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  Index theorem in fermion theory

+ 2 like - 0 dislike

Is it possible to derive the chiral anomaly equation for the axial fermion current (in the form of an index theorem) by using only geometrical arguments, without introducing correlators, path integrals, Dirac seas etc.?

The chiral anomaly has the form of an index theorem, which has a topological origin. On the other hand, in the operator formalism it arises because of the absence of a chiral symmetry preserving regularization. I don't understand how these approaches are connected.

asked Sep 29, 2016 in Theoretical Physics by NAME_XXX (1,060 points) [ revision history ]
edited Oct 9, 2016 by NAME_XXX

You probably need to be more precise about what you want to allow and what you don't want to allow. In order to define the axial fermion current, you need at least a Lagrangian. Then, you also need some information about how to quantize this Lagrangian (a more or less straightforward answer being the path integral / fermion determinant). What do you have in mind for the latter?

@GregGraviton : I have asked rather about principal ability of well-defined chirality of fermions in the presence of the connection - the gauge field.

@NAME_XXX I am afraid, I still don't understand what exactly you are asking. To be able to define the chirality of a fermion, you have to be able to define what a fermion is, and then you are already swimming in the Dirac sea, so to speak. That said, quantum anomalies can be rephrased as questions about the regularized determinant of the Dirac operator on manifolds, which would be a purely mathematical question, but I'm not sure if this is what you want to hear, because regularizing the determinant is essentially a definition of the path integral for fermions.

@GregGraviton : the chiral anomaly has the form of index theorem, which has topological origin. From the other hand, in operator formalism it arises because of absence of chiral symmetry preserving regularization. I don't understand how these approaches are connected.

@GregGraviton : maybe, mathematicians have proved the index theorem for the other space of Dirac operator (not for $R^{4}$), where the all eigenvalues of the Dirac operator are discrete?

@anonymous : If you put the question like that, it makes much more sense. Could you edit the question?

@NAME_XXX : Compact vs non-compact space is actually non-trivial. For open spaces, there is the Callias index theorem .

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