Levi-Civita connection exists and is unique

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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_XY,Z) in terms of X,Y,Z.

Explicitly:

g(\nabla_XY,Z) + g(Y,\nabla_XZ) = Xg(Y,Z) \qquad (1)

g(\nabla_YX,Z) + g(X,\nabla_YZ) = Yg(Z,X) \qquad (2)

g(\nabla_ZX,Y) + g(X,\nabla_ZY) = 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_XY,Z), q = g(\nabla_YZ,X), r = g(\nabla_ZX,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_XY,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_XY,Z) is determined for all Z, as the expression on the right side. Moreover, since the Lie bracket of derivations is \R-bilinear and the metric g is \R-bilinear, the right side is linear in Z. Thus, the function Z \mapsto g(\nabla_XY,Z) is linear. Finally, since g is nondegenerate, there exists a unique vector field \nabla_XY such that g(\nabla_XY,Z) is this function.

The connection exists

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

To check that the connection is C^\infty-linear in X, it suffices to show that the map X \mapsto g(\nabla_XY,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_XY,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_XY,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_XY,Z).

This completes the proof.