$\newcommand{\Dev}{\operatorname{Dev}}$ $\newcommand{\dev}{\operatorname{dev}}$ $\newcommand{\grad}{\operatorname{grad}}$ $\newcommand{\Grad}{\operatorname{Grad}}$ $\newcommand{\divg}{\operatorname{div}}$ $\newcommand{\Ei}{\operatorname{Ei}}$ $\newcommand{\tr}{\operatorname{tr}}$ $\newcommand{\Divg}{\operatorname{Div}}$ $\newcommand{\cay}{\operatorname{cay}}$ $\newcommand{\rot}{\operatorname{rot}}$

Conventionally, the rigid body rotation tensor $\boldsymbol{\Lambda}$ corresponding to a rotation of angle $\chi$ about the unit vector $\mathbf{n}$ may be expressed in terms of the vector $\boldsymbol{\chi}=\chi\mathbf{n}$ as $$\begin{equation} \boldsymbol{\Lambda}\left(\boldsymbol{\chi}\right)=\cos\chi\,\mathbf{I}-\sin\chi\,\boldsymbol{\mathcal{E}}\cdot\mathbf{n}+\left(1-\cos\chi\right)\mathbf{n}\otimes\mathbf{n}\label{eq:rbr-rotation-tensor} \end{equation}$$ where $\boldsymbol{\mathcal{E}}$ is the third-order permutation pseudo-tensor with Cartesian components $\varepsilon_{ijk}$ . Making use of the trigonometric identity, $$\begin{equation} \cos\chi=1-2\sin^{2}\frac{1}{2}\chi\,,\label{eq:rbr-identity-1} \end{equation}$$ this expression may be rearranged as $$\begin{equation} \boldsymbol{\Lambda}\left(\boldsymbol{\chi}\right)=\mathbf{I}-\frac{\sin\chi}{\chi}\,\boldsymbol{\mathcal{E}}\cdot\boldsymbol{\chi}+\frac{2}{\chi^{2}}\sin^{2}\frac{1}{2}\chi\left(\boldsymbol{\mathcal{E}}\cdot\boldsymbol{\chi}\right)^{2}\,,\label{eq:rbr-rotation-tensor-alt-1} \end{equation}$$ where we have made use of the identity $$\begin{equation} \left(\boldsymbol{\mathcal{E}}\cdot\boldsymbol{\chi}\right)^{2}=\boldsymbol{\chi}\otimes\boldsymbol{\chi}-\chi^{2}\mathbf{I}\,.\label{eq:rbr-identity-2} \end{equation}$$ Letting $$\begin{equation} \hat{\boldsymbol{\chi}}=-\boldsymbol{\mathcal{E}}\cdot\boldsymbol{\chi}\label{eq:rbr-antisymmetric-tensor} \end{equation}$$ represent the antisymmetric tensor with axial vector $\boldsymbol{\chi}$ , $\boldsymbol{\Lambda}\left(\boldsymbol{\chi}\right)$ may now be represented as $$\begin{equation} \boldsymbol{\Lambda}\left(\boldsymbol{\chi}\right)\equiv\exp\left[\hat{\boldsymbol{\chi}}\right]=\mathbf{I}+\frac{\sin\chi}{\chi}\hat{\boldsymbol{\chi}}+\frac{2}{\chi^{2}}\sin^{2}\left(\frac{1}{2}\chi\right)\,\hat{\boldsymbol{\chi}}^{2}\,,\label{eq:rbr-rotation-tensor-alt-2} \end{equation}$$ where $\exp\left[\hat{\boldsymbol{\chi}}\right]$ is known as the exponential map. Thus, the exponential map provides the rotation tensor for a rotation $\chi$ about the unit vector $\mathbf{n}$ . Note that $\boldsymbol{\Lambda}\cdot\boldsymbol{\chi}=\boldsymbol{\chi}$ , since $\hat{\boldsymbol{\chi}}\cdot\boldsymbol{\chi}=\boldsymbol{\chi}\times\boldsymbol{\chi}=\mathbf{0}$ .