# $\frac{d}{ds} \langle T, T \rangle = 2\langle \nabla_S T, T \rangle$?

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?

• 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 $(TM,\langle-,-\rangle,\nabla)$. Apr 30, 2021 at 11:57
• @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. May 1, 2021 at 2:30
• (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. May 1, 2021 at 2:50

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.

• 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? May 1, 2021 at 8:01
• 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. May 1, 2021 at 16:27
• 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. May 27, 2021 at 10:55
• (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 $\frac{d}{dt} \langle X, Y \rangle = \langle \nabla_{\sigma'(t)} X, Y \rangle + \langle X, \nabla_{\sigma'(t)} Y \rangle$? May 27, 2021 at 11:48
• I will look into Lee's book. Thank you. May 27, 2021 at 11:50