Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
    
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
add comment

1 Answer

up vote 9 down vote accepted

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.

share|improve this answer
1  
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
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.