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,Zrangle &= frac{1}{2}(Xlangle Y,Zrangle +Ylangle Z,Xrangle -Zlangle X,Yrangle \
&   -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.

Ivo Terek · Accepted Answer

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:

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

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

If we write $Y_flat = langle Y,cdotrangle$, the Koszul formula may be rewritten as $$2langle nabla_XY,Zrangle = (mathcal{L}_Yg)(X,Z) + {rm d}(Y_flat)(X,Z),$$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).

Alice · Answer

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.

Fundamental theorem of Riemannian Geometry question

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