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I'm having a hard time on proving the second Bianchi identity in the case of tangent bundle without choosing of metric. I already know that on a general vector bundle, the second Bianchi identity reads: $$d^{\nabla}F=0$$ where $F$ is the curvature of $\nabla$. I want to use this to prove the second Bianchi identity on tangent bundle: $$0=(\nabla_X R)(Y,Z)+(\nabla_Y R)(Z,X)+(\nabla_Z R)(X,Y)$$ My attempt is to use the formula of $d^{\nabla}$, which will transfer the general result to: $$\nabla_X(R(Y,Z))+\nabla_Y(R(Z,X))+\nabla_Z(R(X,Y))-R([X,Y],Z)-R([Y,Z],X)-R([Z,X],Y)=0$$ But how can I show this implies the above? Note that in both cases, I'm identifying $R$ as a $End(TM)$ valued two form.

GK1202
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    For those who find this, I got the same question answered here: https://math.stackexchange.com/questions/4932614/equivalent-forms-of-second-bianchi-identity-on-tm – Alex Pawelko Jun 15 '24 at 20:08

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