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  Dirac operator in Generalized Geometry

+ 3 like - 0 dislike

I am wondering how the Dirac operator can be built in the context of Hichin's generalized geometry.

In particular, I have the following questions:

  • On a spin manifold, is the conventional spin connection replaced by a generalized one that has two coordinate indices?

  • If so, does this generalized spin connection behave as a non-Abelian $\text{SO}(3,1)$ "B-field" (called also Kalb-Ramond field in physics)?

This post imported from StackExchange MathOverflow at 2015-02-17 11:28 (UTC), posted by SE-user Gian
asked Feb 9, 2015 in Theoretical Physics by Gian (65 points) [ no revision ]
retagged Feb 17, 2015

1 Answer

+ 5 like - 0 dislike

The structure group of $TX \oplus T^*X$ is $SO(n,n)$. Since $w_2(TX) = w_2(T^*X)$, this bundle always admits a spin structure and we lift to $Spin(n,n)$. The sections of the associated spinor bundle are not spinors in the usual way but bispinors like you guessed (they have two spinor indices, one from $TX$ and one from $T^*X$).

A B-field (in the sense of http://arxiv.org/pdf/math/0401221v1 ) determines an extension

$T^*X \to E \to TX$.

By the splitting principle, this bundle also admits a sort of bispinor.

Now it is a bit confusing to me since in physical applications (eg. the wonderful thesis of Yi Li: http://thesis.library.caltech.edu/2098/) only ordinary spinors are used, though I think that the (complex) bispinors are $\psi_+ \otimes \psi_-$ and $\bar\psi_+ \otimes \bar\psi_-$. We have $Spin(n,n) = Spin(n) \times_{\ZZ/2} Spin(n)$ and accordingly the spinor representation splits as a tensor product. A spin structure on $TX$ I think gives a basis of the sections of the bispinor bundle of simple tensors, so we get well-defined $\psi_+$ and $\psi_-$.

answered Feb 19, 2015 by Ryan Thorngren (1,925 points) [ revision history ]
edited Feb 20, 2015 by Ryan Thorngren

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