Torsion is tensorial

This article gives the statement, and possibly proof, that a map constructed in a certain way is tensorial
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Statement

Let $M$ be a differential manifold and $\nabla$ be a linear connection on $M$ (viz., $\nabla$ is a connection on the tangent bundle $TM$ of $M$ ).

Consider the torsion of $\nabla$ , namely:

$\tau(\nabla): \Gamma(TM) \times \Gamma(TM) \to \Gamma(TM)$

given by:

$\tau(\nabla)(X,Y) = \nabla_X Y - \nabla_Y X - [X,Y]$

Then, $\tau(\nabla)$ is a tensorial map in both coordinates. In other words, the value of $\tau(\nabla)$ at a point $p \in M$ depends only on $\nabla, X(p), Y(p)$ and does not depend on the values of the vectors fields $X,Y$ at points other than $p$ .

More explicitly, for any point $p \in M$ , $\tau(\nabla)$ defines a bilinear map:

$\! \tau(\nabla): T_p(M) \times T_p(M) \to T_p(M)$

Further, since in fact torsion is antisymmetric, this is an alternating bilinear map.

Related facts

Facts used

Fact no.	Name	Statement with symbols
1	Any connection is $C^\infty$ -linear in its subscript argument	$\nabla_{fA} = f\nabla_A$ 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 $\nabla_A(fB) = (Af)(B) + f\nabla_A(B)$
3	Corollary of Leibniz rule for Lie bracket (in turn follows from Leibniz rule for derivations	For a function $f$ and vector fields $X,Y$ : $\! f[X,Y] = [fX,Y] + (Yf)X$ $\! f[X,Y] = [X,fY] - (Xf)Y$

Proof

To prove tensoriality in a variable, it suffices to show $C^\infty$ -linearity in that variable. This is because linearity in $C^\infty$ -functions guarantees linearity in a function that is 1 at exactly one point, and zero at others.

The proofs for $X$ and $Y$ are analogous, and rely on manipulation of the Lie bracket $[fX,Y]$ and the property of a connection being $C^\infty$ in the subscript vector.

Tensoriality in the first coordinate

Given: $f:M \to \R$ is $C^\infty$ -function

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

Proof: We start out with the left side:

$\tau(\nabla)(fX,Y)$

Each step below is obtained from the previous one via some manipulation explained along side.

Step no.	Current status of left side	Facts/properties used	Specific rewrites
1	$\nabla_{fX}(Y) - \nabla_Y(fX) - [fX,Y]$	Definition of torsion	whole thing
2	$f \nabla_X Y - \nabla_Y(fX) - [fX,Y]$	Fact (1): $C^\infty$ -linearity of connection in subscript argument	$\nabla_{fX} \mapsto f\nabla_X$
3	$f \nabla_X Y - (f \nabla_Y X + (Yf)(X)) - [fX,Y]$	Fact (2): The Leibniz-like axiom that's part of the definition of a connection	$\nabla_Y(fX) \mapsto f\nabla_YX + (Yf)(X)$
4	$f \nabla_X Y - f \nabla_Y X - ((Yf)(X) + [fX,Y])$	parenthesis rearrangement	--
5	$f \nabla_X Y - f \nabla_Y X - f[X,Y]$	Fact (3)	$(Yf)(X) + [fX,Y] \mapsto f[X,Y]$
6	$f(\nabla_X Y - \nabla_Y X - [X,Y])$	factor out	--
7	$f\tau(\nabla)(X,Y)$	use definition of torsion	$\nabla_X Y - \nabla_Y X - [X,Y] \mapsto \tau(\nabla)(X,Y)$