In the last post, we defined the tangent space to a manifold at a point. That was the first step towards doing calculus on a manifold. In this post, we shall take the second step which is *bundling* the tangent spaces up into a single geometric object called the tangent bundle. We shall describe the tangent bundle in several different ways: As a vector bundle, as a locally free sheaf and as a sheaf of differential operators. Along the way we will see some more definitions from abstract sheaf theory.

Let \(X\) be a manifold. Informally speaking, a vector bundle over \(X\) is a family of vector spaces parameterized smoothly by the points in \(X\). This colloquial language does not suggest a definition. Here is a potential definition (which is incorrect): A vector bundle over \(X\) consists of a manifold \(E\) and a smooth map \(\pi : E \to X\) such that each fiber \(\pi^{-1}(x) = \{ v \in E : \pi(v) = x\}\) is a vector space. The main problem with this definition is that we are not forcing the vector space structure on the fibers to vary smoothly as we move around in \(X\). A second problem is that from this definition, there is no obvious way to describe a specific vector bundle structure. It should be easy to describe a vector bundle you are working with to your friends (assuming that you can describe the base manifold to them). Before we give the correct definition, we need some new terminology. Suppose that \(\pi : E \to X\) is a smooth map. A vector bundle chart on \(E\) consists of an open subset \(U \subseteq X\) and a diffeomorphism \(\pi^{-1}(U) = U \times \mathbb{R}^r\). Now we can give the correct definition: A vector bundle over \(X\) consists of a manifold \(E\) and a smooth map \(\pi : E \to X\) such that

- we can cover \(E\) using vector bundle charts
- given two vector bundle charts \(\pi^{-1}(U) = U \times \mathbb{R}^r\) and \(\pi^{-1}(V) = V \times \mathbb{R}^r\), if \(U \cap V\) is nonempty, then the change of coordinates from \(\pi^{-1}(U)\) to \(\pi^{-1}(V)\) is given by \[(x,v) \mapsto (x,\phi(x)v)\] where \(\phi : U \cap V \to {\rm GL}_r(\mathbb{R})\) is a smooth map (this just means that each matrix entry is a smooth function of \(x \in U \cap V\))

If the fibers are dimension \(r\), we call \(E\) a vector bundle of rank \(r\). Let \(X\) be a manifold. As a set, the tangent bundle is defined by \[TX = \coprod_{x \in X}T_x X\] Write \(\pi : TX \to X\) for the projection map which sends \(T_xX\) to \(x\). Assume that \(x_1,\dots,x_d\) are coordinates on \(U \subseteq X\). Then we have a bundle chart \[ \pi^{-1}(U) \ni \left( x, \sum a_i \frac{\partial}{\partial x_i} \lvert_x \right) \mapsto (x,(a_i)) \in U \times \mathbb{R}^d\] Suppose that \(y_1,\dots,y_d\) are coordinates on \(V \subseteq X\). Then \[\frac{\partial}{\partial x_j}\lvert_x = \sum_i \frac{\partial y_i}{\partial x_j}(x) \frac{\partial}{\partial y_i} \lvert_x\] Therefore the change of coordinates from \(U\) to \(V\), \(\phi : U \cap V \to {\rm GL}_d(\mathbb{R})\) is given by \[\phi(x) = \left( \frac{\partial y_i}{\partial x_j}(x) \right)\] Maybe you are worried that we did not prove that \(TX\) was a manifold before we prove that it was a vector bundle. Everything is OK, because the vector bundle charts we have described on \(TX\) are also plain old coordinate charts. We have defined a new type of mathematical object, so we need to define the morphisms. If \(E,F\) are vector bundles over \(X\) then a map of vector bundles \(f : E \to F\) is a smooth map such that

- \(\pi_F \circ f = \pi_E\) (i.e \(f\) maps fibers into fibers)
- \(f\) is linear on the fibers.

Suppose that \(\pi : E \to X\) is a vector bundle. Let \(X = \cup_{\alpha} U_{\alpha}\) be an open cover such that each \(\pi^{-1}(U_{\alpha})\) is a bundle chart on \(E\). Write \(\phi_{\beta\alpha} : U_{\alpha} \cap U_{\beta} \to {\rm GL}_r(\mathbb{R})\) for the change of coordinates from \(\pi^{-1}(U_{\alpha})\) to \(\pi^{-1}(U_{\beta})\). Then we have

- \(\phi_{\alpha\alpha} = {\rm id}\)
- \(\phi_{\gamma\beta} \phi_{\beta\alpha} = \phi_{\gamma\alpha}\) on \(U_{\alpha} \cap U_{\beta} \cap U_{\gamma}\).

We call \((\phi_{\beta\alpha}, U_{\alpha})\) gluing data for the vector bundle \(E\). From the gluing data we can reconstruct \(E\): The gluing data tells us where the vector bundle charts are and how to change coordinates between them. We can also describe vector bundle maps in terms of gluing data. Suppose that we have two vector bundles \(E\) and \(F\) described by gluing data \((\phi_{\beta\alpha},U_{\alpha})\) and \((\psi_{\beta\alpha},U_{\alpha})\). Then a vector bundle morphism \(f : E \to F\) consists of maps \(f_{\alpha} : U_{\alpha} \to {\rm Mat}_{s \times r}(\mathbb{R})\) such that \[\psi_{\beta\alpha} f_{\alpha} = f_{\beta} \phi_{\beta\alpha}\] on \(U_{\alpha} \cap U_{\beta}\).

This definition is not without flaws. Suppose we write \(U_{\alpha} = \cup_p V_{\alpha, p}\). Then we get new gluing data

\((\phi_{\beta,q,\alpha,p}, V_{\alpha,p})\) defined by

- \(\phi_{\alpha,q,\alpha,p} = {\rm id}\)
- \(\phi_{\beta,q,\alpha,p} = \phi_{\beta,\alpha} \lvert_{V_{\beta,q} \cap V_{\alpha,p}}\).

This gluing data describes the same vector bundle as \((\phi_{\beta\alpha}, U_{\alpha})\), despite being defined over a different open cover of \(X\). Eventually, when we talk about sheaf cohomology, we will fix this problem, but for now we shall just work with the imperfect definition. In practice, this does not cause problems. If \(U \subseteq X\) is a contractible coordinate chart, then \(\pi^{-1}(U)\) is a vector bundle chart on \(E\). This is because every vector bundle over \(\mathbb{R}^d\) is trivial. Therefore, with suitable choices of coordinates on the base, we get gluing data for every vector bundle with the same open cover.

We have seen that every manifold \(X\) has a tangent bundle \(TX\) and written down its gluing data. Let us work through another example to demonstrate how one works with vector bundles. Consider \(\mathbb{RP}^1\), the set of 1-dimensional subspaces of \(\mathbb{R}^2\). We write \([a:b]\) for the line spanned by \((a,b)\). If \(X,Y\) are linear coordinates on \(\mathbb{R}^2\), then \(X/Y\) and \(Y/X\) are smooth partial functions on \(\mathbb{RP}^1\). In high school mathematics, the function \(Y/X\) is often called the gradient. We shall write \(m = Y/X\). We also write \(n = X/Y\). Then \(m = 1/n\) were both functions are defined. Infact, if we set \[ {\rm D}(X) = \{ L \in \mathbb{RP}^1 : X(L) \not= 0 \}\] \[ {\rm D}(Y) = \{ L \in \mathbb{RP}^1 : Y(L) \not= 0 \}\] then \(\mathbb{RP}^1 = {\rm D}(X) \cup {\rm D}(Y)\), \(m = Y/X\) is a coordinate on \({\rm D}(X)\) and \(n = X/Y\) is a coordinate on \({\rm D}(Y)\). Define \[ L = \{ (p,L) \in \mathbb{R}^2 \times \mathbb{RP}^1 : p \in L \} \] Then \(L\) is a rank \(1\) vector bundle (or a line bundle) over \(\mathbb{RP}^1\). Write \(\pi : L \to \mathbb{RP}^1\) for the obvious projection. We have bundle coordinates \[ (\lambda,m) \mapsto ( (\lambda,\lambda m),[1:m]) \quad \text{over ${\rm D}(X)$}\] \[ (\lambda,n) \mapsto ( (\lambda n,\lambda),[n:1]) \quad \text{over ${\rm D}(Y)$} \] The change of coordinates map is \[(\lambda,m) \mapsto ((\lambda,\lambda m),[1:m]) = ((\lambda m \frac{1}{m}, \lambda
m), [1/m:1]) \mapsto (m \lambda, 1/m)\] Therefore the change of coordinate map from \(D(X)\) to \(D(Y)\) is \(Y/X : D(X) \cap D(Y) \to \mathbb{R}^{\times}\). We don't need to worry about the cocycle condition because there are only two coordinate charts. We call \(L\) the *tautological line bundle* over \(\mathbb{RP}^1\). Suppose that \(f : \mathbb{R}^2 \to \mathbb{R}\) is a linear function. Then we have a bundle map \(f : L \to \mathbb{R} \times \mathbb{RP}^1\) defined by \((p,L) \mapsto (f(p),L)\). If \(f = aX + bY\), then we have \[ f = \left(a + b \frac{Y}{X}\right) \text{ over $D(X)$}\] \[ f = \left(a \frac{X}{Y} + b\right) \text{ over $D(Y)$}\]

So far we have been describing vector bundles as manifolds over some base \(X\) where all the fibers are vector spaces. There is also a sheaf theoretic definition of vector bundles which we shall explore in this section.

Let \(X = \mathbb{R}\) and consider the vector bundle \(E = X \times \mathbb{R}\). We have a vector bundle map \(f : E \to E\) defined by \((x,\lambda) \mapsto (x, x \lambda)\). We can think of \(f\) as the matrix \((x)\). If we try and take the kernel of \(f\), something bad happens: The map \(f\) is injective when \(x \not= 0\) and zero when \(x=0\). Therefore the kernel should be like a vector bundle which is zero dimensional over \(\mathbb{R} \backslash \{ 0 \}\) and one dimensional over \(\{ 0 \}\). Obviously, no such vector bundle exists. The reason we need the sheaf theoretic definition of vector bundle is because it gives us a meaningful way to assign kernels and cokernels to all vector bundle maps, not just those with constant rank.

Let \(X\) be a manifold. Recall that \(\mathscr{C}\) is the sheaf of smooth functions on \(X\). A \(\mathscr{C}\)-module is a sheaf of abelian groups \(\mathscr{F}\) such that each \(\mathscr{F}(U)\) is a \(\mathscr{C}(U)\)-module and the restriction maps are compatible with the \(\mathscr{C}\)-action: \((s v) \lvert_V = s \lvert_V v \lvert_V\). Let \(\pi : E \to X\) be a vector bundle. We can build a \(\mathscr{C}\)-module \(\mathscr{E}\) called the *sheaf of sections* as follows. If \(U \subseteq X\) is open, then \(\mathscr{E}(U)\) is the sections of \(E\) over \(U\): \[\mathscr{E}(U) = \{ s : U \to E : \pi s = {\rm id}_U \}\] If \(s,t \in \mathscr{E}(U)\) then we define \((s+t)(x) = s(x) + t(x)\). This makes sense because both \(s(x)\) and \(t(x)\) are vectors in the fiber \(\pi^{-1}(x)\). Therefore \(\mathscr{E}\) is a presheaf of abelian groups. To check that \(\mathscr{E}\) is a sheaf, you just need to unravel the definitions. If \(f \in \mathscr{C}(U)\) and \(s \in \mathscr{E}(U)\), then we define \((fs)(x) = f(x) s(x)\). Therefore \(\mathscr{E}\) is a \(\mathscr{C}\)-module. The sheaf of sections \(\mathscr{E}\) is much nicer than an arbitrary \(\mathscr{C}\)-module. Indeed, \(\mathscr{E}\) is a locally free \(\mathscr{C}\)-module! This means that for each point \(p \in X\), there is a neighborhood \(p \in U\) such that \(\mathscr{E}\lvert_U \cong \mathscr{C}_{U}^{\oplus r}\). To see this, choose a bundle chart \(\pi^{-1}(U) = U \times \mathbb{R}^r\). Write \(e_1,\dots,e_r\) for the standard basis of \(\mathbb{R}^r\). Then \[\mathscr{E}(U) = \{ f_1 e_1 + \dots + f_r e_r : f_i \in \mathscr{C}(U) \} \cong
\mathscr{C}(U)^{\oplus r}.\] To denote this, we shall write \(\mathscr{E}\lvert_U = \mathscr{C} \lvert_U \{ e_1,\dots,e_r \}\). The map \(E \mapsto \mathscr{E}\) sends vector bundles to locally free sheaves. It is an amazing fact that every locally free sheaf arises this way! Suppose that \(\mathscr{F}\) is a locally free sheaf. Suppose that \(\mathscr{F}\lvert_U = \mathscr{C} \lvert_U \{ s_1,\dots,s_r \}\) and \(\mathscr{F} \lvert_V = \mathscr{C} \lvert_U \{ t_1,\dots,t_r \}\). Then the change of basis matrix from \(s_1,\dots,s_r\) to \(t_1,\dots,t_r\) is a smooth map \(\phi : U \cap V \to {\rm GL}_r(\mathbb{R})\). Moreover, if you do this over three open sets, then the cocycle condition is satisfied, so we can construct a vector bundle \(F\) such that \(F \mapsto \mathscr{F}\). We can capture all of this in the following precise statement: The functor \(E \mapsto \mathscr{E}\) from the category of vector bundles to the category of locally free sheaves is an equivalence of categories. Lets explore the map \(f : E \to E\) from the start of the section using our new sheaf theoretic language. The sheaf of sections \(\mathscr{E}\) is just \(\mathscr{C}\), the sheaf of smooth functions on \(X\). The induced map \(\mathscr{C} \to \mathscr{C}\) is given by multiplication by \(x\). If \(f\) is a smooth map and \(fx = 0\) then \(f = 0\) by continuity. Therefore the kernel of \(f\) is the zero sheaf! What the heck? Shouldn't the kernel somehow represent a "vector bundle" which is zero over \(\mathbb{R} \backslash \{ 0 \}\) and one dimensional over \(0\)? The answer is no. We are simply taking the kernel in the category of \(\mathscr{C}\)-modules. If it doesn't behave how we expect, it is not the categories fault, it is our fault for working inside the wrong category. On the bright side, the cokernel of \(f\) will represent a "vector bundle" which is 1-dimensional over \(0\) and zero over \(\mathbb{R} \backslash \{ 0 \}\), but we need to talk about sheafification before we can talk about cokernels (we will do this in a later post).

Let \(X\) be a manifold. Write \(T\) for the tangent bundle and \(\mathscr{T}\) for the associated sheaf of sections. Elements in \(\mathscr{T}(X)\) assign tangent vectors to each point in \(X\), so they are vector fields on \(X\). In the last post, we proved that if \(x_i\) are coordinates on \(U \subseteq X\), then \[\mathscr{T} \lvert_U = \mathscr{C} \lvert_U \left\{ \frac{\partial}{\partial x_1}, \dots, \frac{\partial}{\partial x_d} \right\} \] where \(\partial / \partial x_i\) represents the section which sends \(x\) to \(\partial / \partial x_i \lvert_x\). There is another way to describe the locally free sheaf \(\mathscr{T}\) which is important. We can define a sheaf \({\rm Hom}_{\mathbb{R}}(\mathscr{C},\mathscr{C})\) as follows: The sections over \(U\) are the \(\mathbb{R}\)-linear sheaf homomorphisms \(\mathscr{C} \lvert_U \to \mathscr{C} \lvert_U\). It is routine to check that the sheaf axioms are satisfied. Define \[\mathscr{D} = \{ D \in {\rm Hom}_{\mathbb{R}}(\mathscr{C},\mathscr{C}) : D(fg) = f D(g) + g D(f) \}\] This is the sheaf of first order differential operators. Pointwise multiplication turns \(\mathscr{D}\) into a \(\mathscr{C}\)-module. There is a map \(\mathscr{T} \to \mathscr{D}\) defined by \(V \mapsto f \mapsto (x \mapsto V_x(f))\). Under this map, the vector field \(\partial / \partial x_i\) is sent to partial differentiation with respect to \(x_i\). We can define the inverse map \(\mathscr{D} \to \mathscr{T}\) by \(D \mapsto x \mapsto f \mapsto D(f)(x)\). It is routine to check that these maps establish an isomorphism of \(\mathscr{C}\)-modules.

- write down gluing data for the tautological line bundle on \(\mathbb{RP}^n\).
- Prove that \(E \mapsto \mathscr{E}\) is an equivalence of categories between vector bundles and locally free sheaves.
- Check that \({\rm Hom}(\mathscr{F},\mathscr{G})\) is actually a sheaf.
**(Harder)**Prove that every vector bundle over \(\mathbb{R}^d\) is trivial.