Determinant line bundle

Let X be a paracompact space, then there is a bijection [X,\operatorname{BO}(n)]\xrightarrow\cong\operatorname{Vect}_\mathbb{R}^n(X),[f]\mapsto f^*\gamma_\mathbb{R}^n with the real universal vector bundle \gamma_\mathbb{R}^n . The real determinant \det\colon \operatorname{O}(n)\rightarrow\operatorname{O}(1) is a group homomorphism and hence induces a continuous map \mathcal{B}\det\colon \operatorname{BO}(n)\rightarrow\operatorname{BO}(1)\cong\mathbb{R}P^\infty on the classifying space for O(n). Hence there is a postcomposition: : \det\colon \operatorname{Vect}_\mathbb{R}^n(X) \cong[X,\operatorname{BO}(n)] \xrightarrow{\mathcal{B}\det_*}[X,\operatorname{BO}(1)] \cong\operatorname{Vect}_\mathbb{R}^1(X). Let X be a paracompact space, then there is a bijection [X,\operatorname{BU}(n)]\xrightarrow\cong\operatorname{Vect}_\mathbb{C}^n(X),[f]\mapsto f^*\gamma_\mathbb{C}^n with the complex universal vector bundle \gamma_\mathbb{C}^n . : \det(E) :=\Lambda^{\operatorname{rk}(E)}(E). == Properties ==

• The real determinant line bundle preserves the first Stiefel–Whitney class, which for real line bundles over topological spaces with the homotopy type of a CW complex is a group isomorphism. Since in this case the first Stiefel–Whitney class vanishes if and only if a real line bundle is orientable, both conditions are then equivalent to a trivial determinant line bundle. • The complex determinant line bundle preserves the first Chern class, which for complex line bundles over topological spaces with the homotopy type of a CW complex is a group isomorphism. • The pullback bundle commutes with the determinant line bundle. For a continuous map f\colon X\rightarrow Y between paracompact spaces X and Y as well as a vector bundle E\twoheadrightarrow Y , one has: • : \det(f^*E) \cong f^*\det(E). : Proof: Assume E\twoheadrightarrow Y is a real vector bundle and let g\colon Y\rightarrow\operatorname{BO}(n) be its classifying map with E=g^*\gamma_\mathbb{R}^n , then: :: \det(f^*E) \cong\det(f^*g^*\gamma_\mathbb{R}^n) \cong\det((g\circ f)^*\gamma_\mathbb{R}^n) \cong(\mathcal{B}\det\circ g\circ f)^*\gamma_\mathbb{R}^1 \cong f^*(\mathcal{B}\det\circ g)^*\gamma_\mathbb{R}^1 \cong f^*\det(g^*\gamma_\mathbb{R}^n) \cong f^*\det(E). : For complex vector bundles, the proof is completely analogous. • For vector bundles E,F\twoheadrightarrow X (with the same fields as fibers), one has: • : \det(E\otimes F) \cong\det(E)^{\operatorname{rk}(F)}\otimes\det(F)^{\operatorname{rk}(E)}. == Literature ==