Let $P\rightarrow M$ be a $G$-principal bundle and denote by $\mathrm{at}(P):=TP/G$ the Atiyah Lie algebroid over $M$.
I want to understand how $\mathrm{at}(P)$ is a locally trivial Lie algebroid with $\mathrm{at}(P)\vert_U\cong TU\times\mathfrak{g}$, but I fail to compute the correct transition functions.
Denote by:
- $\theta\colon P\times G\rightarrow P$ the $G$-action on $P$ with $\theta_g\colon P\rightarrow P\colon p\mapsto pg\ $, $\theta^p\colon G\rightarrow P\colon g\mapsto pg$.
- $\hat\xi(p):=T_e\theta^p(\xi)=\frac{d}{dt}p\exp(t\xi)\vert_{t=0}$ the fundamental vector fields.
- $\omega_G\in\Omega^1(G,\mathfrak{g})$ the Maurer-Cartan form of $G$.
Let
$\psi_\alpha\colon P\vert_{U_\alpha}\rightarrow U_\alpha\times G\colon p=s_\alpha(x)g\mapsto\big(\pi(p),\psi^2(p)\big):=(x,g),\ \alpha\in\mathcal{A}$
be local trivializations of $P$ with cocycles
$g_{\alpha\beta}\colon U_\alpha\cap U_\beta\rightarrow G\colon x\mapsto \psi^2_\alpha(s_\beta(x))$
Then the $\psi_\alpha$ induce local trivializations of the Atiyah Lie algebroid:
$\bar{\psi}_\alpha\colon\mathrm{at}(P)\vert_{U_\alpha}\rightarrow TU_\alpha\times\mathfrak{g}\colon[v]\mapsto \Big(T_p\pi(v)\ ,\ T_p(\psi_\alpha^2\circ\theta_{\psi^2_\alpha(p)^{-1}})(v)\Big)$
with inverse
$\bar{\psi}_\alpha^{-1}\colon TU_\alpha\times\mathfrak{g}\rightarrow\mathrm{at}(P)\vert_{U_\alpha}\colon(v,\xi)\mapsto[T_xs_\alpha(v)+\hat{\xi}(s_\alpha(x))]$
I want to compute the cocycles $\bar{g}_{\alpha\beta}\colon TU_\alpha\cap TU_\beta\rightarrow\mathrm{Diff}(\mathfrak{g})$ of the fiber bundle $\mathrm{at}(P)$:
I use the relations:
- $\hat\xi(pg^{-1})=T_p\theta_{g^{-1}}\big((\widehat{\mathrm{Ad}_{g^{-1}}\xi})(p)\big)$
- $T_xs_\beta(v)=T_{s_\alpha(x)}\theta_{g_{\alpha\beta}(x)}\circ T_x s_\alpha(v)+\widehat{g_{\alpha\beta}^*\omega_G(v)}(s_\beta(x))=T_{s_\alpha(x)}\theta_{g_{\alpha\beta}(x)}\circ\big(T_x s_\alpha(v)+(\widehat{\mathrm{Ad}_{g_{\alpha\beta}(x)}g_{\alpha\beta}^*\omega_G(v)})(s_\alpha(x))\big)$
- For $f\colon M\rightarrow G$ consider $f^{-1}\colon M\rightarrow G\colon p\mapsto f(p)^{-1}=\iota\circ f(p)$, then $({f^{-1}}^*\omega_G)\vert_p=-\mathrm{Ad}_{f(p)}(f^*\omega_G)\vert_p$
Then: \begin{align} \bar{\psi}_\alpha\circ\bar{\psi}_\beta^{-1}(v,\xi)&=\bar\psi_\alpha([T_xs_\beta(v)+\hat\xi(s_\beta(x))]) \\ &=\Big(T_{s_\beta(x)}\pi(T_xs_\beta(v)+\hat\xi(s_\beta(x)))\ ,\ T_{s_\beta(x)}\big(\psi^2_\alpha\circ\theta_{\psi^2_\alpha(s_\beta(x))^{-1}}\big)(T_xs_\beta(v)+\hat\xi(s_\beta(x)))\Big) \end{align}
$T_{s_\beta(x)}\pi(T_xs_\beta(v)+\hat\xi(s_\beta(x)))=T_x(\pi\circ s_\beta)(v)+0=T_x id_M v=v\ $ (as expected)
\begin{align} T_{s_\beta(x)}\big(\psi^2_\alpha\circ\theta_{\psi^2_\alpha(s_\beta(x))^{-1}}\big)\circ T_xs_\beta(v)&=T_{s_\beta(x)}\big(\psi^2_\alpha\circ\theta_{g_{\alpha\beta}(x)^{-1}}\big)\circ\Big(T_{s_\alpha(x)}\theta_{g_{\alpha\beta}(x)}\circ\big(T_x s_\alpha(v)+(\widehat{\mathrm{Ad}_{g_{\alpha\beta}(x)}g_{\alpha\beta}^*\omega_G(v)})(s_\alpha(x))\big)\Big) \\ &=T_{s_\alpha(x)}\psi^2_\alpha\big(T_xs_\alpha(v)+(\widehat{\mathrm{Ad}_{g_{\alpha\beta}(x)}g_{\alpha\beta}^*\omega_G(v)})(s_\alpha(x))\big) \\ &=T_x(\psi^2_\alpha\circ s_\alpha)(v)+T_{s_\alpha(x)}\psi^2_\alpha\big((\widehat{\mathrm{Ad}_{g_{\alpha\beta}(x)}g_{\alpha\beta}^*\omega_G(v)})(s_\alpha(x))\big) \\ &=0+\mathrm{Ad}_{g_{\alpha\beta}(x)}g_{\alpha\beta}^*\omega_G(v)=-g_{\beta\alpha}^*\omega_G(v)\hspace{7cm} (!) \end{align}
\begin{align} T_{s_\beta(x)}\big(\psi^2_\alpha\circ\theta_{\psi^2_\alpha(s_\beta(x))^{-1}}\big)\big(\hat\xi(s_\beta(x))\big)&=T_{s_\beta(x)}\big(\psi^2_\alpha\circ\theta_{g_{\alpha\beta}(x)^{-1}}\big)\Big(T_{s_\alpha(x)}\theta_{g_{\alpha\beta}(x)}\big((\widehat{\mathrm{Ad}_{g_{\alpha\beta}(x)}\xi})(s_\alpha(x))\big)\Big) \\ &=T_{s_\alpha(x)}\psi^2_\alpha\big((\widehat{\mathrm{Ad}_{g_{\alpha\beta}(x)}\xi})(s_\alpha(x))\big) \\ &=\mathrm{Ad}_{g_{\alpha\beta}(x)}\xi \end{align}
The result of this computation is:
$\bar{g}_{\alpha\beta}(v)\xi=\mathrm{Ad}_{g_{\alpha\beta}(x)}\xi-g_{\beta\alpha}^*\omega_G(v)$
Thus the transition law on $TU_\alpha\cap TU_\beta$ is:
$(v,\xi_\beta)=(v,\bar{g}_{\beta\alpha}(v)\xi_\alpha)=(v,\mathrm{Ad}_{g_{\beta\alpha}(x)}\xi_\alpha-g_{\alpha\beta}^*\omega_G(v))$
According to the literature the correct equation should be $(v,\xi_\beta)=(v,\mathrm{Ad}_{g_{\beta\alpha}(x)}\xi_\alpha+g_{\alpha\beta}^*\omega_G(v))$