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

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.

• What do you know? I assume you have seen wiki page of Levi-Civita connection. The proof of uniqueness there is not that bad. – Sasha May 4 '12 at 14:35

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.

• I guess i cannot do any better than this. – DanielOfTaebl May 7 '12 at 7:50
• I think if you wanna, for instance, remember the proof, you only need to remember that we derived in a mnemonic order and then sum and subtract. The rest is a matter of using identities listed above =) – matgaio May 7 '12 at 14:49
• @DanielOfTaebl May I know why Koszul formula can uniquely determine $D_YX$? Does it follow from the inner product structure (since the formula holds for all $Z$)? – John Apr 15 '15 at 1:50
• @DanielOfTaebl Also for a connection, we need the target to be a vector field. In this proof, we only define the $D_YX$ point-wisely. How can we guarantee the smoothness? – John Apr 15 '15 at 2:03
• @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. – Kalua Oct 28 '18 at 21:03