1
$\begingroup$

Actually, I am getting confused in the following line "$\nabla_{X_p}Y$" is the directional derivative of the vector field $Y$ along the tangent vector $X_p$.

When we say that the directional derivative of the vector field $Y$ along a curve say $\gamma$ that is $\nabla_{\dot{\gamma(t)}}Y$ so in this case we fix a point $\gamma(t)$ on the curve in the manifold and choose some other point $p$ and then we have a tangent vector $Y_p$ and then we parallel translate the tangent vector at $\gamma (t)$ say $X_{\gamma(t)}$ to point $p$.Then we look for the change and accordingly define the covariant derivative.

My question is in the case of the directional derivative along the curve we moved the vector at $\gamma(t)$ to $Y$ along the curve but what in the case of the directional derivative along the tangent vector $X_p$ mean geometrically? I mean how we are moving our point along what??

$\endgroup$
5
  • $\begingroup$ This should be $\nabla_{\gamma'(t)}Y$, not $\nabla_{\gamma(t)}Y$. $\endgroup$
    – Didier
    Commented Nov 29, 2022 at 17:32
  • $\begingroup$ ohh yes ,thanks. $\endgroup$
    – Andyale
    Commented Nov 29, 2022 at 17:33
  • $\begingroup$ I find your question hard to understand. You are studying how the vector field $Y$ is changing instantaneously at $p$ as you move along any curve $\gamma$ with $\gamma(0)=p$ and $\gamma'(0)=X_p$. One option, if you have the vector field $X$, is to take the integral curve of $X$ through $p$. $\endgroup$ Commented Nov 29, 2022 at 17:40
  • $\begingroup$ Actually, my question is that when we say that directional derivative of the vector field $Y$ along $X_p$ what does it mean I mean are we moving along with the curve parallel from one point to another? I am finding it difficult to visualize. $\endgroup$
    – Andyale
    Commented Nov 29, 2022 at 17:55
  • 1
    $\begingroup$ "along" is not the right word; just like in calculus, we're computing the covariant derivative of the vector field $Y$ at $p$ in the direction of $X_p$. $\endgroup$ Commented Nov 29, 2022 at 18:58

2 Answers 2

2
$\begingroup$

When computing the derivative of a vector field, you want to somehow construct the difference quotient between $Y_p\in T_pM$ and $Y_{γ(t)}\in T_{γ(t)}$.

Now the tangent vector spaces are not the same, they form a bundle over the manifold, which implies some continuity, but that does not allow to embed one into the other uniquely.

So what is needed is some map $A_t:T_{γ(t)}\to T_pM$, so that then the divided quotient $\frac{A_tY_{γ(t)}-Y_p}{t}$ makes sense, has a limit, and that limit does only depend, additionally to $Y$, on $γ'(0)$.

The question is now transformed on how to construct such a map that it is uniquely defined for every path $p$ in some consistent way, independent of the coordinate system used in its construction.

If the maps $A_t$ are isometric towards some Riemannian metric on $M$ and in some sense $t\mapsto A_t$ realizes the "least torsion", one can call it "parallel transport along the path". In extension, also non-metric variants of it are called parallel transport, especially on general vector bundles that have no inherent connection to the metric on the manifold.

$\endgroup$
1
$\begingroup$

Another way to think of it is this. Many books define the tangent space as an equivalence class of paths going through that point. So just pick one of the curves in the equivalence class.

Here is an example reference: https://en.wikipedia.org/wiki/Tangent_space#Definition_via_tangent_curves

$\endgroup$
7
  • $\begingroup$ How is it related to covariant derivatives or affine connections? $\endgroup$
    – Didier
    Commented Dec 6, 2022 at 21:34
  • $\begingroup$ As I understood the question, it was how to intuitively understand the covariant derivative along a vector, when one intuitively understood the covariant derivative along a path. $\endgroup$ Commented Dec 6, 2022 at 21:40
  • $\begingroup$ Sure. But your answer points to the definition of the tangent space (particularly to its definition via equivalent classes of curves). This is not directly related to covariant derivatives. The covariant derivative (along a path) is another object $\endgroup$
    – Didier
    Commented Dec 6, 2022 at 21:42
  • $\begingroup$ But it does answer his question. $\endgroup$ Commented Dec 6, 2022 at 21:44
  • $\begingroup$ I am really not convinced, but well, if you say so $\endgroup$
    – Didier
    Commented Dec 6, 2022 at 21:45

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .