transition functions of a vector bundle

Question

why are the transition functions in the definition of a vector bundle $P:E \rightarrow M$ termed as transition functions? they are $g_{\alpha \beta}:U_{\alpha} \cap U_\beta \rightarrow GL(n,\mathbb{R})$ and defined as $\phi_{\beta} \circ \phi_{\alpha}^{-1} : (U_{\alpha} \cap U_{\beta}) \times \mathbb{R}^n \rightarrow (U_{\alpha} \cap U_{\beta}) \times \mathbb{R}^n$ where $(U_{\alpha}, \phi_{\alpha})$ are charts of $M$.I am not getting how transition functions perform change from one chart of $M$ to another?Please explain with an example if possible

Answer 1

Over $U_\alpha$ and $U_\beta$, you have two different trivializations. When you change coordinates from one trivialization to the other, you have to apply a change-of-basis map to each fiber. The change of coordinate map on the fiber over $x$ is the linear map $g_{\alpha\beta}(x)$.

If this is too general, think of a change-of-coordinates for an ordinary manifold. Its differential is a family of linear change-of-basis maps, one on each tangent space. The $g_{\alpha\beta}$ are defined in the same spirit.

Answer 2

I just learned about vector bundles earlier this year, so perhaps my answer is a bit hand-wavey, but the intuition I get out of it is that the transition maps provide instructions on how to glue the patches $U_\alpha\times\Bbb{R}^n$ and $U_\beta\times\Bbb{R}^n$ together, and these instructions naturally belong to $GL_n(\Bbb{R})$. The image of $x\in U_\alpha\cap U_\beta$ under $g_{\alpha\beta}$ tells you how to identify $\phi_\alpha^{-1}(\pi^{-1}(x))\cong \Bbb{R}^n$ with $\phi_\beta^{-1}(\pi^{-1}(x))\cong \Bbb{R}^n$.

Take the Möbius strip for example (as an $\Bbb{R}$-bundle over $\Bbb{S}^1$). Cover it with three open sets $U_1$, $U_2$, $U_3$ in (sorry for saying this) the most obvious way. On $U_1\cap U_2$ and $U_2\cap U_3$, make your transition function the constant map $x\mapsto$id$_{\Bbb{R}}$, and on $U_3\cap U_1$, the constant map $x\mapsto-$id$_{\Bbb{R}}$. This accounts for the half-twist in the Möbius strip.

Answer 3

Let me answer this question pointwise, i.e. fix $p\in U_\alpha\cap U_\beta\subset \mathcal{M}$. Fix also the easy example of the tangent (vector) bundle, i.e. $P:T\mathcal{M}\rightarrow\mathcal{M}$. Recall that a vector $V\in T_p \mathcal{M}$ at $p$ can be expressed in two different frames like $$V=V^\mu\frac{\partial}{\partial x^\mu}|_p=\bar{V}^\mu\frac{\partial}{\partial y^\mu}|_p\quad (\mu=1,..,n\quad\text{summed}).$$ Note the crucial point that $V$ and $(V^\mu)_{\mu=1,..n}$ and $(\bar{V}^\mu)_{\mu=1,..n}$ all contain the same information about the vector $V$. I.e. we can define bijections $$\phi_{\alpha,p}^{-1}: T_p\mathcal{M}\rightarrow\mathbb{R}^n, V\mapsto (V^\mu)_{\mu=1,..n}\\ \phi_{\beta,p}^{-1}: T_p\mathcal{M}\rightarrow\mathbb{R}^n, V\mapsto (\bar{V}^\mu)_{\mu=1,..n}$$ which (modulo checking diff'ability) are two local trivializations at the fixed point $p$. In the $T\mathcal{M}$ context, they bare the better known name coordinate functions. Coloquially speaking, but rather importantly, they can only tell you something about your geometric structure if you remember which frame they make sense in, namely $\phi_{\alpha,p}^{-1}$ in $x^\mu$ and $\phi_{\beta,p}^{-1}$ in $y^\mu$. (I'm being sloppy about notation of the frames here.) You know how to go from here, I'm convinced: Commutatively completing the triangle $\require{AMScd}$ \begin{CD} T_p\mathcal{M} @>\phi_{\alpha,p}^{-1}>> \mathbb{R}^n\ni(V^1,..V^n)\\ @V \phi_{\beta,p}^{-1} VV\\ \mathbb{R}^n@.\ni(\bar{V}^1,..\bar{V}^n) \end{CD} gives you two (hence equal) maps:

1. the map that translates $V$'s coordinates between frames $x^\mu$ and $y^\mu$
2. $g_{\alpha\beta,p}:\mathbb{R}^n\rightarrow\mathbb{R}^n$. (by your definition - modulo some $\cdot^{-1}$)

Answer 4

I am still learning about vector bundles, so if there is something incomplete, please say! I will give you the example of the most important vector bundle, the tangent bundle!

Let $M$ be a manifold of dimension $n$ and consider an atlas $\{(U_\alpha,\varphi_\alpha)\}$ where the coordinate maps read $\varphi_\alpha : U_\alpha \rightarrow \mathbb{R}^n$. We know that $TM$ is itself a vector bundle, namely of rank $n$ where we can consider the traivializing open cover $\{U_\alpha\}$ together with the trivializations $\{\phi_\alpha\}$ given by $$\phi_\alpha : \pi^{-1}(U_\alpha) \rightarrow U_\alpha \times \mathbb{R}^n,$$where $\pi : TM \rightarrow M$ is the usual projection and $\phi_\alpha(p,v) = (p,d_{\varphi_\alpha}(v))$. Now, it is important to observe that $$d_p\varphi_{\alpha} : T_pM \rightarrow T_{\varphi_\alpha(p)}\mathbb{R}^n \cong \mathbb{R}^n,$$and (by linearity after fixing $p$), is defined as $$d_p\varphi_\alpha(v) = (v_1,\dots,v_n),$$if $v = \sum_{i=1}^nv_1\partial_{x^i}$.

Now, the transition functions, also known as cocycles are given by $g_{\alpha \beta} : U_\alpha \cap U_\beta \rightarrow GL_n(\mathbb{R})$ where $$g_{\alpha \beta}(p) = \phi_\alpha^p \circ (\phi_\beta^p)^{-1} : \mathbb{R}^n \rightarrow \mathbb{R}^n.$$Going back to the trivializing functions of the tangent bundle we notice that fixing $p$ we get a linear isomorphism (since it is a vector bundle) $$\phi_\alpha^p : T_pM \rightarrow \mathbb{R}^n,$$which is given by $\phi_\alpha^p(v) = d_p\varphi_\alpha(v)$ defined previously. But then, for any $p \in U_\alpha \cap U_\beta$ we have $$g_{\alpha \beta}(p) = \phi_\alpha^p \circ (\phi_\beta^p)^{-1} = d_{\varphi_\beta^{-1}}\varphi_\alpha \circ d_p{\varphi_\beta^{-1}} = d_p(\varphi_\alpha \circ \varphi_\beta^{-1}),$$which is indeed a map from $\mathbb{R}^n$ to itself. This shows that the transition functions for $TM$ are given by the Jacobian of the transition functions of our manifold.