Torsion is tensorial

This article gives the statement, and possibly proof, that a map constructed in a certain way is tensorial
View other such statements

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:

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

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

Facts used

• Leibniz rule for derivations: This states that for a vector field $X$ and functions $f,g$, we have:

$X(fg)=(Xf)(g)+f(Xg)$

• Corollary of Leibniz rule for Lie bracket: This states that for a function $f$ and vector fields $X,Y$:

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

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

• The Leibniz rule axiom that's part of the definition of a connection, namely:

$\nabla _{X}(fZ)=(Xf)(Z)+f\nabla _{X}(Z)$

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 on the left side:

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

To prove the equality with $f\tau (\nabla )(X,Y)$, 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 $\tau (\nabla )(X,fY)=f\tau (\nabla )(X,Y)$

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

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

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

Canceling terms, yields the required result.