There is a map $(S^1)^n\to U(n)$, mapping each $n$-tuple to the corresponding diagonal matrix, where $S^1$ is identified with the complex numbers of unit length.
There is an induced map $B((S^1)^n)\to BU(n)$, but $B((S^1)^n)\cong (\mathbb{C}P^\infty)^n$ and $BU(n)\cong G^n(\mathbb{C}^\infty)$, the infinite complex Grassmannian.
So my question is: is there an easy way of thinking about the map $(\mathbb{C}P^\infty)^n\to G^n(\mathbb{C}^\infty)$? (I only care about what this map is up to homotopy)
Just to expand on Mariano Suarez-Alvarez comment. Observe that $B(S^{1})^{n} = BU(1)^{n}$ is the classifying space for $n$-tuples of line bundles. There are $n$ "universal line bundle" over this space, given by pulling back the universal bundle from $n$ distinct factors in the product. Similarly, but more simply, $BU(n)$ is the classifying space for complex vector bundles of rank $n$.
The map $BU(1)^{n} \rightarrow BU(n)$ mentioned by Mariano clearly (?) classifies the sum of the $n$ different "universal line bundles" over $BU(1)^{n}$.
To observe that this is the map you're after, at least up to homotopy, note that $BU(1)$ classifies complex line bundles since transition functions for line bundles are maps into $U(1)$. Now, if $\phi _{i}: V \rightarrow U(1)$, where $V \subseteq X$ is an open subset, are transition functions for some line bundles $\mathcal{L} _{i}$, then $\prod \phi _{i}: V \rightarrow U(1)^{n}$ composed with your map gives you the transition functions of their direct sum.