Question on notations for second covariant differentiation of a tensor I would like to get a better understanding on the concept of $\nabla T$, where $T$ is a tensor field not covector field or vector field.
Could anybody explain why or not the following relation holds or if even makes any sense?
$$
\nabla_X\nabla_YZ = \nabla\nabla Z(X,Y) 
$$
I believe it would be useful to derive the LHS into an equivalent form which unfortunately I can't deduce at the moment.
Thanks!
 A: Let $T$ be an $(r,s)$-type tensor field, then the covariant derivative $\nabla T$ is the $(r,s+1)$-type tensor field is given by
$$\nabla T(X,\vec{Z})=(\nabla_XT)(\vec{Z})=\nabla_X(T(\vec{Z}))-\sum_{i=1}^{s}T(Z_1,\dots,\nabla_XZ_i,\dots,Z_s),$$
where $\vec{Z}=(Z_1,\dots,Z_s)$ is a $s$-tuple of vector fields on $M$.
Now it is easiest to compute the second covariant derivative of a vector field, i.e. when $T$ is a $(1,0)$-type tensor field. We simply have
$$(\nabla\nabla T)(X,Y)=\nabla(\nabla T)(X,Y)=\nabla_X(\nabla T(Y))-\nabla T(\nabla_XY)=\nabla_X\nabla_YT-\nabla_{\nabla_XY}T.$$
Now we try and do the same computation for arbitrary $(r,s)$-type tensor fields.
\begin{align}
(\nabla\nabla T)(X,Y,\vec{Z})=\nabla_X(\nabla T(Y,\vec{Z}))-\nabla T(\nabla_XY,\vec{Z})-\sum_{i=1}^{s}\nabla T(Y,Z_1,\dots,\nabla_XZ_i,\dots,Z_s),
\end{align}
where $\vec{Z}=(Z_1,\dots,Z_s)$ is an $s$-tuple of vectors. We also compute
\begin{align}
\nabla_X(\nabla_YT)(\vec{Z})-(\nabla_{\nabla_XY}T)(\vec{Z}) & = \nabla_X(\nabla_YT(\vec{Z}))-\sum_{i=1}^{s}(\nabla_YT)(Z_1,\dots,\nabla_XZ_i,\dots,Z_s)\\
& \quad - \nabla_{\nabla_XY}(T(\vec{Z}))+\sum_{i=1}^{s}T(Z_1,\dots,\nabla_{\nabla_XY}Z_i,\dots,Z_s) \\
& = \nabla_X(\nabla T(Y,\vec{Z}))-\sum_{i=1}^{s}\nabla T(Y,Z_1,\dots,\nabla_XZ_i,\dots,Z_s)\\
& \quad - \nabla T(\nabla_XY,\vec{Z}).
\end{align}
So we have shown
$$(\nabla\nabla T)(X,Y,\vec{Z})=\nabla_X(\nabla_YT)(\vec{Z})-(\nabla_{\nabla_XY}T)(\vec{Z}).$$
Now we can use the Hessian of the tensor field to simplify things a bit. Define the Hessian of $T$ with respect to $X,Y\in\Gamma(TM)$ to be the $(r,s)$-type tensor field $\nabla\nabla_{(X,Y)}T$ characterized by
$$(\nabla\nabla_{(X,Y)}T)(\vec{Z})=\nabla\nabla T(X,Y,\vec{Z}).$$
Then we see that
$$\nabla\nabla_{(X,Y)}T=\nabla_X\nabla_YT-\nabla_{\nabla_XY}T.$$ (Alternatively you could write by a bit of abuse of noation $\nabla\nabla T(X,Y)=\nabla_X\nabla_YT-\nabla_{\nabla_XY}T$).
An interesting side note that might help with the intuition of $\nabla\nabla T$: if $X,Y,Z\in\Gamma(TM)$ then $\nabla\nabla Z(X,Y)-\nabla\nabla Z(Y,X)=R(X,Y)Z$, where $R$ is the $(1,3)$-type Riemann curvature tensor.
