Torsion and curvature of a linear connection on a parallelizable manifold Could you help me to solve the following problem?

Let $M$ be a parallelizable manifold of dimension $n$, $\{E_1,\dots, E_n\}$ a global frame of $M$. Let $X$, $Y$ be vector fields on $M$ with $Y= \sum_{i=1}^n Y^i E_i$. Let $$\nabla_X Y = \sum_{i=1}^n {X(Y^i) E_i}.$$ Compute the torsion and curvature of $\nabla$.

I tried to compute the curvature:
\begin{align*}
&\ R(X,Y,Z)\\ 
=&\ R_{XY}Z\\
=&\ \nabla_X \nabla_Y Z - \nabla_Y \nabla_X Z -\nabla_{[X,Y]} Z\\
=&\ \nabla_X (Y(Z^j)E_j) - \nabla_Y (X(Z^i)E_i) - [X,Y](Z^k)E_k\\
=&\ X(Y(Z^j))E_j + Y(Z^j) \nabla_X E_j - Y(X(Z^i))E_i - X(Z^i) \nabla_Y E_i -  XY(Z^k)E_k +YX(Z^k)E_k\\
=&\ X(Y(Z^j))E_j - XY(Z^k)E_k - YX(Z^i)E_i + YX(Z^k)E_k + Y(Z^j) \nabla_X E_j  - X(Z^i) \nabla_Y E_i\\
=&\ Y(Z^j) \nabla_X E_j  - X(Z^i) \nabla_Y E_i\\
=&\ 0.
\end{align*}
Is it correct? What about the torsion?
Thanks!
 A: For some reason you seem to expand $\nabla_X Y(Z^j)E_j$ as $X(Y(Z^j))E_j + Y(Z^j)\nabla_XE_j$, while according to the definition, $\nabla_X Y(Z^j)E_j = X(Y(Z^j))E_j$. It follows directly from the definition of $\nabla$ that $\nabla_XE_j = 0$, so you end up with the correct expression for $R(X, Y, Z)$, namely zero, but your computation is more complicated than it needs to be.
You can compute the torsion by using the definition of the connection together with properties of the Lie bracket:
\begin{align*}
T(X, Y) &= \nabla_XY - \nabla_YX - [X, Y]\\
&= \nabla_X(Y^jE_j) - \nabla_Y(X^iE_i) - X(Y^j)E_j + Y_j[X^iE_i, E_j]\\
&= X(Y^j)E_j - Y(X^i)E_i -  X(Y^j)E_j - Y^j[X^iE_i, E_j]\\
&= -Y(X^i)E_i + Y^j[E_j, X^iE_i]\\
&= -Y(X^i)E_i + Y^jE_j(X^i)E_i + Y^jX^i[E_j, E_i]\\
&= -Y(X^i)E_i + Y(X^i)E_i - X^iY^j[E_i, E_j]\\ 
&= -X^iY^j[E_i, E_j].
\end{align*}
An even easier computation is to note that torsion is a tensor, so $T(X, Y) = T(X^iE_i, Y^jE_j) = X^iY^jT(E_i, E_j)$ and
$$T(E_i, E_j) = \nabla_{E_i}E_j - \nabla_{E_j}E_i - [E_i, E_j] = -[E_i, E_j]$$
so $T(X, Y) = -X^iY^j[E_i, E_j]$ as before.
