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

differential-geometryheisenberg-grouplie-algebraslie-groupssymplectic-linear-algebra

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?

Best Answer

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 $g(t) = (tu,\exp it\phi)$.

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