[Math] About the Chern class of the determinant line bundle

algebraic-topologydifferential-topologygeometric-topology

Is it true that the first Chern class of a rank $k$ complex line bundle is equal to the first Chern class of its determinant line bundle?

Best Answer

Yes. You may use the splitting principle and assume that the vector bundle is a direct sum of $k$ line bundles, and note that the first Chern class of direct sum of line bundles is the sum of first Chern classes of the line bundles, which in turn is the Chern class of the tensor product of those bundles, which is isomorphic the top wedge power of the original vector bundle.