Frame Bundle of a Vector Bundle – Related Question

differential-geometrydifferential-topologygroup-actionsprincipal-bundlesvector-bundles

I am currently reading up on principal fiber bundles from a set of lecture notes on the subject, and I am trying to make sense of frame bundle of a vector bundle.

Consider a vector bundle $p:E\to M$ of rank $k$ over a smooth manifold
$M$. For $U\subset M,$ define $E_U:=p^{-1}(U),$ so that $p:E_U\to U$
is a vector bundle over $U$, called the restriction of $E$ to $U.$ A
trivialization of $E$ over an open subset $U\subset M$ is defined to
be an isomorphism $\tau:E_U\to U\times R^k$ of vector bundles. For
$x\in U,$ define the linear isomorphism $\tau_x:E_x\to R^k$ bu
$\tau=(x,\tau_x)$ on $E_x.$ Let $\mathbb{R}^k_M:=M\times
\mathbb{R}^k,$ the trivial bundle over $M$. Then the vector bundle
$\text{Hom}(\mathbb{R}^k_M,E)$ has fiber $\text{Hom}(\mathbb{R}^k,
E_x)$ at the point $x\in M.$ In addition, a trivialization $(U,\tau)$
of $E$ induces a trivialization $\tau'$ of
$\text{Hom}(\mathbb{R}^k,E)$ given by $\tau'(T_x)=\tau_x\circ T_x,$
for $T_x\in \text{Hom}(\mathbb{R}^k,E_x)$

Let $F(E):=\bigcup_{x\in M}F(E_x)$, where $F(E_x)$ is the set of frames in $E_x$ of $\text{Hom}(\mathbb{R}^k_M,E).$ This subset of $\text{Hom}(\mathbb{R}^k_M,E)$ is open and the natural map $F(E)\to M$ defines a sub-fiber bundle of $\text{Hom}(\mathbb{R}^k_M,E).$ A trivialization of $E$ over $U$ induces a trivialization $\tau^{''}$ of $F(E)$ over $U$ given by $\tau_x^{''}(f)=\tau_x \circ f,$ for $f\in F(E_x).$ The group $H=GL(\mathbb{R^k})$ acts from the right on each fiber $F(E_x).$ By looking at the trivializations, we see that these actions together constitute a smooth action of $H$ on $F(E),$ which turns $F(E)$ into a principal fiber bundle with structure group $H.$

Relevant Definition(s): A principal bundle over a smooth manifold $X$ with structure group $H$ is a pair $(P,\pi)$ consisting of a smooth right H-manifold $P$ and a smooth map $\pi:P\to X$ with the following property: For every point $x\in X,$ there exists an open neighborhood $V$ of $x$ in $X$ and a diffeomorphism $\tau:\pi^{-1}(V)\to V\times H$ such that

  1. $\pi=pr_V\circ \tau$ on $\tau^{-1}(V),$ where $pr_V$ denotes the projection $V\times H\to V.$
  2. $\tau$ intertwines the right $H$-actions.

My questions: I was able to make sense of most of it except at the following places.

  1. How is $F(E)$ open?
  2. How does it follow that right action of $H$ on $F(E)$ is smooth? I can see that it is free and transitive, but don't quite see the smoothness of this action.
  3. How does it turn $F(E)$ into a principal fiber bundle? In particular, how does it satisfy the first condition in the definition of PFB?

I am fairly new to this concept and I'm currently stuck on these three questions, so any help in the right direction will be very much appreciated. Thanks in advance.

Best Answer

I would probably reformulate the text you copied by saying first that a trivialization of $E$ over $U$ induces a trivialization of $\text{Hom}(\mathbb R^k_M,E)$ over $U$. This has values in $U\times\text{Hom}(\mathbb R^k,\mathbb R^k)$ and by definition $F(E)$ is mapped to $U\times GL(k,\mathbb R)$ by this trivialization. As stated in the comment by @Echo, this implies that $F(E)$ is open in $\text{Hom}(\mathbb R^k_M,E)$. Thus is is a submanifold and you can use the restriction to define a local trivialization of $F(E)$. Under the chart maps the right action of $H$ corresponds to $(x,A,B)\mapsto (x,AB)$ which implies smoothness of the action. The first statement in the definition of a principal bundle just corresponds to the fact that each fiber $F(E_x)$ by construction is preserved by the right action of $H$.