# Relation between "method of moving frames", spin connection, Cartan forms, and classic rotational kinematics in $\mathbb{E}^n$

I want to know how the "method of moving frames" involving things like connection 1-forms, torsion 2-forms, spin connections, etc. are applied to basic rotational kinematics in flat 3-space (or, $$n$$-space more generally). In particular, how would the angular velocity of a rigid rotating basis be defined and used in the context of differential geometry? I'll give an example of the classic treatment for a rotating basis using basic tensor algebra on the vector space $$\mathbb{E}^3$$:

Classic Rotational Kinematics: Let $$\pmb{\iota}_i\in\mathbb{E}^3$$ (or replace $$3$$ with $$n$$) be some fixed/inertial basis with dual basis $$\pmb{\iota}^i\in\mathbb{E}^{3*}$$. Let $$\pmb{e}_i(t)\in\mathbb{E}^3$$ be some rigid rotating basis with dual basis $$\pmb{e}^i(t)\in\mathbb{E}^{3*}$$. At any time, they are related by a linear transformation:

$$\pmb{e}_i = \pmb{M}\cdot \pmb{\iota}_i = M^j_i \pmb{\iota}_j \quad,\qquad \pmb{e}^i = \pmb{\iota}^i \cdot \pmb{W} = W^i_j \pmb{\iota}^j \quad,\qquad \pmb{M} = \pmb{e}_i\otimes \pmb{\iota}^i \quad,\qquad \pmb{W} = \pmb{\iota}_i\otimes\pmb{e}^i \quad,\qquad M^i_k W^k_j = \delta^i_j$$

where $$\pmb{M}(t)$$, $$\pmb{W}(t)=\pmb{M}^{-1} (t)$$, $$\pmb{e}_{i}(t)$$, and $$\pmb{e}^{i}(t)$$ are some curves in their respective spaces. Further assume that $$\pmb{M}\in SO(\mathbb{E}^3,\pmb{g} )$$ such that the metric components are the same in both bases, $$\pmb{g} ( \pmb{e} _{i} ,\pmb{e} _{j} )= \pmb{g} ( \pmb{\iota} _{i} , \pmb{\iota} _{j} ) = g _{ij}$$, and are constants. We can then define an angular velocity bivector,

$$\pmb{\Omega} := \dot{\pmb{M}} \cdot\pmb{g}^{-1} \cdot \pmb{M}^{\top} = g^{ij} \dot{\pmb{e}}_i\otimes\pmb{e}_j = \tfrac{1}{2}g^{ij} \dot{\pmb{e}}_i\wedge\pmb{e}_j \in \wedge^2 \mathbb{E}^{3} \quad,\qquad \dot{\pmb{e}}_i = {g} _{ij} \pmb{\Omega} \cdot \pmb{e}^j$$

(for $$n=3$$, $$\pmb{\Omega}\cdot\pmb{v}=\pmb{\omega}\times \pmb{v}$$) where $$\pmb{g}^{-1}$$ is the metric on $$\mathbb{E}^{3*}$$. And the angular velocity 2-form,

$$\pmb{\Lambda} := \dot{\pmb{W}}^{\top} \cdot\pmb{g} \cdot \pmb{W} = g_{ij} \dot{\pmb{e}}^i\otimes\pmb{e}^j = \tfrac{1}{2}g_{ij} \dot{\pmb{e}}^i\wedge\pmb{e}^j \in \wedge^2 \mathbb{E}^{3*} \quad,\qquad \dot{\pmb{e}}^i = {g}^{ij} \pmb{\Lambda} \cdot \pmb{e}_j$$

For some curve with displacement vector (relative to some fixed origin), $$\pmb{r}_t= \tilde{r}^i(t)\pmb{\iota}_i = r^i(t) \pmb{e}_i(t) \in\mathbb{E}^3$$, the usual kinematic transport equation for the time-derivative of $$\pmb{r}_t$$ expressed in the rotating basis is then given by (suppressing the argument $$t$$)

$$\dot{\pmb{r}} = \dot{\tilde{r}}^i\pmb{\iota}_i = (\dot{r}^i + r^k g _{kj}\Omega^{ij}) \pmb{e}_i = (\dot{r}^i + r^j\Omega^{i}_j )\pmb{e}_i \in T _{r}\mathbb{E}^3 \qquad,\qquad \dot{\pmb{\rho}} = \dot{\tilde{\rho}}_i\pmb{\iota}^i = ( \dot{\rho} _{i} + {\rho}_k g ^{kj} \Lambda _{ij}) \pmb{e}^i = ( \dot{\rho} _{i} + \rho_j \Lambda^j_i )\pmb{e}^i \in T _{\rho}\mathbb{E}^{3*}$$

where $$r^i$$ and $$\Omega^{ij}$$ are the $$\pmb{e}_i$$-components, and where I've given the equivalent expression for some $$\pmb{\rho}_t\in\mathbb{E}^{3*}$$. I've also used the metric to raise/lower indices in the usual way as $$\Omega^i_j:=g _{jk}\Omega^{ik}$$. Now, if $$\pmb{u} = \tilde{u}^i\pmb{\iota}_i =u^i\pmb{e}_i\in \mathfrak{X}(\mathbb{E}^3)$$ is some vector field then it's time-derivative along the curve is given by

$$\dot{\pmb{u}} = \frac{\partial \tilde{u}^i}{\partial \tilde{r}^j} \dot{\tilde{r}}^j \pmb{\iota}_i = ( \dot{r}^j \frac{\partial u^i}{\partial r^j} + u^j\Omega^{i}_j )\pmb{e}_i \qquad (\text{along } \pmb{r}_t )$$

which has some minor resemblance to the covariant derivative, $${\nabla}_{\dot{\pmb{r}}} \pmb{u}$$, expressed in a coordinate basis.

My Question: If I now treat $$(\mathbb{E}^3,\pmb{g})$$ as a Riemannian manifold, how would I carry out the above ideas — defining an angular velocity tensor for a rotating frame and obtaining the "kinematic transport'' equations for derivatives expressed in that frame — using things like the Levi-Civita connection, connection 1-forms, Cartan's structural equations, etc. Is the concept of a "spin connection" relevant here? I'm familiar with differential geometry and tensor analysis on manifolds, but mostly in the context of symplectic geometry and Hamiltonian mechanics; less so with Riemannian geometry.

First, I’ll preface by saying that once you know what a connection is, you can introduce a local frame, and from that introduce the local connection 1-forms, and that tells you everything you need to know about the frames and how they change from point to point; they’re essentially the angular velocity (in some cases these connection 1-forms are also called the Ricci rotation coefficients). Next, spin connection is a connection on a spinor bundle, i.e not what we’re talking about here. Next, For the situation you have in mind, curvature and torsion forms vanish since you’re dealing with a vector space ($$\Bbb{R}^3$$ or $$\Bbb{R}^n$$) and not a vector bundle. Also, Cartan’s structure equations are of no significance here since you’re studying a curve essentially (1-dimensional), whereas the structure equations talk about 2-forms, so when you pull them back, you get zero. Btw, one thing about Cartan’s formalism is that it is so powerful and quick that the real difficulty is actually recognizing when you have a noteworthy result/quantity worth defining.

• a local frame for a vector space, and the associated connection 1-forms
• $$\mathbf{\Omega}$$ the angular velocity.
• the equation you give for $$\dot{\mathbf{r}}$$
• the equation you give for $$\dot{\mathbf{u}}$$.

The first is the only thing which needs some elaboration; the other three bullet points follow immediately! I’ll briefly finish off with some generalities.

But first, let us get ahead of ourselves and see what E.Cartan himself has to say: in his book Riemannian geometry in an orthogonal frame (free preview available, bottom of page 4):

Thus there are only three independent $$1$$-forms among the forms $$\omega^j_i$$, namely the forms $$\omega^3_2,\omega^1_3$$, and $$\omega^1_2$$. These 1-forms can be called the components of rotation of the trihedron.

1. Local frame for a vector space, and the connection 1-forms.

Let $$V$$ be an $$n$$-dimensional real vector space (no inner products necessary yet). A smooth local frame, also called a moving frame on (an open subset of) $$V$$ is by definition a collection $$\{e_1,\dots, e_n\}$$ of smooth maps from an open subset $$U\subset V$$ into $$V$$ such that for each $$x\in U$$, $$\{e_1(x),\dots, e_n(x)\}$$ is a basis for the vector space $$V$$.

Now, since each $$e_i$$ is a function $$U\to V$$, we can take its exterior derivative $$d e_i$$ to get a $$V$$-valued $$1$$-form on $$U$$ (this is nothing but the usual Frechet derivative, or in less frightening terminology, it’s the total differential of $$e_i$$). Note that the reason we can do this is because our target space $$V$$ is a vector space (if we had a vector bundle as the target, then we’d need a connection to compare vectors in different spaces). Now, there exist unique 1-forms $$\omega^i_{\,j}$$ such that \begin{align} de_j&=\omega^i_{\,j}\,e_i.\tag{*} \end{align} The equation $$(*)$$ is very abridged notation, so in case you don’t fully understand it, here’s the meaning: $$d e_j$$ is a $$V$$-valued $$1$$-form on $$U$$. This means for any vector tangent to $$U$$, it outputs an element of $$V$$. So for any point $$x\in U$$ and any tangent vector $$v_x\in T_xU\cong V$$ (i.e just think of an element $$v\in V$$ as “being attached at the point $$x$$”), we have an output $$(de_j)[v_x]\in V$$. As a side note, just to convince you this is nothing fancy, it is exactly the Frechet derivative $$D(e_j)_x[v]$$. If that is also too daunting, then it is exactly the directional derivative $$\frac{d}{ds}\bigg|_{s=0}e_j(x+sv)$$, of the mapping $$e_j$$ at the point $$x$$ along the vector $$v$$.

Now, every element of $$V$$ can be written as a linear combination of basis vectors, so for the basis $$\{e_1(x),\dots, e_n(x)\}$$ associated with the point $$x$$, we have that for each $$j\in\{1,\dots, n\}$$, the vector $$de_j[v_x]$$ is a linear combination of $$\{e_1(x),\dots, e_n(x)\}$$. So, there exist unique numbers $$\omega^1_{\,j}[v_x],\dots, \omega^n_{\,j}[v_x]\in\Bbb{R}$$ such that \begin{align} (de_j)[v_x]&=\sum_{i=1}^n\omega^i_{\,j}[v_x]\,e_i(x).\tag{**} \end{align} This is now a proper equality of elements of $$V$$: the LHS is an element of $$V$$, and the RHS is a linear combination of basis vectors of $$V$$. Next, since $$de_j$$ depends linearly on $$v_x$$, it follows that $$\omega^i_{\,j}$$ does as well. So look what we just defined: $$\omega^{i}_{\,j}$$ is an object such that for each tangent vector to $$U$$, it assigns a number, and it does so in a linear fashion. This is exactly what a 1-form on $$U$$ is. So, ($$*$$) is just the statement that $$(**)$$ holds for all $$v_x$$. Let’s organize all of this into a definition

Definition.

Let $$V$$ be an $$n$$-dimensional real vector space, $$\{e_1,\dots, e_n\}$$ a smooth local frame for $$V$$ over an open set $$U$$. Then, there is a unique collection, $$\{\omega^i_{\, j}\}_{i,j=1}^n$$, of $$1$$-forms on $$U$$ such that, as an equality of $$V$$-valued $$1$$-forms on $$U$$, we have \begin{align} de_j&=\omega^i_{\,j}\cdot e_i. \end{align} The $$1$$-forms $$\{\omega^i_{\,j}\}_{i,j=1}^n$$ are called the connection 1-forms on $$U$$ (for the standard flat connection on $$V$$) with respect to the local frame $$\{e_1,\dots, e_n\}$$.

Now, if you want to interpret what $$\omega^i_{\,j}$$ is telling us, then from the above equations, we see that it tells us exactly that if we start at a point $$x$$ and move in the direction of a tangent vector $$v_x$$, then the resulting first-order change, $$de_j[v_x]$$, in $$e_i$$, has a component of $$\omega^i_{\,j}[v_x]$$ along the direction $$e_i(x)$$. More succinctly, it tells us how much of the change in $$e_j$$ is occurring along the direction $$e_i$$.

The nice thing about these 1-forms is that they express the changes $$de_j$$ in $$e_j$$ back in terms of the $$e_i$$ themselves! If you think about it, that’s exactly what you’re doing with your rotational mechanics equations as well. Now, in Riemannian geometry, you may have seen the Christoffel symbols, defined with respect to say a frame as the unique functions (not $$1$$-forms) such that $$\nabla_{e_j}(e_k)=\Gamma^i_{jk}\,e_i$$ (and perhaps you may have even specialized to coordinate-induced frames). And here as well, the purpose of the $$\Gamma$$’s is to express the changes in the individual vectors in the frame using the vectors in the frame. But, the key difference is that the $$\omega^{i}_{\,j}$$ have the 1-form nature, meaning we don’t yet plug in the vector field in the lower slot of $$\nabla_{(\cdot)}$$. This may seem like an extra level of abstraction; after all, why leave things “incomplete and unplugged”? Well, working with differential forms is just nicer, and Cartan’s structure equations take a pretty simple format (but Cartan’s equations aren’t important for your purposes, so I won’t elaborate more).

Keeping notation as above, let us now suppose that we have an in addition a smooth curve $$\gamma:I\subset\Bbb{R}\to U$$, and let us define $$\xi_i=\gamma^*e_i=e_i\circ\gamma$$. You can think of $$\gamma(t)\in U$$ as the position of a particle at time $$t$$, and you can think of $$\{\xi_1(t),\dots, \xi_n(t)\}=\{e_1(\gamma(t)),\dots, e_n(\gamma(t))\}$$ as a basis for $$V$$ “carried by the particle at $$\gamma(t)$$”.

Now, we’re interested in how the basis vectors $$\xi_j$$ change with time. Well, we don’t have to do much, we just take equation $$(*)$$, and pullback both sides of the equation using $$\gamma$$, and use that exterior derivatives commute with pullback (essentially the chain rule) \begin{align} d\xi_j&=d(\gamma^*e_j)=\gamma^*(de_j)=\gamma^*(\omega^i_{\,j}\,e_i)=\gamma^*(\omega^i_{\,j})\,\gamma^*e_i=\gamma^*(\omega^i_{\,j})\cdot\xi_i. \end{align} Both sides of this equation are 1-forms on the interval $$I$$ with values in $$V$$. To get something more recognizable, just evaluate both sides at a point $$t\in I$$ on the “unit” tangent vector $$1_t\in T_tI\cong \Bbb{R}$$. This gives \begin{align} \dot{\xi_j}(t)&=(\gamma^*\omega^i_{\,j})[1_t]\cdot\xi_i(t)=\omega^i_{\,j}[\dot{\gamma}(t)]\cdot\xi_i(t).\tag{***} \end{align} With some practice, you can skip these intermediate steps and just jump from $$(*)$$ to $$d\xi_j=\gamma^*(\omega^i_{\,j})\,\xi_i$$. Also, note that in traditional notation, one doesn’t introduce a new letter $$\gamma$$ for the curve, and one doesn’t introduce the notation $$\xi_j=\gamma^*e_j$$, and one does not explicitly write the pullbacks. So, people just say things like “as we move along a curve, we can divide $$(*)$$ by the time interval $$dt$$ to get $$\frac{de_j(t)}{dt}=\frac{\omega^i_{\,j}(t)}{dt}e_i(t)$$”.

With all this in mind:

• The angular velocity at time $$t$$ of the frame $$\{e_1,\dots, e_n\}$$ along the curve $$\gamma$$ is described by the numbers $$\omega^i_{\,j}[\dot{\gamma}(t)]$$. They’re telling you (equation $$(***)$$) how the frame $$\xi_j$$ is changing relative to itself. I should now remark that your equation $$\pmb{\Omega} := \dot{\pmb{M}} \cdot\pmb{g}^{-1} \cdot \pmb{M}^{\top}$$ seems wrong; it should be $$\dot{M}M^{-1}$$, and for orthogonal maps $$M$$, we have $$M^{-1}=M^*$$ where $$M^*$$ is the adjoint endomorphism with respect to the inner product $$g$$; in matrix language you’re missing an extra multiplication by $$\pmb{g}$$ on the right.
• The equation you got for $$\mathbf{\dot{r}}$$ can be interpreted as follows: let $$\mathbf{r}$$ be the identity function on $$V$$. Then, at points of $$x\in U$$, $$\mathbf{r}(x)=x\in V$$ so it can be written as a linear combination of the basis vectors, say $$\mathbf{r}(x)=r^i(x)e_i(x)$$, for some unique smooth functions $$r^1,\dots, r^n:U\to\Bbb{R}$$. Now, we can take the exterior derivative of both sides of the equation and use $$(*)$$ to get \begin{align} d\mathbf{r}&=dr^i\,e_i+r^i\,de_i=dr^i\,e_i+r^j\omega^i_{\, j}\,e_i=(dr^i+\omega^i_{\,j}r^j)e_i. \end{align} Now, pulling back both sides along $$\gamma$$ gives \begin{align} \dot{\gamma}(t)=\left((r^i\circ \gamma)’(t)+\omega^i_{\,j}[\dot{\gamma}(t)]\,(r^j\circ\gamma)(t)\right)\xi_i(t), \end{align} or in classical notation, $$\dot{\mathbf{r}}(t)=\left(\dot{r}^i(t)+\omega^i_{\,j}(t)r^j(t)\right)e_i(t)$$. Now, let me warn you that we are only able to talk about $$\mathbf{r}$$, the identity function, and treat it also as a “position vector” because the target space $$V$$ is a vector space. In more general situations, we do not have this. So, we can’t speak of the component functions $$r^i$$. But, what we can still do is interpret $$d\mathbf{r}$$ accordingly; namely it is simply the identity map on the tangent bundle of the manifold, $$\mathbf{1}_{TM}$$. Afterall, notice what $$d\mathbf{r}$$ is really trying to do: it takes in a tangent vector $$k_x$$ and spits out the same thing $$v_x$$.
• For the equation involving $$\mathbf{\dot{u}}$$, it’s essentially the same idea. Let $$F$$ be any vector field on $$U$$. Then, we can write $$F=F^ie_i$$, and so taking the exterior derivative and using $$(*)$$, \begin{align} dF&=dF^i\,e_i+F^jde_j=dF^i\,e_i+F^j\omega^i_{\,j}e_i=(dF^i+\omega^i_{\,j}F^j)e_i, \end{align} so once again pulling back along $$\gamma$$, we get \begin{align} (F\circ\gamma)’(t)&=\left((F^i\circ\gamma)’(t)+\omega^i_{\,j}[\dot{\gamma}(t)]\cdot (F^j\circ\gamma)(t)\right)\cdot\xi_i(t), \end{align} or in classical notation, $$\dot{F}=\left(\dot{F}^i+\omega^i_{\,j}(t)F^j\right)e_i$$. (If we let $$\nabla$$ be the standard flat connection on $$V$$, then the LHS is also equal to $$\nabla_{\dot{\gamma}(t)}F$$).

One thing you may have noticed is that throughout all my presentation, I did not use any inner products. If we now introduce an (pseudo) inner product $$g$$ on $$V$$, then you can “lower the index“ of the connection 1-forms, to define new $$1$$-forms $$\omega_{ij}=g_{ia}\omega^a_{\,j}$$. These lower-indexed connection 1-forms satisfy $$dg_{ij}=\omega_{ij}+\omega_{ji}$$, so in particular if the moving frame $$\{e_1,\dots, e_n\}$$ is orthonormal (or merely has constant components $$g_{ij}$$) then $$\omega_{ij}+\omega_{ji}=0$$. If we use a positive-definite inner product then for an orthonormal frame, we have that $$\omega^i_{\,j}=\omega_{ij}$$, so $$[\omega^{i}_{\,j}]$$ constitutes a skew-symmetric matrix of $$1$$-forms. In 3-dimensions, this allows us to identify this with an element of $$\Bbb{R}^3$$, the “angular velocity vector”.

3. Some generalities.

Now, say we have a vector bundle $$(E,\pi,M)$$ rather than a single vector space $$V$$. Suppose also we have a connection $$\nabla$$ on $$E$$. Then, we can mimic almost everything above, by replacing exterior derivatives with covariant exterior derivatives (which in the case of a section $$\psi$$ of $$E$$, simply means $$d_{\nabla}\psi[\cdot]=\nabla_{(\cdot)}\psi$$). We can introduce a local frame for the vector bundle $$\{e_1,\dots, e_n\}$$, and the associated connection $$1$$-forms $$d_{\nabla}e_j=\omega^i_{\,j}e_j$$. We can consider a local section $$F:U\to E$$ and expand it as $$F=F^ie_i$$, and write almost analogously to above, \begin{align} d_{\nabla}F&=(dF^i+\omega^i_{\,j}F^j)\,e_i. \end{align} Now we can again pullback along a curve $$\gamma:I\to M$$ and so on. If we want, we can introduce a bundle metric $$g$$ on $$E$$ (compatible with the above connection for convenience), and consider local orthonormal frames.

Like I said above, the only thing we can’t do is talk about a “position vector $$\mathbf{r}$$”, and also in this much generality, we can’t talk about $$d\mathbf{r}$$ either. In the (very important) special case that $$E=TM$$ is the tangent bundle, then we can talk about the identity map $$\mathbf{1}_{TM}$$, and this is a $$TM$$-valued $$1$$-form on $$M$$ (and taking the covariant exterior derivative we get the torsion of the connection $$\tau=d_{\nabla}\mathbf{1}_{TM}$$, but like I said, this is a 2-form (values in $$TM$$), so when pulling back along a curve, it becomes zero, so it’s not important).

One subtlety which I’ll admit to now is that throughout I’ve assumes that the frames are defined on an open set, not just along the curve. If you want to be really proper, you can consider “vector fields and sections along $$\gamma$$”, or more abstractly, work with the pullback bundle $$\gamma^*E$$.

Finally, you can go even more general and talk about principal bundles (here $$G=\text{SO}(V,g)$$ over a base manifold $$B$$ (taken to be $$V$$ in your example), or more generally, consider the bundle of tangent frames to a manifold $$M$$, or even restrict to orthonormal frame bundle). Briefly, the (right (not left, as you’ll see in many sources)) Maurer-Cartan form of a Lie group is $$\omega_R=dg\cdot g^{-1}$$. If you pullback along a curve, you get $$\dot{g}\cdot g^{-1}$$; compare this with the expression $$\dot{M}\cdot M^{-1}$$ that you have (and for the orthogonal group of an inner product space $$M^{-1}=M^*$$ is the adjoint (i.e transpose if in $$\Bbb{R}^n$$ with standard inner product)). So, clearly these ideas are all baked in here.

Summary.

Everything is encoded in the connection 1-forms $$\omega^i_{\,j}$$ associated to a local frame. I could have simply written down the equation $$(*)$$ (or its covariant derivative analogue $$d_{\nabla}e_j=\omega^i_{\,j}e_i$$) and called it a day. With Cartan’s formalism, everything drops out so rapidly that the real difficulty is in recognizing when something important arises.

A final obligatory remark: we’ve barely scratched the surface with Cartan’s method of moving frames. In fact one could even say that we haven’t even started, because Cartan’s approach is to not focus so much on the frames $$e_i$$ themselves, but to focus on the connection 1-forms, and also the dual coframe $$\{\epsilon^1,\dots,\epsilon^n\}$$ to the $$e$$’s. That’s where Cartan’s structural equations come from.

• This is incredibly helpful. I need to re-read it a time or 2 to make sure I get all of the details. You're correct that my angular velocity is better written as $\pmb{\Omega}=\dot{\pmb{M}}\cdot\pmb{M}^{-1}$. I'm confused about how we get from the $\pmb{\omega}^i_j\in\Omega^1(V)$ treated as (0,1)-tensor fields/1-forms to being treated as the tensor \textit{components} (functions), $\omega^i_j\in\mathcal{F}(V)$, of the (1,1)-tensor field $\pmb{\Omega}=\dot{\pmb{M}}\cdot\pmb{W}= \omega^i_j \pmb{e}_i\otimes\pmb{e}^j$. [continued...] Jun 26 at 22:08
• @JPeterson We have for each $t\in I$ and each $j\in\{1,\dots, n\}$, that $\dot{\xi}_j(t)=\bigg(\dot{M}(t)\circ M^{-1}(t)\bigg)(\xi_j(t))$ (evaluation of an endomorphism on a vector). Also, by my $(***)$, we have $\dot{\xi}_j(t)=\sum_{i=1}^n\omega^i_{\,j}[\dot{\gamma}(t)]\cdot \xi_i(t)$. So, equating these two, you see immediately by definition that the matrix representation of the endomorphism $\dot{M}(t)\circ M^{-1}(t)$ with respect to the basis $\{\xi_1(t),\dots,\xi_n(t)\}$ is the $n\times n$ real matrix whose $(i,j)$ entry is the number $\omega^i_{\,j}[\dot{\gamma}(t)]$. Jun 26 at 23:46
• Or, what amounts to the same thing, $(dM)\circ M^{-1}$ gives an $\text{End}(E)$-valued $1$-form on the interval $I$, and its matrix representation with respect to the $\xi_i$’s is $[\gamma^*(\omega^i_{\, j})]$. By plugging in $1_t\in T_tI\cong\Bbb{R}$ on both sides, you get the equality I mentioned in my previous comment. Of course, in this very particular case, considering 1-forms on $I$ is not a very illuminating thing to do. But, if your domain was a more complicated space, then using the differential forms notation is very compact and efficient (compared to loads of partial derivatives). Jun 26 at 23:53
• sorry, $\text{End}(V)$, where $V$ is the vector space we’ve been working with throughout. Jun 27 at 3:29
• btw, you may be interested to read this sort-of-related MSE post. Granted, there OP was asking about more math-stuff about vector-valued forms and structural equations (not really relevant here, but important nonetheless), but perhaps some stuff could be interesting (now or in the future). Also, see the referenced book by Kai Lam (it has some pretty interesting stuff, and turns out explains a bunch of what I said here as well). Also, if you’re merely studying for fun, then certainly read Cartan’s works (Elie and Henri’s). Jun 27 at 3:33