Torsion is tensorial

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.

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:

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

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

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

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:

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

The left hand side is:

<math>(Xf)(Yg) = (Yg)(Xf) = Y(g(Xf))