The necessity of the Lie bracket of vector fields in the definition of the Riemann curvature tensor The Riemann curvature tensor is defined as follows:
Let $\mathscr{X}$ be a set of all vector fields and define the multiply linear map $R$ from $\mathscr{X} \times \mathscr{X} \times \mathscr{X}$ to $\mathscr{X}$ by
$R(X,Y)Z := \nabla_X \nabla_Y Z - \nabla_Y \nabla_X Z - \nabla_{[X,Y]}Z$.
Then Riemann curvature tensor is defined by $(R^t_{k,ij})$, where $R(\frac{\partial}{\partial x^i},\frac{\partial}{\partial x^j})\frac{\partial}{\partial x^k}=\sum_{t=1}^n R^t_{k,ij} \frac{\partial}{\partial x^t}$.
However, even if $R$ is defined by
$R(X,Y)Z := \nabla_X \nabla_Y Z - \nabla_Y \nabla_X Z$,
the $R^t_{k,ij}$ equals.
Since different multiple linear maps should define different tensors, there is some mistake in my thinking.
I would like you to point that out.
 A: Without the $\nabla_{[X,Y]}$ the expression for $R(X,Y)Z$ would not be a tensor because it would depend on derivatives of the $X,Y$ fields. A tensor $T(X,Y)$ depends only on the components of $X$, $Y$ at a point and not on their derivatives.
A: The two definitions of the coefficients ${R^t}_{kij}$ coincide when $[X,Y]=0$ and this is the case for $X= \frac{\partial}{\partial x^i}$ and $Y= \frac{\partial}{\partial x^j}$.
However the definition of the tensor $R_p$ as multilinar map on the tangent space at a given point $p$ concerns generic vector fields $X= X^i\frac{\partial}{\partial x^i}$ and $Y= Y^j\frac{\partial}{\partial x^j}$ such that $[X,Y]\neq 0$.
In this case
$$R(X,Y)Z := \nabla_X \nabla_Y Z - \nabla_Y \nabla_X Z  \quad (wrong)$$
does not define a three-linear map from $T_pM\times T_pM\times T_pM$ to $T_pM$, since the right-hand side depends on the values of the vector fields in a nighborhood of $p$. As a consequence, the right-hand side does not define a tensor at $p$.
Instead, the correct definition
$$R_p(X_p,Y_p)Z_p := (\nabla_X \nabla_Y Z)_p - (\nabla_Y \nabla_X Z)_p - (\nabla_{[X,Y]}Z)_p$$
satisfies that requirement above -- it only depends on the behavior of the vector fields $X,Y,Z$ exactly at $p$ in spite of the presence of derivatives on the right-hand side -- so that it defines a tensor at every  $p$.
