# Covariant derivative in flat space [duplicate]

Is covariant derivative in flat space the same as vector as a derivative, i.e u(v)? If that is correct, doesn’t it also mean that flat space is always torsion-free?

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Within the context of the covariant derivative $$\bf{u}(\bf{v})$$, whether the flow curves generated by $$\bf u$$ and $$\bf v$$ are torsion-free or not depends on the affine connection chosen to define the Christofell symbols involved in the flat space, not the vector fields themselves.

To quickly prove this, the torsion tensor is given by: $$T(\bf{u},\bf{v})=\nabla_{\bf{u}}\bf{v}-\nabla_\bf{v}\bf{u}-[\bf{u},\bf{v}]$$ which upon deriving its tensor components, gives: $$=u^\mu(\partial_\mu v^\sigma+v^\nu\Gamma_{\mu\nu}^\sigma)\partial_\sigma-v^\mu(\partial_\mu u^\sigma+u^\nu\Gamma_{\mu\nu}^\sigma)\partial_\sigma$$ $$=u^\mu v^\nu\Gamma_{\mu\nu}^\sigma\partial_\sigma-v^\mu u^\nu\Gamma_{\mu\nu}^\sigma\partial_\sigma$$ $$T_{\mu\nu}^\sigma=\Gamma_{\mu\nu}^\sigma-\Gamma_{\nu\mu}^\sigma$$ Note: I'm not going through all the steps. For a full derivation see Eigenchris' explanation (19:50)

The property of a covariant derivative being torsion-free means that: $$T(\bf{u},\bf{v})=\nabla_{\bf{u}}\bf{v}-\nabla_\bf{v}\bf{u}-[\bf{u},\bf{v}]=0\implies\nabla_{\bf{u}}\bf{v}-\nabla_\bf{v}\bf{u}=[\bf{u},\bf{v}]$$ and $$\Gamma_{\mu\nu}^\sigma=\Gamma_{\nu\mu}^\sigma$$ Usually, the Levi-Civita connection is the best affine connection to use due to its metric compatibility and torsion-free characteristics. However, there are particular exceptions where a different (not necessarily torsion-free) connection should be used, especially in the case of physical interpretations.

One such example is the Frenet Serret equations (a set of kinematic equations) whose torsion is analogous to the angular momentum experienced by a symmetric "spinning top" pointing tangent to a curve. In this scenario, the connection used (and by extension, the coordinate system) results in a gap between the parallel transport of vectors within a field due to the twisting of reference frames.

In another case where an observer is travelling along a curve (not necessarily a geodesic path) carrying a tangent-pointing rod that traces out a pattern in the flat surface in front of the observer, the observer's parallel transport through the coordinates $$(t,x)$$ is described by: $$\nabla_{\frac{\partial}{\partial t}}\frac{\partial}{\partial x}\bigg|_{x=0}=0$$ and its torsion by the equation: $$T\left(\frac{\partial}{\partial x},\frac{\partial}{\partial t}\right)\bigg|_{x=0}=\nabla_{\frac{\partial}{\partial x}}\frac{\partial}{\partial t}\bigg|_{x=0}$$ Here, $$T\neq 0$$ signifies a helical pattern traced out by the tip of the rod.

The difference between such a connection and the torsion-free Levi-Civita connection applied to the same metric can be computed via the contorsion tensor: $$K_{\mu\nu\sigma}=\frac{1}{2}(T_{\mu\nu\sigma}+T_{\nu\sigma\mu}-T_{\sigma\mu\nu})$$

Note: for this answer, I am not including any type of non-flat Riemannian manifold.

Torsion is a quantity associated to an affine connection $$\nabla$$ defined on the tangent bundle of some manifold $$M$$. Its definition is that $$T(X,Y)=\nabla_XY-\nabla_YX-[X,Y]\tag{1}.$$

That said one must specify what connection one is talking about before discussing its torsion. In particular you ask "Is covariant derivative in flat space the same as vector as a derivative, i.e $$u(v)$$?" but this question has a problem of not realizing that any manifold, even flat space, can be endowed with various possible connections, some of which might have torsion and some of which might not.

What is true is that given a semi-Riemannian manifold $$(M,g)$$ there exists a unique metric-compatible torsion-free connection which is the Levi-Civita connection. In particular, in flat space, the covariant derivative which acts in the way you said in Cartesian coordinates $$\nabla_X Y = (X(Y^1),\dots, X(Y^n))\tag{2}$$

is exactly the Levi-Civita connection and hence torsion-free.

You could endow flat spacetime with other connections which do not take this form and do have torsion. So, no, flat space is not always torsion free, it depends on the affine connection you define on its tangent bundle.