The Chern classes of a rank $n$ vector bundle on $X$ are obtained from composing the associated classifying map in $[X, BU(n)]$ with the maps $BU(n) \to B^{2i} \mathbb{Z}$ corresponding to the cohomology of $BU(n)$, thus giving elements in the cohomology groups $H^{2i}(X,\mathbb{Z})$.

But one could instead write those maps as $BU(n) \to B^{2i-1} U(1)$; and this corresponds to associating a $(2i-1)$-circle bundle to a vector bundle. For $i = 1$, for instance, this corresponds to sending a vector bundle to its top exterior power, the determinant line bundle.

What is then the interpretation of the higher Chern classes in this geometric setting? These should associate to the original rank $n$ vector bundle some circle $(2i-1)$-bundles (e.g. principal $B^{2i-2}U(1)$ $(2i-1)$-bundles), for $1 \leqslant i \leqslant n$.

This should mirror the algebraic side with symmetric polynomials; using $e_1 = x_1 + \ldots + x_n$ we have the line bundle $L = L_1 \otimes \ldots \otimes L_n$ (corresponding to the Chern roots); for other elementary symmetric polynomials $e_i$ we should be able to build a corresponding $(2i-1)$-line bundle directly realising the $i$-th Chern class. (Perhaps it would be more natural to instead consider Schur classes, and hopefully the relationship of these $(2i-1)$-line bundles with Schur functors is elucidated.)

Is there then an obstruction-theoretic picture, when trying to build back up the original vector bundle from these successive $(2i-1)$-line bundles?