Is My Calculation Correct for Topological Action with Veirbein and Levicivita?

[jinbaw]

I'm trying to simplify an action that has the term: levicivita_[a,b,c,d]*levicivita^[mu,nu,rho,sigma]*R^[a,b]_[mu,nu]*R^[c,d]_[rho,sigma]

where a,b,c, and d are flat indices and mu nu rho sigma are curved indices

I got the term: 4*e^mu_a*e^nu_b*e^rho_c*e^sigma_d*R^a,b_mu,nu*R^c,d_rho,sigma

where e i s the veirbein and R is the reimann curvature tensor.
My question is if i have for example a=c levicivita_[a,b,c,d] is 0. however if i have a =c and mu=rho in the answer i got... i won't get a zero. is there some wrong in my computations? thank you

 

[Mentz114]

<br /> \epsilon_{abcd}\epsilon^{\mu\nu\rho\sigma}R{^{ab}}_{\mu\nu}R{^{cd}}_{\rho\sigma}<br />

I'm not sure if your definition of the L-C pseudo-tensor is correct. If any index is repeated it is zero, so the only non-zero indexes are permutations of 0,1,2,3.

0123 : 1
0132 : -1
0213 : -1
0231 : 1
0312 : 1
0321 : -1
1023 : -1
1032 : 1
1203 : 1
1230 : -1
1302 : -1
1320 : 1
2013 : 1
2031 : -1
2103 : -1
2130 : 1
2301 : 1
2310 : -1
3012 : -1
3021 : 1
3102 : 1
3120 : -1
3201 : -1
3210 : 1

 

[jinbaw]

Yes, I have this correctly... but what i used to get my equality is:

LC^mu,nu,rho,sigma*LC_a,b,c,d = e^mu_a*e^nu_b*e^rho_c*e^sigma_d + permutations over mu, nu, rho, sigma.

 

[arkajad]

I think the magic is that, for instance, 4{R^{12}}_{12}{R^{12}}_{12} indeed appears in your term, but it will also appear in those five other terms that you call "permutations" - and they will cancel out. The point of your expansion is to get nice contractions in the formulas, but it is not computationally optimal in the sense that there will be many cancellations of terms. In other words: you are adding and subtracting the same terms in order to get certain nice expressions like square of the scalar curvature etc.

Does it make sense?

 

[jinbaw]

I think it does... there must be some other mistake then.
Actually what I'm trying to do is:
I have a lagrangian of the form: levicivita_a,b,c,d*levicivita^mu,nu,rho,sigma*(lambda^2*e^a_mu*e^b_nu*e^c_rho*e^d_sigma*phi + (8/3)*lambda*e^a_mu*e^b_nu*e^c_rho*R^d,4_sigma,4dot + 2*lambda*a_mu*e^b_nu*phi*R^c,d_rho,sigma + 2*e^a_mu*R^b,c_nu,rho*R^d,4_sigma,4dot + phi*R^a,b_mu,nu*R^c,d_rho,sigma)

where a,b,c,d and 4 represent flat coordinates
and mu,nu,rho,sigma, and 4dot represent curved coordinates

all the veirbeins are diagonal and functions of r only
and phi(r) = e^4_4dot

I already have a solution for these eqs of motion in 5-d... i need to verify that this is still true when i split into 4+1 dimensions (of course 1 is not time here, its the 5th coordinate)

when i plug in my solution into the 4+1 eqs of motion, the solution is not verified... it appears as if I'm getting two extra terms ... so there must be something wrong in my eqs or most probably the way i simplified the lagrangian.

Is there a method to check where the problem comes from?

 

[arkajad]

Can you write your Lagrangian using the tex function available here, and what you have simplified and how? I am not sure if can be of any help, but perhaps I will be able to notice something that will help you.

 

[jinbaw]

My Lagrangian is given by:

\epsilon_{abcd}\epsilon^{\mu\nu\rho\sigma}(\lambda^{2}e^{a}_{\mu}e^{b}_{\nu}e^{c}_{\rho}e^{d}_{\sigma}\phi - \frac{8}{3}\lambda e^{a}_{\mu}e^{b}_{\nu}e^{c}_{\rho}R^{d4}_{\sigma\dot{4}} - 2\lambda e^{a}_{\mu}e^{b}_{\nu}\phi R^{cd}_{\rho\sigma} + 2e^{a}_{\mu}R^{bc}_{\nu\rho}R^{d4}_{\sigma\dot{4}} + \phi R^{ab}_{\mu\nu}R^{cd}_{\rho\sigma})

what i did is simplify each of the 5 terms alone using the fact that:
\epsilon_{abcd}\epsilon^{\mu\nu\rho\sigma} = e^{\mu}_{a} e^{\nu}_{b}e^{\rho}_{c}e^{\sigma}_{d} + permutions over \mu , \nu, \rho, and \sigma

I got:
24 \lambda^{2}\phi - 16\lambda e^{\sigma}_{d} R^{d4}_{\sigma\dot{4}} - 16\lambda e^{\rho}_{c}e^{\sigma}_{d} R^{cd}_{\rho\sigma}\phi + 4 e^{\nu}_{b} e^{\rho}_{c} e^{\sigma}_{d} R^{bc}_{\nu\rho} R^{d4}_{\sigma\dot{4}} + 4 \phi e^{\mu}_{a} e^{\nu}_{b}e^{\rho}_{c}e^{\sigma}_{d} R^{ab}_{\mu\nu} R^{cd}_{\rho\sigma}
Thanks a lot for your help! :):)

 

[arkajad]

Alright, will try to verify. Just give me some time.

 

[jinbaw]

ok sure! thanks again! :)

 

[arkajad]

The first two terms are ok. Will be checking now other terms. The third term I am getting 8 instead of your 16. Factor two from

\epsilon_{abcd}\epsilon^{abuv}=2!\delta^{uv}_{cd}

and another factor two from

\delta^{uv}_{cd}R^{cd}_{\rho\sigma}=2!R^{uv}_{\rho\sigma}.

I stop here until we agree on this term.

 

[jinbaw]

yes I agree with you! Actually, I got the factor of 2 later on while i was replacing indices with numbers. for example, R^{01}_{01} and R^{10}_{10} will add up giving a factor of 2.

 

[arkajad]

From the next term I am getting extra term. Probably in your calculation two terms canceled, but in my calculation they add, so I am getting factor 8 with the term that is missing in your expansion.

 

[jinbaw]

well, i used the fact that in my case the metric is diagonal and function of r only. plus, when i calculate the R's from the spin connections, the only R^{ab}_{\mu\nu} that are non-zero are the ones for which a,b = \mu, \nu. Also R^{d4}_{\sigma\dot{4}} is non zero for d = \sigma only. So i canceled for example terms that do not contain e^{\sigma}_{d} because at the end they will give me a zero.
(terms like 4 e^{\nu}_{b} e^{\rho}_{d} e^{\sigma}_{c} R^{bc}_{\nu\rho} R^{d4}_{\sigma\dot{4}} are always zero either for
d = \sigma the veirbein is zero or for d = \rho then R is zero.)

Am I allowed to do that? or did i oversimplify somewhere?

 

[arkajad]

It seems to me that

<br /> 4 e^{\nu}_{b} e^{\rho}_{d} e^{\sigma}_{c} R^{bc}_{\nu\rho} R^{d4}_{\sigma\dot{4}} <br />

contributes with, for instance:

4e^0_0e^1_1e^1_1R^{01}_{01}R^{14}_{14}

 

[jinbaw]

mmm so in this case i would be taking \rho = \sigma and d = c wouldn't that create sort of a contradiction because the levi civita tensors, in the lagrangian i started with, should give a zero in this case?
I thought i wasn't allowed to repeat the same number for two indices in the same expression because of the presence of the levi civita tensors.

 

[arkajad]

But, as I said before, expanding you are adding and subtracting terms. The term I mentioned appears also in the 4-th term in your expansion. You should be consistent.

 

[jinbaw]

oki I see where the additional term comes from... probably i have the same mistake in the 5th term in my expansion. I'll try to fix that too!

 

[arkajad]

OK. I hope that you will be able to proceed from now on your own. Good luck!
Just one more warning, though I do not know if applicable in your case: It is, in general, not the same to do first dimensional reduction of the Lagrangian and then derive field equations as first derive field equations from the full Lagrangian and then apply dimensional reduction to the equations.

 

[jinbaw]

Let's say I use the initial 5 dimensional lagrangian to get field equations for which i can find a solution. Then I dimensionally reduce this lagrangian taking \phi(r) = e^{4}_{\dot{4}}. I get a new set of field equations from this reduced lagrangian. But, wouldn't the solution be still valid, specially that the metric is independent of the 5th coordinate? Actually, it is a function of r only.

 

[jinbaw]

i did the expansion... and the terms having repeated indices do cancel out as expected.
Just one more question. in 5-d, i calculated the reimann tensor R^{ab}_{\mu\nu} from the spin connections using the formula:

R^{ab}_{\mu\nu} = \partial_{\mu} w_{\nu}^{ab} - \partial_{\nu} w_{\mu}^{ab} + w_{\mu}^{ak}w_{\nu k}^{b} - w_{\nu}^{ak}w_{\mu k}^{b}

where the indices range from 0 to 4.

when i reduced to 4 dimensions + a scalar \phi, I used the formula:

R^{ab}_{\mu\nu} = \partial_{\mu} w_{\nu}^{ab} - \partial_{\nu} w_{\mu}^{ab} + w_{\mu}^{ak}w_{\nu k}^{b} - w_{\nu}^{ak}w_{\mu k}^{b} + w_{\mu}^{a4}w_{\nu 4}^{b} - w_{\nu}^{a4}w_{\mu 4}^{b}

where now the indices range from 0 to 3.

I'm confused if the 2nd expression is correct or not.

+ if the formula is true, how do i calculate the terms: R^{d4}_{\sigma\dot{4}} in my lagrangian? (since now the indices range from 0 to 3... so this is not included in the formula above.

 

[arkajad]

Well, you should calculate again from your formula

<br /> R^{ab}_{\mu\nu} = \partial_{\mu} w_{\nu}^{ab} - \partial_{\nu} w_{\mu}^{ab} + w_{\mu}^{ak}w_{\nu k}^{b} - w_{\nu}^{ak}w_{\mu k}^{b} <br />

using whatever assumptions are in your dimensional reduction.

 

[jinbaw]

with the indices ranging from 0 to 4? so the Reimann curvature components that i have calculated in the original 5-d space will remain the same after dimensional reduction?

 

[arkajad]

I can't say because I do not know the details of your reduction process. In "Riemannian geometry, fiber bundles, Kaluza-Klein theories and all that", p. 246 the "consistency problem" of the dimensional reduction is being mentioned. Field equations from the dimensionally reduced action will, in general, be different from dimensionally reduced full field equations. The devil is in the details.

 

[jinbaw]

I can't find a copy of the book in my university library. I will order a copy asap, but it needs time for shipping. I will have a look at it as soon as it arrives.

Well, the action that i have is from a paper by Chamseddine, entitled: Topological Gravity and Supergravity in Various Dimensions. Reference: Nucl.Phys.B346:213-234,1990

I used the same method that he used to reduce the lagrangian. Actually, the reduced lagrangian is found explicitly in his paper (equation 3.22). but, what i can't understand is how to calculate the reimann curvature in this case.

Another thing that i can't understand and which seems really crucial is this:
there is an action given by: I_{4} = 3k \int_{M_4} \epsilon^{\mu\nu\rho\sigma} \epsilon_{abcde} \phi^{a} R^{bc} R^{de}

where w_{\mu}^{ab}= (e^{\alpha}_{\mu}, w_{\mu}^{\alpha\beta}) and \phi^{a} = (e^{4}_{\dot{4}}, w_{\dot{4}}^{\alpha 4})

I'm not familiar with this type of notation, and i cannot arrive at expanding this action explicitly.

 

[arkajad]

Got the paper. Tomorrow will take a look at it but can't promise anything now (got a cold).

 

[jinbaw]

oki! Get well soon and thank you for your help! :)

 

[arkajad]

Perhaps during the weekend I will try to write a short explanatory note, but to this I will have to refresh my memory. Also Chamseddine paper is rather condensed, so certain details need to be guessed, which makes the task somewhat harder.

 

[jinbaw]

Do you suggest any reference (paper or book) that i could have a look into which will help me a bit until then?

 

[arkajad]

Not really. I am still having problems with understanding what Chamseddine is doing. Lagrangians like his are discussed on p. 216 of the book I mentioned, but still I do not know which group he is gauging (SO(2,4),SO(1,5), SO(2,3) or SO(1,4)) and how is implementing dimensional reduction. The paper is really far from being clear.

Added: Well, in the meantime you can have a look at this paper http://arkadiusz-jadczyk.org/papers/einstein.htm" - but it is unnecessarily complicated for your case.

 

[jinbaw]

well as i understood from his paper, eq 3.8, SO(2,4) is gauged for \lambda =1 and SO(1,5) for \lambda =-1 and ISO(1,4) for \lambda =0. This is how he construted the 5-d action (I_5). Then he obtained I_4 by fixing one of the indices to be 4 and permuting the others.
Please correct me if I'm wrong.

 

[arkajad]

I am trying to understand a relation between your

<br /> \epsilon_{abcd}\epsilon^{\mu\nu\rho\sigma}(\lambda ^{2}e^{a}_{\mu}e^{b}_{\nu}e^{c}_{\rho}e^{d}_{\sigm a}\phi - \frac{8}{3}\lambda e^{a}_{\mu}e^{b}_{\nu}e^{c}_{\rho}R^{d4}_{\sigma\d ot{4}} - 2\lambda e^{a}_{\mu}e^{b}_{\nu}\phi R^{cd}_{\rho\sigma} + 2e^{a}_{\mu}R^{bc}_{\nu\rho}R^{d4}_{\sigma\dot{4}} + \phi R^{ab}_{\mu\nu}R^{cd}_{\rho\sigma})<br />

and (3.22). It seems you have \lambda\mapsto -\lambda and you set
e_{\dot{4}}^\delta=0 and e_\rho^4=0.. But if so, why don't you have the term 2e_\mu^\alpha R^{\beta\gamma}_{\nu\rho}R_{\sigma\dot{4}}^{\delta 4} and the next one?

Moreover, in the paper p. 224 he says "The standard truncation is to set ... e_{\dot{4}}^4 ... to zero. Is that his mistake? I understand this is the scalar field?

 

[jinbaw]

Exactly. I have: \lambda\mapsto -\lambda and e_{\dot{4}}^\delta=0 and e_\rho^4=0.
The term you're talking about is the 4th term in my expansion, but i used latin indices instead of greek, just so i don't get confused between curved and flat indices.
The next term is: 2e_\mu^\alpha R^{\beta 4}_{\nu\rho}R_{\sigma\dot{4}}^{\gamma \delta }
When i calculated the curvature tensor components from the diagonal veirbeins and the spin connections, the R's that contain either 4 or \dot{4} only turned out to be zero.
That's why i ignored this term.

Moreover, in the paper p. 224 he says "The standard truncation is to set ... e^{4}_{\dot{4}} ... to zero. Is that his mistake? I understand this is the scalar field?

Yes, this is supposed to be the scalar field. Plus, if you look at 3.24, e^4_\dot{4} is included. So probably, he’s not using the standard truncation. I can’t exactly understand what he is doing.

 

[arkajad]

His "standard truncation" would give zero for the action!

 

[jinbaw]

yes, exactly.
that's why i suspect i should not use the standard truncation. actually i didn't make any assumptions except that i have a diagonal veirbein from the beginning.

I used this fact in the 5-d action and in the reduced 4-d action. So, I think i should get the same results. That is if i have one solution for 5-d, it should remain valid after reduction. Right?

 

[arkajad]

I don't know. For me it is a dirty business. Moreover, it looks like no referee ever really read the paper. Look at (3.12): it is called an "equation"! Then you substitute these "on-shell" equations whenever it is convenient. I think I give up, unless you find some better paper to base your calculations on.

 

[arkajad]

One more thing, When on p. 224 he writes "If we further relabel..." he probably means:

\omega_\mu^{\alpha 4}=e_\mu^\alpha

\phi^4=e^4_4,\,\phi^\alpha=\omega_4^{\alpha 4}.

There is no "spin connection". He has connection forms with values in the Lie algebra of the appropriate group.

 

[jinbaw]

There is no "spin connection". He has connection forms with values in the Lie algebra of the appropriate group.

Can you explain more please?

 

[arkajad]

Sure, that is easy for me, so if you will have further question - ask. I am not sure at which level to respond, so let me try this way:

Say, we are constructing a gauge theory of SO(3). So we imagine what is called a principal fibre bundle over base manifold M (our space-time). That is locally we take take the product MxSO(3), except that we are forgetting the origin of SO(3) in the fibers. Selecting (in a smooth way) one origin in each fibre is called "choosing a gauge".
Tangent space at a given point to each fibre is essentially tangent space to SO(3) - thus it is isomorphic to the Lie algebra so(3) of SO(3) - that is antisymmetric 3x3 matrices. If we choose a gauge, then to give a connection form is the same as to give a 1-form on M with values in so(3), which we write as

{{\omega_\mu}^A}_B

If we define

{\omega_\mu}^{AB}={{\omega_\mu}^A}_C\eta^{BC}

where \eta is, in our case the Euclidean flat metric, then {\omega_\mu}^{AB}=-{\omega_\mu}^{BA} - thus \omega_\mu have values in so(3).

Spin connection is usually understood in the context of gauging the spin group. In this case that would be gauging SU(2). Its Lie algebra is isomorphic to that of SO(3), but we would represent it normally as 2x2 complex anti-hermitian traceless matrices. And that would be where our connection form would take values for an SU(2) gauge theory.

With this understanding it is not a crime to call an so(3) valued connection form a "spin-connection" (Lie algebras are isomorphic), but it is unnecessarily misleading, because there is no spin around.

If you want me to elaborate more on similar issues - I will try to help as much as I can.

 

[jinbaw]

Actually, I'm very new on this issue. I do understand GR in terms of the metric and affine connection, but when it comes to veirbeins and spin connections i still find difficulties.

What i understand is that the spin connections calculated in chamseddine's paper are actually not the spin connections based on the lie groups, but the "spin connections" based on the lie algebras corresponding to those groups.

Is this correct?

 

[arkajad]

Connections have always values in Lie algebras. Take GR. You are familiar with Levi-Civita connection. It has values in the Lie algebra gl(4), more specifically, it has values in the Lie algebra o(3,1) which is a subset of gl(4). But what it does? At each point of M we gave the set of all orthonormal frames at this point. It is your fiber. The group SO(3,1) acts on these frames (still at the same point of M) - it rotates them. So you have your principal SO(3,1) bundles. What the connection does? It takes an orthonormal frame and transports it along a path in M to an orthonormal frame at a different point of M. How it does it? You solve parallel transport equation using Levi-Civita connection. And so it is in general. You have a fibre, you have group acing on this fibre, you have connection form, with values in the Lie algebra, you have parallel transport equations, you can solve it and transport "rigidly" the whole fibre into the fibre at some other point of the path. Connection form is nothing but a generalization of the Levi-Civita connection.

Normally in GR we use coordinate basis \partial_\mu in the tangent space and define connection coefficients through:

\nabla_\mu\,\partial_\nu=(\partial_\sigma)\,\Gamma^\sigma_{\mu\nu}

I am writing it in this order deliberately, but do not differentiate on the RHS. Then you do not see that you have an o(3,1) valued connection. But if you take instead an orthonormal basis e_a
and define

\nabla_\mu e_a=e_c\,{{\Gamma_\mu}^c}_b

then \Gamma_\mu matrices are in so(3,1).

 

[arkajad]

Suppose you want to do the gauge theory of SO(10) on M. Instead of playing with tangent space, you put a 10-dimensional vector space V_x at each point of M, you endow it with the appropriate SO(10) invariant scalar product (Euclidean in my example), you consider the set of orthonormal bases in each V_x and you suppose you have a covariant derivative allowing you to covariantly differentiate vector valued fields \phi^A,\, A=1\ldots,10. If E_A is a 10-beim, then you define

\nabla_\mu E_A=E_C{{\Gamma_\mu}^C}_A

and because you suppose that your connection preserves the scalar product in the fibers, you find that \Gamma_\mu matrices, \mu=0,1,2,3 are now in so(10) Lie algebra.

Of course at the same time you have the formula

(\nabla_\mu \phi )^A=\partial_\mu\,\phi^A+{{\Gamma_\mu}^A}_C\,\phi^C.

 

[jinbaw]

Yes. Thanks for making the idea clear. I understood a very big part of what u said, though i know i still have lots of gaps because actually i have never taken a real course neither in topology nor in differential geometry.

I still have a question: although it is mentioned in the title of the thread, but i can't really understand what is exactly meant by the term "topological". can u explain this to me?

 

[arkajad]

Topological in the title means, roughly, that the original action integral you start with is a "topological invariant", that is you calculate some expression (a number) using some connection, but the value of the integral (in case of a compact manifold) is, surprise-surprise, the same for every connection. So it does not depend on which connection you put on your manifold. So, what it depends on? It depends on the "shape of your manifold", whether it has holes or handles or such things. Then, of course, you somehow break this independence, because, after all, you want to get field equations for your connection. But I am not well versed in these tricks.