1
$\begingroup$

Let $(M,g)$ be a (Pseudo-)Riemannian manifold. If I perform a transformation on the metric, getting a new metric $\tilde{g}$, which metric should I use to raise and lower indices? As I understand, this leads to a new manifold $(M,\tilde g)$ with the same underlying differentiable manifold but with a different metric. My problem is when I get equations involving both metrics. For example, if I consider a transformation of the type

$ \tilde g = g + T, \tag{1} $

where $T$ is a tensor field (assume that it satisfies everything that has to be satisfied so that $\tilde g$ is a metric indeed). Then I can express the Ricci tensor, for example, in terms of the new metric and I get in coordinates something like

$ R_{ij} = \tilde R_{ij} + B_{ij}, \tag{2} $

where $\tilde R_{ij}$ has the same analytical form of the Ricci curvature but with the new metric $\tilde g$ instead of $g$ and $B_{ij}$ denote collectively other terms depending solely on $\tilde g$ and on $T$. Now, if I want to find $R^{ij}$, should I use $R^{ij}=g^{im}g^{jn}R_{mn}$ or $R^{ij}=\tilde g^{im}\tilde g^{jn}R_{mn}$? What about $\tilde R^{ij}$ and $B^{ij}$? It seems weird to me because the LHS of eq. (2) depends on $g$ while the RHS depends on $\tilde g$.

$\endgroup$
0
$\begingroup$

"Raising and lowering indices using a metric" is only a notional convention and has no real content. In a situation like the one you describe, you can either decide to consider one of the two metrics as being fixed and fix your conventions by using only this metric to raise and lower. This is consistent but potentially dangerous. For example in situation like with Ricci curvatures (which is canonical with lower indices) you get effects like that $R^i_j=g^{ij}R_{ij}$ is the scalar curvature of $g$, whereas $\tilde R^i_j=g^{ij}\tilde R_{ij}$ is not the scalar curvature of $\tilde g$ (which equals $\tilde g^{ij}\tilde R_{ij}$). But it is a convention that works if you are careful enough.

The alternative (which I would probably recommend) is to drop the convention completely and write out the rasing and lowering of indices explicitly all the time.

$\endgroup$
  • $\begingroup$ Thanks, but I have some issues with your answer. First of all I don't think raising/lowering indices has no real content as you said, this is rather a coordinate-dependent way of mapping vectors to their duals and vice-versa and it is what makes explicit the isomorphism between these spaces. Secondly, why are you using indices for the scalar curvature? The indices i and j are summed over, so there's no index left. $\endgroup$ – Mr. K Jun 23 '16 at 9:14
  • $\begingroup$ Now lets go back to my question. Is it consistent to use both metrics at the same time? I mean, can I write $R=g^{ij}R_{ij} = (\tilde g^{ij} + T^{ij})\tilde R_{ij} + (\tilde g^{ij} + T^{ij})B_{ij}$? Then I could relate the old curvature $R$ with the new one $\tilde R$. Note that in this case I am $g^{ij}$ on the LHS to raise the indices, while I am using $\tilde g^{ij}$ on the RHS to raise the indices. $\endgroup$ – Mr. K Jun 23 '16 at 9:16
  • $\begingroup$ If $g^{ij}=\tilde g^{ij}+T^{ij}$ and $R_{ij}=\tilde R^{ij}+B_{ij}$, then of course the equation you propose holds. In this case, you are just writing out an equation. The issue of whom you use to raise indices occurs only if you are using expressions like $R^i_j$, which are then "automatically" interpreted as $g^{ik}R_{kj}$. The expression you use is what I meant by "writing out the raising and lowering of indices explicitly". $\endgroup$ – Andreas Cap Jun 23 '16 at 9:59
  • $\begingroup$ Yes, I know it holds as an equation, but what I actually want to know is if $R=\tilde R + T^{ij}\tilde R_{ij}+(\tilde g^{ij}+T^{ij})B_{ij}$ makes sense since now I have $R=g^{ij}R_{ij}$ and $\tilde R = \tilde g^{ij} \tilde R_{ij}$. That's why I said I am using both metrics to raise/lower indices. $\endgroup$ – Mr. K Jun 23 '16 at 10:13
  • $\begingroup$ Sure it holds, since contractions are linear in both arguments, so $(\tilde g^{ij}+T^{ij})\tilde R^{ij}=\tilde g^{ij}\tilde R_{ij}+T^{ij}\tilde R_{ij}$. $\endgroup$ – Andreas Cap Jun 23 '16 at 10:47

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.