Levi-Civita connection exists and is unique

Statement

Suppose $M$ is a differential manifold and $g$ is a Riemannian metric or a pseudo-Riemannian metric. Then, there is a unique linear connection on $M$ satisfying the following two conditions:

1. It is a metric connection.
2. It is a torsion-free linear connection.

This connection is called the Levi-Civita connection.

Related facts

1. Corollary of Leibniz rule for Lie bracket: This states that:
   • $f[X,Y]=[fX,Y]+(Yf)X$.
   • $f[X,Y]=[X,fY]-(Xf)Y$.

Proof

Formula for the Levi-Civita connection and proof of its uniqueness

Take three vector fields $X,Y,Z$. Now, consider the three equations obtained by cycling $X,Y,Z$ in the first condition. Solving this system of linear equations, we can express $g({\mathrm{\nabla }}_{X}Y,Z)$ in terms of $X,Y,Z$.

Explicitly:

$g({\mathrm{\nabla }}_{X}Y,Z)+g(Y,{\mathrm{\nabla }}_{X}Z)=Xg(Y,Z)(1)$

$g({\mathrm{\nabla }}_{Y}X,Z)+g(X,{\mathrm{\nabla }}_{Y}Z)=Yg(Z,X)(2)$

$g({\mathrm{\nabla }}_{Z}X,Y)+g(X,{\mathrm{\nabla }}_{Z}Y)=Zg(X,Y)(3)$

Now let's choose to focus only on the clockwise cyclic expressions, that is, the three expressions $p=g({\mathrm{\nabla }}_{X}Y,Z),q=g({\mathrm{\nabla }}_{Y}Z,X),r=g({\mathrm{\nabla }}_{Z}X,Y)$.

Writing everything in terms of these three (we here make use of the torsion tensor vanishing):

$p+q=Xg(Y,Z)+g(Y,[Z,X])(4)$

$r+p=Yg(Z,X)+g(Z,[X,Y])(5)$

$q+r=Zg(X,Y)+g(X,[Y,Z])(6)$

Consider (4) + (5) - (6):

$2p=Xg(Y,Z)+Yg(Z,X)-Zg(X,Y)+g(Y,[Z,X])+g(Z,[X,Y])-g(X,[Y,Z])$.

This simplifies to:

$g({\mathrm{\nabla }}_{X}Y,Z)=p=\frac{Xg(Y,Z)+Yg(Z,X)-Zg(X,Y)+g(Y,[Z,X])+g(Z,[X,Y])-g(X,[Y,Z])}{2}$.

Thus, the value $g({\mathrm{\nabla }}_{X}Y,Z)$ is determined for all $Z$, as the expression on the right side. Moreover, since the Lie bracket of derivations is $\mathbb{R}$-bilinear and the metric $g$ is $\mathbb{R}$-bilinear, the right side is linear in $Z$. Thus, the function $Z\mapsto g({\mathrm{\nabla }}_{X}Y,Z)$ is linear. Finally, since $g$ is nondegenerate, there exists a unique vector field ${\mathrm{\nabla }}_{X}Y$ such that $g({\mathrm{\nabla }}_{X}Y,Z)$ is this function.

The connection exists

We now argue that the connection ${\mathrm{\nabla }}_{X}Y$ defined in this way is a connection, i.e., it satisfies the conditions necessary for a connection. First, note that the expression is $\mathbb{R}$-linear in both $X$ and $Y$, so ${\mathrm{\nabla }}_{X}Y$ is $\mathbb{R}$-bilinear.

To check that the connection is $C^{\mathrm{\infty }}$-linear in $X$, it suffices to show that the map $X\mapsto g({\mathrm{\nabla }}_{X}Y,Z)$ is $C^{\mathrm{\infty }}$-linear. We do this:

$g({\mathrm{\nabla }}_{fX}(Y),Z)=\frac{fXg(Y,Z)+Y(fg(Z,X))-Z(fg(X,Y))+g(Y,[Z,fX])+g(Z,[fX,Y])-fg(X,[Y,Z])}{2}$.

We use the Leibniz rule:

$g({\mathrm{\nabla }}_{fX}(Y),Z)=\frac{fXg(Y,Z)+(Yf)g(Z,X)+f(Yg(Z,X))-(Zf)g(X,Y)-f(Zg(X,Y))+g(Y,[Z,fX])+g(Z,[fX,Y])-fg(X,[Y,Z])}{2}$.

We now use fact (1):

$g({\mathrm{\nabla }}_{fX}(Y),Z)=\frac{fXg(Y,Z)+(Yf)g(Z,X)+f(Yg(Z,X))-(Zf)g(X,Y)-f(Zg(X,Y))+fg(Y,[Z,X])+(Zf)g(Y,X)+fg(Z,[X,Y])-(Yf)g(Z,X)-fg(X,[Y,Z])}{2}$.

Cancelling and grouping terms gives:

$g({\mathrm{\nabla }}_{fX}(Y),Z)=\frac{f(Xg(Y,Z)+Yg(Z,X)-Zg(X,Y)+g(Y,[Z,X])+g(Z,[X,Y])-g(X,[Y,Z]))}{2}=fg({\mathrm{\nabla }}_{X}Y,Z)$.

Next, we check the Leibniz rule property. In other words, we need to show that:

${\mathrm{\nabla }}_X(fY)=(Xf)(Y)+f{\mathrm{\nabla }}_X(Y)$.

To do this, it suffices to show that:

$g({\mathrm{\nabla }}_X(fY),Z)=(Xf)g(Y,Z)+fg({\mathrm{\nabla }}_{X}Y,Z)$.

We begin by expanding the left side:

$g({\mathrm{\nabla }}_X(fY),Z)=\frac{X(fg(Y,Z))+fYg(Z,X)-Z(fg(X,Y))+fg(Y,[Z,X])+g(Z,[X,fY])-g(X,[fY,Z])}{2}$

Applying the Leibniz rule:

$g({\mathrm{\nabla }}_X(fY),Z)=\frac{(Xf)g(Y,Z)+f(Xg(Y,Z))+f(Yg(Z,X))-(Zf)g(X,Y)-Z(fg(X,Y))+fg(Y,[Z,X])+g(Z,[X,fY])-g(X,[fY,Z])}{2}$

Using fact (1) now yields:

$g({\mathrm{\nabla }}_X(fY),Z)=\frac{(Xf)g(Y,Z)+f(Xg(Y,Z))+f(Yg(Z,X))-(Zf)g(X,Y)-f(Zg(X,Y))+fg(Y,[Z,X])+fg(Z,[X,Y])+(Xf)g(Z,Y)-fg(X,[Y,Z])+(Zf)g(X,Y)}{2}$

Cancelling and grouping terms gives:

$g({\mathrm{\nabla }}_X(fY),Z)=(Xf)g(Y,Z)+\frac{f(Xg(Y,Z)+Yg(Z,X)-Zg(X,Y)+g(Y,[Z,X])+g(Z,[X,Y])-g(X,[Y,Z]))}{2}=(Xf)g(Y,Z)+fg({\mathrm{\nabla }}_{X}Y,Z)$.

This completes the proof.