When I learned the theory of Vector Bundles at the first time, frankly speaking, I was very afraid about this notion, because I cannot write those ideas down on the drafts, hence not even to make some computations, I still remember how I was afraid about Ralph Cohen’s lecture The Topology of Fiber Bundles, recently, after a series of painful but strictly training and discussing with my classmates, I begun to realize the power of using transition functions in the study of the vector bundles, at least, it can help me do some computation instead of fantasy.

This blog is a summarize of my thoughts on the vector bundles these days.

1. The Transition Functions are ”Cocycles”

We suppose $\pi:E\longrightarrow X$ is a rank $n$ smooth complex (or real, whatever) vector bundle over a smooth manifold $X$ , if we choose an open cover $\mathscr{U}=\{U_i\}_{i\in\Lambda}$ of $X$ , we can find a family of local trivialization of $E$ on each covering member:

$\phi_{i}:\pi^{-1}(U_i):=E_{U_i}\longrightarrow U_i\times \mathbb{C}^n$

each $\phi_i$ must agree with the bundle projection, i.e. $\mathrm{pr}_1\circ\phi_i=\pi$ , and the transition functions:

$\phi_{ij}(z)=\phi_i\circ\phi_i^{-1}(z,\cdot):U_i\cap U_j\longrightarrow \mathrm{GL}_n(\mathbb{C})$

must be $C^\infty$ matrix-valued functions.

People prefer to call these transition functions $\phi_{ij}$ the cocycles, because they satisfy the condition

$\begin{aligned}(1).&\,\phi_{ij}=\phi_{ji}^{-1}\\(2).&\,\phi_{ij}\cdot\phi_{jk}=\phi_{ik}\end{aligned}$

and these two conditions are called cycle conditions, we can see that, two families of transition functions, namely $\phi_{ij}, \psi_{ij}$ (subordinated to the same open cover), defined the same vector bundle, if and only there exists a family of smooth matrix-valued functions $g_i\in\mathrm{GL}_n\left(C^\infty(U_i)\otimes\mathbb{C}\right)$ such that

$\phi_{ij}=g_i\psi_{ij}g_j^{-1}$

Intuitively, these $\phi_{ij}$ form the obstruction of a vector bundle being trivial from local to global, speaking of the obstruction from local to global, there is a very powerful algebraic tool which can help us to investigate, that is the sheaf cohomology, in fact, these cocycles are real cocycles in a cohomology group of a non-Abelian sheaf.

We denote $\mathcal{GL}_n^\infty$ the sheaf of $\mathrm{GL}_n(\mathbb{C})-$ valued smooth functions on a manifold $X$ , we can define the cochain complex as a usual sense as Abelian sheaf:

$\begin{aligned}1\longrightarrow\prod_{i\in\Lambda}\mathcal{GL}_n^\infty(U_i)\stackrel{d^1}{\longrightarrow}\prod_{(i,j)\in\Lambda^2}\mathcal{GL}_n^\infty(U_i\cap U_j)\longrightarrow\cdots\end{aligned}$

But as we are in a non-Abelian case, the subgroup $\mathrm{Im}\,d$ not always a normal subgroup of the kernel, hence we cannot do the quotient, so, in order to define the sheaf cohomology group, we need to redefine an equivalent relation to be quoted. In general, this equivalent relation is very complicated, hence people will only focus on the low-order case, that is the 1st cohomology group $H^1$ only.

We define $(\phi_{ij})\sim(\psi_{ij})$ if and only if there exists $(g_{i})\in C^0$ (the 0-th cochain group) such that $\phi_{ij}=p_i\psi_{ij}p_j^{-1}$ , it is indeed a equivalent relation, and we define the 1st cohomology group as

$H^1\left(X,\mathcal{GL}_n^\infty\right):=\ker d\big/\sim$

Hence by our construction, we can see that the obstructions mentioned above are exactly the cocycles in $H^1\left(X,\mathcal{GL}_n^\infty\right)$ , hence the 1st cohomology group classifies all smooth vector bundles of rank $n$ over $X$ .

As a corollary, if we take $X$ a complex manifold, and we consider the holomorphic line bundle, then our sheaf becomes to $\mathcal{GL}_1^\mathcal{O}$ , which is just $\mathcal{O}_X^*$ , the sheaf of no-where vanishing holomorphic functions on $X$ , and it is an Abelian sheaf. Recall that the isomorphism classes of holomorphic line bundles over $X$ is a group, called the Picard group, hence we have an isomorphism:

$\begin{aligned}\mathrm{Pic}(X)\cong H^1\left(X,\mathcal{O}_X^*\right)\end{aligned}$

Remark: A great different between non-Abelian and the Abelian sheaf cohomology is that, in the non-Abelian case, $H^k$ may loss their functoriality for $k\geqslant 2$ , that is to say, for a short exact sequence

$1\longrightarrow \mathcal{F}\longrightarrow\mathcal{G}\longrightarrow\mathcal{H}\longrightarrow 1$

there is no induced long exact sequence of cohomology groups, but thankfully, $H^1$ will still retain its functoriality, i.e. we have the exact sequence:

$1\longrightarrow H^0(X,\mathcal{F})\longrightarrow H^0(X,\mathcal{G})\longrightarrow H^0(X,\mathcal{H})\longrightarrow H^1(X,\mathcal{F})\longrightarrow H^1(X,\mathcal{G})\longrightarrow H^1(X,\mathcal{H})$

Moreover, the 1st cohomology group of a non-Abelian sheaf can be classified by $\mathcal{G}-$ torsors. There is very good article on Zhihu which shall tell more details.

2. Holomorphic Vector Bundles But Smooth Trivial

This part mainly comes from an answer of a question on Stack Exchange (I was shocked by the flexibility in using tools from complex geometry of Dr. Michael Albanese):

For a compact complex manifold $X$ , does there exist a line bundle $L$ which is holomorphically non-trivial, but trivial as a smooth line bundle?

First, we recall that the topological complex line bundles are classified by the 1st Chern class:

$c_1: \mathrm{Vect}^1_{\mathbb{C}}(X)\longrightarrow H^2(X;\mathbb{Z})$

where $\mathrm{Vect}^1_{\mathbb{C}}(X)$ is the set of isomorphism classes of complex line bundles over $X$ , we need to find the relationship between the Picard group $\mathrm{Pic}(X)$ and the $H^2(X;\mathbb{C})$ . (Recall that from part 1 of this blog, the Picard group is isomorphic to $H^1(X,\mathcal{O}_X^*)$ )

We have the following short exact sequence:

$0\longrightarrow \mathbb{Z}\stackrel{2\pi i}{\longrightarrow}\mathcal{O}_X\stackrel{\exp}{\longrightarrow}\mathcal{O}_X^*\longrightarrow 0$

hence we can gain a long exact sequence:

$\begin{aligned}\cdots\longrightarrow H^1(X,\mathcal{O}_X)\longrightarrow H^1\left(X,\mathcal{O}_X^*\right)\stackrel{\delta}{\longrightarrow }H^2(X,\mathbb{Z})\longrightarrow H^2(X,\mathcal{O}_X)\cdots\end{aligned}$

where the connected morphism $\delta$ is the composition:

$\delta: H^1\left(X,\mathcal{O}_X^*\right)\stackrel{\cong}{\longrightarrow}\mathrm{Pic}(X)\stackrel{c_1}{\longrightarrow}H^2(X,\mathbb{Z})$

Recall that, by Dolbeault’s theorem:

$H^k(X,\mathcal{O}_X)\cong H^{(0,k)}_{\bar{\partial}}(X;\mathbb{C})$

we will use $h^{(0,k)}$ the complex dimension of the group $H^k(X,\mathcal{O}_X)$ . Two interesting cases arise here:

If $h^{(0,1)}=0$ , then $\delta$ is injective, which implies $c_1$ is injective, hence a holomorphic line is topological trivial only if it is holomorphic trivial, for a simple case, consider $X=\mathbb{P}^1$ .
If $h^{(0,2)}=0$ , then $\delta$ is surjective, which implies $c_1$ is surjective, which implies every topological line admits at least 1 holomorphic structure to become a holomorphic line bundle.

In the case $h^{(0,1)}>0$ , the $\ker\delta$ may not trivial, for example, for compact Riemann surface $\mathbb{T}=\mathbb{C}/\Lambda$ with genus $g=1$ , consider the divisor $D=x_1-x_2$ on $\mathbb{T}$ , the line bundle determined by $D$ is topologically trivial, since

$c_1(L_D)=\deg D=0$

but it is not holomorphically trivial, since it has no global holomorphic sections. Indeed, the divisor of a global section of $L_D$ must equal to $D$ , but the divisors of holomorphic sections are always effective, hence it only has trivial global sections.

However, we can show that for $h^{(1,0)}=m>0$ , the $c_1$ does have non-trivial kernel, i.e. there exist some nontrivial holomorphic line bundles but topologically trivial, we denote all such line bundles by $\mathrm{Pic}^0(X):=\ker c_1$ .

Next, we investigate the long exact sequence more carefully, note that $H^0(X,\mathcal{O}_X)\cong\mathbb{C}$ , and $H^0(X,\mathcal{O}_X^*)\cong \mathbb{C}^*$ , hence we have the long exact sequence as follows:

$\cdots\longrightarrow\mathbb{C}\stackrel{\exp}{\longrightarrow}\mathbb{C}^*\stackrel{0}{\longrightarrow} H^1(X,\mathbb{Z})\hookrightarrow H^1(X,\mathcal{O}_X)\stackrel{\varphi}{\longrightarrow} H^1(X,\mathcal{O}_X^*)\stackrel{c_1}{\longrightarrow}H^2(X,\mathbb{Z})\longrightarrow\cdots$

Hence by exactness, we have

$\begin{aligned}\mathrm{Pic}^0(X)&=\ker c_1=\mathrm{Im}\,\varphi\\&\cong H^1(X,\mathcal{O}_X)\big/H^1(X,\mathbb{Z})\\&\cong \mathbb{C}^m/\mathbb{Z}^m\cong\mathbb{T}^m\end{aligned}$

To sum up:

Theorem (Michael Albanese): A topologically trivial line bundle on a compact complex manifold $X$ is holomorphically trivial if and only if $H^{(0,1)}_{\bar{\partial}}(X;\mathbb{C})=0$ . If the complex dimension of the $(0,1)-$ Dolbeault cohomology is $m>0$ , then the collection of isomorphism classes of holomorphic line bundle with zero 1st Chern class, denoted by $\mathrm{Pic}^0(X)$ is isomorphic to an $m-$ dimensional toric group $\mathbb{T}^m$ .