# Proving $d^\nabla( d^\nabla \omega) = F^\nabla \wedge \omega$

I'm going over the exterior covariant derivative $$d^\nabla : \Omega^k(E) \to \Omega^{k+1}(E)$$ of a vector bundle $$E \to M$$ and a connection $$\nabla$$ on $$E$$.

There are some peculiar notions which I'm facing. I'm trying to verify that $$d^\nabla( d^\nabla \omega) = F^\nabla \wedge \omega$$ where $$F^\nabla$$ is the curvature of $$\nabla$$.

The first thing I'm wondering is that can we write $$\omega \in \Omega^k(E)$$ as $$\omega = \alpha \otimes s$$ for $$\alpha \in \Omega^k(M)$$ and $$s \in \Omega^0(E)$$ always or do we need to assume locality or something?

If so, then

\begin{align*} d^\nabla( d^\nabla \omega) &= d^\nabla( d^\nabla \alpha \otimes s) \\ &= d^\nabla( d\alpha \otimes s +(-1)^k\alpha \wedge \nabla s) \\ &= d^\nabla(d\alpha \otimes s) + (-1)^kd^\nabla(\alpha \wedge \nabla s) \end{align*}

but I'm not sure how to proceed from here. I also saw a note that the pairing that the pairing $$F^\nabla \wedge \omega$$ combines the wedge product $$\Lambda^2 M \otimes \Lambda^k M \to \Lambda^{k+2}M$$ with the evaluation map $$\text{End}(E) \otimes E \to E$$ which I did not really understand.

• You most certainly cannot write $\omega\in \Omega^k(E)$ as $\alpha\otimes s$, as pointwise you can't even do that. It does suffice to check on simple tensors by linearity though. Commented Feb 19 at 15:03
• See Ivo Terek’s answer on wedge products with respect to bilinear pairings (and also his online notes for nice short expositions). I’ve written several answers regarding this matter as well e.g here and for a slightly more detailed discussion, here. Commented Feb 19 at 15:17
• Product rule again. Commented Feb 19 at 16:05
• No, @Chris's remark says what you're doing does work. You prove your identity for $\omega = \alpha\otimes s$ first. It then implies the identity for all $\omega\in\Omega^k(E)$ by linearity, because $\omega$ can be written as a sum $$\omega = \sum_{i=1}^N \alpha_i\otimes s_i.$$ Commented Feb 19 at 17:33
• Can you show that $d^\nabla\circ d^\nabla (s)(X,Y,)=F^\nabla(X,Y)s$ for general sections $s\in \Gamma(E)$? Do that first, then you will see it works out here, by noting that $d^\nabla s=\nabla s$. Commented Feb 19 at 17:50

Assuming you went through the links in the comments to understand the following abstract mumbo jumbo, we can give a relatively short answer.

• First, the fact that $$d_{\nabla}$$ satisfies a product rule means that it is a local operator (just like the usual exterior derivative, or any other derivation-like thing you’ve encountered).
• Given $$\omega\in \Omega^k(M;E)$$ and a local coordinate chart $$(U,x=(x^1,\dots, x^n))$$, we can write $$\omega=\sum\omega_I\cdot\,dx^I$$, where the sum is over all increasing multi-indices $$I$$ of length $$k$$, and $$\omega_I:U\to E$$ is a local section and $$dx^I=dx^{i_1}\wedge\cdots\wedge dx^{i_k}$$ is the usual abbreviation (and the $$\cdot$$ is simply the ‘scalar multiplication’, i.e the wedge product of an $$E$$-valued $$0$$-form with a usual $$k$$-form). Notice that since $$d^2=0$$ for usual forms, we have by the product rule that $$d(dx^I)=0$$. Therefore, to prove local formulas, it suffices to work with $$\omega=\psi\cdot \alpha$$ where $$\psi$$ is a local section of $$E$$ and $$\alpha$$ is a closed scalar $$k$$-form.
• If $$\omega=\psi\cdot \alpha$$, with $$\alpha$$ closed, then (by induction and product rule) we see that for all $$p\geq 0$$, \begin{align} d_{\nabla}^p\omega=(d_{\nabla}^p\psi)\wedge \alpha. \end{align} In other words, when you have a ‘basic’ $$E$$-valued $$k$$-form (with $$\alpha$$ closed), if you want to apply exterior covariant derivatives, then you simply let it act on the ‘vector part’, i.e $$\psi$$. In particular, taking $$p=2$$, we get \begin{align} d_{\nabla}^2\omega&=(d_{\nabla}^2\psi)\wedge \alpha =(R\cdot_{\text{ev}}\psi)\wedge \alpha =R\wedge_{\text{ev}}(\psi\cdot\alpha) =R\wedge_{\text{ev}}\omega. \end{align} The second equal sign is essentially by definition of the curvature $$R$$, and the third follows from the associativity of wedges as described in my last link of the comments.
• The equation $$d_{\nabla}^2\omega=R\wedge_{\text{ev}}\omega$$ is linear in $$\omega$$, so because we have proved it for ‘basic’ forms, and because every form is a sum of such guys, we deduce it holds for all forms.

Therefore, we now have a very nice and concise way of writing formulas involving curvature.

What I said above answers your question. But here are some extra good-to-know things.

• In Riemannian geometry, it is often the case that people do not go higher than second derivatives, because ‘everything else is encoded in the curvature’. Of course this is a vague statement, but one way of precisely saying this is that everytime we apply two exterior covariant derivatives, it amounts to wedging by the curvature: \begin{align} d_{\nabla}^4(\omega)=d_{\nabla}^2(d_{\nabla}^2\omega)=R\wedge_{\text{ev}}(R\wedge_{\text{ev}}\omega)=(R\wedge_{\circ}R)\wedge_{\text{ev}}\omega, \end{align} where the final equal sign again uses ‘associativity of wedges’ (in the guise of $$(T\circ S)(x)=T(S(x))$$ for endomorphisms $$T,S$$ and a vector $$x$$, or in words, composing then evaluating is by definition the same as successively evaluating). By induction, you can prove that for all $$p\geq 0$$ (where $$\wedge_{\circ}$$ refers to the wedge product relative to composition on $$\text{End}(E)$$ as the bilinear bundle morphism) \begin{align} d_{\nabla}^{2p}\omega=\underbrace{(R\wedge_{\circ}\cdots\wedge_{\circ}R)}_{\text{p times}}\wedge_{\text{ev}}\omega, \end{align} and so applying the above to $$d_{\nabla}\omega$$, we get \begin{align} d_{\nabla}^{2p+1}\omega=\underbrace{(R\wedge_{\circ}\cdots\wedge_{\circ}R)}_{\text{p times}}\wedge_{\text{ev}}d_{\nabla}\omega. \end{align} Thus, knowledge of $$\omega,d_{\nabla}\omega$$ and $$R$$ allows us to (algebraically) compute all the derivatives $$\{d_{\nabla}^n\omega\}_{n=0}^{\infty}$$.

• The second/differential Bianchi identity $$d_{\nabla^{\text{End}(E)}}R=0$$ also falls out easily (without the tons of index manipulations in GR, or various permuted sums you’ll see explicitly written out in Riemannian geometry books (of course we’re suppressing all of that within our exterior calculus)). For any $$\omega\in \Omega^k(M;E)$$, we have \begin{align} d_{\nabla}^3\omega&=d_{\nabla}^2(d_{\nabla}\omega)=R\wedge_{\text{ev}}d_{\nabla}\omega. \end{align} On the other hand (see the link for the precise statement of this vector-bundle product rule), \begin{align} d_{\nabla}^3\omega&=d_{\nabla}(d_{\nabla}^2\omega)=d_{\nabla}(R\wedge_{\text{ev}}\omega)=d_{\nabla}R\wedge_{{\text{ev}}}\omega+(-1)^2R\wedge_{\text{ev}}d_{\nabla}\omega. \end{align} Comparing these two expressions for $$d_{\nabla}^3\omega$$, we immediately see that $$(d_{\nabla}R)\wedge_{\text{ev}}\omega=0$$ for all $$\omega$$, and thus $$d_{\nabla}R=0$$.

• In the special case of $$E=TM$$, there is a very special $$TM$$-valued $$1$$-form on $$M$$, namely the identity map $$I:TM\to TM$$ (in the classical literature, this is denoted by $$dP$$ or $$d\mathbf{r}$$ to denote the ‘infinitesimal change in position’). If you fix any linear connection $$\nabla$$ on $$TM$$, then the torsion that you learn about in differential geometry is precisely the $$TM$$-valued $$2$$-form on $$M$$ given by $$\tau=d_{\nabla}I$$. We can then investigate the exterior covariant derivative of the torsion: \begin{align} d_{\nabla}\tau&=d_{\nabla}^2I=R\wedge_{\text{ev}}I. \end{align} So, if the LHS vanishes (i.e you have a $$d_{\nabla}$$-closed torsion… which is certainly the case if the connection itself is torsion free) then the RHS vanishes, i.e $$R\wedge_{\text{ev}}I=0$$. This is a $$TM$$-valued $$3$$-form on $$M$$; if we evaluate this on vectors $$x,y,z\in T_pM$$, then the following cyclic sum vanishes: \begin{align} R(x,y)z+R(y,z)x+R(z,x)y&=0. \end{align} This is precisely what the first/algebraic Bianchi identity states. So, $$R\wedge_{\text{ev}}I=0$$ concisely expresses the algebraic Bianchi identity (which is special to $$E=TM$$ and is a trivial consequence of vanishing torsion).

• A local coordinate formula: this is to convince you that, atleast symbolically, $$d_{\nabla}$$ is not that much more complicated than usual exterior derivative $$d$$. You can prove either from your axioms of $$d_{\nabla}$$, or take as a definition the following rule for how $$d_{\nabla}$$ acts on an $$E$$-valued $$k$$-form $$\omega$$ on $$M$$: if we locally write $$\omega=\sum\omega_I\,dx^I$$, then on $$U$$, \begin{align} d_{\nabla}\omega&=\sum_I(\nabla\omega_I)\wedge dx^I=\sum_I\sum_{j=1}^n\left(\nabla_{\frac{\partial}{\partial x^j}}\omega_I\right)\,dx^j\wedge dx^I. \end{align} Notice how this is completely analogous to the usual local coordinate formula for the exterior derivative of scalar forms $$d\omega=\sum_I (d\omega_I)\wedge dx^I=\sum_I\sum_j\frac{\partial\omega_I}{\partial \omega_j}\,dx^j\wedge dx^I$$.

• crap… I just realized you called the curvature $F^{\nabla}$ rather than $R$. Hopefully you can make the mental changes :) Commented Feb 19 at 18:04
• Thanks, yet again for the informative answer. To understand $\wedge_{\text{ev}}$ in $R\wedge_{\text{ev}}\omega$ correctly, is it an element of $\Omega^2(\text{End}(E)) \otimes \Omega^k(M)$ where $\text{ev} : \text{End}(E) \otimes E \to E$ is given by $\varphi \otimes e \mapsto \varphi(e)$? If so what does the notation $\wedge_{\text{ev}}$ exactly mean here? @peek-a-boo Commented Feb 19 at 19:30
• @Tepes $R$ is an element of $\Omega^2(M;\text{End}(E))$, and $\omega$ is an element of $\Omega^k(M;E)$, and $\wedge_{\text{ev}}$ takes these two and produces $R\wedge_{\text{ev}}\omega\in\Omega^{k+2}(M; E)$. And yes, $\text{ev}$ denotes the canonical bilinear bundle morphism $\text{End}(E)\oplus E\to E$, $(\phi,e)\mapsto \phi(e)$… or if you want to descend to a linear bundle morphism out of the tensor product $\text{End}(E)\otimes E\to E$ acting on the pure guys as $(\phi,e)\mapsto \phi(e)$, be my guest. Commented Feb 19 at 19:35
• regarding what exactly $\wedge_{\text{ev}}$ means/ how it is defined, I guess you haven’t read my third link which I left in the comment. Commented Feb 19 at 19:35
• I see, it's the same as $\omega\wedge_{\beta}\eta$, but with $\beta = \text{ev}$. I got a bit carried away with the $k+l$ shuffles there. @peek-a-boo Commented Feb 19 at 19:41