# Problem with vanishing tensor equation and raising all indices

Gold Member
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.

kent davidge

Homework Helper
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.

George Keeling and vanhees71
Mentor
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).

Gold Member
Brilliant Gaussian, thanks.