# version of Bianchi identity

Let $F \to E \to M$ be a smooth fiber bundle with connection $\omega$ and curvature $R$. We can form a (graded) vector bundle by taking the complex of differential forms at each fiber. Call this bundle $A(E) \to M$ where a fiber above $p\in M$ is $\Omega^*(E_p)$ i.e. differential forms on the fiber $E_p$. The connection on the fiber bundle induces a vector bundle connection on $A(E) \to M$, call it $\nabla$. We know that on a local trivialization $U$ we have $\nabla|_U \in \Omega^1(U, End( U \times \Omega^*(F))$. Also, we know that the curvature $R$ of the original fiber bundle is given by a $2$-form $R \in \Omega^2(U, \chi(F))$, i.e. a $2$-form on the base with values on vector fields of $F$.

I want to show that for any vector fields $X,Z,Y$ on $U$ we have that $\nabla|_U(X)\circ \iota_{R(Y,Z)}+ \iota_{R(X,Y)} \circ \nabla|_U(Z)=0$, where $\iota_{R(X,Y)}:\Omega^*(F)\to \Omega^{*-1}(F)$ is contraction with the vector field $R(X,Y)$.

Does this follow from the Bianchi identity?

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I don't think you formula works: first there should be a cyclic sum, and say we add the cyclic sum, then it would only works for $X,Y,Z$ vector fields that commute (plus a small sign issue).

I guess the correct formula is, up to sign issues, something like this: $$\oint_{X,Y,Z}\nabla|_U(X)\circ \iota_{R(Y,Z)}- \iota_{R(X,Y)} \circ \nabla|_U(Z)+\iota_{R([X,Y],Z)}=0.$$ In particular, when $X,Y,Z$ commute, you basically recover what you wanted.

I will try to prove this. First recall that the type of connection you're referring to are called Ehresmann connections, so I will assume that you are familiar with the notion of horizontal lift.

Say we denote $h(X)$ the horizontal lift of some vector field $X$ on $U$ (remember that $h(X)$ is a vector field on $U\times F$ that projects onto $X$) then the operator you denote $\nabla|_U(X)$ coincides with the Lie derivative of the horizontal lift (whether it acts on forms or on vector fields) namely: $$\nabla|_U(X)=\mathcal{L}_{h(X)}.$$

The computation goes like this: \begin{align*} &\oint_{X,Y,Z}\nabla|_U(X)\circ \iota_{R(Y,Z)}- \iota_{R(X,Y)} \circ \nabla|_U(Z)+\iota_{R([X,Y],Z)}=0\\ =&\oint_{X,Y,Z}\nabla|_U(X)\circ \iota_{R(Y,Z)}- \iota_{R(Y,Z)} \circ \nabla|_U(X)+\iota_{R([X,Y],Z)}\\ =&\oint_{X,Y,Z} \mathcal{L}_{h(X)}\circ \iota_{R(Y,Z)}- \iota_{R(Y,Z)} \circ \mathcal{L}_{h(X)}+\iota_{R([X,Y],Z)}\\ =&\oint_{X,Y,Z} \iota_{[\mathcal{L}_{h(X)},{R(Y,Z)}]}+\iota_{R([X,Y],Z)}\\ =&\iota_{\oint_{X,Y,Z} [\mathcal{L}_{h(X)},R(Y,Z)]+R([X,Y],Z)}\\ =&\iota_{\oint_{X,Y,Z} \nabla|_U(X)R(Y,Z)+R([X,Y],Z)}\\ =&0. \end{align*} In the first equality, I used the cyclic sun to rearrange $X,Y$ and $Z$. The second equality follows from the definition that $\nabla|_U(X)=\mathcal{L}_{h(X)}.$ The third equality come from Cartan calculus, namely $\mathcal{L}\circ \iota-\iota\circ\mathcal{L}=\iota_{[\ ,\ ]}$, then the rest is obvious.

Last comment, since I'm not sure you are aware of this, but the (second...) Bianchi identity follows from Jacobi identity for vector fields. More precisely, by the definition, the curvature is given by: $$R(X,Y)=h([X,Y])-[h(X),h(Y)].$$ Then you recover the Bianchi identity $\oint_{X,Y,Z} \nabla|_U(X)R(Y,Z)+R([X,Y],Z)=0$ by replacing $R$ by its formula, using that $\nabla|_U(X)=\mathcal{L}_{h(X)}=[h(X),\ ]$, and then applying Jacobi identity.

Disclaimer: check out my calculations, there may be sign issues ! (especially $+R([X,Y],Z)$ may become $-R([X,Y],Z)$ but my last comment should give you the appropriate sign...)

Hope that helps.

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