Independent components of Christoffel Symbols

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Independent components of Christoffel Symbols

Postby eSOANEM » Fri Feb 27, 2015 3:04 am UTC

I'm interested in this for physical reasons, but I thought that, given the trouble I've had with it (even with the help of my mathmo friends), I'd be more likely to find someone who could help here than in the science board.

Basically, I want to know the number of independent components of the Christoffel symbols in a 3+1 spacetime.

I've recently being doing a research review on classical gauge theories and, in doing so, been learning a bunch of maths I'd never learnt before, in particular about differential forms and fibre bundles. In gauge theories, the connection is a Lie algebra valued 1-form defined on a principle bundle. From this, analogy to the connection I'm familiar with is sufficient to predict the expressions for the gauge covariant derivative and the curvature form which is all great.

The issue is, whilst I know that, geometrically, the Levi-Civita connection and the gauge connection are "the same", I'm nervous about how the fact that the former is defined over a tangent bundle and the latter over a principle bundle affects things.

Assuming the analogy is solid, the Christoffel symbols should also be Lie algebra valued 1-forms and, after some thinking, it seems clear that this should be the Lorentz group which would imply that the Christoffel symbols have 24 free components (4 horizontal components each of which lies in the Lorentz algebra which has 6 dimensions).

To convince myself of this analogy further, I decided to try and derive that this was the number of independent components of the Christoffel symbols from the expressions I knew from GR and this proved difficult.

A general 3-index object has 64 independent components but torsion-free-ness gives us 24 conditions (in terms of the structure group of the tangent bundle, it removes shears from the initial general linear group).

I then thought it should be easy to show how many independent conditions we get from metric compatability but this proved much harder than I anticipated.

Ultimately, the best approach we were able to get was to consider the formula for Γabc, use the fact that (from torsion-free-ness) Γabcacb and consider the parts symmetrised and antisymmetrised in the first two indices for both sides of this equals sign.

This gave us 10 free components of the antisymmetrised version (using torsion-free-ness) and 20 from the symmetrised version.

Normally these would have to be independent (suggesting 30 independent components of the Christoffel symbols which would be incompatible with it being a Lie algebra valued 1-form) but I think that the torsion-free-ness actually enables us to say there might be redundancies between the expression for the symmetrised and antisymmetrised parts derived from opposite sides of the torsion-free-ness equation and thus reduce this number of free components further.

This showed us that it had to have fewer than 30 independent components but, if it's at all possible to consider the connection as a 1-form taking values in some other space, its number of free components should be a multiple of 4 and, by analogy to gauge theories, we'd expect this multiple to be the dimension of the structure group. On physical grounds, this structure group should probably be the Lorentz group (giving the 24 free components I'm expecting) but I'm not entirely sure I can rule out a "dilation", either spatially isotropic, a multiple of a dirac delta or a multiple of η giving 28 components.

So, I was wondering if anyone knew how to show the number of independent components there are to the Christoffel symbols. Or, if not quite that much, provide an argument as to why the dilation can be disregarded.
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Re: Independent components of Christoffel Symbols

Postby doogly » Fri Feb 27, 2015 3:13 pm UTC

To skip your sub-question, which involves work, and answer your real question, yes, connections definitely work in this way. You can treat the tangent bundle as a special case of a principal bundle.

But the Levi Cevita (Christoffel) connection isn't just a connection on the tangent bundle, as you know; it has extra requirements. So counting the degrees of freedom of the Christoffel symbols on the one hand of some Lorentz algebra valued one form on the other would not give you strict agreement.

I'll make a second pot of coffee and try to see how the countings match up if it's important?



But as a general suggestion, I highly, highly recommend Spivak volume 2. It starts with Reimman and Ricci and the debauch of indices, does the Cartan moving frame stuff, the Koszul, the Ehresman, and the Lie algebra principal bundle connections. His goal is to get all these notions of what "connection" means to connect to each other. Which is not the easiest sort of thing. Many books will talk about 'sections on a fiber' right away and make the Christoffel stuff hopelessly obscure.

Morita's Geometry of Differential forms is also very elegant, and has some nice presentation of "what happens when you do connections to the tangent bundle" "what happens when you bundle up a lie group." So much good geometry, and he's doing them together nicely.
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Re: Independent components of Christoffel Symbols

Postby mosgi » Fri Feb 27, 2015 3:32 pm UTC

For completeness: I am one of the people eSOANEM mentioned as helping with this. Specifically with the conclusion that there should be at most 30 independent components to Γ. This part should be fairly standard GR tensor manipulation, and it's probably worth other people checking.

Take the usual expression for Γabc and lower the first index (which just amounts to killing off the gad term and then swapping 'd's with 'a's in the usual expression). That is,

Γabc = 1/2 (gba,c + gac,b - gcb,a) = ga(b,c) - 1/2 gbc,a

Then split it as the sum of two parts:

Γabc = Γ(ab)c + Γ[ab]c

Then we can just count components in teach term: Γ(ab)c is totally symmetric and so has 20 independent components, while Γ[ab]c has some slightly more interesting symmetries (symmetric under b<->c, antisymmetric under a<->b and under a<->c) and ends up with 10 independent components. We counted these in the most obvious way possible: Write out four 4x4 matrices of elements (each matrix being for a fixed value of a) and just cross off terms which were either duplicated or zero due to symmetries.

In theory these two sets of components should be completely independent, but I feel we still could be missing some conditions which would reduce the number of components.

Finally, from googling a fair bit, there are quite a few people claiming there are 40 independent components, but no actual proofs of this fact. At best they seem to be saying "Γ is symmetric in its lower two indices, so has 4x10 = 40 components", and really that's just giving an upper bound. At no point does anyone seem to have used metric compatibility or the actual equation to reduce it any further, or to justify that there are in fact 40.
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Re: Independent components of Christoffel Symbols

Postby z4lis » Fri Feb 27, 2015 10:58 pm UTC

I'm not exactly good with physics or GR or manifolds, so sorry if this is nonsense.I was just bored so figured I'd play around with it.

Here's how I get 40: If I start with a coordinate system that makes the symbols vanish and then do a random coordinate changes and look at how the Christoffel symbols change, it looks like I need 4*4 = 16 numbers to get the partial y/partial x components, and since the only symmetry that random second partials will display is swapping the order, I need 4*(10) = 40 numbers there. So it seems pretty clear to me that the partials derivatives have 56 independent coordinates. Now, if a^i_m and b^m_ij are the partials, the Christoffel symbols themselves form the image of a map f:R^56 ---> R^40 found by taking the appropriate products:

Gammal^k_ij = a^i_m b^m_jk

And this is surjective, since we can take the a^i_m to all be 1 or something.

===

As for your 30, are you sure that \Gamma_{(ab)c} is totally symmetric? I seem to need to know that I can do abc -> cab for Gamma to prove that fact.
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Re: Independent components of Christoffel Symbols

Postby doogly » Sat Feb 28, 2015 12:17 am UTC

Yeah for sure 40. You don't bring in extra conditions from metric compatibility because the christoffel symbol formula you were using had already used that.
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Re: Independent components of Christoffel Symbols

Postby eSOANEM » Sat Feb 28, 2015 9:38 am UTC

You can reduce it to 40 just by assuming lack of torsion though and the fact that there are torsionless metric compatible connections tells you there must be fewer than 40 independent components.

I emailed my relativity lecturer about this who ended up responding with a book recommendation (which is another one from doogly's so I will try to get hold of both and have a look) but also pointed out that metric compatability alone gives 40 conditions (because it has a condition for each partial derivative of g) bringing you to 24 independent components. Clearly there's some redundancy between these conditions and the torsion-free ones because we can show fairly easily that that gives us 24 conditions in general.

From what he's saying (and considering the plausible algebras the connection could lie in), it seems that if we don't eliminate torsion, it is Lorentz algebra valued (giving the 24 components he says) but that, if we assume a lack of torsion then we need to throw away a whole conjugacy class which I think would need to be rotations just leaving the connection being boost-algebra valued giving us 12 free components.
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Re: Independent components of Christoffel Symbols

Postby z4lis » Sat Feb 28, 2015 6:21 pm UTC

Oh, now I see. So I assumed that the relations you mentioned amount the Christoffel symbols must only use real numbers as coefficients, giving my 40. However, you seem to want to also bring in the metric tensor and its partials as coefficients. So which of those do you want to consider when calculating your number of independent components?
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Re: Independent components of Christoffel Symbols

Postby eSOANEM » Sat Feb 28, 2015 6:44 pm UTC

huh? No.

The christoffel symbols have 3 indices. In the absence of any symmetries, it would have 64 components. Torsion-free-ness turns it into 40 free components (because it's 4*10 not 4*16 now). Metric compatability brings that down further. In fact, it imposes 40 conditions but some of those must be redundant with some of the torsion-free conditions or else we'd have no independent components.

This approach doesn't get us anything more specific which is why we (mosgi and I) turned to using the explicit formulae for the christoffel symbols and counting the components moreorless directly but this turned out to still be very hard. This was what gave us the no-more-than-30.
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Re: Independent components of Christoffel Symbols

Postby z4lis » Sat Feb 28, 2015 10:18 pm UTC

The 40 conditions that you impose when you introduce metric compatibility relate the Christoffel symbols to the metric tensor and its partial derivatives. Once you know the metric tensor's components and their partial derivatives, you can specify all of the Christoffel symbols, and this is why you see "no independent components".

What you're doing is like asking how many independent parameters you have in describing a sphere in R3, counting 3 coordinates for the center and the radius (x,y,z,r) as 4 parameters, and then thinking that the volume equation V = 4/3*pi*r^3 should somehow impose additional constraints and lower the number of independent parameters of a sphere to something less than 4. The volume equation imposes constraints on describing spheres using the data (x,y,z,r,V), not (x,y,z,r). A condition that lowers the number of independent components of a set of variables should involve only those variables.
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Re: Independent components of Christoffel Symbols

Postby mosgi » Sun Mar 01, 2015 12:13 pm UTC

z4lis: Ah, that makes so much sense! What I think you're saying is: A priori, we have 64 independent components in each of Γabc and gab,c, but then symmetry (of gamma and g) reduces them to 40 each. Finally, metric compatibility gives 40 equations involving *both* Γabc and gab,c, so fully determines one set of values in terms of the other.

And so, I guess, there *are* 40 independent components in Γabc, and our earlier calculation was wrong (to the person asking why Γ(ab)c would be totally symmetric, I was using the fact that Γ is already symmetric in b,c, and we can build any permutation of a,b,c out of swapping a<->b and b<->c, in case that's what was wrong).

In terms of Γ being a Lie algebra valued 1-form, I guess this means it takes values in a 10-dimensional Lie algebra; possibly the Poincaré group? (so, Lorentz + translations)
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Re: Independent components of Christoffel Symbols

Postby eSOANEM » Sun Mar 01, 2015 2:47 pm UTC

I'm not sure I buy that argument z4lis.

I mean independent components in the same sense as is meant when people say that the Riemann tensor has 20 independent components. That is, what's the minimum number of components of the Christoffel symbol do I need to specify in order to know them all.

If I have an equation linking the Christoffel symbols and the metric's derivatives, it's true that there are 40 independent components to the metric derivatives, but the structure of the equation can reduce the number by making certain terms equivalent.

An example.

gab,c + gc,ab + gbc,a

has significantly fewer than 40 independent components. In fact, because g is itself symmetric, any permutation of indices is equivalent and so we only have 20 independent components (you can count them out yourself).

The expression we're dealing with has less obvious symmetries, because it's

gab,c - gbc,a + gca,b = 2Γabc

but it's structure can still reduce the number of independent components dramatically. This was what we were trying to calculate but, because the expression's symmetries are less obvious, it was a lot harder to work out how many independent components there were and all we were able to do was say was that there were no more than 30.

It occurs to me that this confusion may possibly have come about because I've been somewhat sloppy with using "free components" to mean "independent components". I am aware that the Christoffel symbols are uniquely defined by the metric and so, in that sense, none of the connections components are truly "free", that is, underspecified. What I meant however is that they are independent of other components of the connection; that they're internally free.
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Re: Independent components of Christoffel Symbols

Postby z4lis » Wed Mar 04, 2015 3:57 am UTC

Been too busy to answer anything!

Anyway, the equation you wrote

gthing + gthing - gthing = 2gamma

Gives a big matrix. What's its determinant?
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Re: Independent components of Christoffel Symbols

Postby eSOANEM » Wed Mar 04, 2015 8:48 am UTC

Because it's got 3 indices rather than 2, I'm not quite sure what you mean by it being a matrix or how to calculate its determinant.
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Re: Independent components of Christoffel Symbols

Postby z4lis » Wed Mar 04, 2015 9:55 pm UTC

You've described a linear transformation that takes the 40 g_{ijk} to the 40 \Gamma_{ijk}. This is an invertible transformation. In fact, since I can write:

g_{ab,c} = \Gamma_{bc,a} + \Gamma_{ab,c}

I know the inverse. So any linear relationship you believe holds among the \Gamma would have a corresponding linear relationship for the metric's partial derivatives. But I can cook up a chart that has pretty much anything I want for the partial of the metric tensor, and so that linear relationship can't happen.

On the other hand, the expression

z_{ijk} = g_{ij,k} + g_{jk,i} + g_{ki,j}

takes the 40 g_{ijk} to the 40 z_{ijk}, but as you pointed out, there are symmetries like z_{123} - z_{231} = 0 which mean that the linear transformation taking g's to z's has columns which aren't linearly independent, and so no inverse.
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Re: Independent components of Christoffel Symbols

Postby eSOANEM » Thu Mar 05, 2015 1:07 am UTC

Ah, I see (although, I think you want to remove the commas from the rhs in your first equation).

There are definitely 40 independent components in the metric's partial derivatives so the fact we can invert the expression means it must have exactly 40 independent components as you say.

I'm fairly convinced of this mathematically now.

As I understand it, torsion-free-ness is the only "internal" constraint and so the only one that reduces the number of independent components. Metric compatability then just gives us a way of calculating all these 40 independent components.

In terms of treating it as a Lie algebra valued 1-form, this is interesting then because it'd mean that the structure group for the tangent bundle of a Lorentzian manifold wasn't the Lorentz group and I'm not sure what other 4 translations it would bring in. The two obvious examples are to either upgrade the Lorentz group to the Poincaré group and bring in translations or to bring in dilations, one for each dimension.

Of these two choices, the latter seems more plausible to me as I don't think there's any 4x4 reps of the Poincaré group and, based on the indexing of the Christoffel symbols, the connection form must be taking values in a 4x4 rep.

On the other hand, dilations don't preserve line elements whereas translations do and it seems like this is something that we'd expect the structure group to do (otherwise it would seem to suggest I could construct a spacetime which shrinks you when you circumnavigate it).

Thank you for clearing up my initial confusion about how many components there are. Now I'm mostly confused about what they all correspond to.
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Re: Independent components of Christoffel Symbols

Postby PM 2Ring » Fri Mar 06, 2015 5:44 am UTC

You may find it illuminating and fun to read what John Baez has to say on this topic.

From General Relativity Tutorial - Long Course Outline
John Baez wrote:The CONNECTION is a mathematical gadget that describes "parallel translation along an infinitesimal curve in a given direction". In local coordinates the connection may be described using the components of the CHRISTOFFEL SYMBOL Gammaabc. There is an explicit formula for these components in terms of components gab of the metric, which may be derived from the assumptions 1-3 above. However, this formula is very frightening.
[...]
There is an explicit formula for the components Rabcd in terms of the Christoffel symbols. Together with the aforementioned formula for the Christoffel symbols in terms of the metric, this lets us compute the Riemann tensor of any metric! Thus to do computations in general relativity, these formulas are quite important. However, they are not for the faint of heart, so I will only describe them to readers who have passed certain tests of courage and valor.

See the adventures of Oz and the Wizard for an example of one such daring reader!


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