Understand the definition of covariant derivative for parameterized set

Question

A geometric set $S \subset R^n$ is a set having the property that for each point $p \in S$, there is a vector subspace $T_pS \subset T_p\mathbb R^n$. Moreover, these subspaces should vary smoothly with $p$ and should all have the same dimension. Let $U\subset\mathbb R^k$ be a domain and let $\phi:U\rightarrow\mathbb R^n(k\leq n)$ be a smooth, one-to-one function that is regular for all $p\in U$. A parameterized set $S=\phi(U)$ is defined to be the image of $U$ in $\mathbb R^n$ by $\phi$. The geometric features of the parameterized set $S$ come from “encoding” features of the parameter space $U$ through the function $\phi$.

Then the author (First Steps in Differential Geometry Riemannian, Contact, Symplectic by Andrew McInerney) built intuition like Riemannian metrics, Riemannian Connection and curvature on the geometric set or more specifically the parameterized set.

Proposition 5.1.9. Let $(U, g)$ be a Riemannian space, $\mathcal{X}(U)$ the set of smooth vector fields on $U$, and $\Lambda_1(U)$ the set of smooth one-forms on $U$. Then the map $\gamma: \mathcal{X}(U) \rightarrow \Lambda_1(U)$ given by $\gamma(X)=i(X)$ g for $X \in \mathcal{X}(U)$, i.e., $\gamma(X)$ is the differential one-form such that for any vector field $Y$ on $U$, $$ (\gamma(X))(Y)=g(X, Y), $$ induces a vector space isomorphism $\gamma_p: T_p U \rightarrow T_p^* U$ for all $p \in U$.

Definition 5.3.2. Suppose $X$ and $Y$ are smooth vector fields on a Riemannian space $(U, g)$. Let $\theta_Y=\gamma(Y)$ be the one-form corresponding to the vector field $Y$ under the isomorphism $\gamma$ induced by $g$ defined in Proposition 5.1.9. Construct a new oneform $\theta_{Y, X}$ as follows: $$ \theta_{Y, X}=\frac{1}{2} i(X)\left[\mathcal{L}_Y g+d \theta_Y\right] . $$ The covariant derivative of $Y$ with respect to $X$ (relative to the metric tensor $g$ ), denoted by $\nabla_X Y$, is the vector field $$ \nabla_X Y=\gamma^{-1}\left(\theta_{Y, X}\right) . $$ The assignment $\nabla:(X, Y) \mapsto \nabla_X Y$ is also known as the Riemannian connection corresponding to $g$.

I didn't understand the motivation behind this definition of the covariant derivative, $\nabla_X Y$, as others have explained it as the horizontal (or tangential) component of the directional derivative, $D_X Y$, or as the change of $Y$ in the direction of $X$ using any curve $\gamma$ which goes in the direction of $X$ and uses parallel transport to transport $Y(\gamma(t))$ back to $T_pM$ in order to compare it with $Y(p)$. However, I couldn't connect Definition 5.3.2. with any of those. It would be greatly appreciated if anyone could shed some light on it.

Another question I have is: I didn't come across any examples for computing connections of vector bundles or covariant derivatives on abstract manifolds. Is there a resource where I can find examples or problems to work on? Without seeing these examples, I find it difficult to grasp the complete picture, and I'm starting to forget whatever I've read so far. Thank you in advance.

Answer 1

This is related to the Kozsul formula, also called the fundamental theorem of Riemannian geometry. It gives a formula for computing the unique connection on a Riemannian manifold that preserves the metric and is torsion-free. It is a bit long to write all the details, but see https://en.wikipedia.org/wiki/Fundamental_theorem_of_Riemannian_geometry.

In particular, you get concrete formulas for the Christoffel symbols, and that would allow you to compute many connections and connection forms, in response to your second question.

For the first one, you just need to check that the terms of the Kozsul formula match with the more abstract formula you put there. Let me write out this part.

We want to check that $\nabla_X Y =\gamma^{-1}\theta_{Y, X}$, that is, by definition, for a vector field $Z$, $$ g(\nabla_X Y, Z)=g(\gamma^{-1}\theta_{Y, X}, Z)=\theta_{Y, X}(Z). $$ Now $g(\nabla_X Y, Z)$ is exactly the subject of the Kozsul formula, so we want to check the RHS also produces this result. \begin{align*} \theta_{Y, X}(Z) &= \frac{1}{2} i(X)[{\mathcal L}_Y g + d\theta_Y](Z)\\ &=\frac{1}{2}[({\mathcal L}_Y g)(X, Z) + (d\theta_Y)(X, Z)]\\ &=\frac{1}{2}[({\mathcal L}_Y(g(X, Z))-g(({\mathcal L}_Y X, Z) - g(X, {\mathcal L}_Y Z) + X(\theta_Y(Z)) - Z(\theta_Y (X)) - \theta_Y([X, Z])]\\ &=\frac{1}{2}[Y(g(X, Z)) - g([Y, X], Z) - g(X, [Y, Z]) + X(g(Y, Z)) - Z(g(Y, X)) - g(Y, [X, Z])]\\ &=\frac{1}{2}[X(g(Y, Z)) + Y(g(X, Z)) - Z(g(Y, X)) - g([Y, X], Z) - g( [Y, Z], X)- g([X, Z], Y)], \end{align*} and this exactly the Kozsul formula.

Here we used the definition of the Lie derivative of a tensor, and the formula for exterior differentiation (see https://en.wikipedia.org/wiki/Exterior_derivative).

Answer 2

So if Three aggies answer was too neat for you here's a coordinate calculation that does pretty much the same thing, with added bonus that you get to keep track of indices. Let $Y=Y^i\partial_i$, $X=X^i\partial_i$, and $g=g_{ij}dx^i\otimes dx^j$. We first have that: \begin{align} \theta_{Y}=g_{ij}Y^idx^j \end{align} so: \begin{align} d\theta_Y=(\partial_{k}g_{ij}Y^i+g_{ij}\partial_kY^i)dx^k\wedge dx^j \end{align} Meanwhile, we have that: \begin{align} \mathscr{L}_Yg=(Y^k\partial_kg_{ij}+(\partial_iY^k)g_{kj}+(\partial_jY^k)g_{ik})dx^i\otimes dx^j \end{align} We see that the contraction with $X$ is given by: \begin{align} i(X)\mathscr{L}_Yg=X^i(Y^k\partial_kg_{ij}+(\partial_iY^k)g_{kj}+(\partial_jY^k)g_{ik})dx^j \end{align} meanwhile: \begin{align} i(X)d\theta_Y=&(\partial_kg_{ij}Y^i+g_{ij}\partial_kY^i)X^kdx^j- (\partial_kg_{ij}Y^i+g_{ij}\partial_kY^i)X^jdx^k\\ =&(\partial_ig_{kj}Y^k+g_{kj}\partial_iY^k-\partial_jg_{ki}Y^k-g_{ki}\partial_jY^k)X^idx^j \end{align} When we add the two together we get: \begin{align} \theta_{X,Y}=X^i\frac{1}{2}\left(Y^k\partial_ig_{kj}+g_{kj}\partial_iY^k-Y^k\partial_jg_{ik}+Y^k\partial_kg_{ij}+g_{kj}\partial_iY^k\right)dx^j \end{align} Then $\gamma^{-1}$ of this is: \begin{align} \gamma^{-1}(\theta_{X,Y})=&\frac{1}{2}X^ig^{jl}\left(g_{kj}\partial_iY^k+g_{kj}\partial_iY^k\right)\partial_l+\frac{1}{2}X^iY^kg^{jl}\left(\partial_ig_{kj}+\partial_kg_{ij}-\partial_jg_{ik}\right)\partial_l\\ =&X^i\left(\partial_iY^l+Y^k\Gamma^l_{ik}\right)\partial_l \end{align} which is precisely the Levi-Civita connection in coordinates.

Edit:

So the map $\gamma:\Omega^1(M)\rightarrow \mathfrak{X}(M)$ is induced by the isomorphism $TM\rightarrow T^*M$. Locally, this is just raising and lowering indices via the metric. So if $Z^i\partial_i$ is a vector field written in coordinates, the one form $\gamma^{-1}(Z^i\partial_i)$ is given in coordinates by $Z^ig_{ij}dx^j$. The inverse is just given by using the inverse metric $g^{ij}$. So if we have a one form $\omega_idx^i$, then the vector field associated to it $\omega_ig^{ij}\partial_j$.