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

Could somebody help me to prove the following isomorphism (in particular what is the isomorphism)? $$End(\xi) \cong ad(E_{\xi}) = E_{\xi} \times_{GL(n,\mathbb{R})} \text{Mat}_n(\mathbb{R})$$ 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!

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: $$\varphi_{\beta\displaystyle|\pi_1^{-1}(U_{\alpha\beta})}\circ\varphi^{-1}_{\alpha\displaystyle|\pi_1^{-1}(U_{\alpha\beta})}:(P,M)\in U_{\alpha\beta}\times\mathbb{R}^n_n\to(P,Ad(g_{\alpha\beta}(P)^{-1})(M))\in U_{\alpha\beta}\times\mathbb{R}^n_n$$ 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: $$\psi_{\beta\displaystyle|\pi_2^{-1}(V_{\alpha\beta})}\circ\psi^{-1}_{\alpha\displaystyle|\pi_2^{-1}(V_{\alpha\beta})}:(P,M)\in V_{\alpha\beta}\times\mathbb{R}^n_n\to\left(P,\left(^Tg_{\alpha\beta}^{-1}\otimes g_{\alpha\beta}\right)(P)(M)\right)\in V_{\alpha\beta}\times\mathbb{R}^n_n$$ 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!

• Very clear and readable proof Armando! Many thanks for taking the pain to write down every single step! One last thing to clarify, for the endomorphism bundle $End(F)$ in defining the codomain of the transition function $g_{\alpha\beta}: V_{\alpha\beta} \to GL(n,\mathbb{R})$ you used the fact that $Hom(\mathbb{R}_n^n,\mathbb{R}_n^n) \cong GL(n,\mathbb{R})$ but when you showed the transition function is the same as that of the adjoint vector bundle you switched back to $Hom(\mathbb{R}_n^n,\mathbb{R}_n^n)$ and applied $Hom(V,W) \cong V^{\vee} \otimes W$ instead. Am I right? May 22, 2016 at 18:17
• No: $g_{\alpha\beta}$ are the transition functions of $E$; and I use the fact that $GL(n,\mathbb{R})$ is isomorphic to $Aut(\mathbb{R}^n)$, the group of invertible linear functions of $\mathbb{R}^n$ in itself! Do you remember the construction of the transition functions of a principal bundle? May 23, 2016 at 9:07
• Yes. Sorry for the mistake Armando! But my second statement that you used $Hom(V,W) \cong V^{\vee} \otimes W$ to express the transition function of $End(F)$ in terms of the transition function $g_{\alpha\beta}$ of $E$ is correct i.e. $(g_{\alpha\beta}^T)^{-1} \otimes g_{\alpha\beta}$, am I right? May 23, 2016 at 18:12
• Yes, you are right! ;) May 25, 2016 at 9:36
• Great! That is a perfect ending to this question! Many thanks for your so detailed derivation Armando! May 26, 2016 at 23:52

For a vector bundle $$E$$, here is a "natural" bijection between its endomorphism bundle $$\mathrm{End}(E)$$ and its adjoint bundle $$F(E) \stackrel{\mathrm{GL}_n}{\times} \mathrm{M}_n$$. Each fiber of $$\mathrm{End}(E)$$ above a point $$p$$ consists of linear mappings $$L:E_p\to E_p$$, while the fiber of the adjoint bundle consists of pairs $$(\beta, M)$$, where $$\beta$$ is a basis of $$E_p$$ and $$M$$ is a matrix.

The bijection sends $$L$$ to the pair $$(\beta, [L]_\beta)$$, where $$\beta$$ is any basis and $$[L]_\beta$$ is the matrix of $$L$$ with respect to $$\beta$$. This is well-defined, because change of basis for an endomorphism is given by the (dual) adjoint representation: that is, if we chose any other basis $$A{\cdot} \beta$$ for $$A\in \mathrm{GL}_n$$, we would send $$L$$ to: $$(A{\cdot}\beta, [L]_{A\cdot\beta}) \ =\ (A{\cdot}\beta, A^{-1}[L]_\beta A) \ \sim\ (\beta, AA^{-1}[L]_\beta AA^{-1}) \ =\ (\beta, [L]_\beta),$$ namely the same element as before. The inverse of the bijection evidently sends $$(\beta, M)$$ to the endomorphism $$L$$ defined by the matrix $$M$$ with respect to basis $$\beta$$, and again this is well-defined with respect to the $$\mathrm{GL}_n$$ quotient because of the change-of-basis formula for matrices.

This is a general principle (and the frame bundle is a general principal): any vector space construction with a bundle corresponds to a vector bundle associated to the frame bundle, by realizing the construction with respect to an arbitrary basis.