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 $$c(g_{U,V}(x)) \otimes \nu (g_{U,V}(x))=([c(g_{U,V}(x))]^t)^{-1}.$$


1 Answer 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 $$ S^\ast \cong S \otimes \operatorname{Hom}_{\mathbb{C}\mathrm{l}(E)}(S,S^\ast) \cong S \otimes L_E^\ast. $$

  • $\begingroup$ 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, 2018 at 23:03

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