Pseudo Real Group if anomaly free

  • Thread starter Thread starter robousy
  • Start date Start date
  • Tags Tags
    Anomaly Group
robousy
Messages
332
Reaction score
1
I am working a problem from the Mohapatra textbook unification and supersymmetry and the question is to show that if any rep of a grp is pseudo real then it is automatically anomaly free.


There is not much in the chapter (2) on how to go about this.

All I know is that for a pseudo real group G = G* (the conjugate is the same as the original group). The group is not real though.

I've been searching the web and books for a couple of hours now to no avail so thought I'd leave the question here.
 
Physics news on Phys.org
May I suggest you post this question on the sci.physics.strings board, which you can access through the Strings Branes and LQG subforum. That board is visited by expert string theorists who are likely to have an answer for your question; they are very interested in anomaly cancelation. But they are not likely to visit this Particle Physics subforum.
 
ok, thanks!
 
Mmm, this is a bit of a technical question, you'll want to double check me (say with Weinberg vol 2).

But you basically listed the reason *why*, namely when the left handed fermion fields furnish reps that are equivalent to the complex conjugate rep. Eg the representation of the gauge algebra (take its complex conjugate) is related by a similarity transformation to itself

To see this, remember when you calculate the anomaly from the three point function, you can separate it into symmetric and antisymmetric parts by group index. The anomaly is wholely contained in the symmetric part of this, so you have to do a little bit of algebra (subbing in the similarity condition of the representation into the symmetric part) and you will come out with the required reality or pseudo reality to be anomaly free.

Note that some standard model groups will contain this gauge anomaly, so there you will have to look for cancellations to occur, or else you have an inconsistent theory. Its one of the miracles of SU(3)*SU(2)*U(1) that the required cancellations *do* occur (a deeper reason maybe is that this group is an rep of SO(10) which is by the above anomaly free)
 
Last edited:
Haelfix said:
To see this, remember when you calculate the anomaly from the three point function, you can separate it into symmetric and antisymmetric parts by group index. The anomaly is wholely contained in the symmetric part of this, so you have to do a little bit of algebra (subbing in the similarity condition of the representation into the symmetric part) and you will come out with the required reality or pseudo reality to be anomaly free.


Hey Haelfix. Thanks for your response. I'm still getting used to calculating the anomaly. I've done is so far using young tableux and I've also seen an equation in Mohapatras book that is related to the trace of generators.
Could you possibly give me a reference to the equation you are tallking about - ie regarding the 3 pt function. I have most QFT books - perhaps Weinberg vol 2?

Thanks again.

:smile:
 
Toponium is a hadron which is the bound state of a valance top quark and a valance antitop quark. Oversimplified presentations often state that top quarks don't form hadrons, because they decay to bottom quarks extremely rapidly after they are created, leaving no time to form a hadron. And, the vast majority of the time, this is true. But, the lifetime of a top quark is only an average lifetime. Sometimes it decays faster and sometimes it decays slower. In the highly improbable case that...
I'm following this paper by Kitaev on SL(2,R) representations and I'm having a problem in the normalization of the continuous eigenfunctions (eqs. (67)-(70)), which satisfy \langle f_s | f_{s'} \rangle = \int_{0}^{1} \frac{2}{(1-u)^2} f_s(u)^* f_{s'}(u) \, du. \tag{67} The singular contribution of the integral arises at the endpoint u=1 of the integral, and in the limit u \to 1, the function f_s(u) takes on the form f_s(u) \approx a_s (1-u)^{1/2 + i s} + a_s^* (1-u)^{1/2 - i s}. \tag{70}...
Back
Top