This post was translated from Korean by LLM (Kimi). The translation may contain errors or awkward sentences. The Korean original is the source of truth.
Vector Bundles
Now we need to define vector fields, and to define concepts such as \(C^\infty\) vector fields it is better to first introduce the notion of a vector bundle. Let us begin by defining a vector bundle over a topological space.
Definition 1 A vector bundle over a topological space \(B\) is an object \(\pi:E \rightarrow B\) 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 \(b\in B\), the fiber \(E_b=\pi^{-1}(b)\) 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 \(b\in U\), the map \(x\mapsto h(b,x)\) is an isomorphism.
We call \(k\) the rank of the vector bundle \(E\rightarrow B\). The homomorphism \(h\) in the third condition is called a local trivialization, and if we can take \(U=B\) then we call \(E\) a trivial vector bundle.
Similarly, we can define a vector bundle over a manifold. To do this we require both \(E\) and \(B\) to be manifolds, \(\pi\) to be 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 \(b\in U\), the map \(x\mapsto h(b,x)\) is an isomorphism.
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 map \(\tilde{\varphi}:\pi^{-1}(U)\rightarrow\mathbb{R}^m\times\mathbb{R}^m\) by the formula
\[\tilde{\varphi}(v)=\bigl(x^1(\pi(v)), \ldots, x^m(\pi(v)), dx^1(v),\ldots, dx^m(v)\bigr)\]Then \(\tilde{\varphi}\) is a bijection from \(\pi^{-1}(U)\) onto the open subset \(\varphi(U)\times\mathbb{R}^m\) of \(\mathbb{R}^{2m}\).
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
\[\begin{aligned}(\tilde{\psi}\circ\tilde{\varphi}^{-1})(p^1, \ldots, p^m, v^1, \ldots, v^m)&=\tilde{\psi}\left(\varphi^{-1}(p), \sum v^i\frac{\partial}{\partial x^i}\bigg|_{\varphi^{-1}(p)}\right)\end{aligned}\]Writing \(v=\sum v^i\frac{\partial}{\partial x^i}\) here, the right-hand side can simply be written as
\[\left((\psi\circ\varphi^{-1})(p), dy^1(v), \ldots, dy^m(v)\right)\]Now for arbitrary \(j\) we have
\[dy^j\left(\sum v^i\frac{\partial}{\partial x^i}\bigg|_{\varphi^{-1}(p)}\right)=\sum_{i=1}^m v^i\frac{\partial y^j}{\partial x^i}\bigg|_{\varphi^{-1}(p)}\]Therefore, since each component of the transition map \(\tilde{\psi}\circ\tilde{\varphi}^{-1}\) above is \(C^\infty\), the map \(\tilde{\psi}\circ\tilde{\varphi}^{-1}\) itself is also \(C^\infty\).
Meanwhile, the topology on \(TM\) is generated 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 \(W=\mathbb{R}^{m}\), we can check that the \(\pi^{-1}(U)\) are all open in the topology generated by the above collection, and we can also verify that this topology yields a \(2m\)-dimensional topological manifold.
What remains is the local trivialization on \(TM\). For an arbitrary coordinate system \((U,\varphi)\), define \(\phi:\pi^{-1}(U)\rightarrow U\times\mathbb{R}^m\) this time by the formula
\[v|_p\mapsto (p, dx^1(v),\ldots, dx^m(v))\]That \(\phi\) is an isomorphism between vector spaces on each fixed fiber \(\pi^{-1}(p)\) is obvious, and it is also obvious that for any \(v_x\) we have \((\pi\circ\phi)(x,v)=x\). That \(\phi\) is a diffeomorphism follows from
\[\tilde{\varphi}=(\varphi\times\id_{\mathbb{R}^m})\circ\phi\]since in this formula the two functions other than \(\phi\) are both diffeomorphisms.
In particular, if \(TM\) is a trivial bundle, we call \(M\) a parallelizable manifold.
Smooth Functors
The reason the tangent bundle \(TM\) is important is that most vector bundles defined over a manifold are constructed from \(TM\). For example, the cotangent bundle \(T^\ast M\) is the vector bundle with the cotangent space \(T_p^\ast M\), which is the dual space of the tangent space, attached at each \(p\in M\). Similarly, various vector bundles are defined by algebraic operations at each point \(p\) (Example 5).
Originally, each time we define these we would have to show that they satisfy the conditions of a vector bundle, but [Mil] presents a more fundamental approach.
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 \(\mathbf{FVect}_\text{iso}\) of finite-dimensional \(\mathbb{R}\)-vector spaces whose morphisms are isomorphisms. 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 \((V,W)\overset{(f,g)}{\longrightarrow}(V',W')\) of isomorphisms between finite-dimensional vector spaces.
Thus a functor \(F\) from \(\mathbf{FVect}_\text{iso}\times\mathbf{FVect}_\text{iso}\) to \(\mathbf{FVect}_\text{iso}\) 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 called a smooth functor if \(F(f,g)\) depends smoothly on \(f,g\).
If \(f\in\Hom(V,V'), g\in\Hom(W,W')\), 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
\[\mathbf{FVect}_\text{iso}\times\cdots\times\mathbf{FVect}_\text{iso}\rightarrow \mathbf{FVect}_\text{iso}\]Example 5 \(\Hom(-,-)\) is a smooth functor. Suppose arbitrary isomorphisms \(f:V\rightarrow V'\), \(g:W\rightarrow W'\) are given. Then \(\Hom(f,g)\) is a functor from \(\Hom(V,W)\) to \(\Hom(V',W')\) making the diagram below
commute. In formula, we can write \(\Hom(f,g)(u)=g\circ u\circ f^{-1}\). We can easily check that \(\Hom(f,g)\) depends smoothly on \(g\). Consider the correspondence \(g\mapsto \Hom(f,g)\). Then for a basis \(w_i^j\) of \(\Hom(W,W')\),
\[(g+tw_i^j)\circ u\circ f^{-1}=g\circ u\circ f^{-1}+tw_i^j\circ u\circ f^{-1}\]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, this argument works even if we insert any linear map in place of \(g\), so from this we know that all higher directional derivatives of \(g\mapsto\Hom(f,g)\) are 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 since \(f\) is an isomorphism we can choose \(t\) sufficiently small so that \(f+tw_i^j\) is invertible, and then repeat the above argument.
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 Algebras),
- \(k\)-th symmetric functor \(\mathcal{S}^k(-)\) ([Multilinear Algebra] §Tensor Algebras),
- \(k\)-th exterior functor \(\bigwedge\nolimits^k(-)\) ([Multilinear Algebra] §Tensor Algebras),
- 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 \(n\) vector bundles \(E_i\rightarrow B\) with a common base space \(B\) be given. Then there exists a vector bundle \(E\rightarrow B\) whose fiber at each \(b\in B\) is given by
\[E_b=F((E_1)_b,\ldots,(E_n)_b)\]We denote the vector bundle \(E\) obtained by the above process simply as \(F(E_1,\ldots, E_n)\).
Cotangent Bundle
Applying 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 over a manifold \(M\) means the vector bundle \((TM)^\ast\) obtained by Theorem 6. Following the notation for cotangent space \(T_p^\ast M\), we denote this by \(T^\ast M\).
\(T^\ast M\) is the space with the vector space \(T_p^\ast M\) attached at each point \(p\). 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 of \(T_pM\) and output a real number. We will revisit vector bundles obtained by applying other smooth functors before long.
References
[MS] J.W. Milnor and J.D. Stasheff, Characteristic classes, Princeton university press, 1974.
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