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I am studying a theory with a Lagrangian density that I don't want to write because I would like to try to tackle the problem by myself. The fact is that it is a theory with two fields $\{ A_{\mu\nu},B_\mu \}$ where $A_{\mu\nu}$ is an anti-simmetric field and $B_\mu$ is a vector field.

I found that the action is invariant under a local transformation of the $A_{\mu\nu}$ field and then, to calculate the propagators, I introduce a gauge fixing term that breaks the symmetry which depends on field $A_{\mu\nu}$ and on a lagrange multiplier.

What I have found is that despite the breaking of the local symmetry the Hessian matrix to be inverted to calculate the propagators is still degenerate.

So my question is: is the Hessian matrix degenerate if and only if there is a local symmetry? Or is it possible that it is even if the action has no local symmetries?

Fred
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    Possible duplicate: https://physics.stackexchange.com/q/351104/50583, see in particular part 5 of my answer for the case of gauge-fixing – ACuriousMind Sep 13 '22 at 19:53
  • Thanks, i read your answer and it clarified many things to me. Coming to my example, I noticed that the only way to make the Hessian matrix invertible is to add a gauge fixing term even with field B which has no gauge invariance. How can i justify this? – Fred Sep 14 '22 at 08:23
  • I'm not 100% sure what you mean, but even a non-gauge vector field is a constrained system (just not a gauge theory) because your $B_\mu$ has 4 components but there are only three physical d.o.f. in a massive spin-1 particle. See also this answer of mine – ACuriousMind Sep 14 '22 at 08:33
  • Thank you! I am new to gauge theories and thought constrained systems were only those with local symmetry. – Fred Sep 14 '22 at 08:54

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