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In a comment about variation of metric tensor it was shown that $$\delta g_{\mu\nu}=-g_{\mu\rho}g_{\nu\,\sigma}\delta g^{\rho\,\sigma}$$ which is contrary to the usual rule of lowering indeces of a tensor. That means $\delta g^{\mu\nu}$ is not a tensor, while $g^{\mu\nu}$ is a tensor quantity, so taking a variation leads to loss of tensor nature here.

How often does that happen to other tensor quantities? Are there general rules or guidelines to know if their variations are tensors or not?

Update: Misunderstanding clarified. I should not interpret $\delta T^{\mu\nu}$ as $(\delta T)^{\mu\nu}$ and try to apply tensor rules to indices immediately. Instead, I should take it as $\delta(T^{\mu\nu})$, and if I wish to lower indices, I should write $\delta(g_{\mu\rho}T^{\rho\nu})$ and then apply Leibniz rule to the variation as $\delta(AB)=(\delta A)B+A(\delta B)$. That would generally add some terms in case $\delta g^{\mu\nu}\ne 0$.

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up vote 5 down vote accepted

A variation of a tensor is always a tensor and the formula for the value above doesn't show otherwise.

What you probably find surprising is that $\delta g_{\mu\nu}$ and $\delta g^{\rho\sigma}$ are not related to each other by simply raising the indices $\mu,\nu$ or lowering the indices $\rho,\sigma$. Indeed, they're not related in this way. In this case, there is an extra sign.

The reason why this simple recipe doesn't hold is that when we express $g^{\rho\sigma}$ as $g^{\rho\mu}g^{\sigma\nu}g_{\mu\nu}$, i.e. a formula to raise indices $\mu,\nu$, we mustn't forget to take into account the nonzero variations of the metric tensors $g^{\rho\mu}$ and $g^{\sigma\nu}$ that were used to raise the indices! We're trying to talk about $\delta g^{\rho\sigma}$ so it's clear that $g^{\rho\mu}$ and $g^{\sigma\nu}$ are also nonzero and that's why they give us extra terms from the Leibniz formula for the variation of the product.

But the fact that $\delta g$ with the lower and upper indices aren't related by the simple procedure of raising or lowering indices doesn't mean that these two objects aren't tensors. Being a tensor is not about its relationships to other tensors with different position of indices – about the rules for raising or lowering of the indices. Instead, to judge whether such an object is a tensor, we always keep the indices on the same place. In fact, we could determine whether an object is a tensor even if it were impossible to raise or lower indices, i.e. if there were no metric tensor around to do the job!

The condition that something is a tensor means that if we switch to a different coordinate system, the components transform as $$ T_{\alpha'\beta'\dots}{}^{\lambda\mu\dots} = T_{\alpha\beta\dots}{}^{\lambda\mu\dots}\cdot D_{\alpha'}^\alpha D_{\beta'}^{\beta}\dots $$ where the matrices $D$ are composed of $\partial x^{\alpha'}/\partial x^{\alpha}$ and similar expressions. Note that these $D$ objects have one lower and one upper index and they don't raise or lower anything! The number of covariant and contravariant indices are the same in both coordinate systems when we decide whether something is a tensor!

Note that the formula for the variation of the upper components of the metric may be easily derived from the fact that $$ \delta (\delta_\mu^\alpha ) = 0$$ which holds because the Kronecker delta has constant components $1$ and $0$. You may rewrite $$\delta_\mu^\alpha = g_{\mu\lambda} g^{\alpha\lambda}$$ so its variations, by the Leibniz rule, is $$ \delta (\delta_\mu^\alpha ) = \delta g_{\mu\lambda}\cdot g^{\alpha\lambda}+g_{\mu\lambda}\cdot \delta g^{\alpha\lambda} = 0 $$ which already allows you to calculate the variation of the upper-component metric from the variation of the lower-component metric by a simple renaming and lowering and raising of the indices. Note that the minus sign arises because the sum of the two terms above equals zero.

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So $\delta g^{\mu\nu}$ "ill-behaved" just because it was not independent of the variation of $g_{\mu\nu}$s which are used to lower indices? And while I'm working with some independent variation, I'm totally safe? – firtree May 9 '13 at 6:00
No. Your naive – wrong – prescription for the variation of any tensor (even non-metric tensor) is always wrong as long as $\delta g_{\mu\nu}$ is nonzero. You must use the Leibniz rule $\delta(AB)=\delta A\cdot B+A\cdot\delta B$ to compute the variation. But the variation of a tensor, e.g. $\delta T^{\mu\nu}$, is always a tensor as well. There will be several terms on the right hand side and each of them is actually a tensor by itself! Again, you seem to misunderstand what the word "tensor" means. It doesn't mean that it comes with the naive rules to lower or raise indices. – Luboš Motl May 9 '13 at 6:04
The convention to define the values of tensors with raised or lowered indices is a usual convention we apply to various tensors - we can't apply it to variations - but it has nothing to do with the question whether an object is a tensor! I wrote what the actual condition is above. Could you please reread my answer before you ask the same question again? – Luboš Motl May 9 '13 at 6:05
as long as $\delta g_{\mu\nu}$ is nonzero - I see, thanks. Leibniz rule to rule them all! – firtree May 9 '13 at 7:33
Exactly, @firtree ... The only difference if you consider the upper-index $\delta g$ relatively to the case of $\delta T$ is that you want to compute $\delta g^{ab}$ and the same object appears on the right hand side from the raising and Leibniz rule - so you get a sort of a self-referring equation (what you want to calculate is expressed in terms of the same thing). But this self-referring equation may still be solved - in my answer, it's solved in an easier way using the Kronecker delta. Such subtleties don't arise for $\delta T^{ab}$. – Luboš Motl May 9 '13 at 11:49

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