Raising index of variation

I know how to prove e.g. $$A^{ik}B_{lk}=A_{k}^iB^{k}_l.\tag{1}$$ (Raising and Lowering Indices Question). Today in a book, I find: $$g^{ik}\delta g_{lk}=-g_{kl}\delta g^{ki}.\tag{2}$$

$g^{ik}$ is the metric tensor. There is a negative sign. If I use the regular method to raise/lower the indices, I cannot get the negative sign. I guess this must be due to the variation $\delta$? Do you know how to prove this?

It then says because of the aforementioned equation, therefore:

$$T^{ik}\delta g_{ik}=-T_{ik}\delta g^{ik}.\tag{3}$$

$T^{ik}$ is the energy momentum tensor. Why they have such relationship?

• The crucial observation here is that the object $\delta g^{ij}$ is not the 'raised index version of $\delta g_{ij}$', but rather the variation of the inverse metric $g^{ij}$. – gj255 Oct 30 '16 at 17:37

Remember that the metric tensor $g_{\rho\nu}$ and its inverse $g^{\rho\nu}$ fulfill the relation

$$g^{\mu\rho}g_{\rho\nu}=\delta^\mu_\nu.$$

In the above equation, if you take the variation on both sides you get

$$\delta g^{\mu\rho}g_{\rho\nu}+g^{\mu\rho}\delta g_{\rho\nu}=0.$$

From this you get

$$g^{\mu\rho}\delta g_{\rho\nu}=-g_{\rho\nu}\delta g^{\mu\rho}.$$

Which is what you have in Eq. $(2)$. Equation $(3)$ follows from this.

• Thank you and sorry, I had typo in (3) before. I just corrected. And based on your proof, I can further prove: $$T_{ik}\delta g^{ik}=T^{nm}g_{ni}g_{mk}\delta g^{ik}=-T^{nm}g_{ni}g^{ik}\delta g_{mk}=-T^{nm}\delta^k_n\delta g_{km}=-T^{km}\delta g_{km}=-T^{ik}\delta g_{ik}$$ – HYW Oct 29 '16 at 16:36
• That's it. Very good. – Apogee Oct 29 '16 at 16:51
• Perfect answer. – Nogueira Oct 30 '16 at 17:44