# Calculating the Lie bracket on the Heisenberg algebra of $H=Z\times S^1$

I'm working through Mechanics and Symmetry by Marsden and Ratiu. Let $$(Z,\Omega)$$ be a symplectic vector space and define on $$H:=Z\times S^1$$ the operation $$(u,\exp i\phi)(v,\exp i\psi)=(u+v,\exp i[\phi+\psi+\hbar^{-1} \Omega(u,v)])$$

I need to show that on the Lie algebra of $$H$$, $$\mathfrak{h}=Z\times\mathbb{R}$$, the Lie bracket is defined as follows: $$[(u,\phi),(v,\phi)]=(0,2\hbar^{-1}\Omega(u,v))$$

I'm using the method the book recommends to calculate this bracket, the steps are:

1. Calculate the inner automorphisms: $$I_g:H\to H,\quad\text{where}\quad I_g(h)=ghg^{-1}$$
2. Differentiate $$I_g(h)$$ with respect to $$h$$ at $$h=e$$ to product the adjoint operators $$\textrm{Ad}_g:\mathfrak{h}\to\mathfrak{h};\quad \textrm{Ad}_g\eta=T_eI_g(\eta)$$
3. Then Differentiating $$\textrm{Ad}_g\eta$$ with respect to $$g$$ at the identity $$e$$ in the direction of $$\xi\in\mathfrak{h}$$, since $$T_e(\textrm{Ad}_g\eta)\cdot\xi=[\xi,\eta]$$

So I started by calculating the inner automorphisms. Given $$g=(v,\exp(i\psi))$$ we have $$g^{-1}=(-v,\exp(-i\psi))$$. If we take $$h=(u,\exp i\phi)$$, we have \begin{align} I_g(h)&=(v+u,\exp i(\phi+\psi+\hbar^{-1}\Omega(v,u))(-v,\exp i(-\psi))\\ &=(v+u-v,\exp i(\phi+\psi+-\psi+\hbar^{-1}\Omega(v,u)+\hbar^{-1}\Omega(v+u,-v))\\ &=(u,\exp i(\phi+2\hbar^{-1}\Omega(v,u)) \end{align} After this, I am confused as to how to take the derivative of this, I believe it should be something like: $$T_eI_g(\eta)=(0, 2\hbar^{-1}\Omega(v,u))$$ But I am unsure if this is correct, or how I could justify this step. How am I supposed to compute this derivative?

To be consistent, it would make more sense to take $$g=(u,\exp(i\phi))$$ and $$h=(v,\exp(i\psi))$$, so $$I_g(h) = (v,\exp i(\psi+2\hbar^{-1}\Omega(u,v))).$$ Now one way to calculate $$T_eI_g(\eta)$$ is to calculate $$\frac{d}{dt}\big\vert_{t=0}I_g(h(t))$$, where $$h(t)$$ is a smooth curve such that $$h(0) = e=(0,1)$$, the identity of the Heisenberg group, and $$h'(0) = \eta$$. Taking $$\eta = (v,\psi)$$, a corresponding curve is $$h(t) = (tv,\exp(it\psi))$$. Then $$I_g(h(t)) = (tv,\exp i(t\psi + 2\hbar^{-1}\Omega(u,tv))) \implies T_eI_g(\eta) = \frac{d}{dt}\big\vert_{t=0} I_g(h(t)) = (v,\psi+2\hbar^{-1}\Omega(u,v)).$$ Now repeat this process with $$g(t) = (tu,\exp it\phi)$$.
• So when we take $g(t)=(tu,\exp it\phi)$ then $T_e(\textrm{Ad}_g\eta)\xi=\frac{d}{dt}(v, \psi+2\hbar^{-1}\Omega(tu,v))|_{t=0}=(0,2\hbar^{-1}\Omega(u,v))$? Mar 1, 2021 at 2:02