Solving Vanishing Tensor Eqn & Raising All Indices

[George Keeling]

TL;DR Summary

Problem with vanishing tensor equation with all indices down. Does it still vanish when they are up?

I have an equation $$
\chi_\nu\nabla_\mu\chi_\sigma+\chi_\sigma\nabla_\nu\chi_\mu+\chi_\mu\nabla_\sigma\chi_\nu=0
$$so we also have$$
g_{\nu\rho}g_{\mu\tau}g_{\sigma\lambda}\left(\chi^\rho\nabla^\tau\chi^\lambda+\chi^\lambda\nabla^\rho\chi^\tau+\chi^\tau\nabla^\lambda\chi^\rho\right)=0
$$Does that mean that$$
\chi^\rho\nabla^\tau\chi^\lambda+\chi^\lambda\nabla^\rho\chi^\tau+\chi^\tau\nabla^\lambda\chi^\rho=0
$$as well?

I can prove in two dimensions that $$
x_i=0\Rightarrow g_{ij}x^j=0\Rightarrow x^j=0
$$as long as the metric is not degenerate.

It would be horrendous to extend the proof to four dimensions and three indices. I think there is some more intuitive way to get from the second equation to the third, but the intuition eludes me.

 

[Gaussian97]

Well, actually it is true, the key point as you say is that $g_{\mu\nu}$ has an inverse, so instead of doing what you do in equation 2, why not contract the first equation with $g^{\alpha\nu}g^{\beta\mu}g^{\gamma\sigma}$? What you get from there?

BTW, note that this is true only because $D_{\alpha} g^{\mu\nu} = 0$, if this were not true then an expression true for covariant vectors wouldn't need to be true for contravariant ones.

 

[PeterDonis]

Summary:: Problem with vanishing tensor equation with all indices down. Does it still vanish when they are up?

Raising all the indexes on the LHS of a vanishing tensor equation with all indices down obviously gives a vanishing tensor equation with all indices up (since "raising an index" on $0$ on the RHS just gives $0$ again). As @Gaussian97 says, this will work as long as the metric is not degenerate (so the inverse metric is well-defined).

 

[George Keeling]

Brilliant Gaussian, thanks.