I am looking for a counter example that why the $\mathbb{R}^n$ definition of vector field fail on a manifold.
The following is a summary of what I learnt few years ago.
Start with the idea of differentiating vector fields in Euclidean space: choose a fixed orthonormal basis $\{e_1,\cdots,e_n\}$ for $\mathbb{R}^n$, then any vector field $X$ can be written in the form $\sum_{i=1}^n X^i e_i$ for some smooth function. $X^1,\cdots,X^n$. Then the derivative of $X$ in the direction of a vector $v$ is given by $D_v X=\sum_{i=1}^n v(X^i)e_i$. Essentially, we just differentiate the coeffient functions. The key is to think of vectors $\{e_1,\cdots,e_n\}$ as being constant.
However, on a manifold, this no longer holds: There are no natural vector field which we can take to be `constant'.
Could anyone find a simple explanation of the above sentence? And a counter example?
Best Answer
$\newcommand{\dd}{\partial}\newcommand{\Reals}{\mathbf{R}}\newcommand{\Basis}{\mathbf{e}}$The idea is, just as you say, that in Euclidean space (Cartesian space equipped with the Euclidean metric) there exist "constant" vector fields whose values at each point form a basis of the tangent space, while on a general manifold (including Cartesian space without the Euclidean metric), there do not exist distinguished "constant" vector fields.
For purposes of this question, a vector field $\Basis$ is constant if its "derivative" with respect to an arbitrary vector field $v$ vanishes. Intuitively, a "constant" vector field should have "constant length and direction".
On a general smooth manifold $M$, "length" has no intrinsic meaning; every linear automorphism of a tangent space $T_{p} M$ is effected by some change of coordinates at $p$.
We might say that $\Basis$ is "constant with respect to a coordinate system" if the components of $\Basis$ are constant, but unfortunately this notion depends on the coordinate system, and there exists no (non-zero) vector field constant with respect to an arbitrary coordinate system. With respect to polar coordinates on $\Reals^{2}$, for example, the Cartesian basis is given by $$ \Basis_{1} = (\cos\theta) \dd_{r} - \tfrac{1}{r}(\sin\theta) \dd_{\theta},\qquad \Basis_{2} = (\sin\theta) \dd_{r} + \tfrac{1}{r}(\cos\theta) \dd_{\theta}; $$ the component functions aren't constant.
If the manifold $M$ is equipped with the "extra structure" of a Riemannian metric $g$, however, the notion of an "orthonormal frame" (mutually-perpendicular unit vector fields forming a basis of the tangent space at each point) makes sense, and there is a "more restricted" notion of differentiation in which the metric is used to "detect change of direction" of a vector field along a path. In Euclidean space, it turns out this "covariant derivative" allows us to define an orthonormal frame $(\Basis_{i})_{i=1}^{n}$ that is covariantly constant. Consequently, the formula $$ D_{v} X = \sum_{i=1}^{n} v(X^{i})\, \Basis_{i} $$ holds.
In more detail, with links for reference:
Arbitrary coordinate vector fields on a smooth manifold are "constant with respect to themselves" (in the sense of having vanishing Lie derivative), but are not constant with respect to arbitrary fields.
In the presence of a Riemannian metric, there is a notion of covariant differentiation induced by the Levi-Civita connection of the metric. On a Riemannian manifold, an arbitrary vector can be "carried" along a smooth path $\gamma$ by parallel transport; the resulting vector field is "constant along $\gamma$". Parallel transport does not, however, allow a tangent vector at a point $p$ to be extended to a "constant" vector field in a neighborhood of $p$, because the result of parallel transport generally depends on the path $\gamma$, not merely on the endpoints of the path.
In Euclidean space, parallel transport depends only on the endpoints of a path (because the Euclidean metric is flat and $\Reals^{n}$ is simply-connected). Consequently, each tangent vector at the origin gives rise to a unique "covariantly constant" vector field on $\Reals^{n}$. (Starting with the standard basis at the origin gives the Euclidean vector fields.)