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I am reading Kai Köhler's Differential Geometry and Homogeneous Spaces. In the book, Köhler provides the following second Bianchi identity: $$\nabla^{\mathrm{End}(E)}\Omega = 0,$$ where $E$ is a vector bundle over a smooth connected manifold $M$, $\Omega$ is the curvature of $E$, and $\nabla^{\mathrm{End}(E)}$ is the connection on the endomorphism bundle $\mathrm{End}(E)$ of $E$. Köhler's proof is very concise: for a section $s \in \Gamma(M,E)$, it can be directly calculated that $$(\nabla^{\mathrm{End}(E)}\Omega^E)(s) = \nabla^E(\Omega^Es) - \Omega^E(\nabla^Es) = (\nabla^E)^3s - (\nabla^E)^3s = 0.$$

What confuses me is the first equality in the proof: Köhler has not explicitly stated any properties of $\nabla^{\mathrm{End}(E)}$ in his book before, so this equality seems to have no justification to me. Assuming the proof is correct, this equality must follow from some properties or results that Köhler has mentioned earlier in the book (in fact, there are many such situations in Köhler's book, but I have been able to solve them before, although it may take a lot of time).

I have searched for references for an entire day, but I could not find any explanation of how to naturally derive the first equality without explicitly stating any properties of $\nabla^{\mathrm{End}(E)}$. Therefore, I hope someone can tell me how to naturally derive the first equality in the proof. Here are some results and definitions from Köhler's book that might be used:

  1. A connection on a vector bundle $E \to M$ is an $\mathbb{Z}$-module homomorphism $\nabla^E: \Gamma(M,E) \to \Gamma(M, T^*M \otimes E)$ satisfying the Leibniz rule: $$ \nabla^E(fs) = \mathrm{d}f \otimes s + f\nabla^Es, \qquad \forall f \in C^\infty(M), s \in \Gamma(M,E).$$

  2. A connection $\nabla^E$ on $E$ is extended to an operator $\nabla^E: \mathfrak{A}^q(M,E) \to \mathfrak{A}^{q+1}(M,E)$ by the rule $$ \nabla^E(\alpha \otimes s) = \mathrm{d}\alpha \otimes s + (-1)^{\deg\alpha}\alpha \wedge \nabla^Es$$ for any $\alpha \in \mathfrak{A}^q(M), s \in \Gamma(M,E)$, where $\mathfrak{A}^q(M,E) := \Gamma(M, \Lambda^q T^*M \otimes E)$ is the space of $q$-forms with coefficients in $E$.

  3. The connection induced by $\nabla^E$ and $\nabla^F$ on $E \otimes F$ is defined by $\nabla^{E \otimes F}(s \otimes t) := \nabla^Es \otimes t + s \otimes \nabla^Ft$ for $s \in \Gamma(M,E)$ and $t \in \Gamma(M,F)$.

  4. For vector bundles $E,F$ over $M$ and a map $P: \Gamma(M,E) \to \Gamma(M,F)$, $P$ is a $C^\infty(M)$-module homomorphism if and only if $P \in \Gamma(M, \mathrm{Hom}(E,F))$.

There are almost no calculations in local coordinates in the book, so if possible, I hope to get an explanation that does not use local coordinates. I look forward to any advice or insights you might have. Thank you so much for your time and effort!

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    @QuaereVerum, I prefer to call it the product rule. – Deane Jan 10 '25 at 16:18
  • Definitely not the chain rule. Anyhow, so maybe figure out $\nabla^{E^\otimes E}$ using the criteria you have listed; start with the induced connection $\nabla^{E^}$. – Ted Shifrin Jan 10 '25 at 16:23
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    I have written several answers on this and related topics. See this answer of mine and all the sublinks I mentioned to the basic algebra and calculus of bundle-valued forms. Anyway, almost by definition, given connections on $E,F$, the induced connection on $\text{Hom}(E,F)$ satisfies the following ‘product rule’ (with respect to evaluations): $(\nabla^{\text{Hom}(E,F)}_X\Phi)s=\nabla^F_X(\Phi(s))-\Phi(\nabla_X^Es)$. This then translates into suitable properties of the exterior covariant derivative (see links above). – peek-a-boo Jan 10 '25 at 16:28
  • I dislike the author's notation, so I insist on using my own. It shouldn't be hard to translate between the two. Let $s$ be a section of $E$ and $X,Y,Z$ be tangent vectors at a point on $M$. Observe that $\nabla s$ is an $E$-valued $1$-form. Then \begin{align} d_\nabla(\Omega s)(X,Y,Z) &= (d_\nabla\Omega)(X,Y,Z)s +(\Omega\wedge\nabla s)(X,Y,Z) \end{align} – Deane Jan 10 '25 at 16:30
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    Ahah, I deleted my comment, indeed I should have said "product rule". I haven't been in academia for more than a year now, starting to mix up terminology! – Quaere Verum Jan 10 '25 at 16:33
  • Thanks to everyone's suggestions, I have benefited a lot. My problem is now resolved. By the way, I also found a more abstract method via tensor contraction. I think this was probably Köhler's original approach, as he often skips the specific algebraic details. – MashiroHanser Jan 12 '25 at 04:10

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