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I would be pleased to have some information about an alternate definition for the torsion tensor.

Let us consider a smooth manifold $\mathcal{M}$ together with an arbitrary connection $\nabla$. The first definition for the torsion that I mostly see in the (physics) literature is the following:

${\bf T} ({\bf e}_\alpha, {\bf e}_\beta) := {\nabla}_{{\bf e}_\alpha} {\bf e}_\beta - {\nabla}_{{\bf e}_\beta} {\bf e}_\alpha - [{\bf e}_\alpha,{\bf e}_\beta] \, , $

with $\{ {\bf e}_\alpha \}$ being an arbitrary basis on $\mathcal{M}$. The other definition, always encountered in terms of components in the literature that I have read, is, in its tensorial form,

${\bf T} ({\bf e}_\alpha, {\bf e}_\beta)(f) := \nabla \nabla f ({\bf e}_\alpha,{\bf e}_\beta) - \nabla \nabla f ({\bf e}_\beta,{\bf e}_\alpha) \, , $

where $f$ is a smooth function on $\mathcal{M}$. One can readily check that, for instance in terms of the connection coefficients of $\nabla$, these two definitions are equivalent.

My questions are

(i) As I did not find the second definition in the few mathematics books that I have read, is it indeed a correct definition? (I realize that this question can sound strange as both definitions yield the same result, but not having seen it in a book yet makes me wonder.)

(ii) What's the meaning of the second definition?

(iii) Can the second definition be written in terms of an exterior derivative, and if so, what would be the form?

Thanks in advance for your help.


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Could you explain what is $f$ in the second definition? – abx Jan 24 '14 at 10:57
$f$ is a smooth function on the manifold. I have edited my question to make it clearer. (Also, I have corrected some previous mistakes in the equations). – Xavier Jan 24 '14 at 12:16
Your observation of the equivalence of these two definitions is a standard one. For example, see Milnor's Morse Theory, Definition 8.5 and the attached footnote. – Robert Bryant Jan 24 '14 at 12:20
@RobertBryant Thanks Robert. Does it exist other formulation of the second definition? (Perharps in terms of the exterior derivative?) – Xavier Jan 24 '14 at 12:27
@Xavier: Yes. If you let $\alpha:T^*\otimes T^*\to \Lambda^2(T^*)$ denote antisymmetrization on the cotangent bundle, then consider the operation $\mathsf{T}:T^*\to \Lambda^2(T^*)$ that satisfies $\mathsf{T}(\omega) = \alpha(\nabla\omega)- \mathrm{d}\omega$ for any section $\omega$ of $T^*$ (i.e., $\omega$ is a $1$-form). Then $\mathsf{T}$, regarded as a section of $\Lambda^2(T^*)\otimes T$, is the torsion tensor. Thus, in the special case that $\omega = \mathrm{d} f$, one has your formula $\mathsf{T}(\mathrm{d} f) = \alpha(\nabla\mathrm{d} f)$. – Robert Bryant Jan 24 '14 at 12:32

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