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 $g(\nabla _{X}Y,Z)$ in terms of $X,Y,Z$.

Explicitly:

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

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

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

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

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

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

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

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

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:

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

Thus, the value $g(\nabla _{X}Y,Z)$ 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 $Z\mapsto g(\nabla _{X}Y,Z)$ is linear. Finally, since $g$ is nondegenerate, there exists a unique vector field $\nabla _{X}Y$ such that $g(\nabla _{X}Y,Z)$ is this function.

The connection exists

We now argue that the connection $\nabla _{X}Y$ 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 $\nabla _{X}Y$ is $\mathbb {R}$-bilinear.

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

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

We use the Leibniz rule:

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

We now use fact (1):

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

Cancelling and grouping terms gives:

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

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

$\nabla _{X}(fY)=(Xf)(Y)+f\nabla _{X}(Y)$.

To do this, it suffices to show that:

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

We begin by expanding the left side:

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

Applying the Leibniz rule:

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

Using fact (1) now yields:

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

Cancelling and grouping terms gives:

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

This completes the proof.