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It seems to be that in the context of rigid body dynamics, the moment of inertia is introduced as the quantity that maps the components of the angular velocity into the components of angular momentum.

Well the angular velocity is a generalised velocity, therefore it ought to be an element of the tangent space at some point on the manifold and the angular momentum lives in the cotangent space whose components are defined as gradient components of the Lagrangian i.e.

$$ p_i = \dfrac{\partial\mathcal{L}(q,\dot{q},t)}{\partial\dot{q}^i}$$

Here $p_i$ are the components of the angular momentum and $q^i$ are components of angular velocity. Now is it the case that we assume that these components are related by linear transformation, namely the moment of inertia? If so then it's not at all obvious to me why $L$ ought to depend linearly with $\omega$. Moreover even if it did, how exactly does the moment of inertia qualify as a $(1,1)$ tensor? It seems to be mapping a vector to a covector.

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  1. Moment of inertia is for rotational motion what (inertial) mass is for linear motion. In other words, moment of inertia is a generalized mass.

  2. Let the $n$-dimensional configuration space $M$ have generalized positions $(q^1, \ldots, q^n)$. In point mechanics (as opposed to field theory) the generalized masses are by definition the structure coefficients in the kinetic term $$\frac{1}{2}m_{ij}~\dot{q}^i\dot{q}^j$$ that is quadratic in the velocities $\dot{q}^i$ in the Lagrangian $L$. The generalized mass tensor
    $$m~=~ m_{ij} ~\mathrm{d}q^i \odot \mathrm{d}q^j~\in~ \Gamma\left( {\rm Sym}^2(T^{\ast}M)\right)$$ is a section in the symmetrized tensor product $${\rm Sym}^2(T^{\ast}M)~=~T^{\ast}M\odot T^{\ast}M$$ over the cotangent bundle $T^{\ast}M$. In other words, $m$ is a symmetric $(0,2)$ covariant tensor field. If it is positive definite, it is a metric tensor on the configuration space $M$.

  3. Up until now $m_{ij}$ could in principle depend on the generalized positions $q^i$. If we additionally demand (as one usually does) that $m_{ij}$ is independent of the generalized positions, then we can only allow affine coordinate transformations in the tensor construction.

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