Higher dimensional relation between angular momentum, moment of intertia and angular velocity In 3 dimensions we have the well known relation (summation convention is being used)
$$
L_i = I_{ij} \omega_j
$$
However, as is well known the angular momentum and angular velocity are not vectors but tensors and their Hodge duals are what are used in the above expression. So using the actual 2-forms 
$$
\tilde L_{ij} = \epsilon_{ijk} L_k~,
$$
and likewise for $\tilde \omega$, we get the above relation as
$$
\tilde L_{ij} = \frac{1}{2} ~I_{kl} ~\epsilon_{kij}~\epsilon_{lmn} ~\tilde \omega_{mn}
$$
My question is how do I generalize this to higher dimensions? The angular momentum, moment of inertia and angular velocity will remain second order tensors. However, in higher dimensions the Levi-Civita tensor will be of higher order and I cannot seem to find an unambiguous way to get a $3^rd$ order Levi-Civita tensor from a higher order one.
 A: It turns out the answer is rather simpler in higher dimensions. It gets complicated when specializing to 3-dimensions.
If a rigid body has angular velocity $\omega_{ij}$ then given the location of a point $r_i$ its velocity is given by
$$
v_i = -\omega_{ij} r_j 
$$
and its angular momentum is given by
$$
L_{ij} =\sum m(r) ( -r_i v_j +r_j v_i )\\
= \sum m(r)( r_i r_m \omega_{mj} -r_j r_m \omega_{mi}) \\
$$
Specializing to 3D we get
$$
\tilde L_k = \frac{1}{2} \epsilon_{kij} L_{ij}\\
=\epsilon_{mjt}\epsilon_{ijk} [\sum m(r) r_i r_m] \tilde \omega_t \\
= -\delta^{tm}_{ik} [\sum m(r) r_i r_m ] \tilde \omega_t \\
= [\sum m(r) (-r_k r_i +r^2 \delta_{ki})] \tilde \omega_i \\
= I_{ki} \tilde \omega_i
$$
Thus we see that in arbitrary dimensions the relation between angular momenta, moment of inertia and angular velocity is given by
$$
L_{ij} =  2 \tilde I_{im} \omega_{mj}
$$
where $\tilde I_{ij} = \sum m(r) r_i r_j$.
A: 3D common way
Definition of angular momentum vector ($m = 1$) 
\begin{equation}
L_l = \epsilon_{lij}x^iv^j
\end{equation}
($L_l$ dual to $L^{ij} = x^iv^j - x^jv^i$ tensor)
Rortational velocity of particle 
\begin{equation}
 v^j = \epsilon^{jrk}\omega_rx_k.
\end{equation}
($\omega_r$ dual to $\omega^{jk}$ antisymmetric tensor of angular velocity $ \epsilon^{jrk}\omega_r = \omega^{jk}$) 
Substitute in the angular momentum definition
\begin{equation}
 L_l = \epsilon_{lij}x^i\epsilon^{jrk}\omega_rx_k.
\end{equation}
Let us use the property of the Levi-Civita tensor:
\begin{equation}
 \epsilon^{lij} = -\epsilon^{jli}
\end{equation}
then
\begin{equation}
 L_l = - \epsilon_{jli}\epsilon^{jkr}x^ix_k\omega_r.
\end{equation}
Let us use another property of the Levi-Civita tensor:
\begin{equation}
 \epsilon_{jli}\epsilon^{jkr} = \delta_l^k\delta_i^r - \delta_l^r\delta_i^k.
\end{equation}
\begin{equation}
 L_l = \left( \delta_l^r\delta_i^k - \delta_l^k\delta_i^r \right)  x^ix_k\omega_r.
\end{equation}
Expand the brackets and take into account that $\delta_i^r x^i = x^r$, $\delta_l^k x_k = x_l$ and $\delta_i^k x^i = x^k$, we get
\begin{equation}
 L_l = \left(  \delta_l^r x^kx_k - x^rx_l\right) \omega_r,
\end{equation}
or
\begin{equation}
 L_l = \left(  \delta_l^r x^2 - x^rx_l\right) \omega_r,
\end{equation}
where inertia tensor
\begin{equation}
 I_l^r = \delta_l^r x^2 - x^rx_l
\end{equation}
Arbitrary number of dimensions
For Higher dimensions we can use directly angular momentum tensor 
\begin{equation}
 L^{ij} = x^iv^j - x^jv^i
\end{equation}
and
Rortational velocity of particle we can express via angular momentum tensor 
\begin{equation}
 v^j = \omega^{jm}x_m.
\end{equation}
Then
\begin{equation}
 L^{ij} = x^ix_m\omega^{jm} - x^jx_m\omega^{im}
\end{equation}
Now we can lower indices near $\omega$ with metric tensor:
\begin{align}
 \omega^{jm} = g^{jn}g^{mr}\omega_{nr} \\
 \omega^{im} = g^{in}g^{mr}\omega_{nr} 
\end{align}
So, we get
\begin{equation}
 L^{ij} = \left( x^ix_mg^{jn}g^{mr} - x^jx_m g^{in}g^{mr}\right) \omega_{nr}
\end{equation}
So, we can conclude $L^{ij} = I^{ijnr} \omega_{nr}$, the inertia tensor is
\begin{equation}
 I^{ijnr} = x^ix_mg^{jn}g^{mr} - x^jx_m g^{in}g^{mr}
\end{equation}
A: According to https://arxiv.org/abs/2207.03560, the appropriate generalization is
$$L_{ij} = \frac{1}{2} I_{ijkl} \omega_{kl},$$
where $L$ is the angular momentum two-form (which in three dimensions is the Hodge dual to the angular momentum vector), $\omega$ is the angular velocity two-form (likewise), and $I$ is the rank-four tensor of inertia defined by
$$I_{ijkl} := 4 \int dm \left( x_{[j} \delta_{i][k} x_{l]} \right).$$
Interestingly, the tensor of inertia has the same symmetries and number of degrees of freedom as the Riemann curvature tensor. (This paper (PDF) tries to explain the reason behind this surprising parallel, but it uses some obscure math involving the geometric algebra, and I personally don't find it particularly illuminating.)
