The covariant derivative of any metric tensor is zero. Is this an axiom or something that is derived from other axioms?

robphy
Homework Helper
Gold Member
This is called the "metric compatibility" condition.

HallsofIvy
Homework Helper
It really comes from the definition of "covariant derivative". The difference between the "covariant" derivative and the "naive" derivative is that the covariant discounts changes in a function that are due solely to "non-flatness" in the space. Since the metric tensor just gives that "non-flatness", it is essentially a "constant" and have (covariant) derivative 0.

The covariant derivative of any metric tensor is zero. Is this an axiom or something that is derived from other axioms?
Derived.

Pete

Hurkyl
Staff Emeritus
Gold Member
Is this an axiom or something that is derived from other axioms?
Your question is irrelevant. The only difference between 'something that is derived' and 'an axiom' is the manner in which one decides to present the theory.

Derived.

Pete
Um no Halls of Ivy had it dead on. It's just part of the definition of a covariant derivative, it's not a derivable property. The covariant derivative just one of a few derivative operators that are useful to use.

Thanks.

Is entry #3 the definition of covariant derivative, or only a description of one of its features? If the latter, how exactly is it defined?

Le me rephrase this. It can be derived.
It's just part of the definition of a covariant derivative, it's not a derivable property. The covariant derivative just one of a few derivative operators that are useful to use.
Since when?? I've never seen a treatment of general relativity or tensor analysis when that was taken as a defining property. The reason I said it is derived is that (1) I've derived that relationship myself and (2) most treatements I've seen contain this derivation.

The vanishing of the covariant derivative establishes a relationship between the affine geometry of a manifold and its metric geometry. The derivation proceeds as follows: One requires that the magnitude of the scalar product of two 4-vectors remains unchanged when the 4-vectors are parallel transported. i.e.

$A^\mu B^\nu g{\mu\nu})_{,\beta} = 0[/tex] Since [itex]A^\mu B^\nu g{\mu\nu})_{\beta}$ is a scalar, we can replace the ordinary derivative with the covariant derivative

$A^\mu B^\nu g{\mu\nu})_{;\beta} = 0[/tex] When this is written out there will appear two terms which are the covariant derivatives of the vectors. Since the only changes contributed are those from the parallel transport then it follows that those terms vanish. One the ends up with [itex]A^\mu\nu g{\mu\nu}_{;\beta} = 0[/tex] Since the two vector is arbitrary if follows that [itex]g{\mu\nu}_{;\beta} = 0[/tex] Hence this relationship can be derived. Pete Last edited: nrqed Science Advisor Homework Helper Gold Member The vanishing of the covariant derivative of the metric tensor is neither a defining property of the metric or the covariant derivative. ^ So it is an axiom, right? I mean one must assume that there is no torsion. In that case the conenction is given by the Chrristoffel symbols. But as far as I know, the fact that there is no torsion is an axiom. The reason I said it is derived is that (1) I've derived that relationship myself and (2) most treatements I've seen contain this derivation. Pete I am a bit confused. If one assumes that the connection si given the the Christoffel symbols then it can be derived. But again, this follows from making the assumption that there is no torsion. Sorry nut that 1/2 hour is not enough to get an edit wrinkle free. I had to repost the derivation. So I'll repost the entire post instead - The vanishing of the covariant derivative establishes a relationship between the affine geometry of a manifold and its metric geometry. The derivation proceeds as follows: One requires that the magnitude of the scalar product of two 4-vectors remains unchanged when the 4-vectors are parallel transported. i.e. [itex]A^\mu B^\nu g_{\mu\nu},\beta} = 0[/tex] Since [itex]A^\mu B^\nu g_{\mu\nu;\beta}$ is a scalar, we can replace the ordinary derivative with the covariant derivative

$A^\mu B^\nu g_{\mu\nu;\beta} = 0[/tex] When this is written out there will appear two terms which are the covariant derivatives of the vectors. Since the only changes contributed are those from the parallel transport then it follows that those terms vanish. One the ends up with [itex]A^\mu B\nu g_{\mu\nu;\beta} = 0[/tex] Since the two vector is arbitrary if follows that [itex]g{\mu\nu;\beta} = 0[/tex] Hence this relationship can be derived. So it is an axiom, right? No. I had just explaine why it isn't an axiom. I derived it for you. I mean one must assume that there is no torsion. In that case the conenction is given by the Chrristoffel symbols. But as far as I know, the fact that there is no torsion is an axiom. That is usually assumed when one is working with the affine connection, but yes. Its an assumption that the given that the manifold is torsion free. I am a bit confused. If one assumes that the connection si given the the Christoffel symbols then it can be derived. But again, this follows from making the assumption that there is no torsion. How so? I've never seen that before myself. Pete Hurkyl Staff Emeritus Science Advisor Gold Member Sheesh, does nobody listen to a word I say? Sheesh, does nobody listen to a word I say? Yes. Sometimes we agree with what you post and sometimes it is dismissed as irrelevant. In this case a derivation of the relation starting with basic properties was quite useful, especially since there is no reason to assume that anyone has ever chosen different axioms where the vanishing of the covariant derivative of the metric is zero. Especially since this derivation is found in such GR texts as that of Ohanian and Ruffini. Does that bother you for some reason? I mean one must assume that there is no torsion. In that case the conenction is given by the Chrristoffel symbols. But as far as I know, the fact that there is no torsion is an axiom. Torsion is not an axiom, it is a property of a manifold. In retrospect I don't see any place in the above derivation which assumes zero torsion. Pete Hurkyl Staff Emeritus Science Advisor Gold Member Showing that one set of properties is equivalent to another set of properties: useful. Quibbling about which set of properties is "right" to use as axioms: not useful. Acting as if 'axiomness' is a quality inherent to a set of properties: wrong. Le me rephrase this. It can be derived. Since when?? I've never seen a treatment of general relativity or tensor analysis when that was taken as a defining property. It's in Wald. Showing that one set of properties is equivalent to another set of properties: useful. Quibbling about which set of properties is "right" to use as axioms: not useful. Acting as if 'axiomness' is a quality inherent to a set of properties: wrong. None of which I or the OP seem to care about. A derivation was given where only the typical definitions were used. If that bothers you then ignore it. I never said nor implied that you were wrong so I don't see why you're so concerned about this. In any case I'm not even intersted at this point. DavidWhitbeck - what page in wald? re - [itex]g(\nabla_v(X), Y) = 0$ - Where did this come from? I was referring to the fact that the first two terms in

$$A^\mu_{;\beta}B^\nu g_{\mu\nu} + A^\mu_{;\beta} B^\nu + A^\mu B^\nu g_{\mu\nu;\beta} = 0$$

vanish. I.e. since the only changes in the vectors $A^\mu$ and $B^\mu$ are those contributed by parallel transport then the covariant derivatives of those vectors vanish and the above equation reduces to

$$A^\mu B^\nu g_{\mu\nu;\beta} = 0$$

Pete

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Note: In the derivation I posed above there were supposed to be parentheses around the expression so that it'd be apparent that it was the derivative of the scalar product that the derivative was being taken of. I didn't notice that they were missing until just now.

Due to the recent change in software which allows us only 30 minutes to edit a post you can expect this to continue into the future. As such I'd like to take this opportunity to say that I highly object to such a short limit on the time allowed to edit posts. I use the submit button to view the posts and as such it often takes more than 30 minutes to get a post just right. I stopped using the Preview Post function a very long time ago when I realized that there was a problem with it (I forget exactly what the problem was though).

Pete

Hurkyl - So let me get this straight. You fully believe that, in all possible cases, any theorem that exists, or can exist, in a mathemtical theory can be rephrased as an axiom with appropriate changes in definions and basic axiom set. Is that correct? If so then I find that hard to believe, especially if you expect us to accept it without proof. Do you have or know of a proof of this claim/assertion?

Pete

Let me give you an example which makes this idea of yours hard to believe. If one defines either the metric or the covariant derivative such that the covariant derivative of the metric vanishes then I'd say that your assertion is not possible. This is because such a definition would require all metric spaces to have an affine connection defined on them and all affine spaces to have a metric defined on them. That would require a change in the definition of both metric geometry and affine geometry since as it is now an affine space can exist without an affine connection on it and an affine space can exist without a metric defined on it.

Pete

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Thanks, Pete. I appreciate your efforts.

Hurkyl’s point is well taken, but I think it is always better in math to start with something self-evident and build up from there, rather than simply to declare that it’s all equivalent so it doesn’t matter where one begins. Euclid could have made the Pythagorean Theorem an axiom and used it to prove the parallel postulate, but that would’ve been kind of strange, in my opinion.

Oh, and isn’t the last word in entry #14 supposed to be “priceless”? Hurkyl
Staff Emeritus
Gold Member
Thanks, Pete. I appreciate your efforts.

Hurkyl’s point is well taken, but I think it is always better in math to start with something self-evident and build up from there, rather than simply to declare that it’s all equivalent so it doesn’t matter where one begins. Euclid could have made the Pythagorean Theorem an axiom and used it to prove the parallel postulate, but that would’ve been kind of strange, in my opinion.

Oh, and isn’t the last word in entry #14 supposed to be “priceless”? "Self-evident" is a terrible notion.

Anyways, I wasn't trying to say the axiomatic method was a bad idea -- I'm trying to warn against trying to ascribe any meaning to the axioms beyond "I'm taking this as a starting point". Pedagogically, it's quite useful to start from a core set of assumptions and see what can be derived from them. And formally, axioms are a convenient way to specify theories and work with them, much in the same way a basis provides a convenient way to work with a vector space. (In fact, axioms for a theory and a spanning set for a vector space are both examples of the more general notion of a 'generating set')

And generally speaking, it's a good idea to work from many different starting points and see how they relate. For a relevant example, consider Spivak's Differential Geometry, one of the primary references on the subject. He gives no less than three different definitions of "tangent vector" (by the curve it's tangent to, by the differential operator it defines, and by its components relative to all coordinate charts), and then later gives a fourth abstract axiomatic definition of "tangent bundle", showing that it's unique up to isomorphism and all three of the original definitions really are examples of tangent bundles. The format of http://www.mathpop.com/bookhtms/dg2.htm]Spivak's second book [Broken] is similar; through several of its chapters, he rederives differential geometry from several different starting points.

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Hurkyl
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Gold Member
Hurkyl - So let me get this straight. You fully believe that, in all possible cases, any theorem that exists, or can exist, in a mathemtical theory can be rephrased as an axiom with appropriate changes in definions and basic axiom set. Is that correct? If so then I find that hard to believe, especially if you expect us to accept it without proof. Do you have or know of a proof of this claim/assertion?

Pete

First, there's nothing stopping me from studying a theory by taking as my axioms every single one of its theorems. (Clearly, each of my axioms is true in the theory, and from my axioms, I can (trivially) deduce any of its theorems)

A less over-the-top response would be for me to take as my set of axioms the theorem you selected together with the axioms you were using.

Finally, just reading many math textbooks will provide numerous examples. Look for any theorem that states several different conditions are equivalent; that will yield several different ways to axiomatically define whatever concept they are describing. Also look for theorems describing "necessary and sufficient conditions".

First, there's nothing stopping me from studying a theory by taking as my axioms every single one of its theorems.
That is far from clear. In mathematics one needs to be able to prove assertions like the one you make and not make assumptions such as this one. Although it may appear obvious to you it doesn't mean that its true. Math is too tricky for that.
(Clearly, each of my axioms is true in the theory, and from my axioms, I can (trivially) deduce any of its theorems)
Theorems often have restrictions to which it applies so what one takes as axioms isn't as easy as you make it seem to appear.
Finally, just reading many math textbooks will provide numerous examples. Look for any theorem that states several different conditions are equivalent; that will yield several different ways to axiomatically define whatever concept they are describing. Also look for theorems describing "necessary and sufficient conditions".
Okay. Then address the current problem that I just explained above regarding the unesccesary connection of affine geometries and metric geometries.

Pete

Hurkyl
Staff Emeritus