A Is the variation of the metric ##\delta g_{\mu\nu}## a tensor?

[Kostik]

TL;DR Summary

Two contradictory equations are shown. Where is the flaw?

From $$0 = \delta(g^{\alpha\mu}g_{\mu\nu}) = g^{\alpha\mu} \delta g_{\mu\nu} + g_{\mu\nu} \delta g^{\alpha\mu}$$ we have $$g^{\alpha\mu} \delta g_{\mu\nu} = -g_{\mu\nu} \delta g^{\alpha\mu} \,\, . $$ Multiply both sides by $g_{\alpha\beta}$ to get $$\delta g_{\beta\nu} = -g_{\alpha\beta} g_{\mu\nu} \delta g^{\alpha\mu} \qquad(*)$$ (This is Dirac's eq. (26.9) in "GTR".) On the other hand, the variation $\delta g^{\alpha\mu} = \bar{g}^{\alpha\mu} - g^{\alpha\mu}$ should be a tensor; therefore, $$\delta g_{\beta\nu} = g_{\alpha\beta} g_{\mu\nu} \delta g^{\alpha\mu} \quad\qquad(**)$$ Which equation is correct, $(*)$ or $(**)$?

 

[Kostik]

I think I found the (or an) answer. The relations $$\delta g^{\mu\nu} = -g^{\mu\rho}g^{\nu\sigma}\delta g_{\rho\sigma} \quad , \qquad \delta g_{\mu\nu} = -g_{\mu\rho} g_{\nu\sigma} \delta g^{\rho\sigma} \qquad(*)$$ appear to indicate that $\delta g_{\mu\nu}$ and $\delta g^{\mu\nu}$ are not tensors, since their indices are not raised and lowered in the usual way by the metric tensor. But this is not true; they are tensors, as can be seen by calculating $\delta g'_{\mu\nu}$ using $$\delta g_{\mu\nu} = \bar{g}_{\mu\nu} - g_{\mu\nu} \,\, .$$ $\qquad$ Given the variation $\delta V_\mu$, to find the variation of the contravariant vector $\delta V^\mu$, we must calculate $$\delta V^\mu = \delta (g^{\mu\alpha}V_\alpha) = g^{\mu\alpha}\delta V_\alpha + V_\alpha \delta g^{\mu\alpha} \,\, .$$ Notice that the last term "spoils" the usual rule for raising and lowering indices; the Leibniz rule requires that we account for the variation in $V_\alpha$ and the metric tensor. In the case of the metric tensor $g_{\alpha\beta}$, the Leibnitz rule gives $$\delta g^{\mu\nu} = g^{\mu\alpha} g^{\nu\beta} g_{\alpha\beta} + 2\delta g^{\mu\nu}$$ which gives $(*)$.

$\qquad$ In other words, the Leibniz rule is paramount; the “raising and lowering” rule does not apply across the “$\delta$”.

 

[PeterDonis]

the screenshot below

What reference is this screenshot from?

 

[Kostik]

@PeterDonis These are my own notes.

If I have made an error, please critique it!

 

[PeterDonis]

These are my own notes.

Then you should post them directly using LaTeX for equations, in accordance with the forum rules.

If I have made an error, please critique it!

We can't if all we can quote is an image. We need to be able to quote individual equations and words. That's why we need things posted directly using LaTeX for equations.

 

[Kostik]

That makes sense, I will endeavor to use LaTex in the future. Obviously it’s much faster for me to take a snapshot of my own notes (made with MSWord).

 

[PeterDonis]

I will endeavor to use LaTex in the future.

If you're not going to re-post your notes using LaTeX now, I'm going to close this thread. You asked for feedback; that can only be done if you post your content the way I described.

 

[PeterDonis]

Obviously it’s much faster for me to take a snapshot of my own notes (made with MSWord).

If you can't take the time to post using direct text and LaTeX, why should we take the time to read and answer?

Sorry to be blunt, but that's one of the reasons behind the PF rules on this: you need to be respectful of other posters' time, not just your own.

 

[Kostik]

The narrative in #2 has now been added in LaTex.

 

[anuttarasammyak]

TL;DR Summary: Two contradictory equations are shown. Where is the flaw?

From $$0 = \delta(g^{\alpha\mu}g_{\mu\nu}) = g^{\alpha\mu} \delta g_{\mu\nu} + g_{\mu\nu} \delta g^{\alpha\mu}$$ we have $$g^{\alpha\mu} \delta g_{\mu\nu} = -g_{\mu\nu} \delta g^{\alpha\mu} \,\, . $$ Multiply both sides by $g_{\alpha\beta}$ to get $$\delta g_{\beta\nu} = -g_{\alpha\beta} g_{\mu\nu} \delta g^{\alpha\mu} \qquad(*)$$ (This is Dirac's eq. (26.9) in "GTR".) On the other hand, the variation $\delta g^{\alpha\mu} = \bar{g}^{\alpha\mu} - g^{\alpha\mu}$ should be a tensor; therefore, $$\delta g_{\beta\nu} = g_{\alpha\beta} g_{\mu\nu} \delta g^{\alpha\mu} \quad\qquad(**)$$ Which equation is correct, $(*)$ or $(**)$?

Under variation of metric tensor, $g_{\mu\nu}+\delta g_{\mu\nu}$ is a tensor but its parts $g_{\mu\nu}$ and $\delta g_{\mu\nu}$ are not tensors. Its indeces up-down relation is
$$ g_{\mu\nu}+\delta g_{\mu\nu} =( g_{\alpha\nu}+\delta g_{\alpha\nu}) (g_{\mu\beta}+\delta g_{\mu\beta})( g^{\alpha\beta}+\delta g^{\alpha\beta} )$$
0-th order equation is usual one with no variation. The 1st order equation is
$$ \delta g_{\mu\nu}= g_{\mu\beta} g^{\alpha\beta} \delta g_{\alpha\nu} + g_{\alpha\nu}g^{\alpha\beta}
\delta g_{\mu\beta} + g_{\alpha\nu} g_{\mu\beta}\delta g^{\alpha\beta} $$
which gives $(*)$.

 

[haushofer]

TL;DR Summary: Two contradictory equations are shown. Where is the flaw?

From $$0 = \delta(g^{\alpha\mu}g_{\mu\nu}) = g^{\alpha\mu} \delta g_{\mu\nu} + g_{\mu\nu} \delta g^{\alpha\mu}$$ we have $$g^{\alpha\mu} \delta g_{\mu\nu} = -g_{\mu\nu} \delta g^{\alpha\mu} \,\, . $$ Multiply both sides by $g_{\alpha\beta}$ to get $$\delta g_{\beta\nu} = -g_{\alpha\beta} g_{\mu\nu} \delta g^{\alpha\mu} \qquad(*)$$ (This is Dirac's eq. (26.9) in "GTR".) On the other hand, the variation $\delta g^{\alpha\mu} = \bar{g}^{\alpha\mu} - g^{\alpha\mu}$ should be a tensor; therefore, $$\delta g_{\beta\nu} = g_{\alpha\beta} g_{\mu\nu} \delta g^{\alpha\mu} \quad\qquad(**)$$ Which equation is correct, $(*)$ or $(**)$?

Yes, it's a tensor, as one can easily see by taking the concrete example of the Lie derivative of the metric w.r.t. a vector field written with a covariant derivative.

That extra minus sign doesn't spoil the tensorial transformation properties, does it?

 

[haushofer]

Under variation of metric tensor, $g_{\mu\nu}+\delta g_{\mu\nu}$ is a tensor but its parts $g_{\mu\nu}$ and $\delta g_{\mu\nu}$ are not tensors. Its indeces up-down relation is
$$ g_{\mu\nu}+\delta g_{\mu\nu} =( g_{\alpha\nu}+\delta g_{\alpha\nu}) (g_{\mu\beta}+\delta g_{\mu\beta})( g^{\alpha\beta}+\delta g^{\alpha\beta} )$$
0-th order equation is usual and obvious. The 1st order equation is
$$ \delta g_{\mu\nu}= g_{\mu\beta} g^{\alpha\beta} \delta g_{\alpha\nu} + g_{\alpha\nu}g^{\alpha\beta}
\delta g_{\mu\beta} + g_{\alpha\nu} g_{\mu\beta}\delta g^{\alpha\beta} $$
which gives $(*)$.

I don't understand this comment or the equation.

 

[anuttarasammyak]

@haushofer The equation is the application of the rule for raising and lowering the indices in use of the (varied) metric tensor to the (varied) metric tensor itself.

That extra minus sign doesn't spoil the tensorial transformation properties, does it?

Then how can we distinguish which rule, (*) or (**) in OP, holds for each tensorial quantity of our interest ?

 

[haushofer]

@haushofer The equation is the application of the rule for raising and lowering the indices in use of the (varied) metric tensor to the (varied) metric tensor itself.


Then how can we distinguish which rule, (*) or (**) in OP, holds for each tensorial quantity of our interest ?

Do you know any more tensors which include that minus-sign? You just have to be careful if you vary the very object itself which enables you to lift and lower indices.

 

[anuttarasammyak]

Do you know any more tensors which include that minus-sign?

No, I don't know any case in tensors. That is one of the reasons that I think :

Under variation of metric tensor, gμν+δgμν is a tensor but its parts gμν and δgμν are not tensors.

I am afraid that you are against it in post #11, stating that one of them variation infinitesimal δgμν is a tensor. I should appreciate it if you could explain the validation of this minus-sign for tensor δgμν.

 

[haushofer]

No, I don't know any case in tensors. That is one of the reasons that I think :

I am afraid that you are against it in post #11, stating that one of them variation infinitesimal δgμν is a tensor. I should appreciate it if you could explain the validation of this minus-sign for tensor δgμν.

I don't understand that comment.

The validation of the minus-sign is because you consider the variation of the very metric tensor itself, not just some arbitrary tensor. I also don't understand this comment:

Under variation of metric tensor, $g_{\mu\nu}+\delta g_{\mu\nu}$ is a tensor but its parts $g_{\mu\nu}$ and $\delta g_{\mu\nu}$ are not tensors.

You say that the metric components g_{\mu\nu} don't constitute components of a tensor? And dito for $\delta g_{\mu\nu}$. which is the different of two tensors? Tensor form a vector space, so sums and differences of tensors are tensors. This just doesn't make any sense.

 

[haushofer]

@anuttarasammyak:
By the way, the answer is already given here below.

I think I found the (or an) answer. The relations $$\delta g^{\mu\nu} = -g^{\mu\rho}g^{\nu\sigma}\delta g_{\rho\sigma} \quad , \qquad \delta g_{\mu\nu} = -g_{\mu\rho} g_{\nu\sigma} \delta g^{\rho\sigma} \qquad(*)$$ appear to indicate that $\delta g_{\mu\nu}$ and $\delta g^{\mu\nu}$ are not tensors, since their indices are not raised and lowered in the usual way by the metric tensor. But this is not true; they are tensors, as can be seen by calculating $\delta g'_{\mu\nu}$ using $$\delta g_{\mu\nu} = \bar{g}_{\mu\nu} - g_{\mu\nu} \,\, .$$ $\qquad$ Given the variation $\delta V_\mu$, to find the variation of the contravariant vector $\delta V^\mu$, we must calculate $$\delta V^\mu = \delta (g^{\mu\alpha}V_\alpha) = g^{\mu\alpha}\delta V_\alpha + V_\alpha \delta g^{\mu\alpha} \,\, .$$ Notice that the last term "spoils" the usual rule for raising and lowering indices; the Leibniz rule requires that we account for the variation in $V_\alpha$ and the metric tensor. In the case of the metric tensor $g_{\alpha\beta}$, the Leibnitz rule gives $$\delta g^{\mu\nu} = g^{\mu\alpha} g^{\nu\beta} g_{\alpha\beta} + 2\delta g^{\mu\nu}$$ which gives $(*)$.

$\qquad$ In other words, the Leibniz rule is paramount; the “raising and lowering” rule does not apply across the “$\delta$”.

Yes. I guess that one could be pedantic and distinguish between \delta (V_{\mu}) and (\delta V)_{\mu}. The first one means lowering an index and then vary, while the second one means vary first and then lower an index on this variation. These two operations don't commute.

 

[haushofer]

So, to go back to the OP, if this confuses people, they should be careful and distinguish between \delta(g_{\mu\nu}) and (\delta g)_{\mu\nu}. But I guess this is also confusing, since g is also used for the determinant of the metric.

But I don't see why this raises the question whether these are "tensors". Just see how the components transform under a coordinate transformation. That marks a tensor. Not how we raise and lower indices on it.

 

[anuttarasammyak]

@haushofer Thanks for the teaching.

$$ g_{\mu\nu}+\delta g_{\mu\nu} =( g_{\alpha\nu}+\delta g_{\alpha\nu}) (g_{\mu\beta}+\delta g_{\mu\beta})( g^{\alpha\beta}+\delta g^{\alpha\beta} )$$

@haushofer The equation is the application of the rule for raising and lowering the indices in use of the (varied) metric tensor to the (varied) metric tensor itself.

Is it an incorrect equation though the result coincides with the equation (*) ?

 

[anuttarasammyak]

Under variation of metric tensor, gμν+δgμν is a tensor but its parts gμν and δgμν are not tensors.

@haushofer it is a good chance to learn from your comments.

May I say that in GR when tensor $A_{abc..}^{def...}$ is written arbitrary as $$A_{abc..}^{def...}=X_{abc..}^{def...}+(A_{abc..}^{def...}-X_{abc..}^{def...})$$
, are X and A-X also tensors ?

In GR any linear combination of tensors
$$ \sum_i a(i) T_{bcd...}^{efg..}(i) $$
are tensors ?

I would like to understand that non-linear feature of GR does not spoil these linear algebra relations.

 

[haushofer]

@haushofer it is a good chance to learn from your comments.

May I say that in GR when tensor $A_{abc..}^{def...}$ is written arbitrary as $$A_{abc..}^{def...}=X_{abc..}^{def...}+(A_{abc..}^{def...}-X_{abc..}^{def...})$$
, are X and A-X also tensors ?

In GR any linear combination of tensors
$$ \sum_i a(i) T_{bcd...}^{efg..}(i) $$
are tensors ?

I would like to understand that non-linear feature of GR does not spoil these linear algebra relations.

1) No, not necessarily; look e.g. at the definition of the covariant derivative. A partial derivative is not a tensor under gct's, and neither is the connection, but the inhomogeneous terms of both cancel out such that their sum is a tensor.

2) Yes, as long as you add/subtract tensors of the same type, of course.

 

[haushofer]

@haushofer Thanks for the teaching.

$$ g_{\mu\nu}+\delta g_{\mu\nu} =( g_{\alpha\nu}+\delta g_{\alpha\nu}) (g_{\mu\beta}+\delta g_{\mu\beta})( g^{\alpha\beta}+\delta g^{\alpha\beta} )$$

Is it an incorrect equation though the result coincides with the equation (*) ?

I don't understand what that equation says.

 

[anuttarasammyak]

I don't understand what that equation says.

In the world of metric tensor undertaking variation where the metric tensor is
$$ \bar{g}_{\mu\nu} := g_{\mu\nu}+\delta g_{\mu\nu} $$,
Tensor indeces up-down will be done as
$$ A_{\mu\nu}= \bar{g}_{\alpha\mu}\bar{g}_{\nu\beta}A^{\alpha\beta}$$
I consider the case that
$$ A_{\mu\nu}=\bar{g}_{\mu\nu}$$

[EDIT]
It would be better to wrilte indecies up-down equation as
$$ \bar{A}_{\mu\nu}= \bar{g}_{\alpha\mu}\bar{g}_{\nu\beta}\bar{A}^{\alpha\beta}$$
where with a bar on top means it is a tensor in the world of "new" metric under variation.

 

[anuttarasammyak]

1) No, not necessarily; look e.g. at the definition of the covariant derivative. A partial derivative is not a tensor under gct's, and neither is the connection, but the inhomogeneous terms of both cancel out such that their sum is a tensor.

Thanks. Even with this general caution, may we take it obvious that in tensor division of
$$ \bar{g}_{\mu\nu} := g_{\mu\nu}+\delta g_{\mu\nu} $$
all $\bar{g}_{\mu\nu}$ , $g_{\mu\nu}$ and $\delta g_{\mu\nu}$ are tensors in the world of metric undertaking variation ?

 

[JimWhoKnew]

Thanks. Even with this general caution, may we take it obvious that in tensor division of
$$ \bar{g}_{\mu\nu} := g_{\mu\nu}+\delta g_{\mu\nu} $$
all $\bar{g}_{\mu\nu}$ , $g_{\mu\nu}$ and $\delta g_{\mu\nu}$ are tensors in the world of metric undertaking variation ?

Ordinarily, $~T^{\mu\nu}~$ , $~T_{\mu\nu}~$ and $~T^\mu{}_\nu~$ are coordinate representation of the same coordinate-free object (tensor) $\mathbf{T}$ . In the present case we have the anomaly of 2 metrics. If we regard $g_{\mu\nu}$ as the "basic" metric, then $\bar{g}_{\mu\nu}$ and its reciprocal $\bar{g}^{\mu\nu}$ are not representations of the same coordinate-free object. That is also the case with $\delta g_{\mu\nu}$ and $\delta g^{\mu\nu}$ (and the reason why equation (**) in OP is wrong).

In equations: if $~\bar{g}_{\mu\nu}=(\mathbf{g}_1)_{\mu\nu}~$ then$$(\mathbf{g}_1)^{\mu\nu}=g^{\mu\rho}\bar{g}_{\rho\sigma}g^{\sigma\nu}\neq\bar{g}^{\mu\rho}\bar{g}_{\rho\sigma}\bar{g}^{\sigma\nu}=\bar{g}^{\mu\nu}$$

 

[weirdoguy]

Ordinarily, Tμν , Tμν and Tμν are coordinate representation of the same coordinate-free object (tensor) T

Are they? Tensors are mappings from cartesian products of vector spaces and their duals. What you wrote are coordinate representations of different coordinate-free objects, since they have different domains. Eg. $T^{\mu\nu}e_{\mu}\otimes e_{\nu}$, $T_{\mu\nu}\varepsilon^{\mu}\otimes \varepsilon^{\nu}$ where $e_i$ and $\varepsilon^i$ are bases in $V$ and its dual respectively, are different tensors.

 

[PeterDonis]

Ordinarily, $~T^{\mu\nu}~$ , $~T_{\mu\nu}~$ and $~T^\mu{}_\nu~$ are coordinate representation of the same coordinate-free object (tensor) $\mathbf{T}$

Only in the presence of a metric, so that you can use it to freely raise and lower indexes.

Otherwise those are three different tensors, a (2, 0), a (0, 2), and a (1, 1). Without a metric there is no natural mapping between those different vector spaces.

 

[PeterDonis]

Are they?

I think @JimWhoKnew was assuming the presence of a metric, which gives you a natural mapping between the different vector spaces that allows you to freely raise and lower indexes.

 

[weirdoguy]

which gives you a natural mapping between the different vector spaces

Yes, but still some people wouldn't call those tensors "the same object". Especially most mathematical physicists/mathematicians that I know, including myself. But yes, physicists with their sloppy use of language would

 

[anuttarasammyak]

If we regard gμν as the "basic" metric, then g¯μν and its reciprocal g¯μν are not representations of the same coordinate-free object.

Thank you @JimWhoKnew for the clear and easy-to-understand explanation for the "anomaly". I have preffered to regard "new" $\bar{g}_{\mu\nu}$ as "basic" then old $g_{\mu\nu}$ , $\delta{g}_{\mu\nu}$ and their reciprocals are not representations of the same coordinate free object. That is the resason why I said

Under variation of metric tensor, gμν+δgμν is a tensor but its parts gμν and δgμν are not tensors.

 

[PeterDonis]

physicists with their sloppy use of language would

Yes, and this is the relativity forum, not one of the math forums, so physics sloppiness wins out here.

 

[haushofer]

Thanks. Even with this general caution, may we take it obvious that in tensor division of
$$ \bar{g}_{\mu\nu} := g_{\mu\nu}+\delta g_{\mu\nu} $$
all $\bar{g}_{\mu\nu}$ , $g_{\mu\nu}$ and $\delta g_{\mu\nu}$ are tensors in the world of metric undertaking variation ?

Yes.

 

[anuttarasammyak]

I will summarize what I have written.

The index-raising and -lowering relation for the metric tensor

$$ g_{\mu\nu}=g_{\mu\alpha}g_{\nu\beta}g^{\alpha\beta} ,$$

holds for a varied metric tensor as

$$\bar{g}_{\mu\nu}=\bar{g}_{\mu\alpha}\bar{g}_{\nu\beta}\bar{g}^{\alpha\beta} $$

where

$$ \bar{g}_{\mu\nu} := g_{\mu\nu}+\delta g_{\mu\nu} .$$
$$ \bar{g}^{\mu\nu} := g^{\mu\nu}+\delta g^{\mu\nu} .$$

Equation (*) in the OP follows from this.

I agree with the OP that if $\delta g_{\mu\nu}$ were a tensor, then equation (**) in the OP would hold.

A tensor satisfying (*) remains a mystery to me.

 

[JimWhoKnew]

I will summarize what I have written.

The index-raising and -lowering relation for the metric tensor

$$ g_{\mu\nu}=g_{\mu\alpha}g_{\nu\beta}g^{\alpha\beta} ,$$

holds for a varied metric tensor as

$$\bar{g}_{\mu\nu}=\bar{g}_{\mu\alpha}\bar{g}_{\nu\beta}\bar{g}^{\alpha\beta} $$

where

$$ \bar{g}_{\mu\nu} := g_{\mu\nu}+\delta g_{\mu\nu} .$$
$$ \bar{g}^{\mu\nu} := g^{\mu\nu}+\delta g^{\mu\nu} .$$

Equation (*) in the OP follows from this.

I agree with the OP that if $\delta g_{\mu\nu}$ were a tensor, then equation (**) in the OP would hold.

A tensor satisfying (*) remains a mystery to me.

As @haushofer remarked in #18, it is the behavior under coordinate transformations that determines whether something is a tensor. If Weinberg says $~\delta g_{\mu\nu}~$ qualifies, I'd take it seriously.

I think the notation is misleading. I'll repeat my argument from #25 in a slightly modified form: consider a regular spacetime with a (single!) metric $~g_{\mu\nu}~$ . Suppose we have a smooth symmetric tensor field $~A_{\mu\nu}~$ which has the same signature as the metric at every point. Suppose further that there is also a symmetric tensor field $~B^{\mu\nu}~$ satisfying the same signature condition, such that $~B^{\mu\rho}A_{\rho\nu}=\delta^\mu{}_\nu~$ . Would you necessarily expect $~B^{\mu\nu}=A^{\mu\nu}~~$?

Edit: added "symmetric"

 

[anuttarasammyak]

As @haushofer remarked in #18, it is the behavior under coordinate transformations that determines whether something is a tensor.

Going back to basics, Dirac's textbook defines a tensor as follows:

$$ T^{\alpha'\beta'}_{\gamma'}=x^{\alpha'}_{,\lambda}x^{\beta'}_{,\mu}x^{\nu}_{,\gamma'}T^{\lambda\mu}_{\nu}
\ \ \ (3.6)$$
How exactly should I perform coordinate transformations to investigate the quantity I'm considering? Or is there a more obvious way to check?

 

[anuttarasammyak]

consider a regular spacetime with a (single!) metric gμν . Suppose we have a smooth symmetric tensor field Aμν which has the same signature as the metric at every point. Suppose further that there is also a symmetric tensor field Bμν satisfying the same signature condition, such that BμρAρν=δμν . Would you necessarily expect Bμν=Aμν ?

I am sorry I don't understand "the same signature". If A is g itself, I think that B = A = g.
A=mg, B=1/m g where m is a real number $\neq 0$ satisfy the relation.
I am sorry not to be smart enough to relate A $\neq$ B with the discussion.

 

[JimWhoKnew]

Going back to basics, Dirac's textbook defines a tensor as follows:

$$ T^{\alpha'\beta'}_{\gamma'}=x^{\alpha'}_{,\lambda}x^{\beta'}_{,\mu}x^{\nu}_{,\gamma'}T^{\lambda\mu}_{\nu}
\ \ \ (3.6)$$

So what @haushofer and I said above is in agreement with Dirac's definition. A little nitpicking: if you write $~T^{\lambda\mu}_\nu~$ , then after raising the $\nu$ index, you don't know the order of $~\{\nu,\lambda,\mu\}~$ from left to right.

How exactly should I perform coordinate transformations to investigate the quantity I'm considering? Or is there a more obvious way to check?

In general, you'll need more information. For example: we know that the connections ($\Gamma$'s) are not tensors, because we know how they should be transformed. In the example of #34 we assume that $~A_{\mu\nu}~$ is a tensor field, so it follows that $~A_{\mu\nu}\pm g_{\mu\nu}~$ are also tensor fields.

I am sorry I don't understand "the same signature".

Given a specific inertial frame in Minkowski spacetime, $~C(x)_{\mu\nu}:=\delta_{\mu\nu}~$ is a symmetric invertible smooth tensor field whose signature is different from that of $~\eta_{\mu\nu}~$. So the signature assumption is needed in the example of #34.

If A is g itself, I think that B = A = g.

That's correct, but I want to consider the general case, not only trivial cases. As I said in #34, I find that the notation in OP (following Weinberg and others) obscures the property of "sameness" as pointed in #25. That's why I use A and B instead.

A=mg, B=1/m g where m is a real number $\neq0$ satisfy the relation. I am sorry not to be smart enough to relate A$\neq$B with the discussion.

No need to get personal.

If $~A_{\mu\nu}=m~g_{\mu\nu}~$ then $~A^{\mu\nu}=m~g^{\mu\nu}\neq g^{\mu\nu}/m=B^{\mu\nu} ~~$.
What does this example imply when we write $~\bar{g}_{\mu\nu}:=m~g_{\mu\nu}~$ (and still use g to raise and lower indices)?

Hint: equation (**) in OP is wrong, while (*) is correct to first order in $\delta g~$.

I hoe this study is on a right track.

I don't understand your example. If your manifold is 1D (the ring), it should be charted by a single coordinate. If it is a 2D Euclidean as in the drawing, the line element should contain terms proportional to $~drdr, drd\theta, d\theta d\theta~~$. In 2D Euclidean space, you can easily see that a short arrow tangent to the constant r ring (in your lower drawing) is not necessarily tangent to the constant r' curve (meaning it may have a radial component in the primed coordinates).

 

[anuttarasammyak]

First of all, thanks @JimWhoKnew

In general, you'll need more information.

So in general cases which could include $\delta g^{\mu\nu}$, do we need further information or condition to say it is a tensor or not, though the answer of #32 was affirmative ?

In the example of #34 we assume that Aμν is a tensor field, so it follows that Aμν±gμν are also tensor fields.

I agree that the sum of tensors is a tensor. And sum of no tensors could be sometimes a tensor.

What does this example imply when we write g¯μν:=m gμν (and still use g to raise and lower indices)?

Well, the metric under variation $$\bar{g}^{\mu\nu}=g^{\mu\nu}+(m-1)g^{\mu\nu}=g^{\mu\nu}+\delta g^{\mu\nu}$$
$$ \delta g^{\mu\nu} = (m-1)g^{\mu\nu} $$
from (*)
$$ \delta g_{\mu\nu} = -(m-1)g_{\mu\nu} $$
I may be wrong and/or missing the point. Your further coaching would be appreciated.

 

[JimWhoKnew]

So in general cases which could include $\delta g^{\mu\nu}$, do we need further information or condition to say it is a tensor or not, though the answer of #32 was affirmative ?

@haushofer has already addressed this question in #11. I don't want to get into it here. You'd better consult a textbook (d'Inverno has a nice treatment, despite a little sign inconsistency). Very sloppily: under a coordinate transformation $~T(x)\rightarrow T'(x')~~$. If $~T(x)~$ is a tensor field, so are $~T'(x')~$ , $~T'(x)~$ and $~T'(x)-T(x)~~$. Now, we know that the metric is a tensor field, therefore $~\delta g^{\mu\nu}~$ is a tensor field too (we have to be careful with this economic notation).

Well, the metric under variation $$\bar{g}^{\mu\nu}=g^{\mu\nu}+(m-1)g^{\mu\nu}=g^{\mu\nu}+\delta g^{\mu\nu}$$
$$ \delta g^{\mu\nu} = (m-1)g^{\mu\nu} $$
from (*)
$$ \delta g_{\mu\nu} = -(m-1)g_{\mu\nu} $$
I may be wrong and/or missing the point. Your further coaching would be appreciated.

With $~A_{\mu\nu}=m~g_{\mu\nu}~$ and $~B^{\mu\nu}=g^{\mu\nu}/m~$
1) Find $~\hat{\delta}_{\mu\nu}:=A_{\mu\nu}-g_{\mu\nu}~$ (as a function of g and m).
2) Find $~\tilde{\delta}^{\mu\nu}:=B^{\mu\nu}-g^{\mu\nu}~$.
3) Find $~\hat{\delta}^{\mu\nu}~$.
4) Assume $m=1+\epsilon~$ where $|\epsilon|\ll 1~$. Expand the results of (2) and (3) above to first order in $\epsilon$. Compare with equation (*) in OP.

The important point in my posts: $~A^{\mu\nu}\neq B^{\mu\nu}~$ in general (and therefore equation (**) in OP is wrong).

 

[anuttarasammyak]

With Aμν=m gμν and Bμν=gμν/m
1) Find δ^μν:=Aμν−gμν (as a function of g and m).
2) Find δ~μν:=Bμν−gμν .
3) Find δ^μν .
4) Assume m=1+ϵ where |ϵ|≪1 . Expand the results of (2) and (3) above to first order in ϵ. Compare with equation (*) in OP.

Thanks. I will restate it to confirm my understanding. Here I try not to refer controversial (at least to me) tensor and concentrate on what I found in making use of this example.

The metric
$$ \bar{g}_{\mu\nu}=mg_{\mu\nu} $$
With
$$ \bar{g}_{\mu\alpha} \bar{g}^{\alpha\nu} =\delta^{\nu}_{\mu}$$
$$ \bar{g}^{\mu\nu}=\frac{1}{m}g^{\mu\nu} $$
Say A which satisfies the equation
$$ A_{\mu\nu}=\bar{g}_{\mu\alpha} \bar{g}_{\nu\beta}A^{\alpha\beta} $$,
$\bar{g}_{\mu\nu}$ is A as easily confirmed in this example. The old metric $g_{\mu\nu}$ is not A, actually,
$$ mg_{\mu\nu}=\bar{g}_{\mu\alpha} \bar{g}_{\nu\beta}\frac{1}{m}g^{\alpha\beta} $$
Let us rewrite this equation as
$$ g_{\mu\nu}+\delta g_{\mu\nu}=\bar{g}_{\mu\alpha} \bar{g}_{\nu\beta}(g^{\alpha\beta}+\delta g^{\alpha\beta} )$$
where
$$\delta g_{\mu\nu}=(m-1)g_{\mu\nu}$$
$$\delta g^{\mu\nu}=(\frac{1}{m}-1)g^{\mu\nu}$$
Expecting the variation to be small
$$m=1+\epsilon,\ \ |\epsilon|<<1$$
$$\delta g_{\mu\nu}=\epsilon g_{\mu\nu}$$
$$\delta g^{\mu\nu}=-\epsilon g^{\mu\nu}+0(\epsilon^2)$$

Both $g_{\mu\nu}$ and $\delta g_{\mu\nu}$ are not A. Their sum $g_{\mu\nu}+\delta g_{\mu\nu}$ is A. $\delta g_{\mu\nu}$ and $\delta g^{\mu\nu}$ is introduced to be so. We should not be surprised to see such a conventional quantity satisfies the peculiar equation (*).
Equation (*) holds by changing $g$ to $\bar{g}$. We can regard one metric original and the other varied and vice versa.

The important point in my posts: $~A^{\mu\nu}\neq B^{\mu\nu}~$ in general (and therefore equation (**) in OP is wrong).

In case that A and B are metric, A=B. I used it.

 

[JimWhoKnew]

The metric
$$ \bar{g}_{\mu\nu}=mg_{\mu\nu} $$

I intentionally defined $\hat{\delta}$ and $\tilde{\delta}~$. g is the only metric in this example, and should be used to raise and lower indices. I expected answers to the 4 simple steps as stated.

Edit:
If later on you want to rename $g$ as $\bar{g}$ and regard it as the "new" metric, it doesn't matter, as long as you are consistent and unambiguous. In such a case, the "old" metric will take the form of A and B, which will not represent the same tensor.

 

[anuttarasammyak]

g is the only metric in this example

To me the point is two metrics: original and varied. In varied metric world what kind of things or relations remain and what does not ?

 

[JimWhoKnew]

To me the point is two metrics: original and varied. In varied metric world what kind of things or relations remain and what does not ?

Let's look at it step by step. Point out where you disagree.
We have the original metric $g$ and the varied $\bar{g}~$. Is it legitimate to take the difference$$g_{\mu\nu}=\bar{g}_{\mu\nu}-\delta g_{\mu\nu}\quad ? \tag{1}$$If the answer is "no", the discussion in OP is irrelevant. So let's explore the other option. Now we ask whether there is a meaning to raising the indices of $~\delta g_{\mu\nu}~$. Again, if the answer is "no", the discussion in OP is irrelevant. If "yes", then we are faced with the question how should it be done? We may attempt to preserve symmetry between the metrics by raising one index by $g$ and the other by $\bar{g}~$. But now we have to choose which one raises the left index, which breaks the symmetry. Moreover, the contravariant form will not be symmetric, which is an undesired feature. The other (and more logical) option is to raise both indices with the same metric. Without loss of generality, we may choose $\bar{g}~$ ("the varied world"). But this choice breaks the symmetry between the metrics, making $~\bar{g}~$ the preferred one. With this choice, the RHS of (1) is raised by $~\bar{g}^{\mu\nu}~$ , so to maintain consistency, the LHS must also be raised by $~\bar{g}^{\mu\nu}~$. But now we get the inevitable result that the contravariant form $~\bar{g}^{\mu\rho}g_{\rho\sigma}\bar{g}^{\sigma\nu}~$ is not the inverse of the covariant form $~g_{\mu\nu}~$ (assuming $~\delta g_{\mu\nu}\neq0~~$), and this was my point all along.

Edit:
The OP (following Weinberg) defines $~\delta g^{\mu\nu}:=\bar{g}^{\mu\nu}-g^{\mu\nu}~$ , where $~g^{\mu\nu}~$ is the inverse of $~g_{\mu\nu}~$ . In the framework of the above discussion, where $~\bar{g}~$ was chosen as prefered, this $~\delta g^{\mu\nu}~$ is not the contravariant form of $~\delta g_{\mu\nu}~$ (so equation (**) in OP is inconsistent under the above assumptions, while (*) is satisfied).

 

[anuttarasammyak]

I don't understand your example.

My bad. I withdraw it.

 

[anuttarasammyak]

We may attempt to preserve symmetry between the metrics by raising one index by g and the other by g¯ . But now we have to choose which one raises the left index, which breaks the symmetry. Moreover, the contravariant form will not be symmetric, which is an undesired feature. The other (and more logical) option is to raise both indices with the same metric.

I agree that index raise- lower operation should be done using one same, old or new, metric for all the indexes of all the entities in equations.

 

[anuttarasammyak]

The OP (following Weinberg) defines δgμν:=g¯μν−gμν , where gμν is the inverse of gμν . In the framework of the above discussion, where g¯ was chosen as prefered, this δgμν is not the contravariant form of δgμν (so equation (**) in OP is inconsistent under the above assumptions, while (*) is satisfied).

$$\bar{g}_{\mu\nu} - g_{\mu\nu} = \delta g _{\mu\nu} = - \delta \bar{g}_{\mu\nu} $$
$\bar{g}_{\mu\nu}$ is a tensor in the world where metric is $\bar{g}_{\mu\nu}$ itself.
$g_{\mu\nu}$ is a tensor in the world where metric is $g_{\mu\nu}$ itself.
Their difference $\delta g _{\mu\nu} = - \delta \bar{g}_{\mu\nu}$ belong to neither or both the worlds. I observe that minus sign in (*) , which is shared in both the worlds, comes from this ambivalence.

 

[JimWhoKnew]

$$\bar{g}_{\mu\nu} - g_{\mu\nu} = \delta g _{\mu\nu} = - \delta \bar{g}_{\mu\nu} $$
$\bar{g}_{\mu\nu}$ is a tensor in the world where metric is $\bar{g}_{\mu\nu}$ itself.
$g_{\mu\nu}$ is a tensor in the world where metric is $g_{\mu\nu}$ itself.
Their difference $\delta g _{\mu\nu} = - \delta \bar{g}_{\mu\nu}$ belong to neither or both the worlds. I observe that minus sign in (*) , which is shared in both the worlds, comes from this ambivalence.

The minus sign in (*) follows from the definitions which are used, and from the requirement that $~g^{\mu\nu}~$ and $~\bar{g}^{\mu\nu}~$ are the inverse of $~g_{\mu\nu}~$ and $~\bar{g}_{\mu\nu}~$ respectively (in both worlds, assuming its legitimacy as addressed in #43). Since only terms up to first order in $~\delta g~$ (same as your $~-\delta \bar{g}~~$) are maintained, it is trivial to note that (*) is invariant under "barring".

 

[anuttarasammyak]

Since only terms up to first order in $~\delta g~$ (same as your $~-\delta \bar{g}~~$) are maintained, it is trivial to note that (*) is invariant under "barring".

Only up to first order is the rule on variation though I do not want to argue it is trivial or not.