Verification of Ricci Identity in Riemannian Geometry

Question

Ricci identity:

$\forall X, Y \in \Gamma(TM), T \in\Gamma(\otimes^{r,s}TM)$, we have

$$\nabla^2T(\cdots,X,Y)-\nabla^2T(\cdots,Y,X)=-R(X,Y)T(\cdots)$$

where $R$ is curvature tensor, $R(X,Y):=\nabla_X \nabla_Y – \nabla_Y \nabla_X-\nabla_{[X,Y]}$, $\nabla$ is torsion free connection.

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I want to check Ricci Identity for $T \in\Gamma(\otimes^{1,0}TM)$, i.e.

$$\nabla^2T(\omega,X,Y)-\nabla^2T(\omega,Y,X)=-R(X,Y)T(\omega)$$

My effort:

$\forall X,Y,\in \Gamma(TM), \omega \in \Gamma(T^*M)$, we have

\begin{align}
\nabla^2T(\omega,X,Y)&=\nabla(\nabla T)(\omega,X,Y)=(\nabla_Y(\nabla T))(\omega,X)\\
&=\nabla_Y(\nabla T(\omega,X))-(\nabla T)(\nabla_Y \omega,X)-(\nabla T)(\omega,\nabla_Y X)\\
&=\nabla_Y \nabla_X T(\omega)-\nabla_X T(\nabla_Y \omega)-(\nabla_{\nabla_Y X} T)(\omega)\text{ (**wrong**)}
\end{align}

If we want $\nabla^2T(\omega,X,Y)-\nabla^2T(\omega,Y,X)=-R(X,Y)T(\omega)$

$\quad$ we need $\nabla_X T(\nabla_Y \omega)=\nabla_Y (\nabla_X \omega)$. Are these two things equal?

Update:

As mentioned in answer by Arctic Char, there's a mistake in calculation. The correct one should be:

\begin{align}
\nabla^2T(\omega,X,Y)&=\nabla(\nabla T)(\omega,X,Y)=(\nabla_Y(\nabla T))(\omega,X)\\
&=\nabla_Y(\nabla T(\omega,X))-(\nabla T)(\nabla_Y \omega,X)-(\nabla T)(\omega,\nabla_Y X)\\
&=\nabla_Y (\nabla_X T(\omega))-\nabla_X T(\nabla_Y \omega)-(\nabla_{\nabla_Y X} T)(\omega)\\
&=\nabla_Y \nabla_X T(\omega)+\nabla_X T(\nabla_Y \omega)-\nabla_X T(\nabla_Y \omega)-(\nabla_{\nabla_Y X} T)(\omega)\\
&=\nabla_Y \nabla_X T(\omega)-(\nabla_{\nabla_Y X} T)(\omega)
\end{align}

Answer 1

Note that

\begin{align*} \nabla _Y (\nabla T (\omega, X)) &= \nabla_Y \big((\nabla_X T)(\omega)\big)\\ &= (\nabla_Y \nabla_X T) (\omega) + (\nabla_X T)(\nabla_Y \omega). \end{align*}

So you get

\begin{align} \nabla^2T(\omega,X,Y)= (\nabla_Y \nabla_X T)(\omega)-(\nabla_{\nabla_Y X} T)(\omega) \end{align}