). Writing \((x^i)\) for the coordinate functions of the map \(\varphi\), we can define a map \(\tilde{\varphi} : \pi^{-1}(U) \rightarrow \mathbb{R}^{2n}\) as follows:

\[\tilde{\varphi}\left(v^i \frac{\partial}{\partial x^i}\bigg\rvert_p\right) = (x^1(p),\dots,x^n(p),v^1,\dots,v^n).\]

The image of \(\tilde{\varphi}\) is \(\varphi(U) \times \mathbb{R}^n\), an open subset of \(\mathbb{R}^{2n}\). It maps the pair \(p,v\) to its coordinate representation.

We can define the inverse map as follows:

\[\tilde{\varphi}^{-1}(x^1(p),\dots,x^n(p),v^1,\dots,v^n) = v^i \frac{\partial}{\partial x^i}\bigg\rvert_{\phi^{-1}(x)}.\]

It is a continuous, bijective, open map, and hence a homeomorphism. Thus, we can take \((\pi^{-1}(U),\tilde{\varphi})\) as a chart for \(TM\).

Given any two charts \((U,\varphi)\) and \((V,\varphi)\) for \(M\) with corresponding charts \((\pi^{-1}(U),\tilde{\varphi})\) and \((\pi^{-1}(V),\tilde{\psi})\) for \(TM\), observe that

\[\tilde{\varphi}(\pi^{-1}(U) \cap \pi^{-1}(V)) = \varphi(U \cap V) \times \mathbb{R}^n\]

and

\[\tilde{\psi}(\pi^{-1}(U) \cap \pi^{-1}(V)) = \psi(U \cap V) \times \mathbb{R}^n\]

which are open in \(\mathbb{R}^{2n}\), as required. In the post about the tangent space, we determined that the transition map \(\tilde{\psi} \circ \tilde{\varphi}^{-1} : \varphi(U \cap V) \times \mathbb{R}^n \rightarrow \psi(U \cap V) \times \mathbb{R}^n\) can be written as follows:

\[\tilde{\psi} \circ \tilde{\varphi}^{-1}(x^1,\dots,x^n,v^1,\dots,v^n) = \left(\tilde{x}^1(x),\dots,\tilde{x}^n(x),v^j \frac{\partial \tilde{x}^1}{\partial x^j}(x),\dots,v^j \frac{\partial \tilde{x}^n}{\partial x^j}(x)\right).\]

This transition map is smooth, since it is the composite of smooth functions.

Next, note that every second-countable topological space admits a countable subcover (using an appropriate choice principle). Thus, we may choose a countable cover \(\{(U_i,\varphi_i)\}\) of \(M\) by smooth charts. This induces a corresponding countable cover of \(TM\) by smooth charts \(\{(\pi^{-1}(U_i),\tilde{\varphi}_i)\}\). We have already verified that this cover satisfies most of the required conditions for the chart lemma above; we only need to check the Hausdorff condition. For any two points \((p,v)\) and \((p,w)\) in the same fiber (here \(\pi^{-1}(\{p\})\)), these points are necessarily contained within a single coordinate domain \(\pi^{-1}(U)\) for any open set such that \(p \in U\). For any two points \((p,v)\) and \((q,w)\) in different fibers, since \(M\) is Hausdorff, we can find disjoint open sets \(U,V \subseteq M\) such that \(p \in U\) and \(q \in M\), and thus \(\pi^{-1}(U)\) and \(\pi^{-1}(V)\) are disjoint coordinate domains containing \((p,v)\) and \((q,w)\), respectively.

Next, we want to confirm that the projection map \(\pi\) is smooth. Consider the following diagram:

\begin{CD} TM @>\pi>> M\\ @V\tilde{\varphi}VV @VV{\varphi}V \\ \mathbb{R}^{2n} @>>\varphi\circ\pi\circ\tilde{\varphi}^{-1}> \mathbb{R}^n \end{CD}

The expression of the projection map in coordinates \(\varphi\circ\pi\circ\tilde{\varphi}^{-1}\) is (using \(\pi\) for the name of this map too)

\begin{align}\pi(x^1,\dots,x^n,v^1,\dots,v^n) &= \varphi\circ\pi\circ\tilde{\varphi}^{-1}(x^1,\dots,x^n,v^1,\dots,v^n)\\ &=\varphi(\pi(p,v))\\&= \varphi(p)\\&= x,\end{align}

where \(x\) is the name given to the coordinate representation of \(p\).

This is a smooth function since it is simply a projection.

Thus, we have succeeded in inducing a unique topology on \(TM\).

The Global Differential

We can extend the differential \(dF_p : T_pM \rightarrow T_pN\) of a smooth map \(F : M \rightarrow N\) between smooth manifolds \(M\) and \(N\) at a point \(p \in M\) to the global differential \(dF : TM \rightarrow TN\) as follows for any \(v \in T_pM\):

\[dF(p,v) = (F(p), dF_p(v)).\]

We want to show that this map is smooth. Consider the following diagram:

\begin{CD} TM @>dF>> TN\\ @V\tilde{\varphi}VV @VV\tilde{\psi}V \\ \mathbb{R}^{2n} @>>\tilde{\psi}\circ\pi\circ\tilde{\varphi}^{-1}> \mathbb{R}^{2m} \end{CD}

The expression of the total differential \(dF\) in coordinates \(\tilde{\psi}\circ\pi\circ\tilde{\varphi}^{-1}\) is (using \(dF\) as the name of this map too)

\begin{align}dF(x^1,\dots,x^n,v^1,\dots,v^n) &= \tilde{\psi}\circ dF\circ\tilde{\varphi}^{-1}(x^1,\dots,x^n,v^1,\dots,v^n)\\&= \tilde{\psi}(dF(p, v))\\&= \tilde{\psi}(F(p), dF_p(v))\\&= \tilde{\psi}(F^1(p),\dots,F^m(p),v^i\frac{\partial F^1}{\partial x^i}(x),\dots,v^i\frac{\partial F^m}{\partial x^i}(x)),\end{align}

where the expression for \(dF_p(v)\) in coordinates was calculated in the post about tangent spaces.

This function is the composition of smooth functions, and is a smooth function, so \(dF\) is a smooth map.

In the article about tangent spaces, several important properties of the differential at a point were established; these generalize immediately to the total differential:

• \(d(G \circ F) = dG \circ dF\) (the chain rule).
• \(d(\mathrm{Id}_M) = \mathrm{Id}_{TM}\).
• For any diffeomorphism \(F\), \(dF\) is a diffeomorphism, and \((dF)^{-1} = d(F^{-1})\).

The first two properties establish that there is a functor from the category of manifolds and smooth maps to itself defined such that \(M \mapsto TM\) and \(F : M \rightarrow N \mapsto dF : TM \rightarrow TN\).

Vector Bundles

For the manifold \(\mathbb{R}^n\), since, as we established in the post about tangent spaces, each tangent space \(T_p\mathbb{R}^n\) for each point \(p \in \mathbb{R}^n\) is isomorphic to \(\mathbb{R}^n\), since the tangent bundle consists of one such tangent space per point, it is therefore isomorphic to \(\mathbb{R}^n \times \mathbb{R}^n\):

\[T\mathbb{R}^n = \bigsqcup_{p \in \mathbb{R}^n}T_p\mathbb{R}^n \cong \mathbb{R}^n \times \mathbb{R}^n.\]

Just as an \(n\)-dimensional manifold \(M\) is locally homeomorphic to \(\mathbb{R}^n\) (or, equivalently, to an open subset \(U \subseteq \mathbb{R}^n\)), in the same way, the tangent bundle \(TM\) is, in a certain sense, locally the same as \(\mathbb{R}^n \times \mathbb{R}^n\) (or, equivalently, to \(U \times \mathbb{R}^n\) for some open subset \(U \subseteq \mathbb{R}^n\)). The concept of a vector bundle makes this precise.

For any topological space \(M\), called the base space, a (real) vector bundle of rank \(k\) over \(M\) consists of a topological space \(E\), called the total space, together with a surjective, continuous map \(\pi : E \rightarrow M\), called the projection map, such that the following conditions are satisfied:

• The fiber over \(p\), \(E_p = \pi^{-1}(p)\), is a \(k\)-dimensional real vector space for each point \(p \in M\)
• For each point \(p \in M\), there exist a neighborhood \(U\) of \(p\) and a homeomorphism \(\varphi : \pi^{-1}(U) \rightarrow U \times \mathbb{R}^k\) called the local trivialization of \(E\) over \(U\), such that \(\pi_U \circ \varphi = \pi\) (where \(\pi_U : U \times \mathbb{R}^k \rightarrow U\) is the projection) and the restriction of \(\varphi\) to the fiber \(E_q\) for each point \(q \in M\) is a linear isomorphism from \(E_q\) to \(\mathbb{R}^k \cong \{q\} \times \mathbb{R}^k\).

The tangent bundle satisfies all of these requirements, and is the prime example of a vector bundle. The total space is \(TM\), the base space is \(M\), the projection map is the tangent bundle projection map \(\pi : TM \rightarrow M\), the fiber over any point \(p \in M\) is the tangent space \(T_pM\), the maps \(\tilde{\varphi} : \pi^{-1}(U) \rightarrow U \times \mathbb{R}^n\) for each chart \((U,\varphi)\) on \(M\) witness the local trivialization of \(TM\) over \(U\), and each restriction to \(T_pM\) is a linear isomorphism onto \(\mathbb{R}^n\).