Endomorphism vector bundle isomorphic to the adjoint bundle of its frame bundle?

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Could somebody help me to prove the following isomorphism (in particular what is the isomorphism)? \begin{equation} End(\xi) \cong ad(E_{\xi}) = E_{\xi} \times_{GL(n,\mathbb{R})} \text{Mat}_n(\mathbb{R}) \end{equation} where $E_{\xi}$ is the frame bundle associated with $\xi$ and $GL(n,\mathbb{R})$ acts on its Lie algebra $\text{Mat}_n(\mathbb{R})$ by conjugation

Thank you very much for your attention!

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I change the notations!

If I am not wrong: $E$ is a principal $GL(n,\mathbb{R})$-bundle over a topological space $X$, $Ad(E)$ is the adjoint vector bundle (over $X$) associated to $E$ and $F(E)\equiv F$ is the frame (vector) bundle (over $X$) associated to $E$.

By definition: there exists an open covering $\{U_{\alpha}\}_{\alpha\in A}$ (for $Ad(E)$) of $X$ such that:

  • $\pi_1^{-1}(U_{\alpha})\stackrel{\varphi_{\alpha}}{\cong}U_{\alpha}\times\mathfrak{gl}(n,\mathbb{R})=U_{\alpha}\times\mathbb{R}^n_n$, $\varphi_{\alpha}$ is a homeomorphism;

  • $pr_1\circ\varphi_{\alpha}=\pi_1$ ;

  • getting $U_{\alpha\beta}=U_{\alpha}\cap U_{\beta}\neq\emptyset$, the maps: \begin{equation} \varphi_{\beta\displaystyle|\pi_1^{-1}(U_{\alpha\beta})}\circ\varphi^{-1}_{\alpha\displaystyle|\pi_1^{-1}(U_{\alpha\beta})}:(P,M)\in U_{\alpha\beta}\times\mathbb{R}^n_n\to(P,Ad(g_{\alpha\beta}(P)^{-1})(M))\in U_{\alpha\beta}\times\mathbb{R}^n_n \end{equation} are homeomorphism and the functions $g_{\alpha\beta}:U_{\alpha\beta}\to GL(n,\mathbb{R})$ are the transition functions of $E$.

In the same way: there exists an open covering $\{V_{\alpha}\}_{\alpha\in A}$ (for $End(F)$) of $X$ such that:

  • $\pi_2^{-1}(V_{\alpha})\stackrel{\psi_{\alpha}}{\cong}V_{\alpha}\times End(\mathbb{R}^n)=V_{\alpha}\times\mathbb{R}^n_n$, $\psi_{\alpha}$ is a homeomorphism;

  • $pr_1\circ\psi_{\alpha}=\pi_2$ ;

  • getting $V_{\alpha\beta}=V_{\alpha}\cap V_{\beta}\neq\emptyset$, the maps: \begin{equation} \psi_{\beta\displaystyle|\pi_2^{-1}(V_{\alpha\beta})}\circ\psi^{-1}_{\alpha\displaystyle|\pi_2^{-1}(V_{\alpha\beta})}:(P,M)\in V_{\alpha\beta}\times\mathbb{R}^n_n\to\left(P,\left(^Tg_{\alpha\beta}^{-1}\otimes g_{\alpha\beta}\right)(P)(M)\right)\in V_{\alpha\beta}\times\mathbb{R}^n_n \end{equation} are homeomorphism and the functions $g_{\alpha\beta}:V_{\alpha\beta}\to GL(n,\mathbb{R})$ are the transition functions of $E$ (and of $F$).

Remark. For any pair of vector bundles $V$ and $W$ over $X$: $Hom(V,W)\cong V^{\vee}\otimes W$!, where $V^{\vee}$ is the dual (vector) bundle of $V$.

Whitout loss of generality, we can assume that $Ad(E)$ and $F$ have the same open covering of trivialization $\{U_i\}_{i\in I}$ over $X$!, by a simply computation: \begin{gather} \forall i,j\in I,P\in U_{ij}\neq\emptyset,M\in\mathbb{R}^n_n,\\ Ad(g_{ij}^{-1}(P))(M)=g_{ij}^{-1}(P)\times M\times g_{ij}(P)=\left(^Tg_{ij}^{-1}\otimes g_{ij}\right)(P)(M); \end{gather} in other words, because $Ad(E)$ and $End(F)$ have the same open covering of trivialization and the same transition functions, they are canonically isomorphic!