Vipul: Created page with '==Statement== Suppose $M$ is a differential manifold and $g$ is a fact about::Riemannian metric or a fact about::pseudo-Riemannian metric. Then…'

2009-07-24T18:43:11Z

Created page with '==Statement== Suppose <math>M</math> is a differential manifold and <math>g</math> is a fact about::Riemannian metric or a fact about::pseudo-Riemannian metric. Then…'

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==Statement==

Suppose <math>M</math> is a [[differential manifold]] and <math>g</math> is a [[fact about::Riemannian metric]] or a [[fact about::pseudo-Riemannian metric]]. Then, there is a unique [[linear connection]] on <math>M</math> satisfying the following two conditions:

# It is a [[fact about::metric connection]].
# It is a [[fact about::torsion-free linear connection]].

This connection is called the [[fact about::Levi-Civita connection]].

==Related facts==

# [[uses::Corollary of Leibniz rule for Lie bracket]]: This states that:
#* <math>\! f[X,Y] = [fX,Y] + (Yf)X</math>.
#* <math>\! f[X,Y] = [X,fY] - (Xf)Y</math>.

==Proof==

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

Take three vector fields <math>X, Y, Z</math>. Now, consider the three equations obtained by cycling <math>X, Y, Z</math> in the first condition. Solving this system of linear equations, we can express <math>g(\nabla_XY,Z)</math> in terms of <math>X,Y,Z</math>.

Explicitly:

<math>g(\nabla_XY,Z) + g(Y,\nabla_XZ) = Xg(Y,Z) \qquad (1)</math>

<math>g(\nabla_YX,Z) + g(X,\nabla_YZ) = Yg(Z,X) \qquad (2)</math>

<math>g(\nabla_ZX,Y) + g(X,\nabla_ZY) = Zg(X,Y) \qquad (3)</math>

Now let's choose to focus only on the clockwise cyclic expressions, that is, the three expressions <math>p = g(\nabla_XY,Z), q = g(\nabla_YZ,X), r = g(\nabla_ZX,Y)</math>.

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

<math>p + q = Xg(Y,Z) + g(Y,[Z,X]) \qquad (4)</math>

<math>r + p = Yg(Z,X) + g(Z,[X,Y]) \qquad (5)</math>

<math>q + r = Zg(X,Y) + g(X,[Y,Z]) \qquad (6)</math>

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

<math>\! 2p = Xg(Y,Z) + Yg(Z,X) - Zg(X,Y) + g(Y,[Z,X]) + g(Z,[X,Y]) - g(X,[Y,Z])</math>.

This simplifies to:

<math>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}</math>.

Thus, the value <math>g(\nabla_XY,Z)</math> is determined for all <math>Z</math>, as the expression on the right side. Moreover, since the Lie bracket of derivations is <math>\R</math>-bilinear and the metric <math>g</math> is <math>\R</math>-bilinear, the right side is linear in <math>Z</math>. Thus, the function <math>Z \mapsto g(\nabla_XY,Z)</math> is linear. Finally, since <math>g</math> is nondegenerate, there exists a ''unique'' vector field <math>\nabla_XY</math> such that <math>g(\nabla_XY,Z)</math> is this function.

===The connection exists===

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

To check that the connection is <math>C^\infty</math>-linear in <math>X</math>, it suffices to show that the map <math>X \mapsto g(\nabla_XY,Z)</math> is <math>C^\infty</math>-linear. We do this:

<math>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}</math>.

We use the Leibniz rule:

<math>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}</math>.

We now use fact (1):

<math>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}</math>.

Cancelling and grouping terms gives:

<math>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)</math>.

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

<math>\nabla_X(fY) = (Xf)(Y) + f\nabla_X(Y)</math>.

To do this, it suffices to show that:

<math>g(\nabla_X(fY),Z) = (Xf)g(Y,Z) + fg(\nabla_XY,Z)</math>.

We begin by expanding the left side:

<math>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}</math>

Applying the Leibniz rule:

<math>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}</math>

Using fact (1) now yields:

<math>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}</math>

Cancelling and grouping terms gives:

<math>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)</math>.

This completes the proof.

Levi-Civita connection exists and is unique - Revision history

Vipul: Created page with '==Statement== Suppose M is a differential manifold and g is a fact about::Riemannian metric or a fact about::pseudo-Riemannian metric. Then…'

Vipul: Created page with '==Statement== Suppose $M$ is a differential manifold and $g$ is a fact about::Riemannian metric or a fact about::pseudo-Riemannian metric. Then…'