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Table 3 Elements of the Lie algebra \({\mathfrak t}\) and the associated Lie group T

From: Introducing the Logarithmic finite element method: a geometrically exact planar Bernoulli beam element

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