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]

$$\epsilon_{abcd}\epsilon^{\mu\nu\rho\sigma}R{^{ab}}_{\mu\nu}R{^{cd}}_{\rho\sigma}$$

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

$$4 e^{\nu}_{b} e^{\rho}_{d} e^{\sigma}_{c} R^{bc}_{\nu\rho} R^{d4}_{\sigma\dot{4}}$$

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

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

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

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

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.