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

[anuttarasammyak]

1. We can derive the equation
$$ \delta g^{\mu\nu}=-g^{\mu\alpha}g^{\nu\beta}\delta g_{\alpha\beta} $$
by indeces up down operation. In the world whose metric tensor is g,
$$ g^{\mu\nu}=g^{\mu\alpha}g^{\nu\beta}g_{\alpha\beta} $$
In another new world whose metric tensor is $\bar{g}$,
$$ \bar{g}^{\mu\nu}=\bar{g}^{\mu\alpha}\bar{g}^{\nu\beta}\bar{g}_{\alpha\beta} $$
Say the difference of metric tensors is small
$$ \bar{g}_{\mu\nu}=g_{\mu\nu}+\delta g_{\mu\nu} $$ $$ \bar{g}^{\mu\nu}=g^{\mu\nu}+\delta g^{\mu\nu} $$
we can get this equation in first order of $\delta g$.

2. In the new world where $\delta g \neq 0$, in first order of $\delta g$.
$$ g^{\mu\nu} \neq \bar{g}^{\mu\alpha}\bar{g}^{\nu\beta} g_{\alpha\beta} $$
$$ \delta g^{\mu\nu} \neq \bar{g}^{\mu\alpha}\bar{g}^{\nu\beta} \delta g_{\alpha\beta} $$
Neither old metric nor the differece of new-old satisfies indecies up-down relation but their sum satisfies it. Reciprocally, neither new metric nor the difference of old-new satisfies indecies up-down relation but their sum does it.

3. $$(\delta g)^{\mu\nu}=g^{\mu\alpha}g^{\nu\beta}(\delta g)_{\alpha\beta}$$
satisfies the indeces up down relation. I want concilliation of 2 and 3 for my deeper understanding.

 

[Demystifier]

3. is just a definition of the dual tensor, i.e. the rule how to raise indices in the same world. In 2. you try to compare quantities from two different worlds.

 

[anuttarasammyak]

3. is just a definition of the dual tensor, i.e. the rule how to raise indices in the same world. In 2. you try to compare quantities from two different worlds.

It is a tensor whose covariant and contravariant components are $\delta g_{\mu\nu}$ and $-\delta g^{\mu\nu}$, respectively.

Returning to the original question, does there exist such a quantity whose covariant and contravariant components are $\delta g_{\mu\nu}$ and $\delta g^{\mu\nu}$ ? If so, that is a tensor ? I would like to confirm my assumption that the answer is negative.

 

[Demystifier]

It is a tensor whose covariant and contravariant components are $\delta g_{\mu\nu}$ and $-\delta g^{\mu\nu}$, respectively.

Returning to the original question, does there exist such a quantity whose covariant and contravariant components are $\delta g_{\mu\nu}$ and $\delta g^{\mu\nu}$ ? If so, that is a tensor ? I would like to confirm my assumption that the answer is negative.

I believe there is no such quantity, because you study a variation of the object that defines the covariant and contravariant components.

 

[anuttarasammyak]

Thank you, @Demystifier.

In general, the difference of two tensors is again a tensor. However, in the context of variations of the metric tensor, the quantities
$$ \bar{g}_{\mu\nu}-g_{\mu\nu}$$
$$ \bar{g}^{\mu\nu}-g^{\mu\nu}$$
do not transform as tensors. I have verified this.

 

[Demystifier]

How did you verify this? They do transform as tensors under coordinate transformations, but their indices are not raised and lowered in the usual way. Which is understandable, because when you have two metrics, there is an ambiguity should one raise the indices with $g^{\mu\nu}$ or $\bar{g}^{\mu\nu}$. But raising and lowering indices is not a tensor transformation, and variation of the metric is not a coordinate transformation. So confusion stems from non-distinguishing 3 different notions: (i) tensor transformation under coordinate transformation, (ii) variation of the tensor and (iii) raising and lowering indices.

 

[anuttarasammyak]

They do transform as tensors under coordinate transformations,

I should appreciate it if you could teach me how to verify it.

 

[Demystifier]

I should appreciate it if you could teach me how to verify it.

Under a coordinate transformation $x^{\mu}\to x'^{\mu}=f^{\mu}(x)$, the two metrics transform as
$$g'^{\alpha\beta}=\frac{\partial x'^{\alpha}}{\partial x^{\mu}} \frac{\partial x'^{\beta}}{\partial x^{\nu}} g^{\mu\nu}$$
$$\bar{g}'^{\alpha\beta}=\frac{\partial x'^{\alpha}}{\partial x^{\mu}} \frac{\partial x'^{\beta}}{\partial x^{\nu}} \bar{g}^{\mu\nu}$$
so the quantity $T^{\alpha\beta}\equiv \bar{g}^{\alpha\beta}-g^{\alpha\beta}$ transforms as
$$T'^{\alpha\beta}=\frac{\partial x'^{\alpha}}{\partial x^{\mu}} \frac{\partial x'^{\beta}}{\partial x^{\nu}} T^{\mu\nu}.$$
Similarly for the objects with lower indices.

 

[anuttarasammyak]

Thank you @Demystifier

1. I observe that

a. $\bar{g}_{\mu\nu}-g_{\mu\nu}$ transforms as a covariant tensor under coordinate transformation

b. $\bar{g}^{\mu\nu}-g^{\mu\nu}$ transforms as a contravariant tensor under coordinate transformation.

c. There exists a tensor whose covariant components are $\bar{g}_{\mu\nu}-g_{\mu\nu}$ and contravariant components are $-(\bar{g}^{\mu\nu}-g^{\mu\nu})$ with indeces up-down by metric tensor g or $\bar{g}$.

I hope those are all right.

2. I would like to understand about tensors as united complex of covariant and contravariant components in the context of variation of metric tensor. Which one is right to say ?

a. There exists a tensor whose covariant components are $\bar{g}_{\mu\nu}-g_{\mu\nu}$ and contravariant components are $\bar{g}^{\mu\nu}-g^{\mu\nu}$ with indeces up-down by metric tensor g or $\bar{g}$ but with extra -1 coefficient. Extra minus does not matter this complex of covariant and contravariant components, to be a tensor.

b. There does not exist a complex whose covariant components are $\bar{g}_{\mu\nu}-g_{\mu\nu}$ and contravariant components are $\bar{g}^{\mu\nu}-g^{\mu\nu}$. The only complex possible is the way 1.c.