Torsion tensor of the euclidean space

Question

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.

Answer 1

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

$T(X, Y) = \partial_X Y - \partial_Y X - [X, Y] = 0. \tag{10}$