# Changes

## Torsion is tensorial

, 17:43, 6 January 2012
Proof
==Proof==
To prove tensoriality in a variable, it suffices to show $C^\infty$-linearity in that variable. This is because linearity in $C^\infty$-functions guarantees linearity in a function that is 1 at exactly one point, and zero at others.

The proofs for $X$ and $Y$ are analogous, and rely on manipulation of the Lie bracket $[fX,Y]$ and the property of a connection being $C^\infty$ in the subscript vector.
===Tensoriality in the first coordinate===
We'll use the fact that tensoriality ''Given''': $f:M \to \R$ is equivalent to $C^\infty$-linearity.function
'''To prove''': $\tau(\nabla)(fX,Y) = f\tau(\nabla)(X,Y)$
'''Proof''': We prove this by expanding everything start out on with 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 Each step below is obtained from the equality with $f \tau(\nabla)(X,Y)$, we observe that it reduces to showing:previous one via some manipulation explained along side.
{| class="sortable" border="1"! Step no. !! Current status of left side !! Facts/properties used !! Specific rewrites|-| 1 || $\nabla_{fX}(Y) - \nabla_Y(fX) - [fX,Y]$ || Definition of torsion || whole thing|-| 2 || $f \nabla_X Y - \nabla_Y(fX) - [fX,Y]$ || Fact (1): $C^\infty$-linearity of connection in subscript argument || $\nabla_{fX} \mapsto f\nabla_X$|-| 3 || $f \nabla_X Y - (f \nabla_Y X + (Yf)(X)) - [fX,Y]$ || Fact (2): The Leibniz-like axiom that's part of the definition of a connection || $\nabla_Y(fX) \mapsto f\nabla_YX + (Yf)(X)</math>|-| 4 || [itex]f \nabla_X Y - f \! nabla_Y X - ((Yf)(X) + [fX,Y])$ || parenthesis rearrangement || --|-| 5 || $f \nabla_X Y - f \nabla_Y X - f[X,Y]$ || Fact (3) || $(Yf)(X) = + [fX,Y] \mapsto f[X,Y]$|-| 6 || $f(\nabla_X Y - \nabla_Y X - [fXX,Y])$|| factor out || --|-which is exactly what the corollary | 7 || $f\tau(\nabla)(X,Y)$ || use definition of Leibniz rule above states.torsion || $\nabla_X Y - \nabla_Y X - [X,Y] \mapsto \tau(\nabla)(X,Y)$|}
===Tensoriality in the second coordinate===
''To prove'' $\tau(\nabla)(X,fY) = f \tau(\nabla)(X,Y)$
''Proof'': We prove this by expanding everything out on This is similar to tensoriality in the left side: $\tau(\nabla)(X,fY) = \nabla_X(fY) = \nabla_{fY}(X) - [X,fY] = (Xf)(Y) + f \nabla_XY - f\nabla_YX - f[X,Y] - (Xf)Y$ (the last step uses the corollary of Leibniz rule). Canceling terms, yields the required resultfirst coordinate.