# 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.