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  I cannot solve a contradiction in the second Covariant Derivative of e vector fields

+ 1 like - 0 dislike

I have tried to prove the equation:

\((\nabla_\beta \nabla_ \nu V)^ \mu = {V^\mu}_{;\nu,\beta} + {V^\alpha}_{;\nu}\; {\Gamma^ \mu}_{ \alpha \beta } - {V^\mu}_{;\alpha}\; {\Gamma^ \alpha}_{ \nu \beta }\),

as found in Bernard Schutz book second edition "A first course in general relativity" in the paragraph 6.5 "The curvature tensor" and I found it's correct (here the link to my attempt to prove it: https://feelideas.com/riemann-curvature-tensor-and-covariant-derivative/).

But if I try to develop it in a different way I find a different result:

\((\nabla_{\beta}\nabla_{\nu} \vec{V})^{\mu} = dx^{\mu}(\nabla_{\beta}({V^{\alpha}}_{;\nu}\; \frac{ \partial }{ \partial x^{\alpha}})) = dx^{\mu}( {V^{\alpha}}_{;\nu,\beta}\; \frac{ \partial }{ \partial x^{\alpha}} + {V^{\alpha}}_{;\nu}\; \nabla_{\beta} \frac{ \partial }{ \partial x^{\alpha}} ) =\)

\(dx^{\mu}( {V^{\alpha}}_{;\nu,\beta}\; \frac{ \partial }{ \partial x^{\alpha}} + {V^{\alpha}}_{;\nu}\; {\Gamma^{\sigma}}_{\alpha\beta}\; \frac{ \partial }{ \partial x^{\sigma}} ) = dx^{\mu}( {V^{\alpha}}_{;\nu,\beta}\; \frac{ \partial }{ \partial x^{\alpha}} + {V^{\sigma}}_{;\nu}\; {\Gamma^{\alpha}}_{\sigma\beta}\; \frac{ \partial }{ \partial x^{\alpha}} ) =\)

\(dx^{\mu}( ({V^{\alpha}}_{;\nu,\beta} + {V^{\sigma}}_{;\nu}\; {\Gamma^{\alpha}}_{\sigma\beta})\; \frac{ \partial }{ \partial x^{\alpha}} ) = {V^{\mu}}_{;\nu,\beta} + {V^{\sigma}}_{;\nu}\; {\Gamma^{\mu}}_{\sigma\beta}\)

They differ for the term: 

\(- {V^\mu}_{;\alpha}\; {\Gamma^ \alpha}_{ \nu \beta }\)

Where am I wrong?

asked Aug 19, 2021 in Mathematics by andrea.bali [ revision history ]
edited Aug 19, 2021

Maybe I found the reason of the incongruence.

\((\nabla_\beta \nabla_ \nu V)^ \mu\)  is not a tensor, because it is not linear on the second \(\nabla\).

The correct expression about it is only the last: 

\((\nabla_\beta \nabla_ \nu V)^ \mu = {V^{\mu}}_{;\nu,\beta} + {V^{\sigma}}_{;\nu}\; {\Gamma^{\mu}}_{\sigma\beta}\)

1 Answer

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properties of the covariante derivative of tensor  : https://en.wikipedia.org/wiki/Covariant_derivative

answered Aug 25, 2021 by anonymous [ no revision ]

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