Possibility to fix indices in the model file? about pyrate HOT 4 CLOSED

Hi Tom. Thank you very much for your question and suggestions.
Overall, specifying indices 'by hand' is something that I would prefer to avoid. However, I agree that there might be cases where this is necessary, or at least convenient. In your case, one possible alternative solution would be to express the determinant as a polynomial of traces of the matrix, using the Cayley-Hamilton theorem. In the case of a 3x3 matrix, one has:

For larger matrices however, I expect this 'trick' to have poor performance. Implementing a way of computing the determinant of a 2-index tensor could be a good idea of improvement for future versions. I am also currently working on a possible new feature, allowing the user to implement arbitrarily complicated tensor objects and contractions using Python functions that can be directly called from the model file. With such a feature, the user would be able to either specify indices by hand, or create its own 'determinant' function.

I hope this helps!

Cheers,
Lohan

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Hi Lohan,

yes, that is a very nice trick, I didn't think of that one! A little followup question: since M is bidoublet in a SU(3) x SU(3) gauge group, and your ansatz requires to write down contractions between left and right indices of M. Will such a contraction actually work, since it is between different gauge groups?

Good to know you are looking into a possible enhancement, I'll stay tuned.

cheers,
Tom

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Hi Tom,

Yes. In principle the consistency checks only require indices to have the same range, so the above trick should work. However I didn't anticipate this and must say that I don't really like the idea of indices of different nature being contracted together.

Thinking a bit further about your cubic term, a more elegant solution would be to use the (built-in) 3-index Levi-Civita tensor to produce invariants out of three fundamental SU(3) reps. Namely, the following expression should in principle work :

t : 1/6 * Eps[i,j,k] Eps[a,b,c] H[i,a] H[j,b] H[k,c]

where the 1/6 factor is only needed to match the expression of the determinant.

Cheers,
Lohan

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Ah, very good! I think my student was convinced Eps is only the rank two Levi-Civita symbol, I can convince him otherwise now.

Thanks!

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