Equivalence of two definitions

Question

I have given two definitions of vector fields one is smooth section of tangent bundle and another is we will define vector field of coordinate chart $U_{\alpha}$ by $X_{\alpha}(p)=\sum_{i=1}^n X_{\alpha,i}(p)\frac{\partial}{\partial u_i}$ from $U_{\alpha}$ to $\cup_{p\in U_{\alpha}}{p}\times T_pU_{\alpha}$ such that $X_{\alpha}(p)\in {{p}}\times T_pU_{\alpha}$ and $X_{\alpha,i}:U_{\alpha}\to \mathbb{R}$ are smooth functions. Then vector field of $M$ is collection of $X_{\alpha}$ on $U_{\alpha}$ such that on the intersection $U_{\alpha}\cap U_{\beta}$ the vector fields agree ${X_{\alpha}}_{U_\alpha\cap U_\beta} ={X_{\beta}}_{U_\alpha\cap U_\beta}$.

Now from the definition of section of tangent bundle how to cook up this $U_{\alpha}$'s in the 2 nd definition? And from the 2 nd definition how to get the sections of tangent bundle. Any hints would be very helpful.

Answer 1

The "modern" definition of a vector field on $M$ is that of a smooth section $X : M \to TM$ of the tangent bundle $TM$. For each open $U \subset M$ we get the restriction $X \mid_U : U \to TM\mid_U$ which is also a bundle section. Clearly $(X\mid_U)\mid_{U \cap V} = (X\mid_V)\mid_{U \cap V}$, i.e. the restrictions are compatible with intersection.

Thus for each open cover $\{U_\alpha\}$ of $M$ we get a family $X_\alpha = X \mid_{U_\alpha}: U_\alpha \to TM \mid_{U_\alpha}$ of restrictions which is compatible with intersection. Conversely, given any family $Y_\alpha : U_\alpha \to TM \mid_{U_\alpha}$ of bundle sections which is compatible with intersection ("compatible family"), we clearly get a global bundle section $Y : M \to TM$ such that $Y_\alpha = Y \mid_{U_\alpha}$.

Since $TM\mid_U$ and $TU$ are canonically isomorphic, one frequently writes $TM\mid_U = TU$ which gives a reasonable interpretation to the statement that bundle sections $A : U \to TU$ and $B : V \to TV$ (i.e. vector fields on $U, V$) are compatible on $U \cap V$.

Now let us consider all coordinate charts $(U_\alpha,\phi_\alpha)$ on $M$. By our above considerations a vector field $X$ on $M$ can the be regarded as a compatible family of vector fields $X_\alpha : U_\alpha \to TU_\alpha$. So it remains to be understood how vector fields $X_\alpha : U_\alpha \to TU_\alpha$ can be described. It is well-known that any chart $(U_\alpha,\phi_\alpha)$ gives us a trivialization $\phi_\alpha^* : TU_\alpha \to U_\alpha \times \mathbb R^n$ of the tangent bundle $TU_\alpha$. Under this trivialization the canonical smooth sections $s_i : U_\alpha \to U_\alpha \times \mathbb R^n, s_i(x) = (x,e_i)$, correspond to vector fields on $U_\alpha$ denoted by $\frac{\partial}{\partial x_i}$. Note that the $\frac{\partial}{\partial x_i}$ depend on $\phi_\alpha$ although notation does not really make this transparent. Moreover, each smooth section $s: U_\alpha \to U_\alpha \times \mathbb R^n$ can be uniquely written as $s = \sum_{i=1}^n X_i s_i$ with smooth $X_i : U_\alpha \to \mathbb R$ ($X_i = \pi_i \circ X$ with $\pi_i : U_\alpha \times \mathbb R^n \to \mathbb R, \pi_i(x,t_1,\ldots,t_n) = t_i$). Therefore each vector field $X_\alpha : U_\alpha \to TU_\alpha$ has the form $X_\alpha = \sum_{i=1}^n X_{\alpha,i} \frac{\partial}{\partial x_i}$.

Answer 2

This answer relates to the question where the $U_i$'s in the second definition come from.

Tangent bundle (as any vector bundle -- by definition) is locally trivial. So take any $\{ U_i \}_{i \in I}$ to be a cover such that $TM|_{U_i}$ is trivial for every $i$. Now choose any covering by coordinate charts $\{V_j\}_{j \in J}$. The covering you are looking for can be taken to be $\{U_i \cap V_j\}_{i \in I, j \in J}$. To see this, there are two easy observations:

• Open subset of a coordinate chart is a coordinate chart (coordinate chart here is understood as a set diffeomorphic with an open subset of $\mathbb{R}^n$)
• For an open subset $V$, if $V \subset U$ and $TM|_U$ is trivial, then $TM|_V$ is trivial.

The above reasoning works for any vector bundle. In fact, $TM$ is a little bit special with regard to trivialising covers: if $U \subset M$ is a coordinate chart, then $TM|_U$ is automatically trivial (hence no need to choose a trivialising cover separately for $TM$ -- we could already start with a covering by coordinate charts).