# Non-renormalizable corrections to GUT unification

While writing these answers: Hypercharge for U(1) in SU(2)xU(1) model and Is there a concise-but-thorough statement of the Standard Model? , it occured to me that the unification prediction for Georgi-Glashow is relying on high-scale renormalizability, if you have a nonrenormalizable SU(5) coupling between the SU(5) Higgsing field and the gauge field, you can make the couplings of the three subgroups different.

In general, after breaking, you can make different couplings for SU(2),SU(3) and U(1), depending on the Higgsing details, by allowing nonrenormalizable terms of sufficiently high order. If you Higgs in a natural way, using an SU(5) hermitian tensor field (spacetime scalar) with expectation (a,a,b,b,b), a term of the form $tr(\phi GG)$ where G is the SU(5) gauge field considered as an antihermitian SU(5) matrix, gives an SU(3) coupling correction which goes as (b-a)/M, where M is the higher string or planck scale. Considering that the string scale can be as low as $10^{17} GeV$, while the GUT scale can be as high as $10^{16} GeV$, depending on the details of low energy matter and high energy compactification, is it obvious that the failure of couplings can't be 10%?

Question: Considering that the GUT scale is close enough to the Planck scale that renormalizability is not exact, what are the precise natural errors in coupling unification?

### EDIT: Clarification

Lubos Motl gave an answer which misses the point, so I will give a more detailed description of the question. The question is: how bad can coupling constant unification fail at the GUT scale just from lack of renormalizability, ignoring running.

When you have a big gauge group break into a smaller gauge group, the field strength tensor for the big gauge group is multiplied by $1\over g^2$, and decomposes into the field strength tensors for the smaller gauge groups, which are then also multiplied by 1/g^2 so that the couplings are the same (at the unification scale--- forget about running--- this is a classical question).

But you have a scalar VEV doing the breaking. Normally scalars can't interact with gauge fields to alter the coupling--- the coupling is nondynamical. But the reason for this is renormalizability--- you can't have a renormalizable scalar-scalar gauge-gauge derivative interaction.

But ignore renormalizability, and suppose that there is an additional term in the Lagrangian of the form (the obvious space-time indices on F are suppressed)

$$A \mathrm{tr}(H F F)$$

Where F is the field strength tensor $F^{\mu\nu}$ for SU(5), A is a coupling constant of dimensions inverse mass, and H is an SU(5) adjoint scalar, which acts to Higgs SU(5), whose VEV is diag(a,a,b,b,b) in some gauge. The field H is invariant under SU(2) rotations of the top two components (because the H VEV is $\delta^a_b$ on the upper two-by-two block, and this is an invariant tensor of SU(2)), SU(3) rotations of the bottom 3 components (because the H VEV is $\delta$ on the bottom 3 by 3 block), and any diagonal phase rotations, including U(1) rotations that make up the Georgi-Glashow hypercharge. So this VEV breaks SU(5) to the standard model gauge group.

Now, if you decompose the field strength F into the SU(2) field strength $W$ and the SU(3) field strength $G$, the nonrenormalizable term above decomposes into

$$A a \mathrm{tr}(W^2) + A b \mathrm{tr}(G^2)$$

So that the VEV of the scalar alters the coupling constant of the two gauge groups in the decomposition by the Higgsing, and this happens at the GUT scale, ignoring any running.

This scalar-gauge-gauge nonrenormalizable term is in addition to the ordinary kinetic term, so the full W and G action is

$$({1\over 4g^2} + Aa) \mathrm{tr}(W^2) + ({1\over 4g^2} + A b) \mathrm{tr}(G^2)$$

The coupling constants do not match just from the breaking. The difference in their reciprocal-square coupling at the GUT scale is

$$4A(a-b)$$

Now A is order ${1\over M_\mathrm{PL}}$, and $(a-b)$ is order $M_\mathrm{GUT}$, so you get an order $M_\mathrm{GUT}\over M_\mathrm{PL}$ difference between the coupling constants already at the GUT scale, before any running.

I know that a term like this is ignored in running calculations, since these are RG flows in renormalizable low energy theories only. It is not included in higher loop corrections, or in any threshhold corrections, and to calculate it requires a full string background (and numerical techniques on the string background). The question is whether this term is typically small, or whether it is typically large enough to make the arguments against Georgi-Glashow from failure of unification moot.

### EDIT: more clarification

Unification relies on the coupling constant of a subgroups being equal to the coupling constant of the big group. The reason is that the $FF$ kinetic term breaks up to the subgroup kinetic terms.

But this depends on renormalizability. If you have nonrenormalizable scalar FF interactions, you break the equality of subgroup couplings. These terms can be neglected at ordinary energies, but not at 10^16 GeV. This is for Lubos, who downvoted the question, I am not sure why.

If you downvote this, you might also wish to downvote this: Does the ruling out of TeV scale SUSY breaking disfavor grand unification?

But I urge you to think about it instead, as I am 100% sure that they are both saying correct things.

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funny, i wasn't aware you could use the word 'Higgs' as a verb too –  lurscher Nov 9 '11 at 18:59
Dear @Ron, apologies, I changed my +1 rating of this question to -1 because it is no longer a question and the assertions made in the extensions are incorrect. A scalar-gauge-field interaction doesn't exist? Or doesn't contribute to the RG running? What? Is that a joke? And what does it have to do with your original question, anyway? I also noticed some nonsensical comments you added under my answer – about a gravity-induced failure of unification or whatever. None of these things exists. Gravity gives just a correction to the size that is small as long as the effective QFT is OK. –  Luboš Motl Jun 14 '12 at 11:05
@lubos is it really worth a down vote? Maybe you should change it to 0 because I can think of most questions here, including mine, being brainless in comparison. –  Larry Harson Jun 14 '12 at 16:54
@LubošMotl: I clarified the question--- it is correct, you can see the linked answer. The ordinary renormalizable scalar gauge coupling exists of course, it's the minimal coupling, but this minimal coupling only breaks the group down during Higgsing, it doesn't shift couplings, which are still equal between the broken groups. Only higher order nonrenormalizable terms can alter the equality of couplings, and I gave and analyzed the simplest one in SU(5) explicitly. –  Ron Maimon Jun 14 '12 at 19:03

It seems that somebody has noticed this already, but recently. Here is a reference I found on google: http://arxiv.org/PS_cache/arxiv/pdf/0811/0811.3307v2.pdf

The paper analyzes both dimension 5 and dimension 6 terms together, and this allows you to get whatever you want, since it's a lot of parameters at dimension 6. I like to analyze dimension 5 first, since for true grand unification, these should be the strong corrections. For pure dimension 5 corrections to SU(5) unification, you get the relation here: Does the ruling out of TeV scale SUSY breaking disfavor grand unification? , that the U(1) inverse-coupling-squared at unification is shifted from the SU(3) coupling by a certain fraction of the amount that the SU(2) coupling is shifted. In the coupling graph, you can see that this happens at too low an energy for proton decay.

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Good question. First, for grand unification to make sense, the field theory description must be valid everywhere beneath the GUT scale. It should better be valid at the GUT scale and "a little bit" above it, too. In this regime, one should be able to define the couplings and they should unify above it.

You're right that the naive one-loop calculation of the couplings needed for gauge unification isn't exact. There are two-loop, higher-loop, nonperturbative, and threshold corrections.

http://en.wikipedia.org/wiki/Renormalization_group#Threshold_effect

The latter, see the link above, are the effects of the renormalization group running that take into account the "continuous" change of the slope of the couplings as you're crossing the mass scale of the fields you are integrating out.

The leading corrections to the one-loop calculation are the two-loop perturbative effects – whether or not there are higher-order, nonperturbative, threshold, stringy, or other corrections to the physics nearby. The parameter coming with every loop is generally written as $\alpha/4\pi$ where $\alpha$ is the corresponding fine-structure constant. For the grand unified force, you have $\alpha=1/24$ or so at the GUT scale, so together with $1/4\pi = 1/12$ or so, you get $1/300$. So all these effects should be suppressed by something like multiples of 0.3 percent, not by 10 percent. Of course, the actual detailed numerical factor is model-dependent and one might construct models where it's much greater than one, I guess.

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I wasn't talking about this stuff, which is a rehash of ordinary coupling-unification calculations. I was talking about the possible failure of unification at the GUT scale, different SU(2), SU(3), U(1) couplings at the GUT scale, just from a nonrenormalizable interaction between the SU(5) Higgs field and the SU(5) gauge field of the form detailed above. This might not have been considered before, because I have never seen it adressed. –  Ron Maimon Nov 9 '11 at 8:25
@ron And how many bogus accounts do you have? I mean, who's competent enough to up vote your question, apart from... Lubos... ah that explains it - and he did generously say "good question" :) –  Physiks lover Nov 9 '11 at 22:51
@Physics lover: I like Lubos, and generally he is spot on, always honest, and very accurate, just not this time. The bogus accounts is just a joke, obviously nobody will go through the subterfuge. Is it really that hard to evaluate this question? I gave the precise term in a Lagrangian, and showed you how to decompose it, I even found a reference. This question could be homework in quantum field theory, it only is research because I think that nobody except those people in India noticed before. –  Ron Maimon Nov 10 '11 at 4:21
@Physikslover Ha ha, you are a funny guy (assuming that you are joking and not trolling...) :-P. I have not upvoted this answer so far (it this is what you mean ;-)...) but yes, I like Lumo, Ron and many other people here on physics SE. And of course, I upvote answers, questions, comments, etc if I can get something out and learn something from them. If you have a problem with this, you should read in the FAQ of our site about the purpose of upvotes ;-) –  Dilaton Jun 15 '12 at 7:20
@Physikslover BTW, if you can not get what Lumo, Ron, or other people are talking about in their posts, this does not automatically mean that nobody can understand it ...:-P –  Dilaton Jun 15 '12 at 7:22