Tangent and Cotangent Bundles

Vector Bundles

Now we need to define vector fields, and to define concepts such as \(C^\infty\) vector fields it is better to first define the notion of a vector bundle. First, let us define a vector bundle over a topological space.

Definition 1 A vector bundle over a topological space \(B\), \(\pi:E \rightarrow B\), is an object defined as follows.

• The total space \(E\) and the base space \(B\) are both topological spaces, and \(\pi:E \rightarrow B\) is a continuous surjection.
• For each \(b\in B\), \(E_b=\pi^{-1}(b)\) carries the structure of a \(k\)-dimensional vector space.
• For each \(b_0\in B\) there exists a suitable open neighborhood \(U\subseteq B\) and a homeomorphism \(h:U\times\mathbb{R}^k \rightarrow\pi^{-1}(U)\) such that for every \(b\in U\), \(x\mapsto h(b,x)\) is an isomorphism from \(\mathbb{R}^k\) to \(E_b\).

Here, \(k\) is called the rank of the vector bundle \(E\rightarrow B\). The homeomorphism \(h\) in the third condition is called a local trivialization, and if we can take \(U=B\), then \(E\) is called a trivial vector bundle.

Similarly, we can define a vector bundle over a manifold. To do this, we replace both \(E\) and \(B\) with manifolds, \(\pi\) with a \(C^\infty\) surjection, and replace the third condition with

For each \(b_0\in B\) there exists a suitable coordinate system \(U\subseteq B\) and a diffeomorphism \(h:U\times\mathbb{R}^k\rightarrow\pi^{-1}(U)\) such that for every \(b\in U\), \(x\mapsto h(b,x)\) is an isomorphism from \(\mathbb{R}^k\) to \(E_b\).

Tangent Bundle

A typical example of a vector bundle is the tangent bundle.

Example 2 (Tangent bundle) Define the set \(TM\) by

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

Then there is a natural projection map \(\pi:TM\rightarrow M\). To regard \(TM\) as a vector bundle, we must endow it with a manifold structure.

First, let us define a coordinate system on \(TM\). For an arbitrary coordinate system \((U,\varphi)\), define the function \(\tilde{\varphi}:\pi^{-1}(U)\rightarrow\mathbb{R}^m\times\mathbb{R}^m\) by the formula

\[\tilde{\varphi}(v)=\bigl(x^1(\pi(v)), \ldots, x^m(\pi(v)), \dd{x}^1(v),\ldots, \dd{x}^m(v)\bigr)\]

Then \(\tilde{\varphi}\) is a bijection from \(\pi^{-1}(U)\) onto the open subset of \(\mathbb{R}^{2m}\), \(\varphi(U)\times\mathbb{R}^m\).

These are \(C^\infty\)-compatible. Suppose another coordinate system \((V,\psi)\), \(\psi=(y^j)_{j=1}^m\) is given, and define \(\tilde{\psi}\) as above. Then on \(\pi^{-1}(U)\cap\pi^{-1}(V)=\pi^{-1}(U\cap V)\) we have

\[\begin{aligned}(\tilde{\psi}\circ\tilde{\varphi}^{-1})(p^1, \ldots, p^m, v^1, \ldots, v^m)&=\tilde{\psi}\left(\varphi^{-1}(p), \sum v^i\frac{\partial}{\partial x^i}\bigg\vert_{\varphi^{-1}(p)}\right)\end{aligned}\]

Writing \(v=\sum v^i\frac{\partial}{\partial x^i}\) here, the right-hand side can simply be written as

\[\left((\psi\circ\varphi^{-1})(p), \dd{y}^1(v), \ldots, \dd{y}^m(v)\right)\]

Now for any \(j\),

\[\dd{y}^j\left(\sum v^i\frac{\partial}{\partial x^i}\bigg\vert_{\varphi^{-1}(p)}\right)=\sum_{i=1}^m v^i\frac{\partial y^j}{\partial x^i}\bigg\vert_{\varphi^{-1}(p)}\]

Therefore, since each component of the transition map \(\tilde{\psi}\circ\tilde{\varphi}^{-1}\) above is \(C^\infty\), \(\tilde{\psi}\circ\tilde{\varphi}^{-1}\) is also \(C^\infty\).

Meanwhile, the topology on \(TM\) is obtained by taking the sets

\[\{\tilde{\varphi}^{-1}(W)\mid \text{$W$ open in $\mathbb{R}^{2m}$, $(U,\varphi)\in\mathcal{A}$}\}\]

as a basis. Taking \(W=\mathbb{R}^{2m}\), we can check that the \(\pi^{-1}(U)\) are all open sets in the topology generated by the above collection, and we can also verify that this topology yields a \(2m\)-dimensional topological manifold.

Now all that remains is the local trivialization on \(TM\). For an arbitrary coordinate system \((U,\varphi)\), it suffices this time to define \(\phi:\pi^{-1}(U)\rightarrow U\times\mathbb{R}^m\) by the formula

\[v\vert_p\mapsto (p, \dd{x}^1(v),\ldots, \dd{x}^m(v))\]

On a fixed \(\pi^{-1}(p)\), that \(\phi\) is an isomorphism between vector spaces is obvious, and it is also obvious that for any \((x,v)\in U\times\mathbb{R}^m\) we have \((\pi\circ\phi^{-1})(x,v)=x\). That \(\phi\) is a diffeomorphism holds because

\[\tilde{\varphi}=(\varphi\times\id_{\mathbb{R}^m})\circ\phi\]

and in this formula the two functions other than \(\phi\) are both diffeomorphisms.

In particular, when \(TM\) is a trivial bundle, \(M\) is called a parallelizable manifold.

Smooth functors

The reason the tangent bundle \(TM\) is important is that most vector bundles defined on a manifold are defined from \(TM\). For example, the cotangent bundle \(T^\ast M\) is the vector bundle where at each \(p\in M\), the cotangent space \(T_p^\ast M\), which is the dual space of the tangent space, is attached. Similarly, various vector bundles are defined at each point \(p\) through operations in linear algebra (Example 5).

Ordinarily, each time these are defined, we would have to show that they satisfy the conditions of a vector bundle, but there is a slightly more fundamental approach in [Mil].

Definition 3 Suppose two vector bundles \(E\rightarrow B\), \(E'\rightarrow B'\) are given. Then a bundle map from \(E \rightarrow B\) to \(E' \rightarrow B'\) means, among pairs \(E\rightarrow E', B \rightarrow B'\) making the diagram

commute, those for which \(E_b\rightarrow E'_{b'}\) is an isomorphism.

Now consider the category of finite-dimensional \(\mathbb{R}\)-vector spaces whose morphisms are isomorphisms, \(\mathbf{FVect}_\text{iso}\). Then \(\mathbf{FVect}_\text{iso}\times\mathbf{FVect}_\text{iso}\) is the category whose

• objects are pairs \((V,W)\) of finite-dimensional vector spaces,
• morphisms are pairs \((V,W)\overset{(f,g)}{\rightarrow}(V',W')\) of isomorphisms between finite-dimensional vector spaces.

Therefore, a functor from \(\mathbf{FVect}_\text{iso}\times\mathbf{FVect}_\text{iso}\) to \(\mathbf{FVect}_\text{iso}\), \(F\), must take \((V,W)\) to an \(\mathbb{R}\)-vector space \(F(V,W)\) and \((f,g)\) to an isomorphism \(F(f,g)\).

Definition 4 A functor \(F:\mathbf{FVect}_\text{iso}\times\mathbf{FVect}_\text{iso}\rightarrow \mathbf{FVect}_\text{iso}\) is a smooth functor if \(F(f,g)\) depends smoothly on \(f,g\).

If \(f\in\Hom(V,V'), g\in\Hom(W,W')\) are isomorphisms, then \(F(f,g)\in\Hom(F(V,W),F(V',W'))\). Since these are all vector spaces, they carry the smooth structure described in §Examples of Differentiable Manifolds, ⁋Example 2, and through this we can apply the above definition. Also, it is not difficult to extend this definition to a general \(k\)-fold product

\[\mathbf{FVect}_\text{iso}\times\cdots\times\mathbf{FVect}_\text{iso}\rightarrow \mathbf{FVect}_\text{iso}\]

as well.

Example 5 \(\Hom(-,-)\) is a smooth functor. Suppose two arbitrary isomorphisms \(f:V\rightarrow V'\), \(g:W\rightarrow W'\) are given. Then \(\Hom(f,g)\) is a linear map from \(\Hom(V,W)\) to \(\Hom(V',W')\) making the diagram below

commute. Written as a formula, we can write \(\Hom(f,g)(u)=g\circ u\circ f^{-1}\). We can easily verify that \(\Hom(f,g)\) depends smoothly on \(g\). Consider the correspondence \(g\mapsto \Hom(f,g)\). Then for a basis of \(\Hom(W,W')\), \(w_i^j\),

\[(g+tw_i^j)\circ u\circ f^{-1}=g\circ u\circ f^{-1}+tw_i^j\circ u\circ f^{-1}\]

holds for all \(u\), so the directional derivative of this correspondence in the \(w_i^j\)-direction is \(u\mapsto w_i^j\circ u\circ f^{-1}\), which is continuous. Moreover, since this argument holds no matter what linear map is substituted for \(g\), we know from this that any higher directional derivative of \(g\mapsto\Hom(f,g)\) is always continuous. That is, \(g\mapsto\Hom(f,g)\) is \(C^\infty\). That this correspondence also depends smoothly on \(f\) is somewhat more tedious than for \(g\), but the map assigning \(f^{-1}\) to an isomorphism \(f\) is \(C^\infty\) by Cramer’s rule and \(\Hom(f,g)(u)=g\circ u\circ f^{-1}\) is linear in \(f^{-1}\), so we can repeat the above argument for \(f^{-1}\).

The following are all examples of smooth functors.

• Dual functor \((-)^\ast\) ([Linear Algebra] §Dual Space),
• \(k\)-th tensor functor \(\mathcal{T}^k(-)\) ([Multilinear Algebra] §Tensor Algebra),
• \(k\)-th symmetric functor \(\mathcal{S}^k(-)\) ([Multilinear Algebra] §Tensor Algebra),
• \(k\)-th exterior functor \(\bigwedge\nolimits^k(-)\) ([Multilinear Algebra] §Tensor Algebra),
• Tensor product \(-\otimes -\),
• Direct sum \(-\oplus-\).

The proof of the following theorem can be found in Theorem 3.6 of [MS].

Theorem 6 Let an arbitrary smooth functor \(F:(\mathbf{FVect}_\text{iso})^n\rightarrow \mathbf{FVect}_\text{iso}\) and, with a common base space \(B\), \(n\) vector bundles \(E_i\rightarrow B\) be given. Then, with the fiber at each \(b\in B\) given by

\[E_b=F((E_1)_b,\ldots,(E_n)_b)\]

there exists a vector bundle \(E\rightarrow B\).

We denote the vector bundle \(E\) obtained by the above process simply as \(F(E_1,\ldots, E_n)\).

Cotangent Bundle

Applying the above Theorem 6 to an arbitrary manifold \(M\), the tangent bundle \(E=TM\rightarrow M\), and the dual functor \((-)^\ast\), we obtain the following.

Definition 7 The cotangent bundle defined on a manifold \(M\) is the vector bundle \((TM)^\ast\) obtained by Theorem 6. Following the notation for the cotangent space \(T_p^\ast M\), we denote this by \(T^\ast M\).

\(T^\ast M\) is the space where, at each point \(p\), the vector space \(T_p^\ast M\) is attached. Here, \(T_p^\ast M\) is the dual space of the vector space \(T_pM\), that is, the space of linear maps that take a vector in \(T_pM\) and output a real number. We will revisit vector bundles obtained by applying other smooth functors before long.

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References

[MS] J.W. Milnor and J.D. Stasheff, Characteristic classes, Princeton university press, 1974.

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