Second Covariant Derivative Identity My question is simple. Let $X^{\mu\nu}$ be a rank-2 tensor, and $\nabla_\mu$ be the metric compatible, torsion-free, covariant derivative. Does the identity $(\nabla_\mu \nabla_\nu - \nabla_\nu \nabla_\mu) X^{\mu\nu}=0$ holds ? My proof goes like this


*

*Write the definition of the Riemann $(\nabla_\mu \nabla_\nu - \nabla_\nu \nabla_\mu) X^{\mu\nu} = -\frac{1}{2}(R_{\mu\nu\lambda}^{\phantom{\mu\nu\lambda}\mu}X^{\lambda\nu}+R_{\mu\nu\lambda}^{\phantom{\mu\nu\lambda}\nu}X^{\mu\lambda})$. 

*Rename the indices $\mu\leftrightarrow\nu$ in the first term on the right, and use the antysymmetry $R_{\mu\nu\lambda}^{\phantom{\mu\nu\lambda}\nu}=-R_{\mu\nu\lambda}^{\phantom{\mu\nu\lambda}\mu}$ of the Riemann tensor, to get $(\nabla_\mu \nabla_\nu - \nabla_\nu \nabla_\mu) X^{\mu\nu} = -\frac{1}{2}(-R_{\mu\nu\lambda}^{\phantom{\mu\nu\lambda}\nu}X^{\lambda\mu}+R_{\mu\nu\lambda}^{\phantom{\mu\nu\lambda}\nu}X^{\mu\lambda})$.

*Introduce the symmetric Ricci tensor $R_{\mu\lambda}:=R_{\mu\nu\lambda}^{\phantom{\mu\nu\lambda}\nu}$, to get finally $(\nabla_\mu \nabla_\nu - \nabla_\nu \nabla_\mu)X^{\mu\nu} = \frac{1}{2}R_{\mu\lambda}(X^{\lambda\mu}-X^{\mu\lambda})$ 

*That last term is the contraction of $R_{\mu\lambda}$ (symmetric) and $X^{\lambda\mu}-X^{\mu\lambda}$ (anti-symmetric), therefore vanishes.


This seems to be true for any $X^{\mu\nu}$, but the result seems quite strong, and I've never seen it anywhere. 
Anybody can confirm and/or see a mistake in my reasoning ? 
Thanks
 A: This is correct. Here's another way to see it: Decompose $X^{\mu\nu}=A^{\mu\nu}+B^{\mu\nu}$ into its symmetric and skew symmetric parts $A^{\mu\nu}$ and $B^{\mu\nu}$, respectively.  For your 4th reason, $(\nabla_\mu\nabla_\nu-\nabla_\nu\nabla_\mu)A^{\mu\nu}=0$.  The fact $(\nabla_\mu\nabla_\nu-\nabla_\nu\nabla_\mu)B^{\mu\nu}=0$ follows from the fact $\delta^2=0$, where $\delta$ is the divergence on differential forms; this is the dual of the identity $d^2=0$.  More precisely,
$$ (\nabla_\mu\nabla_\nu-\nabla_\nu\nabla_\mu)B^{\mu\nu}=-2\nabla^\nu\nabla^\mu B_{\mu\nu} = -2\delta^2B = 0 . $$
(I raised and lowered indices just to emphasize that $\delta$ is the divergence.)
A: Edit:It turns I missunderstood the OP and I answered a totaly different question
Thw question I answered is: given a rank-2 tensor A
and two vector fields $\partial_{\mu}$ and $\partial_{\nu}$,
whether or not the identity concerning second
covariant derivatives of $A$ holds
$$(\nabla^2_{\mu,\nu}-\nabla^2_{\nu,\mu})A=0$$
I don't think this holds.
I take any two vector fields $V,\ W$
and a special case for the tensor $A$.
The second covariant derivative of a tensor $A$ is given by the formula
$$
\nabla^2_{V,W}A=\nabla_V(\nabla_WA)-\nabla_{\nabla_VW}A
$$
I give an example where $A$ is the tensor product of two vector fields.
The covariant derivative follows a Leibniz rule with tensor products, 
thus we calculate
\begin{align*}
    \nabla^2_{V,W}X\otimes Y
        &= \nabla_V(\nabla_W(X\otimes Y))-\nabla_{\nabla_VW}(X\otimes Y)\\
    &= \nabla_V(\nabla_WX\otimes Y+X\otimes \nabla_WY)
        -\nabla_{\nabla_VW}X\otimes Y -X\otimes \nabla_{\nabla_VW}Y\\
    &= \nabla_V\nabla_WX\otimes Y + \nabla_WX\otimes \nabla_VY\\
    &\hspace{5mm}+\nabla_VX\otimes \nabla_WY +X\otimes \nabla_V\nabla_WY \\
    &\hspace{5mm}-\nabla_{\nabla_VW}X\otimes Y
        -X\otimes\nabla_{\nabla_VW}Y
\end{align*}
Therefore, after some cancelations,
$(\nabla^2_{V,W}-\nabla^2_{W,V})X\otimes Y$ becomes
\begin{align*}
    (\nabla^2_{V,W}&-\nabla^2_{W,V})X\otimes Y=\\
    &=\nabla_V\nabla_WX\otimes Y -\nabla_W\nabla_VX\otimes Y 
        -\nabla_{\nabla_VW}X\otimes Y+\nabla_{\nabla_WV}X\otimes Y\\
    & \hspace{5mm}+X\otimes \nabla_V\nabla_WY -X\otimes \nabla_W\nabla_VY
        -X\otimes\nabla_{\nabla_VW}Y +X\otimes\nabla_{\nabla_WV}Y\\
    &=(R_{V,W}X)\otimes Y + X\otimes (R_{V,W}Y)
\end{align*}
Where in the last equality, we used the definition of Riemann curvature
with the fact that $\nabla_VW-\nabla_WV=[V,W]$.
Then, for the simple case of the symmetric rank 2 tensor tensor
$X\otimes X$, for some vector field $X$, we have that
\begin{align*}
    (\nabla^2_{V,W} - \nabla^2_{W,V})X\otimes X=0 \text{ iff}\\
    (R_{V,W}X)\otimes X + X\otimes (R_{V,W}X)=0 \text{ iff}\\
    \left\{\begin{array}{ll}
        R_{V,W}X=X &  \text{and}\\
        R_{V,W}X=-X & 
    \end{array}\right.
\end{align*}
But this is clearly impossible for $X\neq0$, thus
$(\nabla^2_{V,W}-\nabla^2_{W,V})X\otimes X$ cannot be zero
Let me know if you notice something I might not have, thanks in advance
