two different version differential Bianchi identity There are two different version of differential Bianchi identities,one use the (0,4) curvature tensor:
$$\nabla Rm(X,Y,Z,V,W) + \nabla Rm(X,Y,V,W,Z) + \nabla Rm(X,Y,W,Z,V) = 0\tag{1}$$
another use (1,3) curvature endomorphism :
$$\left(\nabla_{X} R\right)(Y, Z)W+\left(\nabla_{Y} R\right)(Z, X)W+\left(\nabla_{Z} R\right)(X, Y)W=0\tag{2}$$
I try to show these two are the same.I try to deduce (2) from (1).
first since $Rm = R^\flat$,hence (1) is equivalent to:
$$(\nabla R)^\flat(X,Y,Z,V,W) + (\nabla R)^\flat(X,Y,V,W,Z) + (\nabla R)^\flat(X,Y,W,Z,V) = 0$$
where we use $\nabla$ commute with musical isomorphism.
which implies:
$$\langle(\nabla_VR)(X,Y)Z,W\rangle+\langle(\nabla_WR)(X,Y)V,Z\rangle+\langle(\nabla_ZR)(X,Y)W,V\rangle = 0$$
at this point I don't know how to preceed,maybe we need to use symmetry of curvature tensor at this step?
 A: $\DeclareMathOperator{Rm}{Rm}$
Let us denote by $R$ the Riemann curvature tensor of type $(1,3)$ and $\Rm$ that of type $(0,4)$. They are related by the formula
$
\Rm(X,Y,Z,T) = g(R(X,Y)Z,T),
$
where $g$ denotes the Riemannian metric.
Usual considerations about Leibniz rule and the compatibility between the Riemannian metric and its Levi-Civita connection gives the equality
$$
(\nabla_X\Rm)(Y,Z,V,W) = g((\nabla_XR)(Y,Z)V,W).
$$
In short, it says that the covariant differentiation commutes with the musical isomorphism. Let us sum this last equality over a cyclic permuation of $(X,Y,Z)$:
\begin{multline}
(\nabla_X\Rm)(Y,Z,V,W)+(\nabla_Y\Rm)(Z,X,V,W)+(\nabla_Z\Rm)(X,Y,V,W) = \\ 
g((\nabla_XR)(Y,Z)V+(\nabla_YR)(Z,X)V+ (\nabla_ZR)(X,Y)V,W).
\end{multline}
Hence, (2) being true is equivalent to
$$\tag{3}
(\nabla_X\Rm)(Y,Z,V,W)+(\nabla_Y\Rm)(Z,X,V,W)+(\nabla_Z\Rm)(X,Y,V,W) =0.
$$
Since $\Rm(A,B,C,D) = \Rm(C,D,A,B)$, so does its covariant derivative: $(\nabla_X\Rm)(A,B,C,D) = (\nabla_X\Rm)(C,D,A,B)$ (this is a direct application of Leibniz rule). Hence, (3) is equivalent to
$$
(\nabla_X\Rm)(V,W,Y,Z)+(\nabla_Y\Rm)(V,W,Z,X)+(\nabla_Z\Rm)(V,W,X,Y) =0,
$$
which is now basically (1) with a suitable relabelling of the variables.
If I may make a small comment: I think the notation $\nabla T (X_1,\ldots,X_n)$ for the covariant derivative of a tensor is confusing. I'd rather write $(\nabla_{X_1}T)(X_2,\ldots,X_{n})$ or $(\nabla_{X_n}T)(X_1,\ldots,X_{n-1})$, which leaves no room for doubt. Some authors (e.g Gallot-Hulin-Lafontaine) write $(\nabla T)(X_1;X_2,\ldots,X_n)$ (maybe $(\nabla T)(X_1,\ldots,X_{n-1};X_n)$) to emphasis the difference between the direction of differentiation and the other directions.
