Connect of exchange vector field . $D$ is Levi-Civita connect. $U,V,W,Z$ are vector field. $S$ is (0,4) tensor. 
And $D^2_{X,Y}S=D_XD_YS-D_{D_XY}S$. 
Are there any easy way to compute the below equation ?  
If I unfold it, it is complex to compute ,so I hope there is some easy way.




Assume that $S$ is
$(0,1)$-tensor (Other case are similar). If $E_i$ is coordinate
vector field, first we will enumerate definitions and notations : 
$$ D_k S_m:= (D_kS)_m $$
Here by definition of covariant derivative, $$ (D_kS)_m:=
E_k (S_m) - S_n\Gamma_{km}^n $$
So
$$ D_iD_jS_k
 =
E_i( D_jS_k)- D_jS_m \Gamma_{ik}^m  $$
(Here $T:=D_jS$ is $(0,1)$-tensor That is above is $D_iT_k$)
$$ =
E_iE_j(S_k)- E_i (S_m\Gamma_{jk}^m)- E_j(S_m) \Gamma_{ik}^m+
S_n\Gamma_{jm}^n  \Gamma_{ik}^m   $$
Hence by using symmetry of index $i, \ j$ and direct computation,
$$ D_iD_jS_k - D_jD_i S_k
 = -S_m (E_i\Gamma_{jk}^m-E_j\Gamma_{ik}^m) + S_n(
 \Gamma_{jm}^n  \Gamma_{ik}^m -\Gamma_{im}^n  \Gamma_{jk}^m
) $$
$$ = S_m (-R_{ijk}^m) $$
And note that $ R_{ijk}^m=(D_i D_j E_k-D_jD_i E_k)^m $ Hence we
complete the proof
