6
$\begingroup$

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!

$\endgroup$

2 Answers 2

6
$\begingroup$

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!

$\endgroup$
7
  • $\begingroup$ 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? $\endgroup$ May 22, 2016 at 18:17
  • 1
    $\begingroup$ 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? $\endgroup$ May 23, 2016 at 9:07
  • $\begingroup$ 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? $\endgroup$ May 23, 2016 at 18:12
  • $\begingroup$ Yes, you are right! ;) $\endgroup$ May 25, 2016 at 9:36
  • $\begingroup$ Great! That is a perfect ending to this question! Many thanks for your so detailed derivation Armando! $\endgroup$ May 26, 2016 at 23:52
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.