What kind of object is the (second) exterior derivative of a moving point $R$ in $\mathbb R^n$? What does $dR\wedge\omega \mathbf e$ mean?

Question

TL;DR: Given $f:\mathbb R \to \mathbb R^n$ smooth enough, what are $df$ and $d^2f$ and how are they computed wrt possibly moving bases $\{e_i(t)\}$? I (believe that I) understand the answers in the "reverse" case where $g:\mathbb R^n \to \mathbb R$ - then $dg$ is the second component of the natural extension of $f$ to a fiber bundle homomorphism $T\mathbb R^n\to T\mathbb R$, that is, $dg:\mathbb R^n \to (R^n)^*$, and $d^2g:\mathbb R^n\to(R^n)^*\wedge (R^n)^*$, or $dg\in A^1(T\mathbb R^n)$ and $d^2g\in A^2(T\mathbb R^n)$. But with my $f$, the $df$ and $d^2f$ just don't "type check" because the "star" gives scalar-valued functions, not vector-valued ones.

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Long story: I'm reading a physics book and I just can't get my head around what they're doing. Assume every function here to be $C^\infty$. Using Einstein notation, i.e. every expression $a^i b_i$ is meant to read $\sum_i a^ib_i$.

1. Setup. In a smooth manifold $M\subseteq \mathbb R^n$ we have a rigid body moving within a time interval $I\subseteq \mathbb R$. The body movement is described by a point $\mathbf R:I\to M$ and a coordinate system (basis) $e_i : I\to \mathbb R^n,i=1,\ldots,n$, thought of as "attached" to the body for each $t\in I$ at the point $\mathbf R(t)$. For brevity I'll omit the dependence on time and write $\mathbf R$ even if $\mathbf R(t)$ is meant (and $e_i$ instead of $e_i(t)$).

   Denote the standard basis in $\mathbb R^n$ with $\boldsymbol{\epsilon}_{i}$. Now $\mathbf R=R^i \boldsymbol\epsilon_i$ for some $R^i(t),i=1,\ldots,n$. Since (for each $t\in I$) $\{e_i(t)\}$ forms a basis in $\mathbb R^n$, too, we can express $\boldsymbol{e}_{i}=g_{i}^{j}(\mathbf{R})\boldsymbol{\epsilon}_{j}$ and $\boldsymbol{\epsilon}_{i}=\left(g^{-1}(\mathbf{R})\right)_{i}^{j}\boldsymbol{e}_{j}$ for some invertible matrices $g(\textbf R(t))$.

2. First differential. Next in the book they take the differential of $\mathbf R$ and do some computations: $$ d\mathbf{R} =d\left(R^{i}\boldsymbol{\epsilon}_{i}\right) =\left(dR^{i}\right)\boldsymbol{\epsilon}_{i} =\left(dR^{i}\right)\left(g^{-1}(\mathbf{R})\right)_{i}^{j}\boldsymbol{e}_{j}$$

   I'm uneasy about the second equality $d\left(R^{i}\boldsymbol{\epsilon}_{i}\right) =\left(dR^{i}\right)\boldsymbol{\epsilon}_{i}$ transforming the differential of a vector into a vector multiplied by the differential of a scalar function. Though suspicious, at least symbolically it seems legit because $\boldsymbol{\epsilon}_{i}$ does not depend on the time $t\in I$, so it somehow makes sense to reduce the action of $d$ only to the scalar multiplier $R^i$.

   What is the nature of the product between a differential 1-form $dR^i$ and a vector $\boldsymbol \epsilon_i$?

3. Second differential. Here's where it gets too wild for me to move forward easily. They take the second differential of $\mathbf{R}$ like this: $$0=dd\mathbf{R}=d(\omega^j \boldsymbol{e}_{j})=d\omega^j \boldsymbol{e}_{j}-\omega^j \wedge d\boldsymbol{e}_{j}$$ where the $\omega^j$ are defined s.t. $d\mathbf{R}=\omega^j \boldsymbol{e}_{j}$. We have now a product of a two-form $d\omega^j$ with a vector $\boldsymbol{e}_{j}$, plus a wedge product of a one-form $\omega^j$ with a "vector-form" $d\boldsymbol{e}_{j}$, and both are supposed to be of the same type.

   $\omega^j \wedge d\boldsymbol{e}_{j}$ looks like complete non-sense to me, because a wedge product can take only same-type objects. There is $V\wedge V$ but never $V\wedge W$ for $V\neq W$. And yet, here $\omega^j$ is a "scalar" one-form while $d\boldsymbol{e}_{j}$ is a "vector" one-form, and they're wedged together. And I'm not even sure if I have to write the vector in the second position, it seems if I write $-d\boldsymbol{e}_{j} \wedge \omega^j$ it'd be ok, too.

I suspected something like a "vector-valued form" should exist, and indeed there's a wikipedia article on it. I read through it a few times, but I got too confused with the abstract construction done in two parts - defining a "bundle-valued" form and then taking a tensor product of bundles. Neither could I really understand what are the rules I can use in my more "practical" problem.

My questions are:

• what is the "type signature" of $\mathbf R$ as a (vector-valued) diffferential (0-)form, allowing us take its exterior derivatives $d\mathbf R$ and $d^2\mathbf R$?
• what does it mean to multiply "scalar" and "vector" forms?
• how is such a mixed-type product computed? Why does $d(fg)=df\wedge g-f\wedge dg$ still apply, if $f$ and $g$ are differently typed?

Answer 1

I think you mostly figured out the details, and yes the tensor product allows you to work with things neatly. I just want to emphasize that your questions can all be answered if you have general enough definitions. In what follows, everything is real (vector spaces, vector bundles) and smooth (maps, manifolds, bundles).

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1. Products of Vectors in General.

Definition.

Let $E,F,G$ be vector spaces, and $\beta:E\times F\to G$ a bilinear map. Then we (may if we decide to) call $\beta$ a “product”, and for each $(e,f)\in E\times F$, call the value $\beta(e,f)\in G$ the product of $e$ and $f$ (relative to $\beta$). It is also not uncommon to write $e\cdot_{\beta}f$ or simply $e\cdot f$ or $ef$ if the meaning of $\beta$ is understood from context.

Given vector bundles $E,F,G$ over $M$, a bilinear bundle morphism $\beta:E\oplus F\to G$ is also referred to as a product.

Also, in either case, if nothing else is mentioned and someone just says “product between $E$ and $F$”, then you can just interpret $\beta$ as the canonical bilinear map to the tensor product $\beta\equiv \otimes :E\oplus F\to E\otimes F$

So to go from vector spaces to vector bundles, you simply do things fiberwise (or if you want to think of it the other way, to go from vector bundles to vector spaces, simply take $M$ to be a singleton point).

Here are some common examples:

• multiplication in the field: $\cdot: \Bbb{R}\times\Bbb{R}\to\Bbb{R}$, $(s,t)\mapsto s\cdot t=st$.
• scalar multiplication in a vector space: $\cdot:\Bbb{R}\times E\to E$, $(t,x)\mapsto t\cdot x=tx$.
• evaluation of a linear map on a vector: $\text{ev}:\text{Hom}(E,F)\times E\to F$, $(T,x)\mapsto T(x)$. A special case is when $E=F$, so you get $\text{End}(E)\times E\to E$.
• Duality pairing $E^*\times E\to\Bbb{R}$ (which is a special case of the above when $F=\Bbb{R}$, but is sufficiently important, so I put it as a separate bullet point).
• composition of linear maps: $\circ:\text{Hom}(F,G)\times \text{Hom}(E,F)\to \text{Hom}(E,G)$, $(T,S)\mapsto T\circ S$. A special case is when $E=F=G$.
• An inner product $\langle\cdot,\cdot\rangle:V\times V\to\Bbb{R}$, or more generally, a bilinear non-degenerate map (eg the Lorentzian “inner product” in special relativity)
• A Lie algebra bracket: $[\cdot,\cdot]:\mathfrak{g}\times\mathfrak{g}\to\mathfrak{g}$. Common examples include $\Bbb{R}^3$ with the cross product, or $\text{End}(E)$ with the “commutator”

Doing this fiberwise, you can get relevant vector-bundle examples.

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2. Vector-Valued Alternating Forms, and General Wedge Products.

Here, you can check out Ivo Terek’s answer for some details.

Definition.

Let $V,E$ be two vector spaces, $k\geq 0$ an integer. An E-valued $k$-form on $V$ means a multilinear, alternating map $\omega:\underbrace{V\times\cdots\times V}_{\text{$k$ times}}\to E$, with the understanding that if $k=0$, we just mean an element of $E$.

Notice how this is really not a complicated thing at all, because the usual definition takes $E=\Bbb{R}$. Also, I should remark that there is no differential geometry here, purely abstract algebra; so a $k$-form here is not a $k$-form “field”. Hopefully context will inform you of the intended meaning.

Note that by the universal property for the exterior powers, this amounts to the same thing as a linear map $\omega:\bigwedge^k(V)\to E$ (strictly speaking I should write something like $\tilde{\omega}$ to distinguish from the $\omega$ above, but I’ll run out of letters at this rate), or more simply said, it is an element of $\text{Hom}\left(\bigwedge^k(V),E\right)$. By the usual tensor-product and Hom isomorphisms, this space is isomorphic to $\left(\bigwedge^k(V)\right)^*\otimes E\cong \bigwedge^k(V^*)\otimes E$. Of course, algebraically, it doesn’t matter which concrete realization we work with, but I personally just think of them as alternating multilinear maps with values in $E$, or as a linear mapping $\bigwedge^k(V)\to E$.

Now, we come to a definition of a wedge-product for vector-valued forms:

Definition.

Let $V,E,F,G$ be vector spaces, $\beta:E\times F\to G$ a given bilinear map (a “product”). For each pair of integers $k,l\geq 0$, we define a (obviously bilinear) mapping \begin{align} \wedge_{\beta}:\text{Hom}\left(\bigwedge^k(V),E\right)\times \text{Hom}\left(\bigwedge^l(V),F\right) \to \text{Hom}\left(\bigwedge^{k+l}(V),G\right), \end{align} called the wedge product on $V$ with respect to $\beta$ as follows: $(\omega,\eta)$ is mapped to $\omega\wedge_{\beta}\eta$, which is defined as the mapping \begin{align} (v_1,\dots,v_{k+l})&\mapsto \frac{1}{k!l!}\sum_{\sigma\in S_{k+l}}\text{sgn}(\sigma)\cdot \beta\left(\omega(v_{\sigma(1)},\dots, v_{\sigma(k)}), \eta(v_{\sigma(k+1)}, \dots, v_{\sigma(k+1)})\right)\\ &\equiv \frac{1}{k!l!}\sum_{\sigma\in S_{k+l}}\text{sgn}(\sigma)\cdot \omega(v_{\sigma(1)},\dots, v_{\sigma(k)})\cdot_{\beta} \eta(v_{\sigma(k+1)}, \dots, v_{\sigma(k+1)}). \end{align} We then call $\omega\wedge_{\beta}\eta$ the wedge product of $\omega$ with $\eta$ (relative to the product $\beta$).

One of the comments you raised is:

$\omega^j \wedge d\boldsymbol{e}_{j}$ looks like complete non-sense to me, because a wedge product can take only same-type objects. There is $V\wedge V$ but never $V\wedge W$ for $V\neq W$.

$\omega^j\wedge de_j$ is not nonsense, but you’re entirely correct with the second half. But, if you look at the definition, the domains of these maps are all the same ($\bigwedge^k(V)$ for some $k$), it is only their target space which changes ($E$ or $F$ or $G$). So, for exterior algebra, we can freely mess around with the target as long as we have a product $\beta$. But, what I can’t do is define a wedge product of an element $\text{Hom}\left(\bigwedge^k(V),E\right)$ with an element in $\text{Hom}\left(\bigwedge^klW),F\right)$ if $V\neq W$.

Now, here are some properties easily proved from the definitions (or by opening your favourite book and adapting the proofs (actually just copying line by line and modifying notation appropriately)):

Theorem (“Anti-commutativity”)

Let $V,E,F,G$ be vector spaces, $\beta:E\times F\to G$ a product, and define the reversed product $\beta’:F\times E\to G$ as $\beta’(f,e)=\beta(e,f)$. If $\omega$ is an $E$-valued $k$-form on $V$ and $\eta$ an $F$-valued $l$-form on $V$, then \begin{align} \omega\wedge_{\beta}\eta=(-1)^{kl}\eta\wedge_{\beta’}\omega. \end{align}

Theorem (Associativity)

Let $V,E_1,E_2,E_3,F,\tilde{F},G$ be vector spaces, and suppose we have four products bilinear maps $E_1\times E_2\to F$, and $E_2\times E_3\to \tilde{F}$, and $F\times E_3\to G$ and $E_1\times \tilde{F}\to G$ which are “compatible” in the sense that for all $(e_1,e_2,e_3)\in E_1\times E_2\times E_3$, \begin{align} e_1\cdot(e_2\cdot e_3)=(e_1\cdot e_2)\cdot e_3 \end{align} (I didn’t want to overload notation by introducing so many symbols for the product maps). Suppose that for $i=1,2,3$, we have $E_i$-valued $k_i$-forms $\omega_i$ on $V$. Then with respect to the appropriate products above, the associated wedge product is associative: \begin{align} \omega_1\wedge(\omega_2\wedge \omega_3)=(\omega_1\wedge \omega_2)\wedge \omega_3. \end{align}

One example of such a collection of compatible products is scalar multiplication in the field of course. A non-trivial example is when $E_1$ is a space of linear maps, $E_2$ is another space of linear maps, and $E_3$ is a space of vectors such that composition and evaluation are defined; then obviously $T(S(x))=(T\circ S)(x)$. At the level of forms, this gives \begin{align} \tau\wedge_{\text{ev}}(\sigma\wedge_{\text{ev}}\xi)=(\tau\wedge_{\circ}\sigma)\wedge_{\text{ev}}\xi. \end{align} Here, $\text{ev}$ appears three times, but it’s not necessarily the same evaluation map of course. One important special case is when you consider $\text{ev}:\text{End}(E)\times E\to E$ and $\circ:\text{End}(E)\times\text{End}(E)\to\text{End}(E)$. Then, the same formula holds of course, and the three $\text{ev}$ maps are the same.

Definition(Pullback)

Let $V,W,E$ be vector spaces, and $f:W\to V$ a linear map (which thus induces linear maps between the exterior powers $\bigwedge^kf:\bigwedge^kW\to \bigwedge^kV$). For an $E$-valued $k$-form $\omega$ on $V$, we define the pullback via $f$ to be $f^*\omega=\omega\circ\bigwedge^kf$.

Theorem (Distributivity of pullback)

Let $V,W,E,F,G$ be vector spaces, $\beta:E\times F\to G$ a product, $f:W\to V$ a linear map. If $\omega$ is an $E$-valued $k$-form on $V$ and $\eta$ is an $F$-valued $l$-form on $V$, then we have the following equality of $G$-valued $(k+l)$-forms on $W$: \begin{align} f^*(\omega\wedge_{\beta}\eta)=(f^*\omega)\wedge_{\beta}(f^*\eta). \end{align}

So now, with the bilinearity of $\wedge_{\beta}$, and the theorems above of anti-commutativity, associativity, and distributivity under pullback (extremely important), you can now freely manipulate vector-valued forms as you would with the scalar-valued case. You just need to know which bilinear product is intended at each stage.

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3. Vector-bundle valued differential forms

Definition.

Let $M$ be a smooth manifold, $(E,\pi, M)$ a vector bundle over $M$. A smooth $E$-valued differential $k$-form on $M$ is by definition a smooth bundle morphism $\omega:\bigwedge^k(TM)\to E$. (It is equivalent to define it to be a section of the vector bundle $\bigwedge^k(T^*M)\otimes E$ by virtue of the Hom-tensor product isomorphisms). The set of all such forms is denoted $\Omega^k(M;E)$.

In the case that $E=M\times V$ is a trivial vector bundle ($V$ a given fixed vector space) we abbreviate tbe notation to $\Omega^k(M;V)$. If furthermore $V=\Bbb{R}$ is the scalar field itself, then we write as usual $\Omega^k(M)$.

This is a very simple thing: for each point $x$ in the base, we require $\omega$ to restrict to a linear map $\bigwedge^k(T_xM)\to E_x$. So, we’re simply applying the definition above fiber-by-fiber. There are two noteworthy special cases:

• first, fix a vector space $V$, and take $E$ to be the trivial bundle $M\times V$. Then, applying the definition above to this case simply amounts to providing a smooth map $\omega:\bigwedge^k(TM)\to V$ such that for each $x\in M$, we have a linear map $\bigwedge^k(T_xM)\to V$.
• Next, taking $V=\Bbb{R}$ above recovers the usual definition of a $k$-form on $M$.

One final definition I’ll add is that of the pullback because there’s one subtlety:

Definition(Pullback of bundle-valued forms)

Let $(E,\pi,M)$ be a vector bundle, $N$ a manifold and $f:N\to M$ a smooth map. For an $E$-valued $k$-form $\omega$ on $M$ (which is a bundle morphism $\bigwedge^k(TM)\to E$) we can define $\omega_f:=\omega\circ\bigwedge^k(Tf):T^kN\to E$, which is a bundle morphism over $f$, i.e the diagram below commutes: $\require{AMScd}$ \begin{CD} \bigwedge^k(TN) @>{\omega_f}>> E \\ @V{}VV @VV{}V \\ N @>>{f}> M. \end{CD} We can call $\omega_f$ an *$E$-valued $k$-form on $N$ *along the map $f$**. But if you don’t like considering vector bundles over different bases, you can invoke the universal property of pullback bundles to say this induces a bundle morphism $\tilde{\omega_f}: \bigwedge^k(TN)\to f^*E$. It is up to you which interpretation you prefer, and either can be called $f^*\omega$, the pullback of $\omega$ via $f$.

The reason pullbacks here are slightly more nuanced is because on the target, we don’t just land in a single vector space.

With these definitions in mind, you can generalize all the theorems mentioned in section 2 to the case of differential forms taking values in a vector bundle (and hopefully you can extend the definition of pullback to this case as well; just do it fiber-by-fiber).

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4. Exterior derivatives for vector-space valued differential forms.

When talking about differentiation, we have roughly speaking, three situations:

• You have a map $\psi:M\to N$ with $M,N$ only smooth manifolds. Then, the only notion you have is the tangent map $T\psi:TM\to TN$. This isn’t pleasant in general since the target space is enlarged.
• You have a map $\psi:M\to V$ where $M$ is a smooth manifold and $V$ is a vector space (not bundle). Then, you have of course the notion $T\psi$ above, but by projecting to the second component, you get the exterior derivative $d\psi$ of a vector-space valued map. You can then generalize this notion to define $d$ on all $V$-valued differential $k$-forms on $M$.
• You have a section $\psi:M\to E$ of a vector bundle, and $E$ is equipped with a covariant derivative $\nabla$. I’ll talk about this in the next section.

Here are the properties of the such-defined $d$:

Theorem (Properties of $d$ for vector space-valued differential forms)

Let $M$ be a smooth manifold, $V$ a vector space. Then,

• (linearity): $d:\Omega^k(M;V)\to \Omega^{k+1}(M;V)$ is linear for each $k\geq 0$
• (product rule): For any vector spaces $W,X$ and any bilinear map $\beta:V\times W\to X$ (a product at the vector space level) we have for any $\omega\in \Omega^k(M;V)$ and $\eta\in \Omega^l(M;W)$, that \begin{align} d(\omega\wedge_{\beta}\eta)&=(d\omega)\wedge_{\beta}\eta + (-1)^k\omega\wedge_{\beta}(d\eta). \end{align} To be fully precise I should write $d_{(X)}, d_{(V)},d_{(W)},$ and so on, but I think you get the idea.
• (pullback commutes with $d$) For any smooth map $f:N\to M$ between manifolds, and any $\omega\in \Omega^k(M;V)$, we have \begin{align} f^*(d\omega)=d(f^*\omega). \end{align}

So, this all sounds just as in the scalar case, and you’re right. For the product rule, you just put $\wedge_{\beta}$ and you’re good to go, and this should answer your questions about $(dR^i)\,\mathbf{\epsilon}_i$ and also $\omega^j\wedge d\mathbf{e}_j$.

For example, you asked about the meaning of $(d R^i)\,\mathbf{\epsilon}_i$ in your question. Well, in this case, notice that $R^i$ is a smooth function, meaning $V=\Bbb{R}$ above. Next, $\epsilon_i$ is an element of $\Bbb{R}^n$, so $W=\Bbb{R}^n$. The mapping $\beta$ is thus simply the scalar multiplication of a number with a vector in $\Bbb{R}^n$. To be fully explicit, let us completely evaluate. Given a time $t\in\Bbb{R}$ and a “tangent vector at $t$”, $\xi$, we can get $[(dR^i)\,\mathbf{\epsilon_i}]_t(\xi)= (dR^i)_t[\xi]\cdot \mathbf{\epsilon}_i= (R^i)’(t)\cdot \xi\cdot\mathbf{\epsilon}_i$.

This may be confusing because your $R^i$ is a function $I\to\Bbb{R}$ to it may be too specialized to understand the definitions. If you have a usual $k$-form $\omega$ on a manifold $M$, and you have a vector $v\in V$, then the “product” $\omega\cdot v$ can be thought of as the wedge product of the usual scalar form $\omega$ with the constant map $f:M\to V$, $f(x)=v$ (a $0$-form) and you wedge them with respect to scalar multiplication $\Bbb{R}\times V\to V$. The product nature is more apparent once we eliminate the form-nature: let us evaluate $\omega$ fully: take a point $x\in M$ and tangent vectors $\xi_1,\dots,\xi_1\in T_xM$. Then we get a number $\omega(\xi_1,\dots,\xi_k)$. This number can be multiplied by the vector $v$. So, \begin{align} (\omega\cdot v)(\xi_1,\dots, \xi_k)&=\omega(\xi_1\dots,\xi_k)\cdot v. \end{align} So, the thing to keep in mind is that “the product takes place in the target, not the domain”.

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5. Exterior covariant derivative.

Let $(E,\pi,M)$ be a vector bundle, and suppose $E$ is equipped with a connection $\nabla$. Then, given a section $\psi:M\to E$ (which is also an $E$-valued $0$-form on $M$), you can consider not just the tangent map $T\psi:TM\to TE$, but also the covariant derivative $\nabla\psi:TM\to E$ (which is essentially a “projection” of $T\psi$, but for general vector bundles there’s no canonical way to define this projection, hence we need the connection). Its value on a vector $h_x\in T_xM$ gives you a vector $\nabla_{h_x}\psi\in E_x$. Notice that $\nabla\psi$ is precisely an $E$-valued $1$-form on $M$.

Now, rather than writing $\nabla\psi$, it is sometimes useful to write it as $d_{\nabla}\psi$. This reminds us that $\psi$ is an $E$-valued $0$-form on $M$ and $d_{\nabla}\psi$ is an $E$-valued $1$-form on $M$. One can extend the definition of $d_{\nabla}$ so it acts on all forms, to give a sequence of linear maps $d_{\nabla}:\Omega^k(M;E)\to\Omega^{k+1}(M;E)$. Here are the properties of $d_{\nabla}$:

Theorem (Properties of Exterior covariant derivative)

Let $(E,\pi,M)$ be a vector bundle with a connection $\nabla$. Then, the associated exterior-covariant derivative operators are linear, and

• commute with pullback: if $f:N\to M$ is a smooth map between manifolds, and $\omega\in \Omega^k(M;E)$, then $f^*(d_{\nabla}\omega)=d_{\nabla}(f^*\omega)$, where the $\nabla$ on the right should be thought of as the pullback connection on $f^*E$.
• product rule: Let $E,F,G$ be vector bundles over $M$, with connections $\nabla^E,\nabla^F,\nabla^G$. Let $\beta:E\oplus F\to G$ be a bilinear bundle morphism (a product at the vector bundle level). If $\beta$ is “compatible with the connections” in the sense that for all sections $\psi$ of $E$ and $\phi$ of $F$ and all vector fields $X$ on $M$, we have \begin{align} \nabla^G_X(\beta(\psi,\phi))=\beta(\nabla^E_X\psi,\phi)+\beta(\psi,\nabla^F_X\phi), \end{align} then we have the following product rule: for all $\omega\in \Omega^k(M;E)$ and all $\eta\in \Omega^l(M;F)$, we have \begin{align} d_{\nabla^G}(\omega\wedge_{\beta}\eta)&=(d_{\nabla^F}\omega)\wedge_{\beta}\eta +(-1)^k \omega\wedge_{\beta}(d_{\nabla^F}\eta). \end{align}

In other words, if you omit the superscripts $E,F,G$, then the exterior covariant derivative $d_{\nabla}$ behaves as expected. Note that the compatibility condition is satisfied for many “products” of interest, for example for metric-compatible connections in Riemannian/Lorentzian geometry. Or for example when $\beta$ is evaluation $\text{Hom}(E,F)\oplus E\to F$.

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6. Fully general calculations using all this machinery

Let $f:N\to M$ be a smooth map between smooth manifolds, and $(E,\pi,M)$ a vector bundle equipped with a connection $\nabla$, and consider the pullback bundle $(f^*E,\pi,N)$ with the induced pullback connection, which for convenience I shall continue to denote as $\nabla$.

So, my map $f$ is meant to generalize your $\mathbf{R}:I\to M$. Notice I’m not making any assumptions about my $M$. I don’t care what ambient space it lives in; I simply care that I have a vector bundle with a connection. For example, you can take $E=M\times V$ for a fixed vector space $V$, such as the ambient $\Bbb{R}^n$ in your example, which would then simplify everything below since I no longer need connections. Or you could take $E=TM$, or something fancier if you wish.

Now, suppose I have a local frame $\{e_1,\dots, e_p\}$ for the pullback bundle $f^*E$ over some open set $U\subset N$. Since all my considerations below are local with respect to $N$, I may as well assume $N$ is small enough, and that $U=N$. This essentially means for each point $z\in N$, I have a basis $\{e_1(z),\dots, e_p(z)\}$ for the vector space $E_{f(z)}$. Intuitively, my mapping $f$ maps the point $z\in N$ to $f(z)\in M$ and I now attach a basis at this point for the fiber $E_{f(z)}$. Now, I can introduce the dual frame $\{\sigma^1,\dots,\sigma^p\}$ to the $e$’s. Essentially, this means for each point $z\in N$, $\{\sigma^1(z),\dots, \sigma^p(z)\}$ is the dual basis for $E_{f(z)}^*$ with respect to the basis $\{e_1(z),\dots, e_p(z)\}$ of $E_{f(z)}$.

Ok, so now each $e_j$ is a section of the bundle $f^*E$ over $N$, i.e an $f^*E$-valued $0$-form on $N$. Thus, I can take its covariant exterior derivative $d_{\nabla}(e_j)$ to get an $f^*E$-valued $1$-form on $N$ (again, the $\nabla$ here is really $f^*\nabla$). I can now apply $\sigma^i$ to this to get an honest to god 1-form on $N$, $\omega^i_{\,j}:=\sigma^i(d_{\nabla}e_j)$. If you want to fit this into the above framework, I’m taking $\beta$ to be the duality pairing (which is a form of evaluation) between the $(f^*E)^*$-valued $0$-form $\sigma^i$ and the $f^*E$-valued $1$-form $d_{\nabla}(e_j)$. Anyway, the point of these scalar forms (also called the connection forms) is that I can re-express the exterior covariant derivative in terms of the $e$’s again: \begin{align} d_{\nabla}(e_j)&=\omega^i_{\,j}\cdot e_i, \end{align} where here the $\cdot$ is the product associated to scalar multiplication (note I don’t write $\wedge_{\cdot}$ here because the $e_i$ are ($E$-valued) $0$-forms, but I wouldn’t be wrong if I wrote $d_{\nabla}(e_j)=\omega^i_{\,j}\wedge_{\cdot}e_i$).

If we keep going, then we can apply $d_{\nabla}$ to this equation to get \begin{align} d_{\nabla}^2(e_j)&=d(\omega^i_{\,j})\cdot e_i+ (-1)^1\omega^i_{\,j}\wedge (d_{\nabla}e_j). \end{align}

I leave it to you to carefully figure out what the $E,F,G$ are and what the $\nabla^E,\nabla^F,\nabla^G$ are, in the notation of my theorem above about product rule for vector-bundle valued forms, and why this justifies the equality above. Now, we can simplify further; on the LHS the second exterior covariant derivative is no longer $0$, but actually depends on the curvature. So, if I let $R$ be the curvature of $\nabla$ on $E$ (which is an $\text{End}(E)$-valued $2$-form on $M$), and I let $R_f$ be the curvature of $f^*\nabla$ on $f^*E$, one can show that $R_f=f^*R$ (pull-back of an endomorphism-valued 2-form) and that the LHS of the equation above is $R_f\cdot_{\text{ev}} e_j$, where this $\cdot_{\text{ev}}$ is the product associated to evaluation of an endomorphism on a vector. So, \begin{align} R_f\cdot_{\text{ev}} e_j&= d(\omega^i_{\,j})\cdot e_i -\omega^i_{\,j}\wedge (d_{\nabla}e_j), \end{align} and this is an equality of $f^*E$-valued $2$-forms on $N$.

This is all essentially the beginning of Cartan’s moving frames method, and if you specialize to the tangent bundle and consider Levi-Civita connection, you easily arrive at his two structure equations.

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Edit: A rapid way of introducing the algebraic aspects.

As a matter of principle of illustrating the power of tensor products, let’s now give the rapid definitions. Let $V,W,X$ and $E,F,G$ be vector spaces, and suppose we have two bilinear maps (i.e products) $\alpha:V\times W\to X$ and $\beta:E\times F\to G$. By the universal property for tensor products, you now get linear maps $\tilde{\alpha}:V\otimes W\to X$ and $\tilde{\beta}:E\otimes F\to G$. Now, you can take the tensor product of these linear maps to give you a linear map $\tilde{\alpha}\otimes\tilde{\beta}:(V\otimes W)\otimes (E\otimes F)\to X\otimes G$. Now, by permuting the factors, this linear map is essentially the same as the linear map $(V\otimes E)\otimes (W\otimes F)\to X\otimes G$. And of course such a linear map comes from a bilinear map $(V\otimes E)\times (W\otimes F)\to X\otimes G$. Explicitly, the action on pure tensors is \begin{align} \left((v,e), (w,f)\right)\mapsto \alpha(v,w)\otimes \beta(e,f)\equiv (v\cdot_{\alpha}w)\otimes(e\cdot_{\beta}f). \end{align} In other words, we defined a product between $V\otimes E$ and $W\otimes F$ using $\alpha$ and $\beta$. We can think of $V\otimes E$ as being the space of “$E$-valued vectors in $V$” or “vectors in $V$ of type $E$”.

This is a very general situation, and by taking $V,W,X,E,F,G,\alpha,\beta$ carefully, you can recover so many examples. The most relevant for now is:

• Taking $V$ to be $\bigwedge^k(V^*)$, $W$ to be $\bigwedge^l(V^*)$, and $X$ to be $\bigwedge^{k+l}(V^*)$, and $\alpha=\wedge$ to be the usual wedge product of $k$ and $l$-forms on $V$, this procedure recovers the notion of “wedge-product on $V$ relative to $\beta$”, $\wedge_{\beta}$ as I defined above.

Answer 2

I'd been scratching my head for a week or two when I wrote part of my question, and stubbornly left it as a draft for another week until I (think I) finally understood the situation myself. Still, I decided to self-answer here in hope my issues become more searchable and easier to find.

First, given a p-form $\omega$ and a vector $v$, the most natural way to multiply them - so that the resulting quantity depends linearly on both $\omega$ and $v$, without introducing any other implicit dependencies - is to use the tensor product denoted $\omega \otimes v$ living in the vector space $\bigwedge^p(\mathbb R^n)^* \otimes \mathbb R^n$. It gives the universal way to talk about the two quantities - the form and the vector - as if "juxtaposed" together, somewhat like a pair $(\omega,v)$ but with the freedom to move scalars forward and backwards (and expand linearly): $\lambda\omega \otimes v = \omega\otimes\lambda v$. It is the way to multiply two linear objects of different kinds when they have nothing in common except the underlying field.

Next we attach this form-vector tensor product at each point of the manifold $M$. It is not quite clear how to do this directly - we cannot just use the direct product $M\times\left(\bigwedge^p(\mathbb R^n)^* \otimes \mathbb R^n\right)$ unless $M$ is "straight" (an open subset of $\mathbb R^n$, I believe, but correct me if I'm wrong). Instead (as per wikipedia) we attach separately each part of the tensor product to $M$, getting the vector bundles $A^p(T^*M)$ and $M\times \mathbb R^n$ with fibers $\bigwedge^p (\mathbb R^n)^*$ and $\mathbb R^n$, respectively. The first is a bundle of differential forms and the second bundles the manifold with a vector at each point in the simplest way possible (using a direct product).

Finally, the two bundles have to be glued tensorially, which means that at each point of the manifold we tensor the fibers of both bundles together. We can use the tensor product bundle construction to do this "for us", the resulting bundle denoted $$A^p(T^*M) \otimes (M\times \mathbb R^n).$$ I feel the urge to stress that the $\otimes$ symbol here does not mean tensor product of vectors spaces, but denotes a carefully constructed bundle over the same base manifold $M$ of both operands, with only the fibers of both operands "interleaved" at each point by the standard vector space tensor product.

I'm not familiar with how exactly this new bundle is made, nor do I understand what a bundle-valued form even means (which wikipedia uses for the construction), but for me the important bit is that "on the front" it bundles $M$ with fibers $\bigwedge^p(\mathbb R^n)^* \otimes \mathbb R^n$. That is, it gives me a firm ground to multiply forms with vectors and not be afraid of type checks.

Finally, the exterior derivative is defined on tensors $\omega(x) \otimes v$ only when $v$ is constant, i.e. when it does not depend on the point $x\in M$, naturally by $$d(\omega(x)\otimes v):= (d\omega)\otimes v$$ (here the object $\omega(x) \otimes v$ is not just a tensor but a section of the tensor product bundle, i.e. a function $M\to(\bigwedge^p(\mathbb R^n)^* \otimes \mathbb R^n)$, justifying us speak about "dependence on the point $x$"). This part is not quite clear to me because I'm not sure it correctly defines an action on all $\omega(x)\otimes v(x)$ sections - wikipedia says it does, and it works in my particular problem, so I'm inclined to trust it for now.

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True understanding of what I wrote above easily clears all confusion from my question and answers every point in it. In short, I think I was misguided by a abuse of notation with the missing $\otimes$ symbol to separate the form-part with the vector part. When I put it where necessary, it all becomes clear:

1. The way to interpret $\mathbf R:I\to M$ as something exteriorly differentiable, is to treat $M \cong\mathbb R\otimes M\cong A^0(T^*M)\otimes M.$ The first isomorphism here is trivial - we can just write $1\otimes \mathbf R$ for $\mathbf R$ - but works only for $M\subseteq \mathbb R^n$, which is the case in the book (I'm curious, though, about how it'd work if $M$ was not embedded in $\mathbb R^n$).
2. The exterior derivative now simply "increments the counter" from 0 to 1, and computation-wise we apply $d$ only after expressing $\mathbf R$ in terms of the space-fixed basis $\{\boldsymbol\epsilon_i\}$: $1\otimes\boldsymbol{R}=1\otimes R^{i}\boldsymbol{\epsilon}_{i}=R^{i}\otimes\boldsymbol{\epsilon}_{i}$, like the rule above dictates: $$d\left(R^{i}\otimes\boldsymbol{\epsilon}_{i}\right) =\left(dR^{i}\right)\otimes\boldsymbol{\epsilon}_{i}$$
3. The second time we want to differentiate $\omega^j \otimes \boldsymbol{e}_{j}$, we can't do that directly because the vector part is point-dependent. We can fix this easily, however, if we express the point-dependent vectors in terms of the fixed basis: $\boldsymbol{e}_{i}=g_{i}^{j}(\boldsymbol{R})\boldsymbol{\epsilon}_{j}$. Then we get $$ \begin{align} d\left(\omega^{j}\otimes\boldsymbol{e}_{j}\right)&=d\left(\omega^{j}\otimes g_{j}^{i}\boldsymbol{\epsilon}_{i}\right)=d\left(g_{j}^{i}\omega^{j}\otimes\boldsymbol{\epsilon}_{i}\right)\\ &=d\left(g_{j}^{i}\omega^{j}\right)\otimes\boldsymbol{\epsilon}_{i}\\ &=\left(dg_{j}^{i}\wedge\omega^{j}-g_{j}^{i}d\omega^{j}\right)\otimes\boldsymbol{\epsilon}_{i} \end{align} $$using only "clean" operations. With the two types of product - wedge and tensor - written explicitly, it becomes clear now how the mix of forms and vectors works: we have a tensor product strictly separating the forms from the vectors, and the forms on their own can be expressed via wedge products which can never "leak" to the vector part. Sigh.

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I learnt a lot in my little journey to understanding a page and a half from a classical mechanics book, and I hope my explanation here will help others with similar issues.

And of course, I'm still happy to read anyone else's view on the subject. If you have something to add, please answer!