3
$\begingroup$

Our setting is $(M, g)$, a Riemannian manifold. Let $\Gamma(s,t) \subset M$ be a variation about curve $\gamma(t) = \Gamma(0, t)$ (Let us say that our domain of $\Gamma$ is $(a_0, a_1) \times (b_0, b_1) \subset \mathbb R^2$, and $(a_0, a_1)$ contains $0$.) Define

$$T = \partial_t \Gamma; S = \partial_s \Gamma.$$

My textbook says:

\begin{equation*} \frac{d}{ds} \langle T, T \rangle = 2\langle \nabla_S T, T \rangle.\end{equation*}

If I treat $\frac{\partial}{\partial s}$ as a tangent vector $S$ (or a vector field), then everything makes sense. However, I have a trouble understanding why $\frac{d}{ds}$ is a tangent vector at $T_p M$, where $p = \Gamma(s_0,t_0)$ for some $s_0, t_0$. Note that $\langle T, T \rangle$ is a function $(a_0, a_1) \times (b_0, b_1) \rightarrow \mathbb R$, so it can be treated it as a function from $\mathbb R^2$ to $\mathbb R$. I am merely taking a partial differentiation w.r.t. $s$, and it has nothing to do with tangent vector at $T_p M$. How do I resolve this?

$\endgroup$
3
  • $\begingroup$ It is the same idea when you make $\nabla_{\dot\gamma}\dot\gamma=0$ as the definition of a geodesic. Depending on the author, either you show $\nabla_{\dot\gamma}\dot\gamma$ (on a neighbourhood of the image of $\gamma$) does not depend on how you extend $\dot\gamma$, or you do it via a pullback to interval. The only difference here is that we have a product of interval instead that you need to pull back $(TM,\langle-,-\rangle,\nabla)$. $\endgroup$ Apr 30, 2021 at 11:57
  • $\begingroup$ @user10354138 I have been wondering about what you said too. $\nabla_{\dot \gamma} \dot \gamma$ is interpreted as: taking second derivative on $\gamma$. However, I still don’t understand why it is equivalent to the second derivative. Also, what do you mean by “$\nabla_{\dot\gamma} \dot\gamma$ does not depend on how you extend $\dot\gamma$?” If you have a reference material, I would be happy to take a look at it. $\endgroup$
    – James C
    May 1, 2021 at 2:30
  • $\begingroup$ (Understood what you meant by “$\nabla_{\dot\gamma} \dot\gamma$ does not depend on how you extend $\dot\gamma$.) However, I am still trying to find out why $\nabla_{\dot\gamma} \dot\gamma$ is “equivalent” to the second derivative. $\endgroup$
    – James C
    May 1, 2021 at 2:50

1 Answer 1

2
$\begingroup$

This is because the book uses the notion of covariant derivative along a curve. The covariant derivative along the curve $\sigma : I \to M$ is an operator that sends a vector field along $\sigma$ to another vector field along $\sigma$. If $X : I \to M$ and $Y : I \to M$ are two tangent field along the curve $\sigma$, that is $\forall t, X(t),Y(t) \in T_{\sigma(t)}M$, then $\langle X,Y\rangle$ is a function on $I$, and the covariant derivative $D_t$ along $\sigma$ satisfies:

$$ \frac{\mathrm{d}}{\mathrm{d}t} \langle X,Y\rangle = \langle D_t X,Y\rangle + \langle X,D_t Y\rangle. $$ Here, $\frac{\mathrm{d}}{\mathrm{d}t}$ is not a vector field on $M$ but the usual derivative on $I$.

Moreover, if $X$ is the restriction of a vector field of $M$ on a neighbourhood of $\sigma(t_0)$, then $D_t$ coincides with the usual covariant derivative: $$ D_t X|_{t=t_0} = \nabla_{\sigma'(t)}X|_{\sigma(t_0)}. $$ This should answer your question.

$\endgroup$
5
  • $\begingroup$ So, $\frac{d}{dt} = D_t$, where $D_t$ is covariant derivative along curve $\sigma$, and since $D_t$ commutes with inner product, $D_t\langle X, Y \rangle = \langle D_t X, Y \rangle + \langle X, D_t Y \rangle$. Is this the correct interpretation? $\endgroup$
    – James C
    May 1, 2021 at 8:01
  • $\begingroup$ No, $d/dt$ is the usual differentiation of functions and $t \mapsto \langle X(t),Y(t)\rangle$ is a smooth function on $I$. $D_t$ is the covariant derivative along $\sigma$, and these both different objects are related thanks to the above formula. $\endgroup$
    – Didier
    May 1, 2021 at 16:27
  • $\begingroup$ If $X$ is the restriction of a global vector field $X^0 \in \mathfrak{X}(M)$ along the curve $\sigma$, that is if $X(t) = {X^0}_{\sigma(t)}$, then $D_t X = \nabla_{\sigma'(t)} X^0$: the covariant derivative along $\sigma$ is in some way "the restriction" of the covariant derivative of $M$. The proof can be found in Lee's book, Petersen's book or Gallot, Hullin, Lafontaine's book. $\endgroup$
    – Didier
    May 27, 2021 at 10:55
  • $\begingroup$ (I rewrote my comment for easier legibility) Thank you for a nice answer. I have one last puzzle in my mind: X,Y are covariant derivatives along a curve. However, what does it have to do with the fact that $\frac{d}{dt} \langle X, Y \rangle = \langle \nabla_{\sigma'(t)} X, Y \rangle + \langle X, \nabla_{\sigma'(t)} Y \rangle$? $\endgroup$
    – James C
    May 27, 2021 at 11:48
  • $\begingroup$ I will look into Lee's book. Thank you. $\endgroup$
    – James C
    May 27, 2021 at 11:50

You must log in to answer this question.

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