What is the group of the Lie algebra generated by $\{a(E_{12}-E_{21}),u E_{13},v E_{23}\}$?

Question

I was asked in an exam to find $G$, if $$Lie(G)=\left\{\begin{bmatrix}0 &-a&u\\ a &0&v\\0&0&0\end{bmatrix}; a,u,v\in\mathbb{R}\right\},$$ where $[\cdot,\cdot]$ is the commutator.

My idea was the following: Let $B(t)\in G\forall t\in (-\epsilon,\epsilon)$ and $\epsilon>0$ such that $B(0)=1_{3\times 3}$, then $$\left.\frac{d}{d t}\right|_{t=0}B(t)\in Lie(G).$$ From this I thought that matrices of the following forms could be elements in $G$ $$\begin{bmatrix}1 &-ta&tu\\ ta &1&tv\\0&0&1\end{bmatrix},\begin{bmatrix}0 &-ta&tu\\ ta &0&tv\\0&0&1\end{bmatrix}, \text{etc.}$$ The problem with this is that none of them actually satisfy the properties of a group. Can someone maybe explain how one could find a possible group for the given Lie-Algebra.

Answer 1

Hint: Observe that the matrices in your Lie algebra have a block form with a skew symmetric $2\times 2$-block and a block which is $\mathbb R^2$. Then recall that the skew symmetric $2\times 2$ matrices are the Lie algebra of $O(2)$ and of $SO(2)$.

Answer 2

Calculating the eigenvalues of the matrix $$ M(a,u,v)=\begin{pmatrix} 0 & -a & u\\ a & 0 & v\\ 0 & 0 & 0 \end{pmatrix}\Longrightarrow \det (M-\lambda I)=-\lambda(\lambda-ia)(\lambda+ia) $$ (don't worry too much by the imaginary eigenvalues, everything ends up being rela at the end). Locally the Lie group $G$ is obtained from Lie algebra by exponentiation. Now by Cayley-Hamiltonian theorem, there exists $c,d,e\in \mathbb{C}$ such that $$ X(a,u,v):=\exp(M(a,u,v))= cI + dM + eM^2 $$ Assuming $a\neq 0$, using the eigenvalues (and replacing them instead of $M$) we find $$ \begin{cases} 1=c\\ e^{ia}=c+iad-a^2e\\ e^{-ia} = c-iad-a^2e \end{cases}\Longrightarrow d = \frac{\sin a}{a}, \qquad e=\frac{1-\cos a}{a^2} $$ So, $$ X(a,u,v)= \begin{pmatrix} \cos a & -\sin a& [u\sin a-v(1-\cos a)]/a\\ \sin a &\cos a & [v\sin a+u(1-\cos a)]/a\\ 0& 0& 1 \end{pmatrix} $$ The case $a=0$ can be obtained through taking the limit $a\to 0$. Note that, $$ a^{-1}\begin{pmatrix} u \sin a -v(1-\cos a)\\ u(1-\cos a) + v\sin a \end{pmatrix}= \frac{\sin (\frac{a}{2})}{\frac{a}{2}} \begin{pmatrix} \cos (\frac{a}{2}) & -\sin (\frac{a}{2})\\ \sin (\frac{a}{2}) & \cos (\frac{a}{2}) \end{pmatrix} \begin{pmatrix} u\\ v \end{pmatrix} $$ The matrix on the right hand side is a rotation matrix multiplied by a non-zero number and hence invertible. This means, your $G$ is nothing but the group of matrices of the form $$ \boxed{G=\left\{\begin{pmatrix} \cos a & -\sin a & x\\ \sin a & \cos a & y\\ 0 & 0 & 1 \end{pmatrix}\: :\: a,x,y\in \mathbb{R}\right\}} $$