Is this a correct understanding of the covariant derivative?

The notion of a covariant derivative is important when dealing with curved spaces. We want to be able to find the rate of change in some direction of a vector field. If the space is curved then this will affect the derivative of the vector field in some direction due to how the tangent vector is transported along the curved space to the vector field. So if you have some Riemannian manifold $M$, and a tangent vector $\frac{\partial}{\partial x^i} \rvert_p$ with $p\in M$, then $D_{\frac{\partial}{\partial x^i} \rvert_p} Y$ is essentially a way of giving the derivative of Y along $\frac{\partial}{\partial x^i} \rvert_p$ while taking into account the twists and contractions etc. of $M$. The path along which you transport $\frac{\partial}{\partial x^i} \rvert_p$ to a vector field Y, is loosely determined by the defined connection. Thus the way the manifold twists etc. will be associated with the connection D.


$\newcommand{\Basis}{\mathbf{e}}$Your description isn't exactly wrong, but it's too qualitative/unspecific to say it's right.

For brevity, if $X$ is a tangent vector at some point $p$ of $M$, and if $\gamma$ is a path defined on some neighborhood of $0$ and satisfying $\gamma(0) = p$ and $\gamma'(0) = X$, say $\gamma$ is tangent to $X$.

On a smooth Riemannian manifold $(M, g)$, there is a natural notion of parallel transport (with respect to the metric) of a tangent vector $Y$ along a piecewise-$C^{1}$ path $\gamma$. Qualitatively:

The covariant derivative $D_{X}Y(p)$ measures the failure of $Y$ to agree (near $p$, along a path $\gamma$ tangent to $X$) with the parallel transport of $Y(p)$ along $\gamma$.

Precisely, if $(\Basis_{j})_{j=1}^{n}$ is a frame at $p$, $X$ is a tangent vector at $p$, and $\gamma$ is tangent to $X$, there is a unique covariant-constant frame $(\Basis_{j})_{j=1}^{n}$ along $\gamma$, and consequently there exist uniquely-defined functions $Y^{j}$ such that $$ Y = \sum_{j=1}^{n} Y^{j} \Basis_{j} $$ along $\gamma$. Since $D_{X}\Basis_{j}(p) = 0$, the Leibniz rules gives $$ D_{X}Y = \sum_{j=1}^{n} dY^{j}(X)\, \Basis_{j}(p) = \sum_{j=1}^{n} X(Y^{j})\, \Basis_{j}(p), $$ the vector of derivatives of the components of $Y$ with respect to a covariant-constant frame.

Particularly, $D_{X}Y = 0$ if and only if $X(Y^{j})(p) = 0$ for each $j$, if and only if $Y$ is covariant-constant to first order along $\gamma$ at $p$.

  • $\begingroup$ Thanks for the response. $\endgroup$ – J.Main Jun 14 '17 at 14:04

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.