# Table 3 Elements of the Lie algebra $${\mathfrak t}$$ and the associated Lie group T
Element $$\varvec{\mathbf {T}}_{\varvec{I,r}} \varvec{\in {\mathfrak t}}(3,\mathbb {C})$$ Element $$\varvec{\exp \left( {\mathbf {T}_{I,r}}\right) \in T}(3,\mathbb {C})$$ Subalgebra in $$\varvec{\mathfrak t}(3,\mathbb {C})$$
$$\mathbf {T}_{1,r} := \begin{pmatrix} t_{1,r}\\ t_{1,r}\\ 0\end{pmatrix}$$ $$\exp \left( {\mathbf {T}_{1,r}}\right) = \begin{pmatrix} t_{1,r}\\ t_{1,r}\\ 0\end{pmatrix}$$ $$\mathfrak t_1 = \langle \mathbf {T}_{1,r} \rangle$$, $$r \in \{1,2\}$$
$$\mathbf {T}_{2,r} := \begin{pmatrix} t_{2,r}\\ 0\\ t_{2,r}\end{pmatrix}$$ $$\exp \left( {\mathbf {T}_{2,r}}\right) = \begin{pmatrix} t_{2,r}\\ 0\\ t_{2,r}\end{pmatrix}$$ $$\mathfrak t_2 = \langle \mathbf {T}_{2,r} \rangle$$, $$r \in \{1,2\}$$
$$\mathbf {T}_{3,r} := \begin{pmatrix} t_{3,r}\\ 0\\ 0\end{pmatrix}$$ $$\exp \left( {\mathbf {T}_{3,r}}\right) = \begin{pmatrix} t_{3,r}\\ 0\\ 0\end{pmatrix}$$ $$\mathfrak t_3 = \langle \mathbf {T}_{3,r} \rangle$$, $$r \in \{1,2\}$$
1. $$t_{I,r}$$ assumes the values $$t_{I,1} = 1$$, $$t_{I,2} = \hbox {i}$$ for all $$I\in \{1,2,3\}$$. Note that, for translational lie groups, the representation of the elements of the Lie algebra and their exponentials in the Lie group as column vectors is identical