Torsion tensor of the euclidean space I'm struggling to prove that the euclidean space ($\mathbb R^n$ with the euclidean riemannian metric) is torsion free, i.e.,
$$[X,Y]=\partial_XY-\partial_YX$$
I made all the identifications but I can't finish.
Thank you.
 A: The standard Euclidean Riemannian metric on $\Bbb R^n$ may be written
$ds^2 = \displaystyle \sum_{i, j = 1}^n g_{ij}dx_i dx_j = \sum_{i = 1}^n dx_i^2, \tag 1$
there the $x_i$, $1 \le i \le n$, are the standard global Euclidean coordinates on $\Bbb R^n$.  We see that in this coordinate system, all the metric coefficients $g_{ij}$ are constant; thus, the Christoffel symbols $\Gamma_{jk}^i$ all vanish, since they depend linearly on the first derivatives of the $g_{ij}$.  Therefore covariant derivatives and ordinary derivatives co-incide for this metric.  Then for vector fields
$X = \displaystyle \sum_{i = 1}^n X^i \dfrac{\partial}{\partial x_i}, \tag 2$
$Y = \displaystyle \sum_{i = 1}^n Y^i \dfrac{\partial}{\partial x_i}, \tag 3$
we see that
$\partial_X Y = \left (  \displaystyle \sum_{i = 1}^n X^i \dfrac{\partial}{\partial x_i} \right ) \displaystyle \sum_{j = 1}^n Y^j \dfrac{\partial}{\partial x_j} = \sum_{j = 1}^n \left (\sum_{i = 1}^n X^i \dfrac{\partial Y_j}{\partial x_i} \right ) \dfrac{\partial}{\partial x_j} \tag 4$
and likewise,
$\partial_Y X = \displaystyle \sum_{j = 1}^n \left (\sum_{i = 1}^n Y^i \dfrac{\partial X_j}{\partial x_i} \right ) \dfrac{\partial}{\partial x_j} \tag 5$
and therefore,
$\partial_X Y - \partial_Y X = \displaystyle \sum_{j = 1}^n \left (\sum_{i = 1}^n X^i \dfrac{\partial Y_j}{\partial x_i} \right ) \dfrac{\partial}{\partial x_j} - \displaystyle \sum_{j = 1}^n \left (\sum_{i = 1}^n Y^i \dfrac{\partial X_j}{\partial x_i} \right ) \dfrac{\partial}{\partial x_j}$
$= \displaystyle \sum_{j = 1}^n \sum_{i = 1}^n \left ( X^i \dfrac{\partial Y_j}{\partial x_i} - Y^i \dfrac{\partial X_j}{\partial x_i} \right ) \dfrac{\partial}{\partial x_j}.  \tag 6$
Now for $f$ a differentiable function on $\Bbb R^n$ we may also compute $[X, Y]f$:
$[X, Y]f = X[Y[f]] - Y[X[f]]$
$=  \displaystyle \sum_{i = 1}^n X^i \dfrac{\partial}{\partial x_i}  \left [ \sum_{j = 1}^n  Y^j \dfrac{\partial f}{\partial x_j} \right ] - \sum_{1 = 1}^n Y^i \dfrac{\partial}{\partial x_i} \left [ \sum_{j = 1}^n X^j \dfrac{\partial f}{\partial x_j} \right ]$
$= \displaystyle  \sum_{j = 1}^n  \sum_{i = 1}^n X^i \dfrac{\partial}{\partial x_i}  \left [ Y^j \dfrac{\partial f}{\partial x_j} \right ] - \sum_{j = 1}^n  \sum_{i = 1}^n Y^i \dfrac{\partial}{\partial x_i} \left [X^j \dfrac{\partial f}{\partial x_j} \right ]$
$= \displaystyle \sum_{j = 1}^n  \sum_{i = 1}^n X^i \dfrac{\partial Y^j}{\partial x_i} \dfrac{\partial f}{\partial x_j} + \sum_{j = 1}^n \sum_{i = 1}^n X^i Y^j   \dfrac{\partial^2 f}{\partial x_i \partial x_j}$
$ - \displaystyle \sum_{j = 1}^n  \sum_{i = 1}^n Y^i \dfrac{\partial X^j}{\partial x_i} \dfrac{\partial f}{\partial x_j} - \sum_{j = 1}^n  \sum_{i = 1}^n Y^i X^j \dfrac{\partial^2 f}{\partial x_i\partial x_j}$
$= \displaystyle \sum_{j = 1}^n  \sum_{i = 1}^n X^i \dfrac{\partial Y^j}{\partial x_i} \dfrac{\partial f}{\partial x_j} - \sum_{j = 1}^n  \sum_{i = 1}^n Y^i \dfrac{\partial X^j}{\partial x_i} \dfrac{\partial f}{\partial x_j}; \tag 7$
therefore,
$[X, Y]f = \left [ \displaystyle  \sum_{j = 1}^n  \sum_{i = 1}^n X^i \dfrac{\partial Y^j}{\partial x_i} \dfrac{\partial}{\partial x_j} - \sum_{j = 1}^n  \sum_{i = 1}^n Y^i \dfrac{\partial X^j}{\partial x_i} \dfrac{\partial}{\partial x_j} \right ] f, \tag 8$
since the bracket $[X, Y]$ is uniquely determined by its action on differentiable scalar functions, this shows that the vector field
$[X, Y] =  \displaystyle  \sum_{j = 1}^n  \sum_{i = 1}^n \left ( X^i \dfrac{\partial Y^j}{\partial x_i} \dfrac{\partial}{\partial x_j} - Y^i \dfrac{\partial X^j}{\partial x_i} \right ) \dfrac{\partial}{\partial x_j}  = \partial_X Y - \partial_Y X, \tag 9$
and thus the torsion tensor
$T(X, Y) =  \partial_X Y - \partial_Y X - [X, Y] = 0. \tag{10}$
