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Consider a connection on $E$ which is a vector bundle over $M$ :

$$ \nabla : \Gamma(E) \rightarrow \Omega^1(M)\otimes \Gamma(E),\ s\mapsto \nabla\ s$$

Here $\nabla s =dx^k\otimes \nabla_{\partial_k} s$ where $x^k$ is coordinate function on $M$

Consider $C^\infty(M)$-linear map $d^\nabla$ : $$ d^\nabla (\omega\otimes s) = d\omega \otimes s +\nabla s\wedge \omega $$ where $d$ is an exterior differentiation on $M$.

Then $$ d^\nabla d^\nabla (\omega \otimes s) =\sum_{m<k} dx^m\wedge dx^k\wedge \omega \otimes [ \nabla_{\partial_m},\nabla_{\partial_k}]s $$

Here how can we prove that $d^\nabla d^\nabla d^\nabla (s)=0$ ?

In fact $$ d^\nabla d^\nabla d^\nabla (s)= dx^m\wedge dx^k\wedge dx^t \otimes (\nabla_{\partial_m} \nabla_{\partial_k}(\nabla_{\partial_t}s)).$$

Reference is the book " Einstein manifold - Besse ".

To show this we must have a metric on vector bundle ?

Another Calculation : $\xi$ is a $k$-dimesional vector bundle with a connection $\nabla$

Define $$d^\nabla : \Omega^i(\xi,M) \rightarrow \Omega^{i+1}(\xi,M),\ \omega\otimes s\mapsto d\omega\otimes s + (-1)^i\omega\wedge \nabla s $$ where $\Omega^i(\xi,M)$ is $\xi$-valued $i$-form.

Here $\nabla e_i = A_{ij}e_j $ where $e_i$ is basis on $\xi$ and $A_{ij}$ is $k\times k$-matrix of 1-forms.

Note that $$ d^\nabla \circ \nabla (e_i) = dA_{ij}\otimes e_j - A_{ij}\wedge \nabla e_j =[ dA_{ij} - A_{im}\wedge A_{mj} ]\otimes e_j$$

so that $$ d^\nabla \circ d^\nabla\circ \nabla (e_i) = [ 0 - dA_{im}\wedge A_{mj} + A_{im}\wedge dA_{mj} ]\otimes e_j $$ $$+ [ dA_{is} - A_{im}\wedge A_{ms} ]\wedge (- A_{sj})\otimes e_j$$

$$ = [ -2 dA_{im}\wedge A_{mj} + A_{im}\wedge dA_{mj} + A_{im}\wedge A_{ms} \wedge A_{sj} ]\otimes e_j$$

Why last line is not zero ?

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    $\begingroup$ You mean $\Omega^1(M) \times \Gamma(E)$ as the domain of $\nabla$, no? Also, by $\mathcal{C}M$, you mean $\mathcal{C}^\infty(M)$, right? $\endgroup$ Commented Oct 28, 2013 at 0:52
  • $\begingroup$ You are right yes i missed $\endgroup$
    – HK Lee
    Commented Oct 28, 2013 at 0:56
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    $\begingroup$ Are you sure about the argument of $d^\nabla d^\nabla d^\nabla$? By definition $d^\nabla:\Omega^p(E,M)\to\Omega^{p+1}$, where $\Omega^p(E,M)$ denotes the space of $E$-valued $p$-forms. $\endgroup$
    – gofvonx
    Commented Oct 28, 2013 at 17:45
  • $\begingroup$ I am not sure, since it must be zero. In the above, I write same question. Reference is the book "From calculus to cohomology - Madsen and Tornehave" $\endgroup$
    – HK Lee
    Commented Oct 28, 2013 at 23:53
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    $\begingroup$ Are you sure this is correct? The second Bianchi identity is usually written $d^\nabla \Omega = 0$ for $\Omega \in \Omega^2({\rm End}(E),M)$ the curvature 2-form, which seems to me like it would give something like $(d^\nabla)^3 s = d^\nabla \Omega(s) = \Omega(d^\nabla s)$. Have you tried checking this identity for $E = TM$ with a Levi-Civita connection? $\endgroup$ Commented Oct 29, 2013 at 3:25

1 Answer 1

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As Anthony Carapetis suggests, following the book "From Calculus to cohomology", it said that $d^\nabla F=0$ (Theorem 17.13, .178), where one treat $F \in \Omega^2(M, End(E))$, and $d^\nabla$ the induced connection on $End(E) \cong E^* \otimes E$. So it doesn't really mean $(d^\nabla)^3=0$.

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