Fundamental theorem of Riemannian Geometry question

Question

Theorem:

Let $(M,g)$ be a Riemannian Manifold. Then there exists a unique Riemannian connection on $M$.

My concern is with the existence part. In particular, I am unsure as to why the following holds:

Let $X,Y$ and $Z$ be smooth vector fields on $M$ Define $\nabla_{X}Y$ by

\begin{align} \langle \nabla_{X}Y,Z\rangle &= \frac{1}{2}(X\langle Y,Z\rangle +Y\langle Z,X\rangle -Z\langle X,Y\rangle \\ &\ \ \ -\langle X,[Y,Z]\rangle +\langle Y,[Z,X]\rangle +\langle Z,[X,Y]\rangle ). \end{align} This defines a connection on $M$. Why does this define $\nabla_{X}Y?$ and how do I show that $\nabla_{X}Y$ is a smooth vector field? Please do not use Christoffel symbols.

Answer 1

The actual verification is perhaps annoying but straightforward (people already linked other answers in the comments). But here's a few further comments about the truth behind the Koszul formula:

1. to define $\nabla$, you need to define $\nabla Y$ for all $Y$, where $(\nabla Y)(X) = \nabla_XY$.

2. since the metric is non-degenerate, knowing $\nabla_XY$ is the same as knowing $\langle \nabla_XY,\cdot\rangle$. That is, the same as knowing $\langle \nabla_XY,Z\rangle$ for all $Z$. This is exactly what the Koszul formula does.

3. If we write $Y_\flat = \langle Y,\cdot\rangle$, the Koszul formula may be rewritten as $$2\langle \nabla_XY,Z\rangle = (\mathcal{L}_Yg)(X,Z) + {\rm d}(Y_\flat)(X,Z),$$where $g$ is the metric, $\mathcal{L}$ denotes Lie derivative and ${\rm d}$ is the exterior derivative. Since $\nabla$ will satisfy a Leibniz rule, the above formula makes sense: we have a derivation term $\mathcal{L}_Yg$ and a linear (in $Y$) term ${\rm d}(Y_\flat)$, which should be thought of as the curl of $Y$.

Also, smoothness of $\nabla_XY$ follows from everything else in the Koszul formula being smooth (although smoothness can be shown easier with local computations using Christoffel symbols).

Answer 2

Given a fiber bundle $(E, M, \pi)$, an affine connection is a bilinear form $$\nabla : \mathfrak{X}(M) \times \Gamma(E) \to \Gamma(E) $$ such that

(a.) $\nabla_{fW}V = f \nabla_W V,$

(b.) $\nabla_W fV = f \nabla_W V + W(f) V,$

where $W \in \mathfrak{X}(M), V \in \Gamma(E) $ e $f \in \mathcal{C}^{\infty} (M)$.

A Riemannian connection is defined as an affine connection on the tangent bundle $TM$ of a Riemannian manifold $(M,g)$ which is compatible with the metric and torsion free.

To show that there exists a Riemannian connection you want to show that the Koszul formula defines a Riemannian connection, that is, it is enough to show that:

(1) $\nabla_xY-\nabla_YX = [X,Y],$ (torsion free)

(2) $X g(Y,Z) = g(\nabla_XY,Z) + g(Y,\nabla_X,Z)$, (compatible with the metric)

for all $X,Y,Z \in \mathfrak{X}(M)$.

Note that the space of sections of the tangent bundle is $\mathfrak{X}(M)$. Then a Riemannian connection is a bilinear form $$\nabla : \mathfrak{X}(M) \times \mathfrak{X}(M) \to \mathfrak{X}(M), $$ so $\nabla_X Y \in \mathfrak{X}(M)$.

To prove (1) note that it is equivalent to show that $$ g(\nabla_xY-\nabla_YX, Z) = g([X,Y], Z), \forall Z \in \mathfrak{X}(M),$$ and to prove (2) just use the Koszul formula to calculate the right side of the equation (in (2)). There you go!

I hope this can help you.