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

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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
edited Aug 19

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:

$(\nabla_\beta \nabla_ \nu V)^ \mu = {V^{\mu}}_{;\nu,\beta} + {V^{\sigma}}_{;\nu}\; {\Gamma^{\mu}}_{\sigma\beta}$ answered Aug 25 by anonymous
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