# Levi-Civita connection exists and is unique

## Contents

## Statement

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

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

This connection is called the Levi-Civita connection.

## Related facts

- Corollary of Leibniz rule for Lie bracket: This states that:
- .
- .

## Proof

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

Take three vector fields . Now, consider the three equations obtained by cycling in the first condition. Solving this system of linear equations, we can express in terms of .

Explicitly:

Now let's choose to focus only on the clockwise cyclic expressions, that is, the three expressions .

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

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

.

This simplifies to:

.

Thus, the value is determined for all , as the expression on the right side. Moreover, since the Lie bracket of derivations is -bilinear and the metric is -bilinear, the right side is linear in . Thus, the function is linear. Finally, since is nondegenerate, there exists a *unique* vector field such that is this function.

### The connection exists

We now argue that the connection defined in this way *is* a connection, i.e., it satisfies the conditions necessary for a connection. First, note that the expression is -linear in both and , so is -bilinear.

To check that the connection is -linear in , it suffices to show that the map is -linear. We do this:

.

We use the Leibniz rule:

.

We now use fact (1):

.

Cancelling and grouping terms gives:

.

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

.

To do this, it suffices to show that:

.

We begin by expanding the left side:

Applying the Leibniz rule:

Using fact (1) now yields:

Cancelling and grouping terms gives:

.

This completes the proof.