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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?

math101
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  • Your title asks about vector fields, but the question body seems to ask about differentiating vector fields. It's also not clear to me whether you're interested in Riemannian manifolds or general smooth manifolds. Are you asking why the indicated formula for $D_{v}X$ makes no sense on a smooth manifold, i.e., are you asking why coordinate vector fields on a smooth manifold aren't "constant"? Or are you asking for an invariant/geometric sense in which the Cartesian coordinate fields are "constant" in (flat) Euclidean space? Or something else...? – Andrew D. Hwang Aug 20 '15 at 09:07
  • I am at a beginner level differential geometry. What I really wanted to know here is..Why vector fields are defined differently on Euclidean space and manifold? – math101 Aug 20 '15 at 11:56
  • And also, I wanted to know why one cannot differentiate vector field on a smooth manifold like the way we differentiate vector field on $\mathbb{R}^n$? – math101 Aug 20 '15 at 12:10
  • There is a natural Euclidean structure defined on $R^n$ and you can always transport vectors. So there is a canonical isomorphism between two tangent space at different point. You can compare vectors at different point. You can talk something like "constant" vectors. But for general manifold you can't. – Xipan Xiao Aug 20 '15 at 15:24

1 Answers1

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$\newcommand{\dd}{\partial}\newcommand{\Reals}{\mathbf{R}}\newcommand{\Basis}{\mathbf{e}}$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.)

Andrew D. Hwang
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