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:

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

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

Proof

Tensoriality in the first coordinate

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

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

Proof: We prove this by expanding everything out:

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

To prove the equality with $f\tau (\mathrm{\nabla })(X,Y)$, 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.

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