'picturing' the covariant derivative

Question

Is it correct to say that the covariant derivative cannot be 'pictured' in the same way as other derivatives since the idea of an infinitesimally close neighbourhood of a point p in a manifold $M$ does not make sense i.e. $p+\epsilon$ is ill defined due to the fact that there is no such operation defined on a general Hausdorff topological space? Thus how a vector field Y varies in some direction does not make sense if we think of it as how the vectors of Y vary at points infinitesimally close to p. All we can do is merely create some idea of it in our heads which helps us relate the definition to other concepts like the directional derivative?

Answer 1

I wouldn't go as far as claiming that the covariant derivative can't be pictured. Consider a surface $S$ in $\mathbb{R}^3$ together with the Levi-Civita covariant derivative. Let $\gamma \colon I \rightarrow S$ be a curve in $S$. The curve specifies how to connect a point $p = \gamma(t_0)$ with "infinitesimally close" points $\gamma(t)$ for $t$ near $t_0$ so you don't need to worry about the fact that addition is not defined on $S$ - just work with $\gamma$.

Let $X$ be a vector field along $\gamma$ and think of $X = X(t)$ as a vector field along $\gamma$ in $\mathbb{R}^3$. Then we can take the usual derivative $\dot{X}(t)$ of $X$. The problem is that the resulting vector $\dot{X}(t)$ won't necessarily belong to the tangent space of $S$ at $\gamma(t)$. The covariant derivative $\frac{DX}{dt}(t)$ is then just the orthogonal projection of $\dot{X}(t)$ onto the tangent space of $S$ at $\gamma(t)$.

This is quite geometric. To compute the covariant derivative, you compute the regular derivative which decomposes into two components - the tangential component which is tangent to $S$ and the normal component which is normal to $T_pS$. The covariant derivative gives you only the tangential part. You can hold this picture in your head while learning differential geometry and notice that many things you do don't really require this geometric picture but only the formal properties of the covariant derivative which should give you enough motivation to consider an "abstract" covariant derivative.