Levi-Civita connection in 1d In a one dimensional reimann manifold (topologically equivalent to the real line). with metric tensor $g_{uv}$. what is the Levi-Civita connection? Is it just always equal to the regular derivative $d$ regardless of metric $g_{uv}$?
I suspect this is true, since regardless of the metric tensor, there is only one choice of geodesics.
 A: I'm not sure what you mean by "the regular derivative", but the Levi-Civita connection absolutely does depend on the metric.  Explicitly, let's suppose our manifold is $\mathbb{R}$ and consider the metric tensor as a smooth function $g:\mathbb{R}\to\mathbb{R}_+$ (since the metric tensor has only one entry).  Let $X$ denote the vector field $\frac{d}{dx}$ on $\mathbb{R}$ (i.e., it acts on functions by ordinary differentiation).  The Koszul formula for the Levi-Civita connection $\nabla$ then gives $$\nabla_XX=\frac{g'}{2g}X$$ which determines the connection since every vector field is a smooth function times $X$.  Explicitly, for any smooth function $f$, we get $$\nabla_X(fX)=\left(f'+\frac{fg'}{2g}\right)X$$ using the Leibniz rule.  If you are identifying vector fields with functions via $fX\mapsto f$, this means the "derivative" of a function $f$ (thought of as a vector field) given by the connection is not just the ordinary derivative $f'$ but $f'+\frac{fg'}{2g}$.
Note that while in a $1$-dimensional manifold there is only one possible image that a geodesic can have (independent of the metric), a geodesic must trace out that image with constant speed (with respect to the metric), and so the parametrization of geodesics very much depends on the metric.
