Equation 27 referenced in the quotation is torque equals the time derivative of angular momentum. Equation $\eqref{29}$ is the contraction of a position vector with the angular velocity tensor, giving the linear velocity components of the point
$$ u_{i}=v_{ik}\xi^{k}.\tag{29}\label{29} $$
The discussion pertains to Euclidean space with rectangular Cartesian coordinates where we may toggle indices with impunity.
The following is from Hermann Weyl's Space-Time-Matter beginning on page 74. I find it difficult to follow because the notation used to distinguish between inertial and body coordinates is a bit nebulous. The distinction between what is instantaneously invariant, and what is constant in time is tricky. My questions are at the end.
Euler's Equations for a Spinning Top
As an exercise in tensor calculus, we shall deduce Euler's equations for the motion of a rigid body under no forces about a fixed point $O$. We write the fundamental equations (27) in the co-variant form $$ \frac{dL_{ik}}{dt}=0 $$ and multiply them, for the sake of briefness, by the contra-variant components $w^{ik}$ of an arbitrary skew-symmetrical tensor which is constant (independent of the time), and apply contraction with respect to $i$ and $k$. If we put $H_{ik}$ equal to the sum $$ \sum_{m}mu_{i}\xi_{k} $$ which is to be taken over all the points of mass, we get $$ \tfrac{1}{2}L_{ik}w^{ik}=H_{ik}w^{ik}=H, $$ an invariant, and we can compress our equation into $$ \frac{dH}{dt}=0.\tag{32}\label{32} $$ If we introduce the expressions \eqref{29} for $u_{i}$, and the tensor of inertia $T$, then $$ H_{ik}=v_{ir}T_{k}^{r}.\tag{33}\label{33} $$
We have hitherto assumed that a co-ordinate system which is fixed in space has been used. The components $T$ of inertia then change with the distribution of matter in the course of time. If, however, in place of this we use a co-ordinate system which is fixed in the body, and consider the symbols so far used as referring to the components of the corresponding tensors with respect to this co-ordinate system, whereas we distinguish the components of the same tensors with respect to the co-ordinate system fixed in space by a horizontal bar, the equation \eqref{32} remains valid on account of the invariance of $H$. The $T_{i}^{k}$'s are now constants; on the other hand, however, the $w^{ik}$'s vary with the time. Our equation gives us $$ \frac{dH_{ik}}{dt}w^{ik}+H_{ik}\frac{dw^{ik}}{dt}=0.\tag{34}\label{34} $$
To determine $\dfrac{dw^{ik}}{dt}$, we choose two arbitrary vectors fixed in the body, of which the co-variant components in the co-ordinate system attached to the body are $\xi_{i}$ and $\eta_{i}$ respectively. These quantities are thus constants, but their components $\bar{\xi}_{i}$, $\bar{\eta}_{i}$ in the space co-ordinate system are functions of the time. Now, $$ w^{ik}\xi_{i}\eta_{k}=\bar{w}^{ik}\bar{\xi}_{i}\bar{\eta}_{k}, $$ and hence, differentiating with respect to the time $$ \frac{dw^{ik}}{dt}\xi_{i}\eta_{k}=\bar{w}^{ik}\left(\frac{d\bar{\xi}_{i}}{dt}\bar{\eta}_{k}+\bar{\xi}_{i}\frac{d\bar{\eta}_{k}}{dt}\right).\tag{35}\label{35} $$ By formula \eqref{29} $$ \frac{d\bar{\xi}_{i}}{dt}=\bar{v}_{ir}\bar{\xi}^{r}=\bar{v}_{i}^{r}\bar{\xi}_{r}. $$ We thus get for the right-hand side of \eqref{35} $$ \bar{w}^{ik}\left(\bar{v}_{i}^{r}\bar{\xi}_{r}\bar{\eta}_{k}+\bar{v}_{k}^{r}\bar{\xi}_{i}\bar{\eta}_{r}\right), $$ and as this is an invariant, we may remove the bars, obtaining $$ \xi_{i}\eta_{k}\frac{dw^{ik}}{dt}=w^{ik}\left(\xi_{r}\eta_{k}v_{i}^{r}+\xi_{i}\eta_{r}v_{k}^{r}\right). $$ This holds identically in $\xi$ and $\eta$; thus if the $H^{ik}$ are arbitrary numbers, $$ H_{ik}\frac{dw^{ik}}{dt}=w^{ik}\left(v_{i}^{r}H_{rk}+v_{k}^{r}H_{ir}\right). $$ If we take the $H_{ik}$'s to be the quantities which we denoted above by this symbol, the second term of \eqref{34} is determined, and our equation becomes $$ \left\{ \frac{dH_{ik}}{dt}+\left(v_{i}^{r}H_{rk}+v_{k}^{r}H_{ir}\right)\right\} w^{ik}=0, $$ which is an identity in the skew-symmetrical tensor $w^{ik}$; hence $$ \frac{d\left(H_{ik}-H_{ki}\right)}{dt}+\left[v_{i}^{r}H_{rk}+v_{k}^{r}H_{ir}-v_{k}^{r}H_{ri}-v_{i}^{r}H_{kr}\right]=0. $$ We shall now substitute the expression \eqref{33} for $H_{ik}$. Since, on account of the symmetry of $T_{ik}$, $$ v_{k}^{r}H_{ir}\left(=v_{k}^{r}v_{i}^{s}T_{rs}\right) $$ is also symmetrical in $i$ and $k$, the two last terms of the sum in the square brackets destroy one another. If we now put the symmetrical tensor $$ v_{i}^{r}v_{kr}=g_{rs}v_{i}^{r}v_{k}^{s}=\left(v,v\right){}_{ik} $$ we finally get our equations into the form $$ \frac{d}{dt}\left(v_{ir}T_{k}^{r}-v_{kr}T_{i}^{r}\right)=\left(v,v\right){}_{ir}T_{k}^{r}-\left(v,v\right){}_{kr}T_{i}^{r}. $$
It is well known that we may introduce a Cartesian co-ordinate system composed of the three principal axes of inertia, so that in these $$ g_{ik}=\begin{cases} 1 & (i=k),\\ 0 & (i\neq k), \end{cases}\quad\text{and}\quad T_{ik}=0\quad\text{(for } i\neq k\text{).} $$ If we then write $T_{1}$ in place of $T_{1}^{1}$, and do the same for the remaining indices, our equations in this co-ordinate system assume the simple form $$ \left(T_{i}+T_{k}\right)\frac{dv_{ik}}{dt}=\left(T_{k}-T_{i}\right)\left(v,v\right){}_{ik}. $$ These are the differential equations for the components $v_{ik}$ of the unknown angular velocity---equations which, as is known, may be solved in elliptic functions of $t$. The principal moments of inertia $T_{i}$ which occur here are connected with those, $T_{i}^{*}$, given in accordance with the usual definitions by the equations $$ T_{1}^{*}=T_{2}+T_{3},\qquad T_{2}^{*}=T_{3}+T_{1},\qquad T_{3}^{*}=T_{1}+T_{2}. $$
The above treatment of the problem of rotation may, in contra-distinction to the usual method, be transposed, word for word, from three-dimensional space to multi-dimensional spaces. This is, indeed, irrelevant in practice. On the other hand, the fact that we have freed ourselves from the limitation to a definite dimensional number and that we have formulated physical laws in such a way that the dimensional number appears accidental in them, gives us an assurance that we have succeeded fully in grasping them mathematically.
Questions
The first term on the left-hand side of \eqref{34} appears to mean
$$ \frac{dH_{ik}}{dt}w^{ik}=\frac{dv_{i}^{j}}{dt}\sum_{m}m\xi_{j}\xi_{k}w^{ik}, $$
because the position vectors are constant over time relative the coordinate system fixed in the body. Thus the components of the angular velocity tensor relative to the body's coordinate system are (in general) time-variable. Is this correct?
The statement (following \eqref{35}) that
$$ \bar{w}^{ik}\left(\bar{v}_{i}^{r}\bar{\xi}_{r}\bar{\eta}_{k}+\bar{v}_{k}^{r}\bar{\xi}_{i}\bar{\eta}_{r}\right) $$
is invariant appears to mean that the value is the same in both the inertial and body coordinates at a given instant in time, but not necessarily constant over time. Again, is this correct?
Finally, is it correct that all expressions following "we may remove the bars" are in terms of body coordinates?
