Torsion is tensorial: Difference between revisions

Revision as of 17:36, 6 January 2012

This article gives the statement, and possibly proof, that a map constructed in a certain way is tensorial
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Let $M$ be a differential manifold and $\nabla$ be a linear connection on $M$ (viz., $\nabla$ is a connection on the tangent bundle $T M$ of $M$ ).

Consider the torsion of $\nabla$ , namely:

$τ (\nabla) : Γ (T M) \times Γ (T M) \to Γ (T M)$

given by:

$τ (\nabla) (X, Y) = \nabla_{X} Y - \nabla_{Y} X - [X, Y]$

Then, $τ (\nabla)$ is a tensorial map in both coordinates.

Fact no.	Name	Statement with symbols
1	Any connection is $C^{\infty}$ -linear in its subscript argument	$\nabla_{f A} = f \nabla_{A}$ for any $C^{\infty}$ -function $f$ and vector field $A$ .
2	The Leibniz-like axiom that is part of the definition of a connection	For a function $f$ and vector fields $A, B$ , and a connection $\nabla$ , we have $\nabla_{A} (f B) = (A f) (B) + f \nabla_{A} (B)$
3	Corollary of Leibniz rule for Lie bracket (in turn follows from Leibniz rule for derivations	For a function $f$ and vector fields $X, Y$ : $f [X, Y] = [f X, Y] + (Y f) X$ $f [X, Y] = [X, f Y] - (X f) Y$

We'll use the fact that tensoriality is equivalent to $C^{\infty}$ -linearity.

To prove: $τ (\nabla) (f X, Y) = f τ (\nabla) (X, Y)$

Proof: We prove this by expanding everything out on the left side:

$τ (\nabla) (f X, Y) = \nabla_{f X} (Y) - \nabla_{Y} (f X) - [f X, Y] = f \nabla_{X} Y - f \nabla_{Y} X - (Y f) (X) - [f X, Y]$

To prove the equality with $f τ (\nabla) (X, Y)$ , we observe that it reduces to showing:

$(Y f) (X) = f [X, Y] - [f X, Y]$

which is exactly what the corollary of Leibniz rule above states.

The proof is analogous to that for the first coordinate.

To prove $τ (\nabla) (X, f Y) = f τ (\nabla) (X, Y)$

Proof: We prove this by expanding everything out on the left side:

$τ (\nabla) (X, f Y) = \nabla_{X} (f Y) = \nabla_{f Y} (X) - [X, f Y] = (X f) (Y) + f \nabla_{X} Y - f \nabla_{Y} X - f [X, Y] - (X f) Y$

(the last step uses the corollary of Leibniz rule).

Canceling terms, yields the required result.

@@ Line 17: / Line 17: @@
 ==Facts used==
-* [[Leibniz rule for derivations]]: This states that for a vector field <math>X</math> and functions <math>f,g</math>, we have:
-<math>\! X(fg) = (Xf)(g) + f(Xg)</math>
+{| class="sortable" border="1"
+! Fact no. !! Name !! Statement with symbols
-* [[Corollary of Leibniz rule for Lie bracket]]: This states that for a function <math>f</math> and vector fields <math>X,Y</math>:
+|-
+| 1 || Any connection is <math>C^\infty</math>-linear in its subscript argument || <math>\nabla_{fA} = f\nabla_A</math> for any <math>C^\infty</math>-function <math>f</math> and vector field <math>A</math>.
-<math>\! f[X,Y] = [fX,Y] + (Yf)X</math>
+|-
+| 2 || The Leibniz-like axiom that is part of the definition of a connection || For a function <math>f</math> and vector fields <math>A,B</math>, and a connection <math>\nabla</math>, we have <math>\nabla_A(fB) = (Af)(B) + f\nabla_A(B)</math>
-<math>\! f[X,Y] = [X,fY] - (Xf)Y</math>
+|-
+| 3 || [[uses::Corollary of Leibniz rule for Lie bracket]] (in turn follows from [[uses::Leibniz rule for derivations]]|| For a function <math>f</math> and vector fields <math>X,Y</math>:
-* The Leibniz rule axiom that's part of the definition of a [[connection]], namely:
+<br><math>\! f[X,Y] = [fX,Y] + (Yf)X</math><br><math>\! f[X,Y] = [X,fY] - (Xf)Y</math>
+|}
-<math>\! \nabla_X(fZ) = (Xf)(Z) + f\nabla_X(Z)</math>
 ==Proof==