Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $M$ be a Riemannian manifold with metric tensor $g$ and Levi-Civita connection $\nabla$.

Also, let $u: \mathbb{R}\to T_pM$ be a smooth curve in $T_pM$. In a proof, my course notes assure that

$$\frac{d}{dt} \left(g_p(u(t),u(t))\right)=2g_p\left(\frac{d}{dt} u(t),u(t)\right)\text{ (1).}$$

This is giving me a few problems. First of all, $\frac{d}{dt} u(t)$ is an element of $T(T_pM)$, but we identify it with $T_pM$, is that right ?

But then, for me, by definition of the Levi-Civita connection,

$$\frac{d}{dt} \left(g_p(u(t),u(t))\right)=2g_p\left(\frac{D}{dt}u(t),u(t)\right)\text{ (2)},$$ where $\frac{D}{dt}$ is the covariant derivative, giving here $$\frac{D}{dt} u(t)=\dot u_i(t)\frac{\partial}{\partial x_i}+u_i(t)\nabla_{\dot u(t)}\frac{\partial}{\partial x_i}$$ with $\frac{\partial}{\partial x_i}$ a basis of $T_pM$.

Are the two expressions equal (which doesn't seem because of the second terme of the last equation) or does one of them contain a mistake ?

share|cite|improve this question
up vote 3 down vote accepted

You got a small mistake in your last formula. However, you can see the result in two different way.

The first one, simplest one, is the following.

If you fix $p\in M$ then the map $(X,Y)\mapsto g_p(X,Y)$ is a bilinear map from $T_pM\times T_pM$ to $\mathbb R$. But $g_p$ gives you a norm on the vector space $T_pM$.

The curve $u:\mathbb R \rightarrow T_pM$ can then be seen as a smooth map on the vector space $T_pM \cong \mathbb R^n$. So everything is about derivative in a fixed vector space and you know the formula of the derivative of a bilinear map, a composition of two smooth maps, etc...

Finally, you get that the derivative of the map $h:t\mapsto g_p(u(t),u(t))$ is exactly $$h'(t)=g_p(u'(t),u(t))+g_p(u(t),u'(t))=2g_p(u'(t),u(t)).$$

The second one, using the covariant derivative along a curve.

Let me remind you that a vector field along a curve $\gamma:\mathbb R \rightarrow M$ is a smooth map $X:\mathbb R \rightarrow TM$ such that $\pi\circ X=\gamma$ where $\pi:TM\rightarrow M$ is the canonical projection.

If we fix coordinates $(x_1,\cdots,x_n)$ on $M$, and write $X=\sum_{i=1}^nX_i\dfrac{\partial}{\partial x_i}$ then $$\dfrac{DX}{dt}=\sum_{i=1}^n\dfrac{dX_i}{dt}\dfrac{\partial}{\partial x_i}+\sum_{i=1}^nX_i\nabla_{\gamma'}\dfrac{\partial}{\partial x_i}$$

Now, one can see your map $u$ as a smooth map : $X:\mathbb R\rightarrow TM, t\mapsto (p,u(t))$ i.e. $\forall t\in \mathbb R, X(t)\in T_pM$ and its value is $u(t)$.

Since $\forall t\in\mathbb R, \pi\circ X(t)=p$ is a constant map, you can say that $X$ is a vector field along the constant curve $\gamma:\mathbb R \rightarrow M, t\mapsto p$.

Hence $\gamma'(t)=0$ and the computation of $\dfrac{DX}{dt}$ gives only one term: $$\dfrac{DX}{dt}=\sum_{i=1}^n\dfrac{dX_i}{dt}\dfrac{\partial}{\partial x_i}=\sum_{i=1}^n \dot{u}_i(t)\dfrac{\partial }{\partial x_i}=\dfrac{dX}{dt}$$ You finally recover the formula : $$\dfrac{d}{dt}g_p(u(t),u(t))=\dfrac{d}{dt}g_{\gamma(t)}(X,X)=2g_{\gamma(t)}(\dfrac{DX}{dt},X)=2g_p(\dfrac{dX}{dt},X)=2g_p(u'(t),u(t)).$$

share|cite|improve this answer
I wish I could upvote your answer twice, it really clarified everything! Thanks a lot! – Klaus Dec 9 '12 at 13:24

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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