Let $E$ be a $spin^c$ bundle and $L_E$ be a (complex) line bundle defined using transition functions $\nu \circ g_{U,V}$ where $\nu:spin^c(n) \to \mathbb{T}$ is map such that $\ker \nu=spin(n)$ and $g_{U,V}$ are transition functions for $spin^c(n)$-principial bundle $spin(E)$ (see also here). Let $S$ be a spinor bundle i.e. vector bundle constructed from the system of transition functions $c \circ g_{U,V}$ where $c$ is irreducible representation of Clifford algebra.

How to prove that the bundle $L_E$ satsifies the following $S \otimes L_E \cong S^*$ (where $V^*$ denotes the dual bundle).

I tried to show this using tranistion functions: the best situation would be if transition functions from both sides exactly coincide. This is equivalent to $$c(g_{U,V}(x)) \otimes \nu (g_{U,V}(x))=([c(g_{U,V}(x))]^t)^{-1}.$$


Let $c : \mathbb{C}\mathrm{l}_n \to \operatorname{End}(S_n)$ be your irrep, so that the “dual” irrep $c^\ast : \mathbb{C}\mathrm{l}_n \to \operatorname{End}(S_n^\ast)$ is defined by $$ \forall x \in \mathbb{C}\mathrm{l}_n, \quad c^\ast(x) := c(x^!)^t. $$ Thus, if $\pi := c\vert_{\operatorname{Spin}^\mathbb{C}(n)} : \operatorname{Spin}^\mathbb{C}(n) \to U(S_n)$, then $\pi^\ast : \operatorname{Spin}^\mathbb{C}(n) \to U(S_n^\ast)$ is given by $$ \forall \lambda \in \mathbb{T}, \; \forall x \in \operatorname{Spin}(n), \quad \pi^\ast(\lambda x) = c((\lambda x)^{-1})^t = c^\ast(\lambda^{-1}x), $$ so that the induced representation $$\operatorname{Hom}(\pi,\pi^\ast) : \operatorname{Spin}^\mathbb{C}(n) \to U(\operatorname{Hom}_{\mathbb{C}\mathrm{l}_n}(S_n,S_n^\ast)) = \mathbb{T}\operatorname{id}_{\operatorname{Hom}_{\mathbb{C}\mathrm{l}_n}(S_n,S_n^\ast)}$$ is given by $$ \forall \lambda \in \mathbb{T}, \; \forall x \in \operatorname{Spin}(n), \; \forall T \in \operatorname{Hom}_{\mathbb{C}\mathrm{l}_n}(S_n,S_n^\ast), \\ \operatorname{Hom}(\pi,\pi^\ast)(\lambda x)T := \pi^\ast(\lambda x) \circ T \circ \pi(\lambda x)^{-1} = c^\ast(\lambda^{-1} x) \circ T \circ c(\lambda^{-1}x^!) = c^\ast\left(\nu((\lambda x)^{-1})\right) \circ T. $$ Since $$ S \cong \operatorname{Spin}(E) \times_\pi S_n, \quad S^\ast \cong \operatorname{Spin}(E) \times_{\pi^\ast} S_n^\ast, $$ you can now use local trivialisations of $\operatorname{Spin}(E)$ to show that $$ \operatorname{Hom}_{\mathbb{C}\mathrm{l}(E)}(S,S^\ast) \cong \operatorname{Spin}(E) \times_{\nu^{-1}} \mathbb{C} \cong L_E^\ast, $$ and hence, by the natural isomorphism $S_n^\ast \cong S_n \otimes \operatorname{Hom}_{\mathbb{C}\mathrm{l}_n}(S_n,S_n^\ast)$, that $$ S^\ast \cong S \otimes \operatorname{Hom}_{\mathbb{C}\mathrm{l}(E)}(S,S^\ast) \cong S \otimes L_E^\ast. $$

| cite | improve this answer | |
  • $\begingroup$ Thank you for your answer. Still I have a couple of questions: you computed $Hom(\pi,\pi^*)$ and this calculation was needed to identify $Hom_{\mathbb{C}l(E)}(S,S^*)$ with $Spin^c(E) \times_{\nu ^{-1}} \mathbb{C}$: how you obtain $\nu ^{-1}$ since you computed that $Hom(\pi,\pi^*)(\lambda x)$ acts as the composition with $c^*(\nu((\lambda x)^{-1}))$? $\endgroup$ – truebaran Feb 17 '18 at 23:03

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.