Understanding and Expressing the Definition of Inertia Tensor in the Language of Differential Geometry

I am confused about translating the definition of the inertia tensor I know into the language of differential geometry. Part of this confusion arises because in every physics textbook I read, the term "coordinates" is used in different places with different meaning (even within the same book).

Anyway, let me try to illustrate what I am familiar with first, and then try to explain my doubts. For example, the definitions I am familiar with are:

Given a (say real) vector space $$V$$, an $$(r,s)$$ tensor over $$V$$ is a multi-linear map $$T: \underbrace{V^* \times \dots \times V^*}_{\text{r times}} \times \underbrace{V \times \dots \times V}_{\text{s times}} \to \Bbb{R}$$.

and

Given a smooth manifold $$M$$, a smooth $$(r,s)$$ tensor field on $$M$$ is a smooth section $$\xi : M \to T^r_s(M)$$ of the $$(r,s)$$ tensor bundle over $$M$$. (i.e at each point $$p$$ of the manifold, we have an $$(r,s)$$ tensor $$\xi(p)$$ over the tangent space $$T_pM$$, such that the association $$p \mapsto \xi(p)$$ is smooth).

Of course, I know a few examples of tensors and tensor fields from basic linear algebra. For example, if $$V$$ is a finite-dimensional vector space, we can always equip it with an inner product (a $$(0,2)$$ tensor over $$V$$). Also, a typical example of a tensor field is a metric tensor field $$g$$ on a smooth manifold (a $$(0,2)$$ tensor field). I'm also "acquainted"(i.e I've seen them but haven't had any practice with them) with other examples of tensor fields from physics, such as the stress-energy tensor field in the context of electromagnetism.

Now, the reason I'm not so confused in these cases is because I know precisely what the spaces $$V$$ and $$M$$ are, and I know the exact definition (i.e the rule for the map). When it comes to the inertia tensor however, I'm not so sure.

Now, given a rigid body, here's the definition I know (from Landau and Lifshitz, Volume $$1$$, $$\S 32$$).

We take a "moving system of coordinates $$x_1, x_2, x_3$$, which is supposed to be rigidly fixed in the body, and to participate in its motion", and in these coordinates, we define \begin{align} I_{ij} &= \int_{\text{Body}}(\delta_{ij} \lVert x\rVert^2 - x_i x_j) \cdot \rho \, dV. \end{align} (this discussion occurs over a few pages, so I'm summarizing the essentials).

I understood their motivation for such a definition: namely by defining $$I_{ij}$$ like this, the rotational kinetic energy can be expressed as $$T_{\text{rot}} = \dfrac{1}{2}I_{ij} \Omega^i \Omega^j$$ ($$\vec{\Omega}$$ being the angular velocity of the rigid body). However, here are some things I am unclear about:

$$1$$. What does it mean in a precise and technical manner to have a system of coordinates attached to a point in the body? I understand the intuitive notion that I'm supposed to think of myself as being anchored to a point in the rigid body and "describe how I see things". But I'm having trouble precisely formulating this "simple" idea as a precise mathematical definition. I'm hoping someone can fill in something along the lines of "A coordinate system (____ on a certain space ____ )is a (_____ type of object ____). And a coordinate system attached to a point is (____ something_____)"

$$2$$. My next doubt is whether the inertia tensor is actually a tensor over a fixed vector space $$V$$ (if so which vector space? is it $$\Bbb{R}^3$$? the tangent space at a point of the rigid body? ) or whether it is actually a tensor field over a certain manifold $$M$$ (if so which one?). The reason I ask this is because I'm aware that in the physics literature, it's not uncommon to leave out the term "field", because it's usually clear from context... but unfortunately it isn't clear to me $$\ddot{\frown}$$.

$$3.$$ Following up with ($$2$$), what is the type/rank of the inertia tensor(field?) (i.e what are $$r$$ and $$s$$)? My guess is that based purely on the way it is written, it is a $$(0,2)$$ tensor (field?) based on the index structure. But I'm not sure because this is only Volume $$1$$ of Landau and Lifshitz, and from my understanding, at this point they do not make any distinction between upper vs lower placement of indices. Another reason I ask this is because people usually identify $$(2,0), (1,1), (0,2)$$ tensor (fields) all together using an inner product/ Riemannian metric tensor field. So, I'm wondering, which "type" is the most natural to begin with.

$$4.$$ Is it possible to define the inertia tensor in a manner which manifestly makes it clear that it is actually a tensor (field?). For example, if we consider $$M = \Bbb{R}^4$$ as a smooth manifold, then with the identity chart $$(\Bbb{R}^4, \text{id}_{\Bbb{R}^4})$$, where we denote its four component functions as $$\text{id}_{\Bbb{R}^4}(\cdot) = \left( t(\cdot), x^1(\cdot), x^2(\cdot), x^3(\cdot)\right)$$, we can define $$g := -dt \otimes dt + \delta_{ij}dx^i \otimes dx^j$$. Written in this manner, although we have used the component functions of the identity chart, all the operations used (exterior derivative, tensor product) etc are all clearly chart-independent and purely geometric operations. So, the result is very clearly chart-independent and is indeed easily seen to be a (symmetric) $$(0,2)$$-tensor field on the manifold $$M$$. So, my question is if we can describe the inertia tensor (field?) in a similar terminology.

So, really, my issue is one of translating between terminology in the math books I'm familiar with and the physics texts which I also read simultaneously, and one about the geomtric way of defining such objects.

Nice question you got here! Let's go through this step by step:

1) Embedding of the body

To understand the coordinates "rigidly fixed" on the rotating body, think of the object as a manifold $$B$$ embedded in space $$M=\mathbb{R}^3$$, i.e. an injective immersion $$\imath:B\rightarrow M.$$ Since we want to discribe a motion of this body, we let this mapping be dependent on time: $$\imath:T\times B\rightarrow M,$$ where $$T\subseteq \mathbb{R}$$ is a time interval. Now we can introduce those "rigidly fixed" coordinates as coordinates on the manifold $$B$$. Since the dynamics of this object are fully containt in the embedding map $$\imath$$, these coordinates are fixed on the body.

2) Moment of inertia

The moment of inertia is a tensor, but of which vector space? We will see that it is a tensor on the tangent space $$T_p M$$ at the point $$p\in M$$ around that the body rotates. More specifically, it is a mapping $$I:T_p M \otimes T_p M \rightarrow \mathbb{R}.$$ At this level, this is not a tensor field, but rather a tensor on the tangent space of a specific point in space. However, it can be made one by mapping each point $$p\in M$$ to the inertia tensor $$I_p$$ of a rotation around this point. Nevertheless, I'm not sure wheter it is of some use to consider this as a tensor field.

Now we want to connect this to the stuff above. Choose a submanifold $$B$$ of $$\mathbb{R}^3$$, then the motion of $$B$$ around the point $$p\in\mathbb{R}^3$$ and an angular velocity vector $$\omega\in T_p\mathbb{R}^3 \cong \mathbb{R}^3$$ is described by the embedding \begin{align} \imath: T\times B &\rightarrow \mathbb{R}^3 \\ (t,x) &\mapsto R[t\omega](x-p) + p. \end{align} Here, $$R[v]:\mathbb{R}^3\rightarrow\mathbb{R}^3$$ is the rotation map by angle $$|v|$$ around the axis $$v$$ through the origin of $$\mathbb{R}^3$$.

To motivate the definition of the moment of inertia, we will take a look at the kinetic energy $$E=\frac{1}{2} \int_B dV \rho(x) \langle \partial_t \imath(x),\partial_t \imath(x)\rangle$$ of the system, where $$\rho:B\rightarrow \mathbb{R^+}$$ is the mass density of the body and $$\partial_t \imath(t,x)$$ its "local velocity". Note that $$\partial_t \imath(t,x) = \omega \times (x-p)$$ with $$\bullet\times\bullet$$ the vector product on $$\mathbb{R}^3$$. Using the identity $$\langle a\times b,a\times b\rangle = |a|^2|b|^2 - \langle a,b\rangle^2$$ then gives rise to $$E=\int_B dV \rho(x) (|\omega|^2 |x-p|^2 - \langle \omega, x-p \rangle^2).$$ Motivated by this expression, we define the moment of inertia tensor as $$I_p(v,w):=\int_B dV \rho(x) (|x-p|^2 \langle v,w\rangle - \langle v,x-p \rangle \langle w,x-p \rangle)$$ and, thus, $$E=\frac{1}{2} I_p(\omega,\omega)$$. As we see, the moment of inertia is a type $$(2,0)$$ tensor on the tangent space $$T_p M$$ and, thus, depends on the point $$p$$ in space, the body is rotated around. As mentioned above, you can consider the map $$I: p\in M \mapsto I_p$$ as a tensor field, but I am not sure whether this brings any advantages. I hope this answers your question.

Cheers!

• This was very helpful, but I'm afraid I still don't fully understand point 1. So I understand that the map $\iota: T \times B \to M$ allows us to think of for each $t \in T$, $\iota[\{t\} \times B] \subset M$ as "how the rigid body $B$ is situated in the ambient space $M$ at time $t$". But I don't understand how the map $\iota$ allows us to introduce coordinates (charts?) on the manifold $B$. Also, I recently had this question on regularity: is it really appropriate to model rigid bodies as smooth manifolds? Because there are several examples where they're clearly not, like pointy cones Commented Apr 3, 2020 at 18:04
• and also cylinders (their edges are not smooth), and I'm sure these are not the only examples. Also, for the most part of your section ($2$), it seems that the only key properties of $B$ that are needed is that it is a subset of an ambient manifold $M$ (in our case $\Bbb{R}^3$), and that we can integrate over it. So it seems more fitting to model a rigid body as a Lebesgue-measurable subset of $M = \Bbb{R}^3$ (perhaps you considered the case $B$ is a smooth submanifold to simplify the discussion?). I appreciate your input on this. Commented Apr 3, 2020 at 18:09
• You don't have to restrict yourself to $B$ being a manifold, however, if you want to talk about coordinates it is an assumption making live easy. You don't need the map $\imath$ to introduce coordinates on $B$. If it is a manifold (not necessary smooth), it is possible to introduce coordinates. The point is that you don't need coordinates to define the inertia tensor. If you want to calculate $I$, you choose appropriate coordinates on $B$ and calculate the components $I_{ij}$ of its coordinate representation. Landau and Lifshitz defines the tensor by this components in a certain coordinate rep Commented Apr 4, 2020 at 11:41
• I think using a Lebesgue-messurable subset of $\mathbb{R}^n$ instead is possible, however, from a physical perspective describing a body by a topological manifold seems more reasonably to me. Commented Apr 4, 2020 at 12:03

$$\newcommand{\R}{\mathbb{R}} \newcommand{\E}{\mathcal{E}} \renewcommand{\O}{\mathrm{O}} \newcommand{\SO}{\mathrm{SO}} \newcommand{\so}{\mathfrak{so}} \renewcommand{\d}{\mathrm{d}} \renewcommand{\r}{\mathrm{r}} \newcommand{\inv}{^{-1}} \newcommand{\w}{\omega}$$

You have posed a significant and intriguing question. Providing a complete answer reveals much about the underlying mathematics of rigid body motion.

The inertia tensor is a $$\binom{0}{2}$$-tensor on the vector space of angular velocities. When applied to the angular velocity $$\omega$$ in both slots, it yields a scalar — twice the kinetic energy of a rotating body. Written in components, $$I(\omega,\omega) = I_{ij} \omega^i \omega^j = 2K$$.

The answer $$T_p\R^3$$ is not generally correct. Consider a rotating rigid body not in a 3-dimensional space, but on a 2-dimensional plane. Here, the angular velocity is a vector in a 1-dimensional vector space, and the inertia tensor (consisting of just one moment of inertia) is a tensor on this 1-dimensional space, and not on the 2-dimensional tangent space to the place. It is merely coincidental that in three dimensions, the dimension of angular velocities equals the dimension of the space itself. Thus, the vector space on which the inertia tensor is defined is not $$T_p\R^3$$.

Problem's setting. Ambient space

We will now rigorously define the problem's setting. To get to the notion of the abstract inertia tensor, not its component matrix, we need to work in a coordinate-free setting. A rigid body is embedded in an ambient space most accurately modeled as a Euclidean space of dimension $$n$$, denoted by $$\E(n)$$. It is instructive to keep the setting general and use $$n$$, but one can always plug in $$n=3$$. A Euclidean space is an affine space with a metric. For a detailed definition I refer to this Wikipedia page. In brief, an affine space is a vector space without an origin. Its elements are only points, not vectors. However, differences between points (translations) $$p-q$$ are vectors. In a Euclidean space, the vector space of translations $$V$$ is additionally equipped with an inner product $$g$$, which allows measuring distances between points as the norm of the difference vector. This associated vector space of translations $$(V,g)$$ is called a Euclidean vector space. This is why inner product spaces are often called Euclidean vector spaces.

The inertia tensor describes the inertial properties of a rigid body undergoing rotational motion, i.e., a rigid motion about a fixed point. Therefore, there exists a fixed point $$o\in\E(n)$$ about which the body rotates, or a reference point about which we want to calculate the inertia tensor should it rotate around this point.

A choice of a different reference point yields another inertia tensor. This is not merely a change of coordinates, but a different tensor itself. For example, consider a point-mass $$p$$ with mass $$m$$ on a plane ($$n=2$$). Its moment of inertia about a point $$o$$ is $$I = m\cdot||p-o||^2$$, which is non-zero if $$o\neq p$$ and zero if $$o = p$$. This is clearly not a linear transformation. Therefore, the inertia tensor is not just a property of a body, but also of the chosen reference point $$o\in\E(n)$$.

The existence of such a fixed point allows identifying the Euclidean space $$\E(n)$$ with the $$n$$-dimensional Euclidean vector space $$(V,g)$$ by subtracting the point $$o$$: \begin{align*} \E(n) &\to V\\ p &\mapsto p-o \end{align*} Under the identification, body is now embedded into an inner product space $$(V,g)$$.

Space of rotations

A rotation of the rigid body about the origin of $$V$$ is a transformation of $$V$$ that preserves the distances and the origin. It is, therefore, a linear isometry of the inner product space $$(V,g)$$, i.e., a linear map $$r\colon V \to V$$ that preserves the inner product: $$g\bigl(r(v),r(w)\bigr) = g(v,w) \quad \forall\, v,w\in V.$$ The set of all linear isometries $$r$$ of $$(V,g)$$ together with a composition of transformations forms a group — the so-called orthogonal group of the inner product space $$(V,g)$$, denoted by $$\O(V, g) := \{ \text{linear } r\colon V \to V \mid g\bigl(r(v),r(w)\bigr) = g(v,w) \quad \forall\, v,w\in V\}.$$ This group is a subgroup of the group of linear isomorphisms $$\mathrm{GL}(V)$$ of $$V$$.

Note that we have used no notion of coordinates so far. Now, let us choose an orthonormal basis $$(e_i)_{i=1}^n$$ of $$(V,g)$$ (the inner product allows to do so). In the initial Euclidean space $$\E(n)$$, this basis would form an orthonormal frame centred at $$o$$.

Answer to your 1st question: a coordinate frame (system) in $$\E(n)$$ is a tuple $$\Psi = (o, e_1, \dots, e_n)$$ consisting of a point $$o\in\E(n)$$ (the origin) and a basis $$(e_1, \dots, e_n)$$ of the associated Euclidean vector space $$V$$. It allows identifying $$\E(n)$$ with $$\R^n$$ by first subtracting $$o$$ and obtaining vectors $$p-o$$ in $$V$$ and then taking their coordinates with respect to the basis.

The components of an element $$r\in\O(V,g)$$ of the orthogonal group in an orthonormal basis form an orthogonal matrix $$R$$, hence the name of the group. The set of all orthogonal matrices, together with the matrix multiplication, forms a group: $$\O(n) := \{R \in \R^{n\times n} \mid R^\top = R\inv\}$$ — the orthogonal group in dimension $$n$$. This group consists of coordinate representations of linear isometries $$r$$ in an orthonormal basis, or alternatively, of linear isometries of $$\R^n$$ with the standard inner product.

Linear isometries and corresponding orthogonal matrices have determinants of $$\pm1$$. This includes both rotations ($$+1$$) and reflections ($$-1$$). When studying the motion of a rigid body, we are only interested in rotations. Therefore, we restrict our groups to determinants of $$+1$$ only, i.e., to rotations. The resulting group is called the special orthogonal group and is denoted b \begin{align*} \SO(V, g) &:= \{r \in \O(V, g) \mid \det r = +1\}\\ \SO(n) &:= \{R \in \O(n) \mid \det R = +1\} \end{align*} in the coordinate-free and coordinate-based cases, respectively. Elements $$r\in\SO(V, g)$$ are called rotations, and elements $$R\in\SO(n)$$ are called rotation matrices. $$\SO(V, g)$$ and $$\SO(n)$$ are isomorphic under a choice of an orthonormal basis of $$V$$.

One can equip these groups with the structure of a smooth manifold, making them Lie groups. A Lie group is a group that forms a smooth manifold, where the group operations of multiplication and inversion are smooth. For example, in 2D, $$\SO(2)$$ is a one-dimensional manifold homeomorphic to a circle. In 3D, it is 3-dimensional (which is coincidental). In general, the dimension of $$\SO(n)$$ is $$\frac{n(n-1)}{2}$$ (the number of rotational degrees of freedom).

Angular velocity

Now consider an initial configuration of the rigid body and identify it with the identity of the group $$\SO(V, g)$$. The Lie group represents the configuration space of the body rotating around a point. A continuous and smooth rotation of the body corresponds to a smooth curve $$r(t)$$ on the group parametrised by time — a trajectory on the configuration space. The velocity of the curve at some point $$r\in\SO(V, g)$$ is a tangent vector to the Lie group denoted by $$\dot{r}\in T_r\SO(V, g)$$.

In coordinates (orthonormal basis of $$V$$), $$\SO(n)$$ is used, and the configuration evolution is represented by rotation matrices $$R(t)$$ parametrised by time. The velocity of the trajectory, in this case, can be written as the time derivative of the rotation matrix $$\dot{R}$$ (this is a property of matrix Lie groups). However, this is not yet an angular velocity.

From classical mechanics, one may be familiar with the definition of the angular velocity given by a skew-symmetric matrix $$\Omega:= \dot{R}R\inv.$$ The reason it is defined this way comes from the structure of a Lie group, particularly its Lie algebra. The tangent space $$T_e G$$ at the identity $$e$$ of a Lie group $$G$$ has a special name — the Lie algebra of the Lie group. In the case of special orthogonal group, it is denoted with $$\so(V,g)$$ and $$\so(n)$$. The coordinate-based Lie algebra $$\so(n)$$ of the special orthogonal group $$\SO(n)$$ can be shown to be the set of all skew-symmetric $$n\times n$$ matrices, i.e. $$\so(n) = \{\Omega \in \R^{n\times n} \mid \Omega^\top = - \Omega \}.$$

A powerful property of Lie groups allows translating any tangent vector to a Lie group into the tangent space at the identity, i.e., into the Lie algebra. To do this, consider the right multiplication by $$h\in G$$: \begin{align*} \mathrm{r}_h\colon G &\to G\\ g &\mapsto g\cdot h, \end{align*} which is a diffeomorphism. Its differential or push-forward at $$g\in G$$ is a linear map between the tangent spaces $$\d_g \r_h \equiv (\r_h)_{*,g} \colon T_g G \to T_{g\cdot h} G.$$

Now, given a trajectory $$r(t)$$ and its velocity $$\dot{r}$$ at some point $$r$$, the push-forward of the map $$\r_{r\inv}$$ translates the velocity to the tangent space at the identity, i.e., to the Lie algebra $$\so(V,g)$$: \begin{align*} (\r_{r\inv})_{*,r} \colon T_r \SO(V,g) &\to T_e \SO(V,g) = \so(V,g) \\ \dot{r} &\mapsto (\r_{r\inv})_{*,r} \dot{r} =: \omega. \end{align*} The resulting element of $$\so(V,g)$$ is called the angular velocity of the body, and thus, $$\so(V,g)$$ is the space of angular velocities in question.

In coordinates, the same operation appears as matrix multiplication by $$R\inv$$ from the right: \begin{align*} (\r_{R\inv})_{*,R} \colon T_R \SO(n) &\to T_e \SO(n) = \so(n)\\ \dot{R} &\mapsto (\r_{R\inv})_{*,R} \dot{R} = \dot{R}R\inv =: \Omega. \end{align*}

The dimension of the Lie algebra as a vector space is the dimension of the Lie group as a manifold. $$\so(n)$$ can be canonically isomorphically identified with $$\R^{\frac{n(n-1)}{2}}$$. In dimensions 2 and 3 this is given by \begin{aligned} \so(2) &\to \R\\ \begin{pmatrix} 0 & -\w\\ \w & 0 \end{pmatrix} &\mapsto \w \end{aligned} \qquad\text{and}\qquad \begin{aligned} \so(3) &\to \R^3\\ \begin{pmatrix} 0 & -\w_z & \w_y\\ \w_z & 0 & -\w_x\\ -\w_y & \w_x & 0 \end{pmatrix} &\mapsto \begin{pmatrix} \w_x\\ \w_y\\ \w_z \end{pmatrix} \end{aligned} respectively. These are equivalent representations of the angular velocity in coordinates in matrix and column form.

The choice of an orthonormal frame in $$\E(n)$$ centred at $$o$$, which is equivalent to the choice of an orthonormal basis of $$V$$, induces a choice of a basis in the Lie algebra $$\so(V,g)$$, identifying it isomorphically with $$\so(n)$$ and $$\R^{\frac{n(n-1)}{2}}$$.

Inertia tensor

Finally, the inertia tensor $$I$$ relative to a point $$o\in\E(n)$$ is a symmetric $$\binom{0}{2}$$-tensor on the vector space $$\so(V,g)$$ — the Lie algebra of the special orthogonal group $$\SO(V,g)$$: $$I\colon \so(V,g) \times \so(V,g) \to \R.$$

If one chooses an orthonormal frame in $$\E(n)$$ centred at $$o$$, the components of the inertia tensor (with respect to the induced basis of $$\so(V,g)$$) form a symmetric $$\frac{n(n-1)}{2}\times\frac{n(n-1)}{2}$$ matrix. Alternatively, this matrix is a symmetric $$\binom{0}{2}$$-tensor on $$\so(n)$$ or $$\R^{\frac{n(n-1)}{2}}$$: $$[I_{ij}]\colon \R^{\frac{n(n-1)}{2}} \times \R^{\frac{n(n-1)}{2}} \to \R.$$ This provides answers to your 2nd and 3rd questions.

Bonus: The angular momentum is obtained using the flat map of the inertia tensor: \begin{align*} I^\flat \colon \so(V,g) &\to \so^*(V,g)\\ \w &\mapsto I^\flat(\w) := I(\w, \cdot), \end{align*} and it is an element of the dual space of the Lie algebra $$L := I^\flat (\w) \in \so^*(V,g).$$ Angular momentum can be naturally paired with angular velocity yielding a scalar: $$\langle L, \w \rangle = I(\w, \w) = 2K,$$ which illustrates that the angular momentum is a covector (an element of the dual space).

In dimensions 2 and 3, one determines the inertia tensor about a point $$o$$ of a particular rigid body with a specified mass distribution using the formulas. Let $$B$$ be a measurable subset of $$\E$$ of material points of the body, and $$\rho\colon B \to \R$$ be its mass density. For points $$p\in B$$, let us denote the difference vector by $$p-o$$, and its components in a chosen orthonormal basis $$(e_i)_{i=1}^n$$ of $$V$$ by $$x^i := (p-o)^i$$ forming a vector $$\mathbf{x}\in\R^n$$. Then in dimension 2, the only component of the inertia in the induced basis of $$\so(V,g)$$ tensor is: $$I_{11} = \int_B ||\mathbf{x}||^2 \cdot \rho \mathrm{d}V,$$ and in dimension 3, the components of the inertia tensor are given by the matrix: $$[I_{ij}] = \int_B (||\mathbf{x}||^2 \cdot \mathbb{I}_{3\times3} - \begin{pmatrix} x^1x^1 & x^1x^2 & x^1x^3\\ x^2x^1 & x^2x^2 & x^2x^3\\ x^3x^1 & x^3x^2 & x^3x^3 \end{pmatrix}) \cdot \rho \mathrm{d}V.$$ A change of the basis of $$V$$ induces a corresponding change of the basis of $$\so(V,g)$$ and yields a transformation of the components of $$I$$ (which will transform as components of a $$\binom{0}{2}$$-tensor). $$I$$ itself is an abstract, basis-independent tensor.

Possible further directions

• Euler equations of a rigid body $$\dot{L} = 0$$, principal inertia frame.
• The concept of the action of the Lie group on the configuration space. In general, Lie groups are used do define symmetries in physics, whose significance is hard to overstate.
• What is the induced basis of $$\so(V,g)$$ given a basis of $$V$$? Transformation of the inertia tensor under a rotation of the coordinate frame and adjoint representation $$\mathrm{Ad}$$ of a Lie group on its Lie algebra.
• From the Lie algebra $$\so(V,g)$$, the inertia tensor can be extended to a left- or right-invariant tensor field on the entire Lie group $$\SO(V,g)$$, establishing a metric on the configuration space. (It contains the same information as just $$I$$.) This determines the dynamics of a rotating rigid body: the time evolution of the configuration is a geodesic on $$\SO(V,g)$$ with respect to this metric.
• Why the right translation to the Lie algebra in this case in more natural than the left translation? This can be understood by eliminating the reference configuration and considering an abstract configuration space on which $$\SO(V,g)$$ acts from the left freely and transitively.
• To eliminate dependence on the reference point, consider the group of all rigid transformations of $$\E(n)$$ (rotations and translations) — the so-called, special Euclidean group $$\mathrm{SE}(\E)$$. A choice of an orthonormal coordinate frame in $$\E(n)$$ (including a choice of the origin) induces its coordinate representation — $$\mathrm{SE}(n)$$ consisting of homogeneous matrices. This is the configurations space of a rigid body under undergoing general motion in space. Then the Lie algebra $$\mathfrak{se}(\E)$$ is the vector space of generalised velocities (angular and translational) (also called twists). For $$n=3$$, its dimension is 6. Mass distribution and the total mass of the body determine the full inertia tensor, which is now a symmetric $$\binom{0}{2}$$-tensor on $$\mathfrak{se}(\E)$$. Applied to a generalised velocity, it produces a generalised momentum (angular and translational). In components in dimension 3, it is a $$6\times6$$ matrix.
• Since writing this question ~4 years ago I had figured most of this stuff out along the way but thanks for writing it up regardless. There are some small typos/slips of the tongue (e.g you say right in the beginning of the section on rotations that they are linear isometries, but in fact are orthogonal transformations, but ok later on the usage seems to be consistent) Commented Aug 25 at 18:40