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If you define a "generalized" Christoffel tensor as the following: $Chris(\omega; u, v) := (\nabla_u(\omega))(v) - (\tilde{\nabla}_u(\omega))(v)$ where $\omega$ is a dual vector, $u,v$ are vectors, and $\nabla$ and $\tilde{\nabla}$ are two covariant derivatives, then is $Chris$ a tensor field if you take $nabla$ to be the metric covariant derivative and $\tilde{\nabla}=\mathcal{L}$ to be the lie derivative? (as opposed to the Christoffel symbol not being a tensor when $\tilde{\nabla}$ is the ordinary partial derivative $\partial$)

I guess also with $\tilde{\nabla}=\partial$, $Chris$ is a tensor (a multinear map from duals and vectors to the reals), but some people insist on defining a tensor as something that transforms in a certain way under a change of charts. I guess my question is then will it "transform like a tensor" in this case?

PPR
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1 Answers1

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If $\nabla$ and $\tilde{\nabla}$ are two connections, then the difference $\nabla - \tilde{\nabla}$ acts tensorially, so gives rise to a tensor, called the difference tensor (or contorsion) of $\nabla$ and $\tilde{\nabla}$.

The Lie derivative is not a connection, so the difference $\nabla - \mathcal{L}$ makes little sense.

The Christoffel symbol of $\nabla$ is defined only with respect to a local chart as the difference $\nabla - \mathrm{d}$ where $\mathrm{d}$ is the "trivial" connection in the chart (takes directional derivatives of components).

The details see, for instance, in R.Wald's "General relativity".

Yuri Vyatkin
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  • Thanks! I'm a bit confused by the terminology though, and I allow myself to insist because the terminology I know is exactly from Wald. There, $\nabla$ and $\mathcal{L}$ are not "connections", but rather derivative operations, and as such they stand on equal footing. What sets $\nabla$ apart is the requirement that $\nabla(g)=0$. I suppose you already answered: the contorsion tensor is a tensor. Though I still don't get why the contorsion tensor for $\nabla-\mathcal{L}$ would not be a tensor (though still, not presuming it makes any sense to talk about it...) – PPR Jan 29 '14 at 14:31
  • @Psycho_pr I should have clarified that the difference $\nabla_X - \mathcal{L}_X$ is not tensorial because $\mathcal{L}_X$ is not linear in the slot $X$ while $\nabla_X$ is. Therefore there is no "contorsion tensor for $\nabla_X - \mathcal{L}_X$". – Yuri Vyatkin Jan 29 '14 at 19:42
  • Notice also, that there is no need to restrict to metric connections (those that satisfy $\nabla(g)=0$ for a metric $g$). Everything works for arbitrary linear connections. – Yuri Vyatkin Jan 29 '14 at 19:47