# Issues with the variation of Christoffel symbols

• I
• JuanC97

#### JuanC97

Hello everyone,
I'm sure a lot of you know that the Christoffel symbols are not tensors by themselves but, their variation is a tensor.
I want to revive a post that was made in 2016 about this: The Variation of Christoffel Symbol and ask again "How is that you can calculate ∇ρδgμν if δ{gμν} is not a tensor?"

You know that δgμν can be written as (- gμα gνβ δgαβ) since δ{δμν}=0 so... at first glance, you could say that δgμν is not a tensor since you can't lower its indices just by using two metrics BUT the truth is that those indices are not the indices of the variation so you can't just think like that, it is, in fact, a matter of notation as I will explain now:

As far as I've tought, δgμν stands for "the variation of the components" of the metric ( i.e. δ{gμν} )
and that's really different from "the components of the variation" ( i.e. (δg)μν - wich, of course, would be tensorial since that's just the difference of two metrics ), take this into account:

δg = δ{gμνdxμ⊗dxν} = δ{gμν}dxμ⊗dxν + gμνδ{dxμ⊗dxν} so (δg)μν is, in general different from δ{gμν}.

Now...maybe the problem is just about notation...
The issue that really makes me think twice about it is... What does it really mean that covariant derivative of δgμν that you found in the formula for the variation of the Christoffel symbols? and, even more, how can you interpret that equation in the proper form?

Could you please rephrase ∇ρδgμν in a different notation or in words, or maybe interpreting as multilinear maps?.

Thanks btw.

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https://anhngq.wordpress.com/2010/0...ariant-derivative-of-tensors-in-a-short-form/.

First of all, by ∇ρδgμν we refer to the scalar that results from the evaluation of (∇ρδg)(∂μ,∂ν).
Then, you should interpret (∇ρδg)(∂μ,∂ν) as the covariant derivative of the metric tensor along ∂ρ evaluated in the basis given by ∂μ and ∂ν .

Using the properties of the covariant derivative you can find that (∇ρδg)(∂μ,∂ν) is given by
ρδg(∂μ,∂ν) - δg(∇ρμ,∂ν) - δg(∂μ,∇ρν)
= ∂ρδg(∂μ,∂ν) - δg( Γλρμλ,∂ν) - δg(∂μ, Γλρνλ)
= ∂ρδg(∂μ,∂ν) - Γλρμδg( ∂λ,∂ν) - Γλρνδg(∂μ,∂λ)
= ∂ρδgμν - Γλρμδgλν - Γλρνδgμλ

The conclusion is that ∇ρδgμν = ∂ρδgμν - Γλρμδgλν - Γλρνδgμλ has to be interpreted as the components of ∇ρδg so... this is a tensor (?)

Hmmm, and... what would happen if I use the same reasoning with δg(∂μ,∂ν) = δgμν ?
As I see it, δg(∂μ,∂ν) stands for the components of δg so δgμν is a tensor (?).

Just my 2 cents: the variation of the metric is the difference between two metric tensors in the same coordinate system and hence a tensor. Take e.g. the Lie derivative of the metric as explicit example.

Just my 2 cents: the variation of the metric is the difference between two metric tensors in the same coordinate system and hence a tensor. Take e.g. the Lie derivative of the metric as explicit example.

I agree with you when you said that (δg) is a tensor but we have to remember that we also have this two equations from the literature that are always used when computing variations:

(i) (∇ρδgμν) = ∂ρδgμν - Γλρμδgλν - Γλρνδgμλ
states that δgμν is the component of a tensor and this relation can be deduced from its interpretation as (δg)μν.
(ii) δgμν = - gμα gνβ δgαβ
states that δgμν is not the component of a tensor and this relation can be deduced from its interpretation as δ{gμν}.

Also, taking into account that
δg = δ{gμνdxμ⊗dxν} = δ{gμν}dxμ⊗dxν + gμνδ{dxμ⊗dxν}
and the second term in the right-hand side is, in general, different from 0, it's clear that (δg)μν ≠ δ{gμν} so both interpretations are, in principle, uncompatible. The question is then... Why we usually use both of this formulas?, maybe I am misreading something?.

It seems to me that there is no real distinction between ##(\delta \boldsymbol{g})_{\mu \nu}## and ##\delta g_{\mu \nu}##. As you say,

##\boldsymbol{g} = g_{\mu \nu} dx^\mu \otimes dx^\nu##

So if we vary ##\boldsymbol{g}##, you get:

##\delta \boldsymbol{g} = \delta g_{\mu \nu} dx^\mu \otimes dx^\nu + g_{\mu \nu} \delta(dx^\mu \otimes dx^\nu)##

But you're not varying your coordinate system, so ##dx^\mu## isn't varying. So the last term is zero.

• JuanC97
I think the confusion is that ##g^{\alpha \beta}## should really be written as ##(g^{-1})^{\alpha \beta}##. So both ##\delta \boldsymbol{g}## and ##\delta (\boldsymbol {g}^{-1})## are tensors, but they are different tensors. They are related by:

##\delta (\boldsymbol{g}^{-1})^{\mu \nu} = - g^{\mu \alpha} g^{\nu \beta} \delta \boldsymbol {g}_{\alpha \beta}##

In other words, ##\delta (\boldsymbol{g}^{-1})_{\mu \nu} \neq \delta \boldsymbol {g}_{\mu \nu}##. They are actually the negatives of each other.

• JuanC97
In other words, ##\delta (\boldsymbol{g}^{-1})_{\mu \nu} \neq \delta \boldsymbol {g}_{\mu \nu}##. They are actually the negatives of each other.

Did you mean to write the indexes on ##\delta (\boldsymbol{g}^{-1})_{\mu \nu}## as lower indexes here? They were upper indexes in your earlier formula.

Did you mean to write the indexes on ##\delta (\boldsymbol{g}^{-1})_{\mu \nu}## as lower indexes here? They were upper indexes in your earlier formula.

Peter, he is saying that ##\, (\delta\boldsymbol{g^{-1}}) \,## is a tensor, hence:
## (\delta\boldsymbol{g^{-1}})^{\mu \nu} \equiv g^{\mu \alpha} g^{\nu \beta} (\delta\boldsymbol{g^{-1}})_{\alpha \beta}##

but we've found that
## \delta\{ \delta_\mu^\nu \} = \delta\{ g_{\mu\alpha} (\boldsymbol{g^{-1}})^{\alpha\nu} \} = 0
\;\Leftrightarrow\;
(\delta\boldsymbol{g^{-1}})^{\mu \nu} = - g^{\mu \alpha} g^{\nu \beta} \delta g_{\alpha \beta} \hspace{0.3cm}##, so ##\,\, (\delta\boldsymbol{g^{-1}})_{\alpha \beta} = - \delta g_{\alpha \beta}##

I agree with this point of view but I'm still dealing with the argument of ## \delta\{ dx^\mu \} = 0 ## and ## \delta\{ \partial_\mu \}=0 ## since, I think about ## \delta ## as a general variation (it could be a variation due to a change in coordinates or anything else) which makes it difficult to me to understand why the basis (co-)vectors would remain unchanged under any kind of variation.

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Did you mean to write the indexes on ##\delta (\boldsymbol{g}^{-1})_{\mu \nu}## as lower indexes here? They were upper indexes in your earlier formula.

No, I meant for them to be lowered. if ##\delta(\boldsymbol g^{-1})## is a tensor, then we can use ##g## to raise and lower its indices. So

##(\delta(\boldsymbol g^{-1}))_{\mu \nu} \equiv g_{\mu \alpha} g_{\nu \beta} (\delta(\boldsymbol g^{-1}))^{\alpha \beta} = - (\delta \boldsymbol{g})_{\mu \nu}##

I agree with this point of view but I'm still dealing with the argument of ## \delta\{ dx^\mu \} = 0 ## and ## \delta\{ \partial_\mu \}=0 ## since, I think about ## \delta ## as a general variation (it could be a variation due to a change in coordinates or anything else) which makes it difficult to me to understand why the basis (co-)vectors would remaining unchanged under any kind of variation.

You can certainly allow more general variations. But in the context of variational principles in deriving Einstein's field equations, the way that I think of it is this:
• You have a candidate metric, ##\boldsymbol{g}##
• You consider a second tensor, ##\boldsymbol{\bar{g}}## that is infinitesimally different than ##\boldsymbol{g}##
• You subtract them to get ##\delta \boldsymbol{g} \equiv \boldsymbol{\bar{g}} - \boldsymbol{g}##
• You calculate the effect on the action to first-order in ##\delta \boldsymbol{g}##.
• Your candidate ##\boldsymbol{g}## is the real metric if the first-order variation vanishes.

• JuanC97
In the euler lagrange eqns you only consider functional changes, not coordinate ones. That's why variations and partial derivatives commute. In the Lie derivative you consider the difference between two tensors in the same coordinate system. I can't think of an explicit example of more general ones which I have encountered.