Let $$M=\begin{pmatrix} u_1 & u_2 & \ldots & u_n \\ v_1 & v_2 & \ldots & v_n \\ \end{pmatrix}$$ be a $2 \times n$ matrix. Define $\nu(M)$ to be the $k \times n$ matrix $$\nu(M) = \begin{pmatrix} u_1^{k-1} & u_2^{k-1} & \ldots & u_n^{k-1} \\ u_1^{k-2} v_1 & u_2^{k-2} v_2 & \ldots & u_n^{k-2} v_n \\ \ldots \\ v_1^{k-1} & v_2^{k-1} & \ldots & u_n^{k-1}\\ \end{pmatrix}.$$ This immediately descends to a map $\nu: G(2,n) \to G(k,n)$, by thinking of a matrix as its rowspan.

**EDIT** Actually, as Sasha points out below, it only gives a rational map. The matrix $\nu(M)$ has full rank if and only if $M$ has at least $k$ pairwise linearly independent columns.

I was talking to a bunch of physicists today who are interested in the cohomology class of the image of this map. Does anyone know of papers which look at this?

In particular, if I understood them correctly, they have been computing the intersection of $\nu(G(2,n))$ with $G(k-2, n-4)$ embedded in the more normal way. (That is to say, we embed $G(k-2, n-4)$ into $G(k,n)$ as the space of $k$ planes containing a given $2$-plane $A$, and contained in a given $(n-2)$-plane $B$, with $A \subseteq B$.) Experimentally, the number of intersection points is coming out to be the Eulerian number $\left\langle \begin{matrix} n-3 \\ k-1 \end{matrix} \right\rangle$. For example, the intersection $\nu(G(2,6)) \cap G(1,2)$ inside $G(3,6)$ gives $4$.

Has anyone seen this?