How to define the Symplectic form on the cotangent bundle of a Lie group?

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.

• It is just cross evaluation: $\omega((v,\phi),(w,\psi))=\phi(w)-\psi(v)$ or minus that Commented Oct 10, 2022 at 4:27
• @MarianoSuárez-Álvarez That 2-form is not closed for non-abelian Lie groups Commented Oct 10, 2022 at 5:46

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 $$\begin{equation*} \omega_{(g,\lambda)}(X,\phi),(Y,\psi)) = \phi(Y) - \psi(X) - \lambda[X,Y]. \end{equation*}$$

• Thank you Eero, it's a wonderful computation！Besides, what interesting here is that the Liouville 1-form on the $T^*G$ coincides with the 1-form induced by the Maurer-Cartan form on $G$. Commented Oct 10, 2022 at 11:14

Well, I'd love to share some of my thoughts.

There is a Maurer-Cartan form defined on a Lie group $$G$$, that is

$$\alpha_g=(L_{g^{-1}})_*: T_gG\longrightarrow\mathfrak{g}$$

where $$(L_{g^{-1}})_*$$ is the differential of left-multiplication $$L_{g^{-1}}$$, and in particular, the $$\alpha$$ is a $$\mathfrak{g}$$-valued 1-form, i.e. $$\alpha\in \Omega^1(G,\mathfrak{g})$$.

This 1-form can induce a $$\mathfrak{g}$$-valued one form on $$T^*G\cong G\times \mathfrak{g} ^*$$, by sending $$(X, A')$$ to $$\alpha_g(X)= (L_{g^{-1}})_* X$$, where $$(X,A')\in T_{(g,A)}T^*G=T_gG\times\mathfrak{g}^*$$, which is just the differential of the induced left action on the cotangent bundle: $$\tilde{L_{g^{-1}}}(g, A)=(e, A)$$, notice that, if we identify $$T_gG$$ with $$\mathfrak{g}$$, that $$\alpha_g$$ is in fact the identity ！

To gain a "real" 1-form on $$T^*G$$, we can take the pairing $$A(X)$$, let's denote this one form by $$\theta$$, and by our construction:

$$\theta_{(g,A)}(X,A')=A(X)$$

Maybe by differentiating this $$1-$$form (Maurer-Cartan equation maybe involved) $$d\theta$$, one can probably get a desired Symplectic form.