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.
Most succinct proof of the uniqueness and existence of the Levi-Civita connection.
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1What do you know? I assume you have seen wiki page o$f$ [Levi-Civita connection](http://en.wikipedia.org/wiki/Levi-Civita_connection). The proo$f$ of uniqueness there is not that bad. – 2012-05-04
1 Answers
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:
Symmetry: $\nabla_XY-\nabla_YX=[X,Y]$
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.
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0@John Rearranging the terms in the Koszul formula, we get $2\langle\nabla_YX,Z\rangle=K(X,Y,Z)$ where $K(X,Y,Z)$ denotes the remaining terms. Assuming $\nabla_YX$ defines a vector field, one can show with some lengthy but straight forward computations, that $\nabla$ in fact is a metric torsion free connection by using $K(X,Y,Z)$ in the computations. The fact that $\nabla_YX$ is a vector field is more tricky. One can show that $T(Z):=K(X,Y,Z)$ is $C^\infty(M)$-linear in $Z$, and conclude the unique existence of some vector field $X_0$ with $\langle X_0,\cdot\rangle=T$ using coordinate patches. – 2018-10-28