Let $G$ be a Lie group, for the convenience of computation, you can assume $G=\mathrm{GL}_n(\mathbb{R})$, its cotangent bundle $T^*G$, by the left trivialisation, is actually a trivial bundle $T^*G\cong G\times \mathfrak{g}^*$.
It is well known that there is a (cannonical) Symplectic form on every cotangent bundle (via the Liouville 1-form), so a natural question is, how to find the Symplectic form on the case of cotangent bundle over a Lie group $T^*G$?
That is to find a nondegenerate skew-symmetric bilinear form:
$$\omega_{(g,A)}: (\mathfrak{g}\times\mathfrak{g}^*)\times (\mathfrak{g}\times\mathfrak{g}^*) \longrightarrow\mathbb{R}$$
for every $(g,A)\in T^*G\cong G\times\mathfrak{g}^*$, where $T_{(g,A)}T^*G\cong T_gG\times T_A\mathfrak{g}^*\cong \mathfrak{g}\times\mathfrak{g}^*$, and what does the Liouville 1-form look like in this case?
If for a matrix group, $\mathrm{GL}_n(\mathbb{R}) $ for instance, the Symplectic form maybe given by some operations of matrices, so thinking on this special example maybe a little easier for it may provide some concrete computation, but I cannot find it.
Sincerely looking forward to your answer, Thanks.
Best Answer
The Liouville form or tautological one-form $\sigma$ on $T^*G$ is given at any point $\lambda\in T^*G$ by $\sigma_{\lambda} = \lambda\circ \pi_*$, where $\pi\colon T^*G\to G$ is the bundle projection and $\pi_*\colon T(T^*G)\to TG$ its differential. After left-trivializing everything we obtain the left-trivialized tautological form as the one-form \begin{equation*} \sigma\colon G\times\mathfrak{g}^*\to (\mathfrak{g}\times\mathfrak{g}^*\to\mathbb{R}),\quad \sigma_{g,\lambda}(X,\phi) = \lambda\circ\pi_*(X,\phi) = \lambda(X). \end{equation*}
In general, the differential of a smooth one-form $\alpha\in \Omega^1(T^*M)$ can be evaluated along two smooth vector fields $X,Y\in\operatorname{Vec}(M)$ with the formula \begin{equation*} d\alpha(V,W) = V(\alpha(W))-W(\alpha(V))-\alpha[V,W]. \end{equation*}
In the case of the left-trivialized tautological form, we take two vectors $(X,\phi)$ and $(Y,\psi)$ in $\mathfrak{g}\times\mathfrak{g}^*$ and consider these two as left-invariant vector fields on $G\times\mathfrak{g}^*$. Then the above formula gives the left-trivialized symplectic form at any point $(g,\lambda)\in G\times\mathfrak{g}^*$ as \begin{equation*} \omega_{g,\lambda}((X,\phi),(Y,\psi)) = (X,\phi)_{g,\lambda}(\sigma(Y,\psi)) - (Y,\psi)_{g,\lambda}(\sigma(X,\phi)) - \sigma_{g,\lambda}[(X,\phi),(Y,\psi)] \end{equation*} The last term is easy to evaluate as $\pi_*[(X,\phi),(Y,\psi)]=[X,Y]$. For the first term observe that the curve $t\mapsto (g\exp(tX),\lambda+t\phi)$ is a smooth curve in $G\times\mathfrak{g}^*$ passing through $(g,\lambda)$ with (note the left-trivialization again!) derivative $(X,\phi)$ at $t=0$, so \begin{equation*} (X,\phi)_{g,\lambda}(\sigma(Y,\psi)) = \frac{d}{dt}\Big|_{t=0} \sigma_{(g\exp(tX),\lambda+t\phi)}(Y,\psi) = \frac{d}{dt}\Big|_{t=0} (\lambda+t\phi)(Y) = \phi(Y). \end{equation*} With a similar computation for the middle term, we obtain an explicit formula for the left-trivialized symplectic form at a point $(g,\lambda)\in G\times\mathfrak{g}^*$ as \begin{equation*} \omega_{(g,\lambda)}(X,\phi),(Y,\psi)) = \phi(Y) - \psi(X) - \lambda[X,Y]. \end{equation*}