Most succinct proof of the uniqueness and existence of the Levi-Civita connection.

Question

Seeing as proving the existence and/or uniqueness of the Levi-Civita connection seems to crop up in every single exam in Geometry and General Relativity, what is the most succinct proof of this, to memorize.

Answer 1

It's just a matter of remembering the order of things. The most practical way of proving the existence of Levi-Civita connection in a Riemannian manifold $(M,\langle\cdot,\cdot\rangle)$ is using its desirable properties, say:

1. Symmetry: $\nabla_XY-\nabla_YX=[X,Y]$

2. Compatibility: $X\langle Y,Z\rangle=\langle\nabla_XY,Z\rangle+\langle Y,\nabla_XZ\rangle$

We construct the connection based on the behavior over the fields: lets $X,Y,Z$ be fields over $M$. By 2. and 1.,

a) $X\langle Y,Z\rangle=\langle\nabla_XY,Z\rangle+\langle Y,\nabla_XZ\rangle=\langle [X,Y],Z\rangle+\langle \nabla_YX,Z\rangle+\langle Y,\nabla_XZ\rangle$

b) $Y\langle Z,X\rangle=\langle\nabla_YZ,X\rangle+\langle Z,\nabla_YX\rangle$

c) $Z\langle X,Y\rangle=\langle\nabla_ZX,Y\rangle+\langle X,\nabla_ZY\rangle$

Notice that in first and second equations above we have $\langle \nabla_YX,Z\rangle$ appearing two times. Calculating "a) $+$ b) $-$ c)" we get, putting similar terms togheter,

\begin{eqnarray} X\langle Y,Z\rangle+Y\langle Z,X\rangle-Z\langle X,Y\rangle&=&\langle [X,Y],Z\rangle+2\langle \nabla_YX,Z\rangle\\ &+&\langle Y,\nabla_XZ-\nabla_ZX\rangle+\langle\nabla_YZ-\nabla_ZY,X\rangle\\ &=&2\langle\nabla_YX,Z\rangle+\langle [X,Y],Z\rangle+\langle [Y,Z],X\rangle+\langle [X,Z],Y\rangle \end{eqnarray}

(this equation is known as the Koszul formula) Isolating $\langle\nabla_YX,Z\rangle$, you get a formulae and hence can define point to point the value $\nabla_YX$. You just have to remember the (natural) order of derivations in a), b) and c) and "a) $+$ b) $-$ c)". The rest is straightforward.