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Let $(V,\Omega)$ be a symplectic vector space, and consider $H:=V\times \mathbb{R}=\{(v,t)\}$. This space $H$ with the multiplication $$(v_1,t_1)\cdot(v_2,t_2)=\left( v_1+v_2,\frac{1}{2}\Omega(v_1,v_2)+t_1+t_2 \right) $$ is a Lie group called the Heisenberg group. Verify that $\operatorname{ad}_{(Y,s)}(X,r)=(0,\Omega(Y,X))$ for $(Y,s),(X,r)\in\operatorname{Lie}(H)\cong V\times \mathbb{R}$.

It is straightforward to compute that $$(Y,s)\cdot(X,r)-(Y,s)\cdot(X,r)=(0,\Omega(Y,X)),$$ but to show that indeed $$\operatorname{ad}_{(Y,s)}(X,r)=(Y,s)\cdot(X,r)-(Y,s)\cdot(X,r)$$ I should "differentiate" $\operatorname{Ad}$. I think this "derivative" is what Lee, Intro. to Smooth Manifolds calls the "induced morphism on Lie algebras". (Perhaps the equation above is immediate from definition but I'm kind of lost on how to argue.)

In a previous question of this exercise I showed that $\operatorname{Ad}_{(v,t)}(X,r)=(X,r+\Omega(v,X))$, for $(v,t)\in H$ and $(X,r)\in\mathsf{Lie}(H)$.

danimalabares
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    Consider a curve $u \mapsto (v(u), t(u))$ passing at $u=0$ by $(0,0)$ (the identity of the Heisenberg group) with velocity $(Y, s)$. Evaluate your formula for $Ad$ along this curve, differentiate it with respect to $u$ and evaluate the derivative at $u = 0$: that will give you $ad_{(Y,s)}(X, r)$. – Jordan Payette Oct 16 '24 at 21:12

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