[Math] Global Trivializing Sections of Tangent bundle

complex-geometrydifferential-geometrytangent-bundle

I quote from Complex Geometry- An Introduction by Daniel Huybrechts.

Let $U\subseteq \mathbb{C}^n$ be an open subset. Thus, $U$ can in particular be considered as a $2n$-dimensional real manifold. For $x\in U$ we have real tangent space $T_xU$ at the point $x$ which is of real dimension $2n$. A canonical basis of $T_xU$ is given by tangent vectors $$\frac{\partial}{\partial x_1},\frac{\partial}{\partial x_2},\ldots,\frac{\partial}{\partial x_n},\frac{\partial}{\partial y_1},\frac{\partial}{\partial y_2},\ldots,\frac{\partial}{\partial y_n}$$where $z_1=x_1+iy_1,\cdots,z_n=x_n+iy_n$ are the standard coordinates on $\mathbb{C}^n$. Moreover, the vectors $\frac{\partial}{\partial x_1},\frac{\partial}{\partial x_2},\ldots,\frac{\partial}{\partial y_n}$ are global trivializing sections of $TU$.

I understand almost everything but the last line. What does it mean to say tangent vectors are global trivializing sections of $TU$?

We have tangent bundle $TU$ with map $\pi:TU\rightarrow U$ where $\pi(p,\partial/\partial x_i)=p=\pi(p,\partial/ \partial y_i)$. By section of a tangent bundle we mean $\sigma: U\rightarrow TU$ such that $\pi\circ\sigma=Id_U$.

We have $\partial/\partial x_i: U\rightarrow TU$ is given by $p\mapsto (p,\partial/\partial x_i|_p)=(p,\partial/\partial x_i)$ and $\pi\circ \partial/\partial x_i= Id_U$.

These sections almost behaves likes identity maps. Is this the trivialization of sections they are referring to?

Any reference for definition of global trivializing sections is welcome.

Best Answer

Let $\pi \colon E \longrightarrow M$ be a (real) vector bundle of rank $n$. A trivialization of this bundle over an open subset $U \subset M$ consists of a diffeomorphism $$\phi \colon {\pi}^{-1}(U) \longrightarrow U\times \mathbb{R}^n,$$ satisfying:

  1. $\textrm{p}_1 \circ \phi (p) = \pi(p)$ for every $p \in {\pi}^{-1}(U)$ (here $\textrm{p}_1$ denotes projection onto the first factor on $U\times\mathbb{R}^n$);
  2. $\phi$ acts as a linear isomorphism on each fiber, i.e., ${\phi}|_{{\pi}^{-1}(x)} \colon E_x \longrightarrow \mathbb{R}^n$ is an isomorphism of vector spaces.

Such maps allow you to choose "nice" coordinates on the total space of the bundle, adapted to the vector space structure on the fibers. In general, this can only be done locally. An additional condition on the definition of vector bundles requires such local trivializations to interplay in a controlled way on overlaps (see section 2.2 on Huybrechts' book, for instance).

Once a trivialization $\phi$ over $U$ is chosen, one can obtain a basis of $\textit{local sections}$ over $U$ (i.e., a collection of sections over $U$ which at each point $x \in U$ is a basis of the fibers). This can be done by choosing ${\sigma}_k \colon U \longrightarrow {\pi}^{-1}(U)$ given by $$\sigma_k(x)={\phi}^{-1}(x,e_k),$$ where $\{e_k\}$ denotes a basis on $\mathbb{R}^n$.

Similarly, conditions 1. and 2. above imply if one has a basis $\{\sigma_k\}_{k=1}^n$ of local sections, then one can collect such sections to define a trivialization as follows: given $p \in {\pi}^{-1}(x)$, we define $$\phi(p)=(x,a^1,a^2,\cdots,a^k),$$ where $a^j$ are the coordinates of the vector $p \in E_x$ with respect to the basis $\{\sigma_k\}$. Such a collection of sections is then called a set of trivializing sections, according to Huybrechts.

The quote from your post says that over an open subset $U$ of $\mathbb{C}^n$, the tangent bundle $TU$ can be trivialized globally, i.e. over the whole set $U$, by the collection of sections induced by a system of coordinates on $U$.

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