Equivalence for Christoffel symbol and Koszul formula

Question

I am trying to show to define a Levi-civita connection, it's equivalent to define Christoffel symbols or define Koszul formula.

$$ 2g(\nabla_XY, Z) = \partial_X (g(Y,Z)) + \partial_Y (g(X,Z)) - \partial_Z (g(X,Y))+ g([X,Y],Z) - g([X,Z],Y) - g([Y,Z],X)$$

One direction is easy, given Koszul formula, take $X=\frac{\partial}{\partial x^i},Y=\frac{\partial}{\partial x^j}, Z=\frac{\partial}{\partial x^k}$ and compute.

For the other direction, I set $X=x^i\partial_i,Y=y^j\partial_i,Z=z^k\partial_k$ and try to show LHS and RHS of Koszul formula are equal. After expressing both sides explicitly, I think I need to use some other conditions to cancel some terms before I can plug in the definition of Christoffel symbols. Could you give some reference or a detailed illustration? Thanks.

Answer 1

This is proved in my Riemannian Manifolds book. (See the proof of Theorem 5.4.)

Answer 2

Actually, you should add $\frac{1}{2}$ in RHS of formula, and it should have been $$\langle\nabla_XY,Z\rangle = \frac{1}{2}(X\langle Y,Z\rangle+Y\langle X,Z\rangle-Z\langle X,Y\rangle+\langle[X,Y],Z\rangle-\langle[X,Z],Y\rangle-\langle[Y,Z],X\rangle)$$

• $\langle\cdot,\cdot\rangle=g(\cdot,\cdot)$

• You can find it on http://en.wikipedia.org/wiki/Levi-Civita_connection

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It is accessible to use the local frame to prove the formula. However, we can prove it by the axioms, which also can be founded on http://en.wikipedia.org/wiki/Levi-Civita_connection.

Below is the proof by local coordinate(sketch)

Assume $X=X_i\frac{\partial}{\partial x_i},~~Y=Y_i\frac{\partial}{\partial x_i},~~Z=Z_i\frac{\partial}{\partial x_i}$, then $$[X,Y]=(X_j\frac{\partial Y_i}{\partial x_j}-Y_j\frac{\partial X_i}{\partial x_j})\frac{\partial}{\partial x_i}$$ $$[X,Z]=(X_j\frac{\partial Z_i}{\partial x_j}-Z_j\frac{\partial X_i}{\partial x_j})\frac{\partial}{\partial x_i}$$ $$[Y,Z]=(Y_j\frac{\partial Z_i}{\partial x_j}-Z_j\frac{\partial Y_i}{\partial x_j})\frac{\partial}{\partial x_i}$$ and $$\nabla_XY=(X_iY_j\Gamma_{ij}^k+X_i\frac{\partial Y_k}{\partial x_i})\frac{\partial}{\partial x_k}$$

Noticing there are three facts

• $\langle X,Y\rangle=g_{ij}X_iY_j$;

• $\Gamma_{ij}^k=\frac{1}{2}g^{kl}(\frac{\partial g_{il}}{\partial x_j}+\frac{\partial g_{jl}}{\partial x_i}-\frac{\partial g_{ij}}{\partial x_l})$

• $(g_{ij})(g^{kl})=I$

Then take the local presentation into formula, use the three facts, and you will find $$\text{LHS=RHS}$$

Because the work needs patience and carefulness, you know $\cdots$.