How is the Index Raised or Lowered on a Conformally Related Metric?

[Silviu]

Hello! I have a manifold with a metric $g$ and another metric on M is $\bar{g}=e^{2\sigma(p)}g$, where $\sigma$ is a function on manifold (0-form). After some tedious calculations one reaches the formula $\bar{Ric}_{\mu \nu} = Ric_{\mu \nu}- g_{\mu \nu}B_\lambda^\lambda-2(m-1)B_{\nu \mu}$, where $B$ is a symmetric tensor previously defined (not important for my question), m the dimension of the manifold and $Ric$ the Ricci tensor. My first question is quite general. If we want to rise/lower indices on a tensor, we need to apply the metric according to which it is considered a tensor, right? I.e. $\bar{g}$ for $\bar{Ric}$ and $g$ for $Ric$. So something like $g^\mu_\nu \bar{Ric}_{\mu \nu}$, would rise the index, right? Then, in order to get the scalar curvature, I wanted to multiply the above relation by $g^\mu_\nu$. On the right everything would act normal, but on the left I expected to get $g^\mu_\nu \bar{Ric}_{\mu \nu} = e^{-2\sigma} e^{2\sigma} g^\mu_\nu {Ric}_{\mu \nu} = e^{-2\sigma} \bar{R}$, where $R$ is the scalar curvature. However the results in the book is $e^{2\sigma}\bar{R}=R-2(m-1)B_\lambda^\lambda$. How do they get $e^{+2\sigma}$ on the left? Also, I am confused about the scalar curvature now. I thought that being a scalar means that it stays the same under a change of coordinates, so why does it change here? Thank you!

 

[andrewkirk]

something like $g^\mu_\nu \bar{Ric}_{\mu \nu}$, would rise the index, right?

That would raise the index, but in a purely arithmetical manner. The result would not have any meaning as a tensor because it is mixing the two metrics. To raise the index of a tensor in the bar metric, the bar metric tensor needs to be used. Secondly, to raise an index the metric must be written with both indices as superscripts, so for the above we'd need something like
$$\bar g^{\mu\xi}\ \overline{Ric}_{\xi\nu}\quad\textrm{giving}\ \overline{Ric}^\mu{}_\nu$$

To get the scalar curvature of $\overline{Ric}_{\mu\nu}$, we have to multiply by $\bar g^{\mu\nu}$ and perform the double-summing that is implied by that.

I am confused about the scalar curvature now. I thought that being a scalar means that it stays the same under a change of coordinates, so why does it change here? Thank you!

Because changing the metric tensor is much more than just changing the coordinates. If we change the metric then we have a completely different differentiable manifold. The differential structure (geometry) of the manifold has changed. Only its topology has remained constant, and curvature is not defined by topology.

 

[Silviu]

That would raise the index, but in a purely arithmetical manner. The result would not have any meaning as a tensor because it is mixing the two metrics. To raise the index of a tensor in the bar metric, the bar metric tensor needs to be used. Secondly, to raise an index the metric must be written with both indices as superscripts, so for the above we'd need something like
$$\bar g^{\mu\xi}\ \overline{Ric}_{\xi\nu}\quad\textrm{giving}\ \overline{Ric}^\mu{}_\nu$$

To get the scalar curvature of $\overline{Ric}_{\mu\nu}$, we have to multiply by $\bar g^{\mu\nu}$ and perform the double-summing that is implied by that.

Because changing the metric tensor is much more than just changing the coordinates. If we change the metric then we have a completely different differentiable manifold. The differential structure (geometry) of the manifold has changed. Only its topology has remained constant, and curvature is not defined by topology.

Thank you for your reply! How about my derivation above? What is it wrong with it and how do they get $e^{2\simga}$ on the left? Also, about the scalar curvature, isn't a change in coordinate a change in the metric? Like going in euclidean space from cartesian to polar, changes the metric, but the curvature of the space is still 0.

 

[andrewkirk]

Thank you for your reply! How about my derivation above? What is it wrong with it and how do they get $e^{2\simga}$ on the left? Also, about the scalar curvature, isn't a change in coordinate a change in the metric? Like going in euclidean space from cartesian to polar, changes the metric, but the curvature of the space is still 0.

Changing coordinates doesn't change the metric, which is a coordinate-independent object. The block of numbers used to represent a metric in a given coordinate system - eg a square matrix for a second order tensor, and a column of numbers for a vector - is just a representation of the object, not the object itself.

It's a good idea to familiarise yourself with the idea of coordinate-free representations of tensors, as understanding that can make a lot of things simpler. A tensor is a function that takes one or more vectors or dual vectors as inputs and gives a scalar as result. Changing coordinates doesn't change the function, just as changing from inches to centimetres doesn't change the length of a stick.

 

[martinbn]

You wrote it in an index free notation and may be that confuses you. In index notation the equation $\bar{g}=e^{2\sigma(p)}g$ is $\bar{g}_{\mu\nu}=e^{2\sigma(p)}g_{\mu\nu}$. So for the inverse metrics, which are $g^{\mu\nu}$ and $\bar{g}^{\mu\nu}$, the relation is $\bar{g}^{\mu\nu}=e^{-2\sigma(p)}g^{\mu\nu}$. For the Ricci scalor relation you multiply both sides by $g^{\mu\nu}$ and sum, as already written above.

$g^{\mu\nu}\overline{Ric}_{\mu \nu} = g^{\mu\nu}Ric_{\mu \nu}+\cdots$

then

$e^{2\sigma(p)}\bar{g}^{\mu\nu}\overline{Ric}_{\mu \nu} = R+\cdots$

and

$e^{2\sigma(p)}\bar{R} = R+\cdots$