I'm currently working through Introduction to Mechanics and Symmetry's chapter on Lie Group actions and I want to make sure I've calculated the Adjoint action of the Galilean $G$ on its Lie algebra $\mathfrak{g}$ correctly.
We'll take $G$ to be the set of matrices with the block structure: $$(R,v,a,\tau)=\begin{pmatrix} R&v&a \\ 0&1&\tau \\ 0&0&1\end{pmatrix}$$ With $R\in SO(3)$, $v,a\in \mathbb{R}^3$ and $\tau\in \mathbb{R}$.
We'll start by calculating the the inverses in $G$. Let $g=(R,v,a,\tau)$ and $\tilde{g}=(\tilde{R},\tilde{v},\tilde{a},\tilde{\tau})$ such that $g\tilde{g}=\mathrm{Id}_{G}=(I_3,0,0,0)$ We have $$g\tilde{g}=\begin{pmatrix} R&v&a \\ 0&1&\tau \\ 0&0&1\end{pmatrix}\begin{pmatrix} \tilde{R}&\tilde{v}&\tilde{a} \\ 0&1&\tilde{\tau} \\ 0&0&1\end{pmatrix}=\begin{pmatrix}R\tilde{R}&R\tilde{v}+v&R\tilde{a}+\tilde{\tau} v+a\\0&1&\tau+\tilde{\tau}\\0&0&1\end{pmatrix}$$ We can then conclude that $\tilde{R}=R^{-1}$, $\tilde{v}=-R^{-1}v$ and $\tilde{\tau}=-\tau$. From these we have $R\tilde{a}-\tau v+a=0$ hence: $\tilde{a}=R^{-1}(\tau v-a)$. So $g^{-1}=(R^{-1}, -R^{-1}v, R^{-1}(\tau v-a), -\tau)$.
Now we will calculate the inner automorphisms of $g$. Take $g=(R,v,a,\tau), h=(R',v',a',\tau')$. We have \begin{align}ghg^{-1}&=\begin{pmatrix} R&v&a\\ 0&1&\tau\\ 0&0&1\end{pmatrix}\begin{pmatrix} R'&v'&a'\\ 0&1&\tau'\\ 0&0&1\end{pmatrix} \begin{pmatrix} R^{-1}&-R^{-1}v&R^{-1}(\tau v-a)\\ 0&1&-\tau\\ 0&0&1\end{pmatrix}\\ &=\begin{pmatrix} RR'&Rv'+v&Ra'+\tau'v+a\\ 0&1&\tau+\tau'\\0&0&1 \end{pmatrix}\begin{pmatrix} R^{-1}&-R^{-1}v&R^{-1}(\tau v-a)\\ 0&1&-\tau\\ 0&0&1\end{pmatrix}\\ &=\begin{pmatrix}RR'R^{-1}&Rv'+v-RR'R^{-1}v&RR'R^{-1}(\tau v-a)-\tau Rv'-\tau v+Ra'+\tau'v+a\\ 0&1&\tau'\\0&0&1 \end{pmatrix} \end{align} Assuming that this previous calculation is correct, we'll parameterize a curve $h(t)$ such that $h(0)=(\mathrm{id}_{3},0,0,0)$ and $\dot{h}(0)=\begin{pmatrix} \hat{\omega}&u&\alpha\\ 0&0& \theta\\ 0&0&0\end{pmatrix}=\xi\in\mathfrak{g}$, where $\hat{}$ denotes the identification of $\mathbb{R}^3$ with $\mathfrak{so}(3)$. Differentiating $I_g(h(t))$ at $t=0$ will yield $\mathrm{Ad}_g(\xi)$. Let $$h(t)=(R'(t), tu, t\alpha, t\theta)$$ for some differentiable curve $R(t)$ in $SO(3)$ with $R'(0)=I_3$ and $\dot{R}'(0)=\hat{\omega}$. Then $$I_g(h(t))=\begin{pmatrix} RR'(t)R^{-1}& Rtu+v-RR'(t)R^{-1}v& RR'(t)R^{-1}(\tau v-a)-\tau Rtu-\tau v+Rt\alpha+t\theta v+a\\ 0&1&t\theta\\ 0&0&1\end{pmatrix}$$ Since $\frac{d}{dt}|_{t=0}(RR'(t)R^{1})=\hat{[R\omega]}$ from the adjoint operator on $SO(3)$ We end up with the following:
$$\mathrm{Ad}_g(\xi)=\begin{pmatrix}\hat{[R\omega]}& Ru-\hat{[R\omega]}v& \hat{[R\omega]}(\tau v-a)-\tau Ru+R\alpha+\theta v\\ 0&0&\theta\\0&0&0 \end{pmatrix}$$