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: $f:M\to \mathbb {R}$ 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	$\nabla _{Y}(fX)\mapsto f\nabla _{Y}X+(Yf)(X)$
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	$\nabla _{X}Y-\nabla _{Y}X-[X,Y]\mapsto \tau (\nabla )(X,Y)$