# Verification of Ricci Identity in Riemannian Geometry

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.

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}

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