Vector Bundles

Posted on July 6, 2021 by Parth
(geometry)
Among the first ideas one encounters in the study of homotopy groups is to consider covering spaces– things that locally look like products with some discrete topological space $$\Delta$$ but can have vastly different global properties than a simple product space. In fact the number of non-trivial covering spaces is a measure of the topological complexity of the space, in a sense made precise by the Galois correspondence of fundamental groups. It is then natural to ask what happens if one replaces $$\Delta$$ with more a complicated topological space $$F$$: what results is a richer analogue of a covering space called a fiber bundle, or a vector bundle if $$F$$ is a vector space.

# Locally product-like spaces

Recall that we say $$E\xrightarrow{p} B$$ is a covering space with fibers $$F$$ (where $$F$$ is a discrete topological space) if $$B$$ has an open cover $$\{U_\alpha\}$$ such that for each $$\alpha$$, there is a homeomorphism $$\phi_\alpha: U_\alpha \times F \rightarrow p^{-1}(U_\alpha)$$ which makes the following diagram commute (here $$\pi_i$$ can any of the projection maps which the product comes with.) This motivates the definition of a vector bundle:

Definition For a topological space $$B$$, we say $$E\xrightarrow{p} B$$ is a (real) vector bundle of rank $$n$$ if $$B$$ has an open cover $$\{U_\alpha\}$$ such that for each $$\alpha$$, we have a fiber-preserving homeomorphism $$\phi_\alpha: U_\alpha\times \mathbb{R}^n \rightarrow p^{-1}(U_\alpha)$$ which is a linear isomorphism on each fiber (i.e. for each $$x\in U_\alpha$$, $$\phi_\alpha$$ restricts to a linear isomorphism $$x\times \mathbb{R}^n\rightarrow p^{-1}(x)$$ of affine spaces).

Vector bundles of rank $$1$$ are termed line bundles.

$$E$$ is termed the total space, while $$B$$ is the base space. The collection of charts $$\{(\phi_\alpha, U_\alpha)\}$$ is called a trivialization.

Note that one may replace $$\mathbb{R}$$ with any other field, and strengthen ‘homeomorphism’ in the above definition to ‘diffeomorphism’ or ‘isomorphism’ depending on what kind of geometry is involved.

The condition we ask for is local trivialization: to each $$U_\alpha$$, the map $$p$$ must look just like a projection $$U_\alpha \times \mathbb{R}^n \rightarrow U_\alpha$$. As with covering spaces, this says nothing about the global structure that $$E$$ might possess. $$E$$ has carrier set $$\coprod_{x\in B} \{x\}\times \mathbb{R}^n$$ and $$p:E\rightarrow B$$ acts as the natural map on sets, but $$E$$ has additional structure induced by both the topology on $$B$$ and the vector space structure on $$\mathbb{R}^n$$. The vector bundle then is a $$B$$-parametrised family of vector spaces which varies ‘smoothly’.

Example What is the simplest thing that locally looks like a product? A product, of course. Accordingly, we call $$E= B\times \mathbb{R}^n\rightarrow B$$ the trivial bundle of rank $$n$$. Note that a bundle is trivial if and only if $$\{(\text{id}, B)\}$$ is a trivialization.

A slightly more involved example is that of a tangent bundle: given an $$n$$-dimensional differentiable manifold $$B$$, to each $$x\in B$$ we can associate the tangent space $$T_xB$$ ($$\cong \mathbb{R}^n$$). Then the tangent bundle has carrier set \begin{align*} E &=\coprod_{x\in B} \{x\}\times T_xB \\ &= \{(x,y)\;|\;x\in B, \; y\in T_xB\} \end{align*} and there is a natural map $$p: E\rightarrow B$$ given by $$(x,y)\mapsto x$$. We can turn this into a vector bundle– in fact, a manifold in its own right. Recall that $$B$$ has an open cover $$\{U_\alpha\}$$ such that each $$U_\alpha$$ is diffeomorphic to $$\mathbb{R}^n$$ via some chart $$\theta_\alpha$$. Identifying each $$T_xB$$ with $$\mathbb{R}^n$$, we can define functions (of sets) $$\phi_\alpha: U_\alpha \times \mathbb{R}^n \cong \mathbb{R}^{2n}\rightarrow p^{-1}(U_\alpha)$$. Giving $$E$$ the weak topology induced by these, we have an atlas of charts on $$E$$ whose transition functions can be checked to be smooth. Then in addition to being a vector bundle, $$E$$ is also a $$2n$$-dimensional manifold.

What does the tangent bundle of $$S^1$$ look like? If $$(x,y)\in S^1$$, we have that $$T_{(x,y)}S^1 = \{(u,v)\;|\; ux+vy=0\}\subset \mathbb{R}^2$$ hence the tangent bundle $$TS^1$$ is precisely the manifold $$\{(x,y,u,v)\;|\;x^2+y^2=1, \; ux+vy=0\}\subset \mathbb{R}^4$$, isomorphic to the annulus $$\{(x,y,u)\;|\;x^2+y^2=1\}\subset \mathbb{R}^3$$. But this is the trivial bundle $$S^1\times \mathbb{R}$$.

What about other spheres? It is a fact that $$S^1, S^3$$ and $$S^7$$ are the only spheres whose tangent bundle is trivial. We will be able to address the case for $$S^2$$ once we understand what sections are.

## Linearity

In the definition of a vector bundle we insist that the charts are linear isomorphisms on fibers, not just diffeomorphisms. This is a strong constraint, but also a helpful one since vector spaces are rich with structure and we understand them well. Firstly, observe that associating a vector space to each point is morally the right thing to do when dealing with objects like tangent bundles of a manifold. And all our favourite vector space operations extend directly to vector bundles– in essence, all operations are ‘pointwise’:

• Direct sums: If $$E_1\xrightarrow{p_1}B$$ and $$E_2\xrightarrow{p_2}B$$ are two vector bundles on $$B$$, then $$E_1\oplus E_2$$ has carrier set $$\{(v_1,v_2)\;|\; p_1(v_1)=p_2(v_2)\}$$ and associated projection $$p(v_1,v_2)=p_1(v_1)=p_2(v_2)$$. After checking local triviality, we can see that this does what we want: for any $$x\in B$$, $$p^{-1}(x)=p_1^{-1}(x)\oplus p_2^{-1}(x)$$.

• Tensor products: For $$E_1$$ and $$E_2$$ as above, we define $$E_1\otimes E_2$$ to have carrier set the disjoint union of $$p_1^{-1}(x)\otimes p_2^{-1}(x)$$ for $$x\in B$$, and the natural projection map. The topology is given by taking open sets which trivialize both $$E_1$$ and $$E_2$$, and then letting the fiberwise tensor products of the charts on these be homeomorphisms.

• Linear Duals: If $$E\xrightarrow{p}B$$ is a vector bundle, then we can define the dual bundle $$E^\ast$$ to have carrier set given by the disjoint union of $$p^{-1}(x)^\ast$$ for each $$x\in B$$. This can be given a topology by explicitly gluing the pullbacks of open sets along transition functions, a method of construction we shall meet next.

# Twisted functions

If $$f:X\rightarrow Y$$ is a morphism, we can replace it with a morphism $$\tilde{f}: X\rightarrow X\times Y$$ given by $$\tilde{f}(x)=(x,f(y))$$. $$\tilde{f}$$ thus given is called the graph of $$f$$. Moreover, the correspondence is one-to-one in that if $$\tilde{f}:X\rightarrow X\times Y$$ is any morphism such that $$\pi_1\circ \tilde{f}=\text{id}_X$$, then there is a unique morphism $$f$$ which $$\tilde{f}$$ is the graph of.

Now suppose $$B$$ is a topological space, and $$\{U_\alpha\}$$ is an open cover of $$B$$. If $$f:B\rightarrow \mathbb{R}^n$$ is any continuous map, then its graph $$\tilde{f}$$ is a map from $$B$$ to the trivial vector bundle $$B\times \mathbb{R}^n$$ satisfying $$\tilde{f}\circ \pi_1 = \text{id}_B$$. In a trivial way, it induces maps $$U_\alpha\rightarrow U_\alpha\times \mathbb{R}^n$$ which when ‘stitched’ together, give a map $$B\rightarrow B\times \mathbb{R}^n$$. Then a function $$f: B\rightarrow \mathbb{R}^n$$ is what we get by ‘stitching’ locally defined functions into a global function. We can generalise this construction by not restricting ourselves to trivial stitching.

Definition If $$E\xrightarrow{p} B$$ is a vector bundle, we say $$f:B\rightarrow E$$ is a section if $$p\circ f=\text{id}_B$$.

A particularly nice one is the zero section, which assigns to each $$x\in B$$ the point $$(x,0)\in E$$. Since vector bundle isomorphisms act like linear isomorphisms on fibres, zero sections are preserved.

Since vector bundles are locally trivial, the restrictions of $$f$$ to trivialising open sets are functions. It is immediate from the preceding discussion that $$\mathbb{R}^n$$-valued functions are precisely sections of the trivial bundle of rank $$n$$.

Example The space $$\mathbb{RP}^n$$ is defined as the space of lines of $$\mathbb{R}^{n+1}$$ through the origin. Assigning to every ‘point’ of $$\mathbb{RP}^n$$ the line that it actually is gives us the tautological bundle $$E$$, whose carrier set is $$\{(\ell,v)\;|\; \ell \in \mathbb{RP}^n, v\in \mathbb{R}^{n+1}, v\in \ell\}$$. The simplest case is obtained for $$n=1$$, when $$\mathbb{RP}^1$$ is homeomorphic to $$\mathbb{S}^1$$. The tautological bundle $$E$$ in this case is called the Mobius bundle, since it is homeomorphic to the interior of a Mobius strip: $$\mathbb{RP}^1\cong[0,\pi]/\{0,\pi\}$$ is swept by a line rotating through $$\pi$$, with $$\{0\}\times\mathbb{R}$$ identified with $$\{\pi\}\times \mathbb{R}$$ by $$(0,t)\sim (\pi,-t)$$.

What do sections of the Mobius bundle look like? We can trivialise $$E$$ by taking charts $$\mathbb{RP}^n \setminus \{x\} \cong (0,1)$$ for each $$x$$– then the restriction of any section $$f$$ to this chart gives a function $$g:(0,1)\rightarrow \mathbb{R}$$ such that $$\lim_{x\rightarrow 0}g(x)=-\lim_{x\rightarrow 1} g(x)$$. This is a family of partial functions on $$\mathbb{RP}^n$$ that carries the same information up to some transition function: it is morally the right thing to put them together into one packet. However, there is no total function that does the job since restrictions of total functions on $$\mathbb{RP}^n (\cong S^1)$$ to the chosen charts agree at end points. The construction we need is precisely that of a Mobius bundle, and the sections of this bundle are the ‘twisted function’ we want.

Then is the Mobius bundle isomorphic to the trivial bundle on $$S^1$$? No: a simple way to see this is that every section on the Mobius bundle must meet the zero section at some point (think intermediate value theorem) while the trivial bundle has a nowhere-vanishing section.

We will now make precise the idea of gluing these partial functions with appropriate twists involved.

## Transition functions and structure groups

What does it mean to ‘stitch’ these partial functions? If $$\{U_\alpha\}$$ is the trivializing cover of the vector bundle $$E\xrightarrow{p}B$$, then for any $$\alpha,\beta$$ writing $$U_{\alpha\beta}=U_\alpha\cap U_\beta$$ we see that $$p^{-1}(U_{\alpha\beta})$$ sits inside both $$\phi_\alpha(U_\alpha\times \mathbb{R}^n)$$ and $$\phi_\beta(U_\beta\times \mathbb{R}^n)$$. Identifying these copies gives us a map $$\phi_\alpha \circ \phi_\beta^{-1}:U_{\alpha\beta}\times \mathbb{R}^n \rightarrow U_{\alpha\beta}\times \mathbb{R}^n$$ that is an automorphism of $$\mathbb{R}^n$$ on each fiber (i.e. for each fixed $$x\in U_{\alpha\beta}$$). In other words, for each $$x\in U_{\alpha\beta}$$ we have a linear map $$g_{\alpha\beta}(x)\in GL_n(\mathbb{R})$$. We can check that these maps satisfy the cocycle condition: $\begin{gather*} g_{\alpha\alpha}=\text{id} \\ g_{\alpha\beta}\cdot g_{\beta\gamma}=g_{\alpha\gamma} \end{gather*}$

This method of using explicit transition functions can be used to construct new vector bundles from scratch: given $$B$$ with a collection of charts $$U_\alpha$$ and functions $$g_{\alpha\beta}:U_{\alpha}\cap U_\beta \rightarrow GL_n(\mathbb{R})$$ which satisfy the cocycle conditions, there is a fiber bundle $$E$$ that is trivializable over $$\{U_\alpha\}$$ with transition functions $$g_{ij}$$.

Now it may happen that the maps $$\{g_{\alpha\beta}\}$$ take values in a subgroup $$G\subset GL_n(\mathbb{R})$$, which then captures the symmetries of how the different charts overlap. This group $$G$$ is then called a structure group and the charts $$\{U_\alpha\}$$ define a $$G$$-structure on the vector bundle $$E\rightarrow B$$. While the cocycles $$\{g_{\alpha\beta}\}$$ depend on the choice of trivialization, structure groups are an invariant of the bundle.

Proposition For a smooth manifold $$B$$, the vector bundle $$E\rightarrow B$$ is trivial if and only if it has a trivial structure group.
Proof Suppose $$E$$ is the trivial bundle of rank $$n$$ on $$B$$, then $$\{(\text{id}, B)\}$$ is a trivialization with cocycles $$\{\text{id}\}\subset GLn(\mathbb{R})$$. Conversely, suppose $$E\rightarrow B$$ is a vector bundle with trivial structure-group, and $$\{(\phi_\alpha,U_\alpha)\}$$ is a choice of trivialization. Then for any $$\alpha,\beta$$, $$\phi_\alpha\circ \phi_{\beta}^{-1}$$ is the identity on $$\{x\} \times \mathbb{R}^n$$ for any $$x\in U_\alpha\cap U_\beta$$, so in fact $$\phi_\alpha^{-1}$$ and $$\phi_\beta^{-1}$$ agree on $$p^{-1}(U_\alpha)\cap p^{-1}(U_\beta)$$. Then by the gluing lemma for smooth maps [CITATION NEEDED] we find a smooth map $$\phi^{-1}: \bigcup_\alpha p^{-1}(U_\alpha) = E \rightarrow E\times \mathbb{R}^n$$ that we can check is a diffeomorphism. $$\square$$

What is the structure group of the Mobius bundle $$\mathcal{M}$$ on $$S^1= [0,1]/\{0,1\}$$? We can triviale the bundle as $$\{(\phi_0, U_0), (\phi_1,U_1\})\}$$ where $$U_0 = S^1\setminus\{0\}$$ and $$U_1=S^1\setminus\{\frac{1}{2}\}$$. Then $$U_0\cap U_1=(0,\frac{1}{2})\cap (\frac{1}{2}, 1)$$, so the transition function is $g_{01}(x): t\mapsto \begin{cases} t, &x<\frac{1}{2} \\ -t, &x>\frac{1}{2}.\end{cases}$ The structure group can be seen to be $$\mathbb{Z}/2\mathbb{Z}\cong O(1)\subset GL_1(\mathbb{R})$$. In particular, the Mobius bundle is non-trivial.

Note that the structure group of the Mobius bundle contains elements of negative determinant– this corresponds to the transition function reversing the orientation of the fibers when gluing. Accordingly, vector bundles who have a structure group with only elements of positive determinant are termed orientable. Hence the Mobius bundle is non-orientable.

## Vector fields

This idea of writing functions as sections provides a useful tool when you naturally care about maps whose codomain depends on the argument– for instance, a vector field $$v$$ on a smooth manifold $$B$$ which continuously assigns to each $$x\in B$$ a vector $$v(x)\in T_xB$$. While using classical terminology would require us to consider an embedding of $$B$$ in a larger affine space, we can now define vector fields as sections of the tangent bundle $$TB$$. While there are ways to embed the manifold with its tangent bundles in a larger space to avoid variable codomains, the tool becomes indispensible when considering cotangent bundles (the linear dual to the tangent bundle), where a section $$f$$ gives, for each $$x\in B$$, a linear functional $$f(x)\in (T_xB)^\ast$$.

The additional linear structure that vector bundles contain over fiber bundles gives us a lot more to talk about. For instance, we say a collection of vector fields $$\{V_\alpha\}$$ on smooth manifold $$B$$ is linearly independent if $$\{V_\alpha(x)\}$$ is a linearly independent subset of $$T_xB$$ for each $$x\in B$$. The vector fields form a basis of the tangent bundle if the corresponding vectors at each point form a basis of the tangent space. Then if an $$n$$ dimensional manifold has a basis of $$n$$ vector fields, we can decompose the tangent bundle into linear combinations of these fields and construct an isomorphism with the trivial bundle. These definitions and results are true for general vector bundles:

Theorem If $$E\xrightarrow{p}B$$ is a vector bundle of rank $$n$$, then $$E$$ is trivial if and only if it admits a basis of $$n$$ sections.

This, for instance, immediately shows that the tangent bundle of $$S^1$$ is trivial since there is a non-vanishing vector field on $$S^1$$. On the other hand, the hairy ball theorem asserts that even dimensional spheres admit no nowhere-vanishing continuous vector fields– in particular, there cannot be a basis of $$n$$ vector fields hence the tangent bundles are nontrivial. In a similar vein, the fact that the Mobius bundle has no non-vanishing sections (a fact that follows from the intermediate value theorem) shows that it is a non-trivial bundle.

To conclude, we shall give a construction of ‘twisted functions’ that naturally arises in algebraic geometry.

## Homogeneous polynomials on $$\mathbb{P}^n$$

The complex projective space $$\mathbb{P}^n$$ (with homogeneous coordinates $$(X_0:...:X_n)$$) is covered by $$n+1$$ affine patches $$\{U_0,...,U_n\}$$, where $$U_i = \{X_i \neq 0\}$$. Each patch is a copy of $$\mathbb{A}^n$$, and polynomials in coordinate functions are morphisms, or sections of $$U_i\times \mathbb{C}$$. For $$n=1$$, consider $$1+\frac{X_1}{X_0}$$ on the patch $$U_0$$ and $$1+\frac{X_0}{X_1}$$ on the patch $$U_1$$. Up to multiplication with $$\frac{X_1}{X_0}$$, these associate a well-defined scalar to each point of $$\mathbb{P}^1$$. It however, does not come from a function (a section of the trivial bundle) since the transition functions are not identity– but what we have here is a perfect situation for a vector bundle.

Recall that $$\mathbb{P}^n$$ is the space of all (complex) ‘lines’ through the origin in $$\mathbb{C}^{n+1}$$, so has an associated tautological line bundle which we call $$\mathcal{O}(-1)\rightarrow \mathbb{P}^n$$. Let $$\mathcal{O}(1)\xrightarrow{p} \mathbb{P}^n$$ denote the linear dual of this bundle, i.e. $$p^{-1}(x_0:...:x_n)$$ is the space of linear functionals on $$(x_0,...,x_n)\mathbb{C}\subset \mathbb{C}^{n+1}$$. This carries the structure of a quasi-projective variety: writing $$(X_0:...:X_{n+1})$$ for the homogeneous coordinates on $$\mathbb{P}^{n+1}$$, consider the quasi-projective variety given by \begin{align*} V &= \mathbb{P}^{n+1}\setminus\{(0:...:0:1)\} \\ &= \{(x_0:...:x_n:\lambda)\;|\; (x_0:...:x_n)\in \mathbb{P}^n, \; \lambda \in \mathbb{C}\}. \end{align*} Then we can give an isomorphism $$V\rightarrow \mathcal{O}(1)$$ by mapping $$(x_0:...:x_n:\lambda)$$ to the linear map $$\lambda\in \mathcal{O}(1)_{(x_0:...:x_n)}$$. Henceforth we shall use the two realisations of $$\mathcal{O}(1)$$ interchangeably.

Equipped with the natural projection map $$p$$, we can explicitly show that this is a (complex) line bundle over $$\mathbb{P}^n$$ trivialized by the affine patches: on $$U_i$$ we have the holomorphic chart $$(x_0:x_1:...:x_n:\lambda)\mapsto ((x_0:x_1:...:x_n), \frac{\lambda}{x_i})\subset \mathbb{P}^n\times \mathbb{C}$$ and for a fixed $$(x_0:...:x_n)\in U_i\cap U_j$$, the transition function between charts is the linear isomorphism given by $$\frac{x_i}{x_j}\in \text{GL}(1,\mathbb{C})$$.

If $$f:\mathbb{P}^n\rightarrow V$$ is a holomorphic section of $$\mathcal{O}(1)=\mathcal{O}(-1)^\ast$$, then for each $$(x_0:...:x_n)\in \mathbb{P}^n$$, The linear map $$f(x_0:...:x_n)$$ maps $$(x_0,...,x_n)\in \mathcal{O}(-1)_{(x_0:...:x_n)}$$ to some $$F(x_0,...,x_n)\in \mathbb{C}$$. Then $$F$$ is a well-defined holomorphic function $$\mathbb{C}^{n+1}\setminus \{0\} \rightarrow \mathbb{C}$$. From the description it is clear that $$F(\lambda x_0,...,\lambda x_n) = \lambda F(x_0,...,x_n)$$, so $$F$$ is in fact linear and homogeneous. In particular, the singularity at $$0$$ is removable, and we have a homogeneous polynomial of degree $$1$$.

On the other hand if $$F\in \mathbb{C}[X_0,...,X_n]$$ is any homogeneous polynomial of degree $$1$$, then $$F$$ corresponds to the section $$z\mapsto (z:F(z))$$ of $$V\subset \mathbb{P}^{n+1}$$ (the degree $$1$$ condition ensures this section is well-defined). Thus the holomorphic sections of (_{^n}(1)) are precisely homogeneous degree 1 polynomials.

Writing $$\mathcal{O}(k)=\mathcal{O}(1)^{\otimes k}$$, sections of $$\mathcal{O}(k)\rightarrow \mathbb{P}^n$$ would correspond to formal products of $$k$$ sections of $$\mathcal{O}(1)\rightarrow \mathbb{P}^n$$, i.e. of the form $$F_1\otimes F_2 \otimes ... \otimes F_k$$ for homogeneous polynomials $$F_1,...,F_k$$. For instance, if $$n=1, k=2$$ then the restriction of $$\mathcal{O}(k)$$ to the patch $$Z\neq 0$$ is isomorphic to the trivial bundle, and moreover the sections $$\frac{X}{Z}\otimes \frac{Y}{Z}$$ and $$\frac{Y}{Z}\otimes \frac{X}{Z}$$ are both equal to $$(\frac{XY}{Z^2})(1\otimes 1)$$ since smooth functions can be treated like scalars. This reasoning can be extended to show that the space of holomorphic sections of $$\mathcal{O}_{\mathbb{P}^n}(k)$$ has basis $$X_{i_1}\otimes ... \otimes X_{i_n}$$ for $$0\leq i_1\leq i_2 \leq ... \leq i_k\leq n$$. In other words, holomorphic sections of $$\mathcal{O}(k)$$ are precisely degree $$k$$ homogeneous polynomials.

# A lens to view the base space

Going back to the initial motivation, we study covering spaces because the isomorphism classes of covering spaces on a space are a measure of its complexity (roughly the same measure as the fundamental group). We can play similar games with vector bundles.

Observe that $$\mathbb{R}\setminus \{0\}$$ is homotopy equivalent to the discrete two-point space $$S^0$$, so if $$E\rightarrow B$$ is a real line bundle with zero-section $$Z$$ then $$E\setminus Z$$ has the homotopy type of a fiber bundle with fibers $$S^0$$ i.e. a double-cover of $$B$$. It is an easy check that the trivial line bundle gives the trivial double cover (i.e. $$B\times S^0$$). This gives us another way to see that the Mobius bundle is non-trivial: removing the zero-section leaves us with (up to homotopy) the non-trivial double-cover of $$S^1$$.

Just like covering spaces, the classes of vector bundles on a compact Hausdorff base space is a homotopy invariant. In particular, we have the following result

Theorem Every vector bundle over a contractible compact Hausdorff space is trivial.

Of course, the whole article was an elaborate proof of the fact that $$S^1$$ is not contractible.