## Enhanced Stiffness Model of the 3 DJM
### Displacement and Load of the 3 DJM on SO(3)  
According to the exponential formula [31], the trajectory curve of the 3 DJM \(\mathbf{R}(t)\) yields  
\[\mathbf{R}(t)=e^{\hat{\mathbf{\zeta}}(t)}, \tag{1}\]  
where \(\hat{\mathbf{\zeta}}\in so(3)\) is an element of Lie algebra \(so(3)\) and satisfies  
\[\mathbf{\zeta}=\left(\begin{array}{c}\zeta_{1}\\ \zeta_{2}\\ \zeta_{3}\end{array}\right)\rightarrow\hat{\mathbf{\zeta}}=\left(\begin{array}{ ccc}0&-\zeta_{3}&\zeta_{2}\\ \zeta_{3}&0&-\zeta_{1}\\ -\zeta_{2}&\zeta_{1}&0\end{array}\right). \tag{2}\]  
Here, \(\mathbf{\zeta}\in\mathbb{R}^{3}\) is called the coordinates of \(\hat{\mathbf{\zeta}}\in so(3)\). The derivative of \(\mathbf{R}(t)\) to time \(t\), i.e., \(\hat{\mathbf{R}}(t)\), is an element of the tangent space of \(SO(3)\) at the point \(\mathbf{R}(t)\), denoted as \(T_{\mathbf{R}(t)}SO(3)\). According to (1), \(\hat{\mathbf{R}}(t)\) satisfies the following equation  
\[\hat{\mathbf{R}}(t)=\mathbf{R}(t)\hat{\mathbf{\zeta}}(t)=\mathbf{R}(t)\hat{\mathbf{\omega}}(t), \tag{3}\]  
where \(\mathbf{\omega}(t)=\hat{\mathbf{\zeta}}(t)\in so(3)\) is the angular velocity of the 3 DJM. Given \(\hat{\mathbf{e}}_{j}\)\((j=1,2,3)\) as the basis of \(so(3)\), where \(\mathbf{e}_{1}=(1,0,0)^{T}\), \(\mathbf{e}_{2}=(0,1,0)^{T}\) and \(\mathbf{e}_{3}=(0,0,1)^{T}\), then \(\hat{\mathbf{R}}(t)\) can be expressed as  
\[\hat{\mathbf{R}}(t)=\mathbf{R}(t)\hat{\mathbf{\omega}}(t)=\mathbf{R}(t)\sum_{j=1}^{3}\omega^{ j}\hat{\mathbf{e}}_{j}=\sum_{j=1}^{3}\omega^{j}\mathbf{R}(t)\hat{\mathbf{e}}_{j}=\sum_{j=1}^{3} \omega^{j}\mathbf{L}_{j}|_{\mathbf{R}(t)}, \tag{4}\]  
where \(\mathbf{L}_{j}|_{\mathbf{R}(t)}=\mathbf{R}(t)\hat{\mathbf{e}}_{j}\)\((j=1,2,3)\) are taken as the basis of \(T_{\mathbf{R}(t)}SO(3)\) and \(\mathbf{L}_{j}|_{\mathbf{R}(t)}\) are always written as \(\mathbf{L}_{j}\) for short. From (4), it can be concluded that \(T_{\mathbf{R}(t)}SO(3)\) is isomorphic to \(so(3)\).  
Since \(\hat{\mathbf{R}}(t)\) describes the velocity of the 3 DJM, the instantaneous displacement of the 3 DJM can be represented by \(\delta\mathbf{R}\), yielding  
\[\delta\mathbf{R}(t)=\hat{\mathbf{R}}(t)\delta t. \tag{5}\]  
That means the instantaneous displacement of the 3 DJM can be described on the tangent space of \(SO(3)\). Similarly, since the moment load \(\hat{\mathbf{\tau}}\) is an element of \(so^{*}(3)\), the dual space of \(so(3)\), by analogy with \(\hat{\mathbf{\omega}}\in so(3)\) and \(\hat{\mathbf{R}}(t)\in T_{\mathbf{R}(t)}SO(3)\), we define \(\mathbf{F}(t)=\mathbf{R}(t)\hat{\mathbf{\tau}}(t)\) as an element of the dual space of the tangent space \(T_{\mathbf{R}(t)}SO(3)\) at \(\mathbf{R}(t)\), called _cotangent space_ and denoted as \(T^{*}_{\mathbf{R}(t)}SO(3)\). The element of the cotangent space is named as the cotangent vector. Given \(\hat{\mathbf{\lambda}}^{k}\)\((k=1,2,3)\) as the basis of \(so^{*}(3)\) and \(\mathbf{\Lambda}^{k}|_{\mathbf{R}(t)}=\mathbf{R}(t)\hat{\mathbf{\lambda}}^{k}\)\((k=1,2,3)\) as the basis of \(T^{*}_{\mathbf{R}(t)}SO(3)\), then \(\mathbf{F}(t)\in T^{*}_{\mathbf{R}(t)}SO(3)\) satisfies  
\[\mathbf{F}(t)=\mathbf{R}(t)\hat{\mathbf{\tau}}(t)=\sum_{k=1}^{3}\tau_{k}\mathbf{R}(t)\hat\mathbf{{\lambda}}^{k}=\sum_{k=1}^{3}\tau_{k}\mathbf{\Lambda}^{k}|_{\mathbf{R}(t)}. \tag{6}\]  
It shows that \(T^{*}_{\mathbf{R}(t)}SO(3)\) is isomorphic to \(so^{*}(3)\). Thus, \(\mathbf{F}(t)\) can be employed to represent the load applied on the 3 DJM. That means the load applied on the 3 DJM can be described on the cotangent space of \(SO(3)\).