# Twisting Spinor Bundles with Line Bundles

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?

• The Clifford action on $S \otimes E$ is defined by tensoring $c$ with the identity on $E$. The Dirac operator is defined using the "tensor product connection" $\nabla_S \otimes 1 + 1 \otimes \nabla_E$ where $\nabla_E$ is a connection on the line bundle $E$ compatible with the metric. – Michael Mar 21 '11 at 16:12
• So if I've understood correctly, what you've said gives the Dirac operator I proposed. No? – Janos Erdmann Mar 21 '11 at 16:25
• Yes, it's the same thing; except that your notation is perhaps a little unusual. The usual notation is for $c$ to be the map $\Omega^1(M) \to End(S)$, whence $$D(s\otimes e) = Ds \otimes e + \sum_i c(\omega^i) s \otimes \nabla_i e$$ The answer to your second question is also affirmative. – José Figueroa-O'Farrill Mar 21 '11 at 18:18
• Michael and José: at least one of you should fill in an answer. – S. Carnahan Mar 23 '11 at 6:51

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
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.