Importance of Closing Algebra in Wess-Zumino Lagrangian: Simple SUSY Question

[nrqed]

This is surely a stupid question but here it goes...

An auxiliary field is needed in the Wess-Zumino Lagrangian in order to have the algebra close off-shell. I have a couple of questions about this but let me ask the first one only for now.

Why is this required? The variation of the Lagrangian density gives a total derivative without using the equations of motion, even before the auxiliary fields are introduced. Why do we need to in addition impose that the algebra of the SUSY charges close? I guess I am asking: what would go wrong if the algrebra of charges would not close? This seems to be an issue that is totally unrelated to the invariance of the action but nobody explains what problem would happen if the algebra did not close.
Actually, this is not a purely SUSY question. Why is it important for the algebra of the charges of *any* symmetry to close?

Thanks in advance

 

[Alwi]

Dear nrqed,

If the algebra does not close - then you have to introduce new degrees of freedom so that the symmetry can be preserved with the new set of fields. Otherwise you lose the symmetry because you miss out on degrees of freedom which must be part of the symmetry. There would be symmetry operations generated by the left-out degrees of freedom and how will you account for it if you ignore those left out degrees of freedom ?

Best Regards
Alwi

 

[Demystifier]

I think that a closed algebra is sufficient but not necessary property to guarantee that the required symmetry is really there. If the algebra closes, then you are sure that everything is OK. If it does not close, then it might be consistent as well, but it is not so easy to prove it.

 

[Alwi]

Dear Demystifier,

I think the biggest problem from a Physics point of view - is that you would have left out some degrees of freedom that could possibly generate a symmetry operation. In that sense - an algebra that is not closed is not very satisfactory from a Physics perspective. The notion that there are left out degrees of freedom that can generate a symmetry operation is - not nice !

Best Regards
Alwi

 

[nrqed]

I think that a closed algebra is sufficient but not necessary property to guarantee that the required symmetry is really there. If the algebra closes, then you are sure that everything is OK. If it does not close, then it might be consistent as well, but it is not so easy to prove it.

Thanks..Yes, from what I have read, this seems to be the situation. Mots references on SUSY makes it sound as if it's a disaster if the algebra does not close off-shell (butthey don't say exactly why), but it turns out that the situation is more murky than that.

Where would a possible problem arise if it does not close off-shell?

 

[Alwi]

Dear nrqed,

Can you tolerate a symmetry operation generated by degrees of freedom that are NOT found in your Lagrangian ? My answer is NO ! I think that lies at the heart of the question. How do you deal with such a symmetry ? What about the Noether currents and so on ? Conserved charges ? I hope that answers your question.

Best Regards
Alwi