I was doing an exercise in Schutz (A First course in General Relativity). The exercise wanted the double covariant derivative calculated for a vector $V^\mu$ i.e. $\nabla_\alpha \nabla_\beta V^\mu$ i.e. $(V^\mu_{;\alpha})_{;\beta}$. This basically amounts to calculate the covariant derivative of a mixed (1,1) tensor $T^\alpha_\beta$. I was able to calculate the covariant derivative of this mixed tensor by converting it into $T^\alpha_\beta = A^\alpha B_\beta$ and it worked out.

But I originally tried to calculate it by lowering the index on a (2,0) tensor $T^{\alpha\beta}$ using a metric tensor and then calculating the covariant derivative, since the covariant derivative of metric tensor is zero. But it did not work out. Here is how I did it.

Given that, $$ T^{\alpha \beta}_{;\gamma}= T^{\alpha \beta}_{,\gamma} + \Gamma^\alpha_{\mu\gamma}T^{\mu\beta}+\Gamma^\beta_{\mu\gamma}T^{\alpha\mu} $$ I lowered the index, $$ (T^{\alpha}_\lambda g^{\lambda\beta})_{;\gamma} =(T^{\alpha}_\lambda )_{;\gamma}g^{\lambda\beta}= (T^{\alpha \beta}_{,\gamma} + \Gamma^\alpha_{\mu\gamma}T^{\mu\beta}+\Gamma^\beta_{\mu\gamma}T^{\alpha\mu})g^{\lambda\beta} $$ $$ (T^{\alpha}_\lambda)_{;\gamma} = g_{\lambda\beta}T^{\alpha \beta}_{,\gamma} + g_{\lambda\beta}\Gamma^\alpha_{\mu\gamma}T^{\mu\beta}+g_{\lambda\beta}\Gamma^\beta_{\mu\gamma}T^{\alpha\mu} $$ The confusion is that $g_{\lambda\beta}\Gamma^\beta_{\mu\gamma}$ in the last term, after the contraction on $\beta$ becomes, $\Gamma_{\lambda\mu\gamma}$ which is not the correct term for the covariant derivative of $T^{\alpha}_\lambda$.

So, I am missing something in the contraction of $\Gamma^\beta_{\mu\gamma}$ or may be that is not a valid operation(?). I have a hunch that the reason has something to do with $\Gamma^\beta_{\mu\gamma}$ not being a valid tensor but I cant place what it is mathematically/physically(?). Or may be there is an identity of $\Gamma^\beta_{\mu\gamma}$ that I am missing here(?).

  • $\begingroup$ Which term would you expect? Note that you cannot contract the first term either, because of the not covariant derivative in $T^{\alpha\beta}_{\,,\gamma}$. $\endgroup$
    – Photon
    Commented Jul 13, 2018 at 9:52
  • $\begingroup$ Of course! Thanks for point that out. So in general contraction isn't the way to go with covariant derivatives. The Christoffel symbol isn't the only problem here! What is the physical meaning of this thought? If you can shed some light on it. $\endgroup$
    – Shaz
    Commented Jul 13, 2018 at 10:26
  • $\begingroup$ The physical meaning of what exactly? $\endgroup$
    – Photon
    Commented Jul 13, 2018 at 13:04

1 Answer 1


Your hunch is basically correct. Raising and lowering indices really means that you are exploiting a canonical isomorphism between tangent and contangent spaces in the presence of a metric: $v^\mu=v_\nu g^{\nu\mu}$ is just the map $$\begin{align*} V &\longrightarrow V^*\\v&\longmapsto g(v,\cdot)\end{align*}$$ expressed in a basis. The same applies to tensors with more indices, i.e. elements of tensor products of $V$ and $V^*$.

This implies that raising and lowering is only sensible for proper tensors (elements of $V\otimes\dotsm \otimes V\otimes V^*\otimes\dots \otimes V^*$). Partial derivatives and Christoffel symbols are not such tensors, and so you should not raise/lower the indices here. (Of course, the covariant derivative combines $\partial_\mu$ and $\Gamma_{\mu\nu}^\rho$ in the right way to be a tensor, hence the above iosomrphism applies, and you can freely raise/lower indices her.)

  • $\begingroup$ There is an interesting angle to this question in the Math forum. math.stackexchange.com/questions/2849440/… $\endgroup$
    – Shaz
    Commented Jul 14, 2018 at 17:18
  • $\begingroup$ can you raise index on christopher symbols? for computing the partial derivative of contravariant metric tensor? $\endgroup$ Commented Feb 18, 2019 at 9:09
  • $\begingroup$ @vtd-xml-author: As the answer says, "Christoffel symbols are not such tensors, and so you should not raise/lower the indices here" $\endgroup$
    – Toffomat
    Commented Feb 18, 2019 at 9:17
  • $\begingroup$ Thanks for the confirmation... but in many text books, christoffel symbols have first kind and second kind, both of which are related by lowering /raising indexes... en.wikipedia.org/wiki/… $\endgroup$ Commented Feb 19, 2019 at 4:02
  • $\begingroup$ this could simply be abuse of notion... $\endgroup$ Commented Feb 19, 2019 at 4:02

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