Euler class of dual bundle Let $L^*$ be the dual bundle of complex line bundle $L$. Since the bundle $L\otimes L^* = \text{Hom}(L,L^*)$ has nowhere vanishing section given by the identity map, the first Chern class $c_1(L\otimes L^*) = c_1(L) + c_1(L^*) = 0$, and the $c_1(L) =-c_1(L^*)$. The first Chern class $c_1(L)$ it is Euler class $e(L_\mathbb{R})$. So, $e(L_\mathbb{R}) = -e(L^{*}_{\mathbb{R}})$. There exist an isomorphis between vector bundle and their dual bundle. So, their Euler classes should be equal. Where is the mistake?
 A: A complex vector bundle and its dual bundle are in general not isomorphic. As your argument shows, if $E \longrightarrow M$ is a complex vector bundle of odd rank $n$, then $c_n(E) = -c_n(E^\ast)$ and $e(E_{\Bbb R}) = - e((E^\ast)_{\Bbb R})$, so that both $E \not\cong E^\ast$ and $E_{\Bbb R} \not\cong (E^\ast)_{\Bbb R}$ (if $c_n(E) \neq 0 \neq e(E)$).
Note that this is in contrast to the case of real vector bundles. A Euclidean metric on a real vector bundle $F \longrightarrow M$ gives an isomorphism
$$F \cong F^\ast = \mathrm{Hom}(F, \underline{\Bbb R}).$$
The difference in the complex case is that a Hermitian metric on a complex vector bundle $E \longrightarrow M$ gives an isomorphism
$$\bar{E} \cong E^\ast = \mathrm{Hom}(E, \underline{\Bbb C}),$$
where $\bar{E}$ is the conjugate bundle of $E$. The conjugate bundle is not necessarily isomorphic to the original bundle.
A: I think you're confusing the dual, over the real numbers, of the underlying real bundle $L_{\mathbb R}$ of $L$ with the underlying real bundle of the complex dual of $L$.
