# Index position when varying an action with respect to the metric

I'm confused about where we should put tensor indices when we vary an action wrt the metric. For example, if I have in the Lagrangian a term such as $$A_{\mu\nu}B^{\mu\nu},$$ do I necessarily have to write it as $$g^{\rho\mu}g^{\nu\sigma}A_{\mu\nu}B_{\rho\sigma}$$ before performing the variation? My guess is that if (prior to the definition of the Lagrangian) A is defined as a (0,2) tensor and B as a (2,0) tensor, then $$A_{\mu\nu}B^{\mu\nu}$$ doesn't depend on the metric, whereas things like $$F_{\mu\nu}F^{\mu\nu} = g^{\rho\mu}g^{\nu\sigma}F_{\mu\nu}F_{\rho\sigma}$$ do depend on the metric as one needs to map $$F_{\mu\nu}$$ to its dual space to be able to contract it with itself.

1. Is the above correct?

2. What if the quantities being contracted aren't tensors (thus cannot be mapped back and forth from/to tangent/cotangent space), eg terms formed by contracting the Christoffel symbol $$\Gamma^\alpha_{\mu\nu}$$ with itself? I'd assume there wouldn't be any metric dependence of the form $$\Gamma^\alpha_{\mu\nu} = g^{\alpha\beta}\Gamma_{\beta\mu\nu}$$.

• The point is that once you specify a set of independent variables, you just vary as normal. You can take these variables to be, for example, A with lowered indices and B with raised indices, or both with raised indices. In the former case their contraction does not depend on the metric. However, with any choice you’ll end up with the same equations of motion. – knzhou Jan 23 '19 at 11:33
• Why I end up with the same eom? For example, suppose $A_\mu$ is a covector and $B^\mu$ a vector, both constants. In this case I don't need the metric to contract them, so as you say $A_\mu B^\mu$ is metric-independent, thus $\frac{\delta}{\delta g^{\mu\nu}}(A_\lambda B^\lambda)=0$. On the other hand, if I am given two constant vectors $A^\mu$ and $B^\mu$, then I do need the metric and $\frac{\delta}{\delta g^{\mu\nu}}\ (A_\lambda B^\lambda) = \frac{\delta}{\delta g^{\mu\nu}}\ (g_{\lambda\sigma}A^\lambda B^\sigma)=A^\lambda B^\sigma\frac{\delta}{\delta g^{\mu\nu}}\ g_{\lambda\sigma}\neq 0$. – Mr. K Jan 23 '19 at 13:03
• Right, some details will look slightly different, but the equations of motion will be the same. Try and find an example where they don't match! – knzhou Jan 23 '19 at 14:04
• Just showed you one... the Lagrangian $\mathcal{L}=\sqrt{g}A(B)$, where $A$ is a covector and $B$ a vector, is different than the Lagrangian $\mathcal{L}=\sqrt{g}g(A,B)$, where A and B are vectors. They produce different EOMs and different physics. – Mr. K Jan 23 '19 at 14:40

In case of the Einstein–Hilbert action, the action is extremized against an infinitesimal variation of the components of the metric tensor $$g_{\mu \nu}$$, even if for convenience it is used the inverse metric $$g^{\mu \nu}$$. That means that you have to express the Lagrangian showing the components of the metric tensor. For instance if you have the composition of two rank-2 tensors, you show:
$$A_{\mu \nu} B^{\mu \nu} = g^{\mu \rho} g^{\nu \sigma} A_{\mu \nu} B_{\rho \sigma}$$
As for objects which are not tensors, as the physical laws are stated in terms of a tensorial formalism, there are other terms that together allow for a tensorial description. For instance the covariant derivative is made up of the partial derivative and the $$\Gamma's$$: each piece separately is not a tensor, but together they are.