Twisting Spinor Bundles with Line Bundles

Question

In a paper I am reading, the following framework was given: Let $S$ be a spinor bundle, over a Riemannian manifold $M$, with Clifford action $$ c:S \otimes \Omega^1(M) \to S. $$ Moreover, let $E$ be a line bundle over $M$ with connection $\nabla$.

The author then speaks of the canonical Dirac operator $D$ on $S \otimes E$. What does he mean by this? My guess is as follows: Let $s \in S$ and $e \in E$, such that $\nabla(e) = \sum_i e_i \otimes \omega_i$, for $\omega_i \in \Omega^1(M)$. Moreover, let $D_S$ be the Dirac operator on $S$. I would define $D$ by $$ D(s \otimes e) = D_S(s) \otimes e + \sum_i c(s \otimes \omega_i) \otimes e_i. $$ Is this correct? If so, how does one define the Clifford action for $S \otimes E$. Finally, does this work for a twisting by any vector bundle?

Answer 1

As was already pointed out in the comments, one can always tensor a Cliffordmodul with another bundle to obtain a new twisted Clifformodul. Of course the Clifford action is only on the first factor. Given a connection on the twisting bundle one obtains a new twisted Dirac operator (formula as in the comments). One thing which is very intressting is the Weizenboeck formula in that context: the curvature remainder $K$ in $$D^2=\nabla^*\nabla+K$$ consists of the raw term for the spinor bundle, which is the scalar curvature up to an factor (of course, this is only true on spin manifolds), and an curvature term from the twisting bundle. These observations are also the first steps into an heat equation proof of an index theorem for Dirac operators. FOr more on that, see Roe's book on elliptic operators, or for much more, see Lawson & Michaelson

Answer 2

I think you might make a mistake, the connection form $\omega_i$ is not a form.

And the twisted Dirac is the twisted on the connection part and the action structure becomes the $spin^{\mathbb C}$, i.e. $$D^L=c(e_i)\nabla^{S\otimes L}_{e_i}.$$ Of course when $L$ is trivial, is equals to the normal Dirac.