Covariant derivative do Carmo help clarification concept

Question

Let $M$ be a Riemannian manifold and let $\nabla$ be a connection compatible with the metric. Then for any vector fields $V,W$ along the smooth curve $c:I\rightarrow M$ , we have

$\frac{d}{dt}\langle V, W\rangle =\langle \frac{DV}{dt},W\rangle+ \langle V, \frac{DW}{dt}\rangle$

However my question is: Why does $\frac{d}{dt} \langle V, W \rangle$ make sense?

$V$ and $W$ are vector fields along a curve, so I suppose $\langle V, W\rangle$ is viewed as a map $I\rightarrow \mathbb{R}$ , where $\langle V,W \rangle(t)=\langle V_t, W_t \rangle$ . But how do we show that it is smooth? Also compatibility of the metric is defined on smooth vector fields.

Answer 1

If $M$ is a smooth manifold then $TM$ also has a natural smooth structure, hence, does $E=TM\times TM$. The projection $\pi: TM\to M$ defines a smooth map $$ \pi\times \pi: E\to M^2. $$ I will leave it to you to verify that the preimage of the diagonal under $\pi\times \pi$ is a smooth submanifold $A$ in $E$. Explicitly: $$ A=\{(u,v)\in TM\times TM: \pi(u)=\pi(v)\}, $$ i.e. $A$ consists of pairs of vectors in the same tangent space $T_pM$, $p\in M$.

A Riemannian metric $g$ on $M$ is a smooth function $g: A\to {\mathbb R}$ (satisfying further linearity conditions which are irrelevant for us). Given a curve $c: I\to M$, a vector field $X$ along $c$ is a smooth map $X: I\to TM$ such that $\pi\circ X=c$. A pair of such vector fields $X, Y$ defines a smooth map $P: I\to A$. Hence, the composition of two smooth maps $g\circ P$ is again smooth. This composition is nothing but the function $$ t\mapsto g(X,Y)(t), t\in I. $$

Answer 2

So here's the story without using pullback bundles: A vector field $V$ along the curve $c: I \rightarrow M$ is defined to be a smooth map $V: I \rightarrow T_*M$, where $V(t) \in T_{c(t)}M$. In local coordinates, the covariant derivative of $V = V^k\partial_k$ along $c$ is defined to be $$ \frac{DV}{dt} = \left(\frac{dV^i}{dt} + \Gamma^i_{jk}\dot{c}^j V^k\right)\partial_i, $$ where $\dot{c}$ is the derivative of $c$ and $\Gamma^i_{jk}$ are the Christoffel symbols. You can check that this definition is co-ordinate-independent and satisfies all the expected properties.

If you want the story using pullback bundles, see, for example, https://www.math.nyu.edu/~yangd/papers/pullback.pdf

Answer 3

Indeed, $\langle V,W \rangle: I\rightarrow \mathbb{R}$ given by $\langle V,W \rangle (t)=\langle V_t,W_t \rangle_{c(t)}$ is indeed smooth.

Let $M$ be your manifold. Let $t\in I$, So, $c(t)\in M$. Choose a frame $(U,e_1,.....,e_n)$ of $c(t)$ such that $\{$ $e_1,.....,e_n$ $\}$ forms an orthonormal local frame $U\rightarrow TM$ (Note $\langle e_i, e_j \rangle$ is smooth on $U$ for each $i,j$). Therefore $V_t=V^j(t)e_j$ and $W=W^k(t)e_k$ for smooth functions $V^j$ and $W^k$ on $I$. Hence, $\langle V_t , W_t\rangle= \sum_iV^i(t)W^i(t)$.