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

Symbolic statement

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

Consider the torsion of $\mathrm{\nabla }$, namely:

$\tau (\mathrm{\nabla }):\mathrm{\Gamma }(TM)\times \mathrm{\Gamma }(TM)\to \mathrm{\Gamma }(TM)$

given by:

$\tau (\mathrm{\nabla })(X,Y)={\mathrm{\nabla }}_{X}Y-{\mathrm{\nabla }}_{Y}X-[X,Y]$

Then, $\tau (\mathrm{\nabla })$ is a tensorial map in both coordinates.

Facts used

Fact no.	Name	Statement with symbols
1	Any connection is $C^{\mathrm{\infty }}$-linear in its subscript argument	${\mathrm{\nabla }}_{fA}=f{\mathrm{\nabla }}_A$ for any $C^{\mathrm{\infty }}$-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 $\mathrm{\nabla }$, we have ${\mathrm{\nabla }}_A(fB)=(Af)(B)+f{\mathrm{\nabla }}_A(B)$
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

Tensoriality in the first coordinate

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

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

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

$\tau (\mathrm{\nabla })(fX,Y)={\mathrm{\nabla }}_{fX}(Y)-{\mathrm{\nabla }}_Y(fX)-[fX,Y]=f{\mathrm{\nabla }}_{X}Y-f{\mathrm{\nabla }}_{Y}X-(Yf)(X)-[fX,Y]$

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

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

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

Tensoriality in the second coordinate

The proof is analogous to that for the first coordinate.

To prove $\tau (\mathrm{\nabla })(X,fY)=f\tau (\mathrm{\nabla })(X,Y)$

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

$\tau (\mathrm{\nabla })(X,fY)={\mathrm{\nabla }}_X(fY)={\mathrm{\nabla }}_{fY}(X)-[X,fY]=(Xf)(Y)+f{\mathrm{\nabla }}_{X}Y-f{\mathrm{\nabla }}_{Y}X-f[X,Y]-(Xf)Y$

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

Canceling terms, yields the required result.