First Chern class of a specific line bundle

Question

Let $E$ be a spin$^c$ bundle and $spin^c(E)$ the corresponding $spin^c(n)$-principial bundle. Let $g_{U,V}: U \cap V \to spin^c(n)$ denote transition functions for this principial bundle and consider the map $\nu:spin^c(n) \to \mathbb{T}$ defined by $\nu(w)=w^!w$ where $(v_1 \cdot ... \cdot v_r)^!=v_r \cdot ... \cdot v_1$. We can form the composition $\nu \circ g_{U,V}:U \cap V \to \mathbb{T}$. As this satisfies cocycle property we can form the line bundle $L_E$.

Let us assume that $E$ is $spin^c$ but is not spin.

How to prove that the first Chern class of $L_E$ is odd (in the sense that $j^*(c_1(L_E)) \neq 0 $ where $j^*$ is mod 2 reduction of coefficients)?

Answer 1

One way to prove this is to first check that it is true when $E$ is the spin$^c$ bundle canonically associated to a complex vector bundle $F$ and then argue that complex and spin$^c$ bundles have the essentially same characteristic classes in degree up to 4.

To associate a spin$^c$-bundle to a complex vector bundle, consider the inclusion $\rho: U \hookrightarrow Spin^c$ obtained by lifting $i \times \det : U \hookrightarrow SO \times U(1)$ by the double cover $Spin^c \to SO \times U(1)$ (where $i : U \hookrightarrow SO$ is the obvious inclusion). Given a principal $U$-bundle $F$, we can thus consider $E = F \times_\rho Spin^c$, i.e. the quotient $(F \times Spin^c)/U$ where $h \in U$ acts by $(f,g) \mapsto (f\cdot h, \,\rho(h^{-1})g)$. Then we get a well-defined action of $Spin^c$ on $E$ by setting $[(f,g)]\cdot h \mapsto [(f, g\rho(h))]$ for $h \in Spin^c$, and this makes $E$ a principal $Spin^c$-bundle.

In a similar way, the fundamental line bundle $L_E$ of any spin$^c$ bundle $E$ is the $U(1)$-bundle associated to $E$ by the homomorphism $\psi : Spin^c \to U(1)$ defined by composing the double cover of $SO \times U(1)$ with projection to the right factor. Since $\psi \circ \rho : U \to U(1)$ is precisely $\det$, the fundamental line bundle of the spin$^c$ bundle associated to a complex vector bundle is simply $\det F$. Hence $c_1(L_E) = c_1(\det F) = c_1(F)$ in this case, and it is well-known that $w_2(E) = c_1(F)$ mod 2.

For the final part, note that the classifying map $f : BU \to BSpin^c$ for the universal complex bundle $EU$ as a spin$^c$ bundle can be represented as a fibre bundle with fibres $Spin^c/U$. For if $ESpin^c$ is a contractible space with a free $Spin^c$ action, then the subgroup $U$ also acts freely. Hence we can take $ESpin^c/U$ as a representative of $BU$, and then $f$ is simply the projection $ESpin^c/U \to ESpin^c/Spin^c$, whose fibres are $Spin^c/U$.

This fibre has a transitive action by $Spin$ with stabiliser $SU$, so it's homeomorphic to $Spin/SU$. In degree $0 < i \leq 5$, the homotopy groups are $$ \pi_i(Spin/SU) = \pi_i(Spin(6)/SU(3)) = \pi_i(SU(4)/SU(3) = \pi_i(S^7) = 0, $$ implying that $f$ is a 5-equivalence (i.e. $f_* : \pi_i(BU) \to \pi_i(BSpin^c)$ is an isomorphism for $i < 5$, and surjective for $i = 5$). Therefore $f^* : H^i(BSpin^c) \to H^i(BU)$ is an isomorphism for $i \leq 4$. Hence any relation between spin$^c$ characteristic classes of degree up to $4$ that is valid for complex vector bundles is valid for any spin$^c$ bundle.