I am trying to understand the map between vector bundles and principal fibre bundles and would like to work out the following example explicitly. Moreover in this example I am only currently looking at one direction, I would like to understand this well to attempt the other direction after this.
Given a vector bundle $ \pi:E\rightarrow M$ of rank $n$, we can consider the frame bundle $F(E)$. Since $F(E)$ is a principle fibre bundle we can consider the vector bundle associated to $F(E)$ which is $F(E)\times \mathbb{R}^n/GL(n,\mathbb{R})$. Show that $\mathcal{P}=F(E)\times \mathbb{R}^n/GL(n,\mathbb{R})\cong E$.
Intuitively this makes sense but to fully understand this I should be able to produce a rigour argument.
As I understand we can think elements of $F(E)$ fibrewise as maps $p: \mathbb{R}^n\rightarrow E_x$. For $\mathcal{P}$ an element is an equivalence class $[p,v]$ where $p$ is a map as above, $v\in E_x$ and $(p,v)\sim(pg,\rho(g^{-1})v)$ (I'm not entirely sure what $\rho(g^{-1})v$ is, I know $\rho$ is the "standard representation of $GL(n,\mathbb{R})$ in $\mathbb{R}^n$ but I don't know what this representation is).
Taking inspiration from the wikipedia page, I claim that the isomorphism is given by $[p,v]\mapsto p(v)$. First we show this is well defined. By definition of this map $[pg,\rho(g^{-1})v]\mapsto pg(\rho(g^{-1})v)$? If I understood the representation It would probably make sense why the $g$ and $\rho(g^{-1})$ cancel.
It seemed like surjectivety should be clear, as for any $v\in E$ we can just take $[id,v]$ but this doesn't make sense. Also should the elements in $E$ be pairs $(v,p)$ with $p$ a point in the base manifold?
Let $x\in M$ and $E_x$ its fibre. A frame at $x$ is a linear isomorphism $u:\mathbb{R}^n\rightarrow E_x$. There exists a morphism $h:F(E)\times \mathbb{R}^n\rightarrow E$ defined by $h(u,v)=u(v)$. This morphism is invariant by $Gl(n,\mathbb{R})$ since $h(g.(u,v))=h(ug^{-1},g(v))=(ug^{-1})(gv)=u(v)$. (you have to take $\rho=Id$) Remark that the action of $Gl(n,\mathbb{R})$ on $F(E)$ is defined by $u.g=ug^{-1}$ (It acts on the right) and induces an isomorphism $F(E)\times \mathbb{R}^n/Gl(n,\mathbb{R}^n)\rightarrow E$.