2
$\begingroup$

I consider the spinor vector bundle $\Sigma$ over a spin manifold $M$. Next, I take the tensor product $\bigotimes^2 \Sigma$. I consider now the following Dirac operator acting on the tensors: $$ {\cal D}(\psi \otimes \psi')= \sum_i e_i .\nabla_{e_i} (\psi \otimes \psi') $$ with $e_i$ an orthonormal basis, $\nabla (\psi \otimes \psi')= (\nabla \psi)\otimes \psi' +\psi \otimes (\nabla \psi')$ and $e_i (\psi \otimes \psi')= (e_i \psi)\otimes \psi'$ . Have I the Lischnerowicz formula and the inequalities over the first proper value of the Dirac operator? $${\cal D}^2 = \Delta \otimes 1 + 1 \otimes \Delta + 2 \sum_i (\nabla_{e_i} \otimes \nabla_{e_i}) + \frac {r}{4}$$ with $\Delta$ the Laplacian.

$\endgroup$
  • $\begingroup$ Depending on which spinor bundle you take and the dimension of $M$, $\Sigma\otimes\Sigma$ is some variant of $\Lambda^*TM$, and $\mathcal D = d + d^*$. There is a Lichnerowicz-type formula for this "twisted" Dirac operator, you can find it in the book "Heat kernels and Dirac operators" by Berline, Getzler and Vergne (Theorem 3.52). In particular they compute the "Weitzenböck identity" you are after (Equation 3.16). $\endgroup$ – Bertram Arnold Oct 11 '16 at 13:23

Your Answer

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

Browse other questions tagged or ask your own question.