Geodesics, isometries and connections.

I am trying to prove with the differential definition that isometries preserve geodesics. That is: Let $c(t)$ be a geodesic in a $n$-dimensional semirriemannian manifold $M$ and $g$ an isometry of $M$ on itself. Then $g\circ c$ will be a geodesic if $$\frac{d^2y^k}{dt^2}+\Gamma^k_{ij}\frac{dy^i}{dt}\frac{dy^j}{dt}=0,\ \ k=1,...,n,\ (\ast)$$ where $(\phi\circ g\circ c)(t)=(y^1(t),...,y^n(t)$ for a chart $\phi$ of $M$.

Doing some research online I found Isometries preserve geodesics . Here they talk about connections. I've looked for a definition (that allowed me to understand them) of connection for quite a while but so far the whole concept makes no sense to me.

Why is it necessary to use connections to prove $(\ast)$? Can $(\ast)$ be proved without using them? What is a connection? Is a concrete mathematical object like a set with a concrete structure or it is just a "meta-object" just to say that "we transport data along a curve"? I am really confused about the connection topic but I'm just a beginner in differential geometry so I guess I'm missing some obvious fact.

Thank u in advance. Any help would be appreciated.

• connections or covariant derivatives are generalizations of the directional derivative concept suitable to determine how vector and tensor fields vary among them – janmarqz Jun 11 '16 at 15:48
• take a look at math.stackexchange.com/questions/1099619/… – janmarqz Jun 11 '16 at 15:59
• IMHO, connections are a quite nice tool in discussing such stuff. They give you a coordniate free way of thinking about a lot of interesting geometric things such as geodesics, parallel transport... But I'm sure you can solve your problem as well in index notation. If you take a chart $\psi$ than $g \circ \psi$ will also be a chart. Now how are $\psi \circ c$ and $g \circ \psi \circ c$ related? How are the cristoffelsymbols wrt $\psi$ and wrt to $g \circ \psi$ related? – hase_olaf Jun 11 '16 at 19:32
• @janmarqz Are you saying that a connection is just another name for a covariant derivative? It was my understanding that the covariant derivative is a type of connection. Sorry for the insistence but I still don't get it. – user335721 Jun 11 '16 at 22:35
• @hase_olaf $c(t)$ is supposed to be a curve from $\mathbb{R}$ to $M$ and your chart $\psi$ seems to be a function from $\mathbb{R}^n$ to $M$ (I deduce that because of the composition $g\circ\psi$), so what sense has the composition $\psi\circ c$ or $g\circ\psi\circ c$ ?? – user335721 Jun 11 '16 at 22:40