# 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!

• You can easily see that the relation cannot be correct by observing that the right hand side is linear over smooth functions in $Y$ (since $\nabla\nabla Z$ is a tensor field), while the left hand side does not have this property. – Andreas Cap Aug 18 '17 at 10:51

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.