Levi Civita connection intuition and motivation Can someone explain why need we the Levi-Civita connection and what it does intuitively?
 A: I fix a Riemannian manifold $(M,g)$ of dimension $n$, a (linear) connection $\nabla$ on $M$.
I recall you that we can write, in a local chart $(U,\varphi)$ of $M$:
\begin{gather*}
\forall i,j\in\{1,\dots,n\},\,g\left(\frac{\partial}{\partial x^i},\frac{\partial}{\partial x^j}\right)=g_{ij},\\
\forall P\in U,v,w\in T_PM,\,g_P(v,w)=g_{ij}(P)[dx^i\otimes dx^j(v,w)]\equiv\langle v,w\rangle_P;
\end{gather*}
where $\{x^1,\dots,x^n\}$ are the local coordinates of $M$ in $U$, and $\left\{\textstyle\frac{\partial}{\partial x^1},\dots,\frac{\partial}{\partial x^n}\right\}$ is the corresponding local frame in $TM$ (the tangent bundle of $M$).
Then $\langle\cdot,\cdot\rangle$ is a vector field on $M$ defined by $g$, and it makes sense compute $\nabla_X\langle Y,Z\rangle$; where $X$, $Y$ and $Z$ are vector fields on $M$.
I say that:


*

*$\nabla$ is compatible with the metric induced by $g$ if
\begin{equation*}
\forall X,Y,Z\in\mathfrak{X}(M),\,\nabla_X\langle Y,Z\rangle=\langle\nabla_XY,Z\rangle+\langle Y,\nabla_XZ\rangle;
\end{equation*}

*$\nabla$ is symmetric if in any local chart $(U,\varphi)$ of $M$, the Christoffel symbols are symmetric; I mean that
\begin{equation*}
\forall i,j,k\in\{1,\dots,n\},\,\Gamma_{ij}^k=\Gamma_{ji}^k;
\end{equation*}

*$\nabla$ is the Levi-Civita connection if it is symmetric and compatible with the metric induced from $g$.


Let $\sigma:[0,1]\equiv I\to M$ be a smooth curve over $M$; I remember you that (a generic connection) $\nabla$ induces a unique covariant derivation $\nabla^{\sigma}$ along $\sigma$, and via $\nabla^{\sigma}$ you can define the parallel transport $\widetilde{\sigma}$ of the vector fields on $M$ along $\sigma$.
One can prove that $\nabla$ is compatible with the metric induced from $g$, if and only if for any smooth curve $\sigma$ on $M$, $\widetilde{\sigma}$ is an isometry! Moreover, if $\nabla$ is the Levi-Civita connection on $M$, one can prove that:
\begin{equation*}
\forall V\in\mathfrak{X}(\sigma),\,d(\nabla^{\sigma}(V))=\nabla^{\sigma}(dV(f))
\end{equation*}
where $\mathfrak{X}(\sigma)$ is the vector space of vector fields on $M$ along $\sigma$. Or in a general setting:
\begin{equation*}
\forall X,Y\in\mathfrak{X}(M),\,d(\nabla_XY)=\nabla_{dX}(dY).
\end{equation*}
Roughly speaking, the Levi-Civita connection and the connections induced on the smooth curves on $M$ commute with the differential!
However, in general $\widetilde{\sigma}$ is a linear isomorphism between $E_{\sigma(0)}$ and $E_{\sigma(1)}$, whenever $\nabla$ is a connection on a vector bundle $E$ over $M$.
Some reference:


*

*Kobayashi and Nomizu - Foundations od Differential Geometry; volume 1, chapters 2, 3 and 4; I have a doubt about the chapter 5.

*Kolář, Michor and Slovák - Natural Operations in Differential Geometry; chapters 3 and 10; available at http://www.emis.de/monographs/KSM/kmsbookh.pdf.

*Spivak - A Comprehensive Introduction to Differential Geometry; volume II, chapter 6.


If you would like, I can explain a Lagrangian mechanical application of the Levi-Civita connection...
Is it all clear?
