The matrix \(\boldsymbol{\dot B} - 2C\) is skew symmetric
\(C\) is an \(n \times n\) matrix such that its elements \(C_{ij}\)
satisfy (see Equations of motion):
\[\sum_{j=1}^{n} C_{ij}\,\dot{q}_{j}
= \sum_{j=1}^{n} \sum_{k=1}^{n} h_{ijk}\,\dot{q}_{k}\,\dot{q}_{j}\]
\[\text{where:}\]
\[h_{ijk}
= \frac{\partial b_{ij}}{\partial q_{k}}
- \tfrac{1}{2}\,\frac{\partial b_{jk}}{\partial q_{i}}\]
Hence:
\[\sum_{j=1}^n C_{ij}\,\dot q_j
= \sum_{j=1}^n \sum_{k=1}^n
\frac{\partial b_{ij}}{\partial q_k}\,\dot q_k\,\dot q_j
\;-\;\frac12
\sum_{j=1}^n \sum_{k=1}^n
\frac{\partial b_{jk}}{\partial q_i}\,\dot q_k\,\dot q_j\]
\[\sum_{j=1}^n C_{ij}\,\dot q_j
=
\underbrace{\sum_{j=1}^n \sum_{k=1}^n
\frac{\partial b_{ij}}{\partial q_k}\,\dot q_k\,\dot q_j}_{\text{this is the same as:}}
\;-\;\tfrac12
\sum_{j=1}^n \sum_{k=1}^n
\frac{\partial b_{jk}}{\partial q_i}\,\dot q_k\,\dot q_j\]
\[\quad\quad\quad\quad\quad\quad\sum_{j=1}^n\sum_{k=1}^n \frac{\partial b_{ik}}{\partial q_j}\,\dot q_k\,\dot q_j
\quad\text{or equivalently}\quad
\sum_{j=1}^n\sum_{k=1}^n \tfrac12
\Bigl[
\frac{\partial b_{ij}}{\partial q_k}
+ \frac{\partial b_{ik}}{\partial q_j}
\Bigr]
\dot q_k\,\dot q_j\]
Then:
\[\sum_{j=1}^n C_{ij}\,\dot q_j
= \sum_{j=1}^n
\underbrace{\sum_{k=1}^n\tfrac12
\Bigl[
\frac{\partial b_{ij}}{\partial q_k}
+ \frac{\partial b_{ik}}{\partial q_j}
- \frac{\partial b_{jk}}{\partial q_i}
\Bigr]
\dot q_k}_{C_{ij}}\,\dot q_j\]
That is:
\[C_{ij} = \sum_{k=1}^n C_{ijk}\,\dot q_k,\]
\[C_{ijk}
= \tfrac12
\Bigl(
\frac{\partial b_{ij}}{\partial q_k}
+ \frac{\partial b_{ik}}{\partial q_j}
- \frac{\partial b_{jk}}{\partial q_i}
\Bigr)\]
\[\text{which are the Christoffel symbols of the first type } (C_{ijk} = C_{ikj}).\]
Let’s rewrite \(C_{ij}\) as:
\[ \begin{align}\begin{aligned}C_{ij}
= \tfrac12 \sum_{k=1}^n \frac{\partial b_{ij}}{\partial q_k}\,\dot q_k
+ \tfrac12 \sum_{k=1}^n
\Bigl(
\frac{\partial b_{ik}}{\partial q_j}
- \frac{\partial b_{jk}}{\partial q_i}
\Bigr)
\dot q_k\\= \tfrac12\,\dot b_{ij}
+ \tfrac12 \sum_{k=1}^n
\Bigl(
\frac{\partial b_{ik}}{\partial q_j}
- \frac{\partial b_{jk}}{\partial q_i}
\Bigr)
\dot q_k\end{aligned}\end{align} \]
Let us now define:
\[N(\boldsymbol q,\boldsymbol{\dot{q}})
= \dot{B}(\boldsymbol q)\;-\;2\,C(\boldsymbol q,\dot{\boldsymbol q})\]
\[n_{ij}
= \dot{b}_{ij}\;-\;2\,C_{ij}
= \sum_{k=1}^{n}
\Bigl(
\frac{\partial b_{ik}}{\partial q_{j}}
- \frac{\partial b_{jk}}{\partial q_{i}}
\Bigr)
\dot q_{k}\]
We observe that \(n_{ij} = -n_{ji}\), i.e. \(N(\boldsymbol q,\dot{\boldsymbol q})\)
is skew-symmetric.
Then:
\[\boxed{\boldsymbol \omega^{T}\,N(\boldsymbol q,\dot{\boldsymbol q})\,\boldsymbol{\omega} \;=\; 0}\]
In particular we will use:
\[\boxed{\boldsymbol{\dot q^{T}}\,N(\boldsymbol q,\boldsymbol{\dot{q}})\,\boldsymbol{\dot{q}} \;=\; 0}\]
Linearity with the dynamic parameters
The Lagrangian \(\mathcal{L}(\boldsymbol q,\boldsymbol{\dot{q}}) = \mathcal{T}(\boldsymbol q,\boldsymbol{\dot{q}}) - \mathcal{U}(\boldsymbol q)\)
can be rewritten as
\[\mathcal{L}(\boldsymbol q,\boldsymbol{\dot{q}})
= \sum_{j=1}^{n}
\bigl(
\beta_{T_j}^{T}(\boldsymbol q,\boldsymbol{\dot{q}})
- \beta_{U_j}^{T}(\boldsymbol q)
\bigr)\,\boldsymbol{\pi}_{j}\]
where
\(\boldsymbol{\pi}_{j}\) is a \((p\times 1)\) vector of parameters of link \(i\) (mass, position of center of mass, inertia).
\(\beta_{T_j}\) and \(\beta_{U_j}\) are \((p\times 1)\) vectors of the coefficients that multiply the
dynamic parameters of \(\boldsymbol{\pi}_{j}\).
Then the Lagrange equations:
\[\frac{d}{dt}\Bigl(\frac{\partial \mathcal{L}}{\partial \dot q_i}\Bigr)
\;-\;\frac{\partial \mathcal{L}}{\partial q_i}
= \xi_i,\quad i=1,\dots,n\]
can be rewritten as:
\[\xi_i
= \sum_{j=1}^n \boldsymbol y_{ij}^{T}\,\boldsymbol{\pi}_j,\quad i=1,\dots,n\]
where
\[\boldsymbol y_{ij}
= \frac{d}{dt}\Bigl(\frac{\partial \beta_{T_j}}{\partial \dot q_i}\Bigr)
- \Bigl(\frac{\partial \beta_{T_j}}{\partial q_i}\Bigr)
+ \Bigl(\frac{\partial \beta_{U_j}}{\partial q_i}\Bigr)\]
The derivatives appearing in \(\boldsymbol y_{ij}\) are null when \(j<i\). Then:
\[\begin{split}\begin{bmatrix}
\xi_1 \\[6pt]
\vdots \\[3pt]
\xi_n
\end{bmatrix}
=
\begin{bmatrix}
\boldsymbol y_{11}^{T} & \boldsymbol y_{12}^{T} & \cdots & \boldsymbol y_{1n}^{T} \\[3pt]
0 & \boldsymbol y_{22}^{T} & \cdots & \boldsymbol y_{2n}^{T} \\[3pt]
\vdots & \ddots & \ddots & \vdots \\[3pt]
0 & \cdots & 0 & \boldsymbol y_{nn}^{T}
\end{bmatrix}
\begin{bmatrix}
\boldsymbol{\pi}_1 \\[3pt] \vdots \\[3pt] \boldsymbol{\pi}_n
\end{bmatrix}\,.\end{split}\]
If we add the friction coefficients in vector \(\boldsymbol \pi_i\), then this can be rewritten in matrix form as:
\[\boxed{\boldsymbol \tau = y(\boldsymbol q,\dot{\boldsymbol q},\ddot{\boldsymbol q})\,\boldsymbol \pi}\]
and \(y(\boldsymbol q,\boldsymbol{\dot q},\boldsymbol{\ddot q})\) is called the regressor.