Torsion is tensorial

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Revision as of 01:18, 24 July 2009 by Vipul (talk | contribs) (Proof)
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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_X Y - \nabla_Y X - [X,Y]

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

Facts used

\! X(fg) = (Xf)(g) + f(Xg)

\! 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_XY - f\nabla_YX - f[X,Y] - (Xf)Y

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

Canceling terms, yields the required result.