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?


1 Answer 1


Let $V = k^n$. The map in question is the composition of the canonical map $$ f:G(2,V) \to G(k,S^{k-1}V) $$ given by the $(k-1)$-th symmetric power of the tautological bundle, and the (noncanonical) linear map $$ g:G(k,S^{k-1}V) \to G(k,V) $$ given by the choice of the basis in $V$. For both maps it is easy to compute the pullbacks of the Chern classes of the tautological bundles. Since the Chow ring is generated by those classes you can compute the pullback of any Schuber class and this determines the class of the image.

In fact, the map $g$ (as well as the initial map) is not regular but only rational. But it has a simple resolution. Let $m = \dim S^{k-1}V - (n - k)$ and $K = Ker(S^{k-1}V \to V)$. Consider the partial flag variety $F(k,m;S^{k-1}V)$ and its subscheme $Z$ consisting of all flags $U_k \subset U_m \subset S^{k-1}V$ such that $K \subset U_m$. Then the projection $Z \to G(k,S^{k-1}V)$ is birational (for generic $U_k$ we have $U_m = U_k \oplus K$) and the projection to $G(k,S^{k-1}V/K) = G(k,V)$ is regular. On the other hand, by definition $Z$ in $F(k,m;S^{k-1}V)$ is the zero locus of the vector bundle $K^*\otimes Q_m$, where $Q_m$ is the tautological quotient bundle (with fiber $S^{k-1}V/U_m$). This description allows to compute the pullback for $g$.

Addition. The precise formula in case $n = 6$, $k = 3$ for the map is the following. Let $U$ denote the tautological rank 2 subbundle on $G(2,6)$ and let $Q$ denote the tautological rank 3 quotient bundle both on $G(3,6)$ and $G(18,21)$. Consider the product $X = G(2,6)\times G(18,21)$. Let $z = c_9(S^2U^*\otimes Q)$ on $X$. Let $p_*$ denote the map on the Chow rings induced by the projection $X \to G(2,6)$. Then the map on the Chow rings induced by the composition $g\circ f$ takes $$ \prod c_i(Q)^{k_i} \mapsto p_*(z\cdot c_3(Q)^{15}\cdot \prod c_i(Q)^{k_i}) $$ (in the LHS $Q$ is considered as a bundle on $G(3,6)$ and in the RHS as a bundle on $G(18,21)$).

  • $\begingroup$ Thanks! I'm having trouble pulling back along $g$, though. Let's look at $G(2,6) \to G(3,15)$, followed by the rational map to $G(3,6)$. I know that the image is a hypersurface in $G(3,6)$ of degree $4$. In other words, if I pull back the Schubert class $(3,3,2)$, I should get $4$ times the class of a point. I tried pulling back $(3,3,2)$ from $G(3,15)$ and got $128$ times a point instead. (I can write out the details if you'd like.) I am clearly not understanding how to use your resolution. $\endgroup$ Nov 12, 2011 at 13:29
  • $\begingroup$ Minor correction: 15 should read 21 $\endgroup$ Nov 12, 2011 at 14:52
  • $\begingroup$ I wrote a more detailed explanation. Is it all right now? $\endgroup$
    – Sasha
    Nov 12, 2011 at 16:17
  • $\begingroup$ You are describing how to compute the map $H^{\ast}(G(18,21)) \to H^{\ast}(G(2,6))$. I understand that. But how do you compute the cohomology class of the image of $G(2,6)$ in $G(3,6)$? Note that the image of $G(2,6)$ in $G(18,21)$ is NOT transverse to the exceptional locus of the rational map. $\endgroup$ Nov 12, 2011 at 19:31
  • $\begingroup$ The class of the image is determined by its intersections with classes of $G(3,6)$. Those are generated by Chern classes of $Q$, and the intersection with products of these classes are given by the above formula. $\endgroup$
    – Sasha
    Nov 12, 2011 at 21:58

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