9
$\begingroup$

So far I met three definitions of the so called generalized Dirac operator(or Dirac type operators. Everything takes place over Riemannian manifols $M$ and we have smooth hermitian vector bundle $S \to M$ over $M$.
First definition The first order differential operator $D$ is called Dirac type if its symbol $\sigma$ has the property: $$\sigma(x,\xi)^2u=-\|\xi\|^2u$$ for $x \in M, \xi \in T^*_xM, u \in S_x$.
Second definition $D$ is called Dirac type operator if $D^2$ is of the form $$\sum_{i,j}g^{ij}\partial_i\partial_j$$ modulo the lower order terms (here $g^{ij}$ are components of Riemannian metric on cotangent bundle).
Third definition $D$ is called Dirac type if there is a Clifford action $c$ and a Clifford connection $\nabla$ (i.e. metric connection compatible with this Clifford action) such that $D=c \circ \nabla$ (some authors put $i$ in front of this operator).

Are these definitions equivalent? If so, why is it true?

The one implication which is evident for me is that the second definition implies the first

$\endgroup$
7
$\begingroup$

A very good place to read about this is the 3. chapter of the book "Heat kernels and Dirac operators" by Nicole, Getzler & Vergne.

Up to the wrong sign in your second definition, 1. and 2. are equivalent, as the symbol just collects the highest order terms. Also you obtain from the basic properties of the symbol map that the bundle which admits a Dirac operator of the 1. definition admits a Clifford representation of the cotangent (and via the Riemannian metric by the tangent) bundle. The action is just given by the symbol of $D$. A slightly tricky computation (carried out in the aforementioned book) gives you a metric connection (if $D$ is self-adjoint) on your bundle, such that the corresponding Dirac operator differs from $D$ only by a endomorphism. As you cannot get rid of this zeroth order term in general, you see that the 3. definition is a little bit more special than 1. & 2.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.