1
$\begingroup$

Let $(M,g)$ be a pseudo-Riemannian spin manifold and let us denote by $S$ the spinor bundle, i.e. the associated vector bundle with respect to the spin representation. Usually, the "gamma matrices" are defined to be maps of the form $$\gamma:TM\to\mathrm{End}(S)$$ such that the Clifford relations are fulfilled, i.e. $\gamma(v)^{2}=g(v,v)\mathrm{id}$ or equivalently, $\gamma(v)\gamma(w)+\gamma(w)\gamma(v)=2g(v,w)\mathrm{id}$. I have to questions related to this:

  • Is it usuall assumed that $\gamma$ is linear at each fibre, i.e. that $\gamma(v_{p}+\lambda w_{p})=\gamma(v_{p})+\lambda \gamma(w_{p})$? I checked in the literature and it seems to be me that it somehow used sometimes but never explicitely mentioned. In particular, is it true that $\gamma$ can be viewed as sections of the bundle $\mathrm{End}(S)\otimes T^{\ast}M$.
  • I have seen somewhere the claim that $\nabla\gamma=0$. My question, how to I interbret this? If the answer to my first question is positive, then I would guess it is supposed to mean $\nabla^{\mathrm{End}(S)\otimes T^{\ast}M}\gamma=0$ where $\nabla^{\mathrm{End}(S)}$ is the connection induced by the spin connection $\nabla^{S}$ and $\nabla^{T^{\ast}M}$ the usual Levi-Civita connection. If I choose a local chart $x^{\mu}$ on $U\subset M$, set $\gamma_{\mu}:=\gamma(\partial_{\mu})\in\Gamma(\mathrm{End}(E)\vert_{U})$ and write $\gamma=\gamma_{\mu}dx^{\mu}$, then this would mean that $\nabla^{\mathrm{End}(S)\otimes T^{\ast}M}\gamma=0$ means $$\nabla^{\mathrm{End}(S)}_{\partial_{\mu}}\gamma_{\nu}-\Gamma_{\mu\nu}^{\alpha}\gamma_{\alpha}=0$$ Is this correct?

Any answer and/or reference is appreciated.

$\endgroup$
1
  • 1
    $\begingroup$ 1) The so-called Clifford action $\gamma$ is always taken to be bundle morphism, so that for each $p \in M$, $\gamma$ restricts to a linear map $\gamma_p : T_p(M) \to \operatorname{End}(S_p)$. 2) Your interpretation of $\nabla \gamma = 0$ is correct. In general, this is a condition you can look for in a connection on a spinor bundle, which makes it a Clifford connection. $\endgroup$ Commented Dec 12 at 14:54

1 Answer 1

0
$\begingroup$

The following affirmative answer was not given yet by Branimir Cacic: Yes you can also see Clifford multiplication as a section of $\mathrm{End}(S)\otimes T^∗M$.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .