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 $\nabla$ 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:

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

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

Proof

Tensoriality in the first coordinate

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

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

Proof: We prove this by expanding everything out:

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

To prove the equality with $f\tau (\nabla )(X,Y)$, we need to show:

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

To prove this, we need to show that both sides evaluate to the same expression for any function $g$. Plugging a function $g$, we see that the right side becomes:

$f(X(Yg)-Y(Xg))-f(X(Yg))+Y((fX)g)=Y(f(Xg))-f(Y(Xg))$

Applying the Leibniz rule for $Y$ on the product of functions $f$ and $Xg$, this simplifies to:

$(Yf)(Xg)+Y(Xg)(f)-f(Y(Xg))=(Yf)(Xg)$

which is precisely equal to the left side.

Tensoriality in the second coordinate

The proof is analogous to that for the first coordinate.

Fill this in later