The q-Vandermonde identity reads:

$$ \binom{m + n}{k}_{\!\!q} =\sum_{j} \binom{m}{k - j}_{\!\!q} \binom{n}{j}_{\!\!q} q^{j(m-k+j)} $$

The q-binomial coefficients: $$ \binom{ a }{ b}_{\!\!q} $$

are known to count the number of points over the Grassmannians $Gr(a,b)$ over $F_q$, and $q^{j(m-k+j)}$ is the number of points over $A^{j(m-k+j)}$ over $F_q$.

So the identity above states that number of points of $Gr(k, m+n)$ can be written as the sum of numbers of points of smaller Grassmannians and $A^l$. That would be naturally explained if there would be some geometric relation between these manifolds, like fiber bundles.

Question 1 What could be the geometric interpretation of the identity above, if any? If yes, what about other fields, e.g. $C$?

Question 2 Is the identity above true on the level of motives?

Question 2b If yes, then is it true for arbitrary fields?

For the case of q=1 - the hypothetical "field with one element", everything is "Okay": the $Gr(a,b,F_1)$ are just the sets of all a-combinations out of b and so the usual Vandermonde identity implies that $Gr(m+n,k)$ can be factorized in terms of smaller Grassmannians. That is, however, a kind of cheating, since over $F_1$, geometry disappears and counting is enough to give a "geometric" identity.

  • $\begingroup$ I remember trying to prove something like this identity in a comment on math.stackexchange using exact sequences, but I didn't have the time to go through the grueling details. But now I'm thinking that maybe it's easiest to forget about the coordinate-free Grassmannian and think of RREMs (= reduced row echelon matrices). A RREM with $n+m$ columns and $k$ rows can be regarded as a RREM with $n$ columns using only the first $j$ rows (for some $j \leq k$) attached to a RREM with $m$ columns using the remaining rows and an arbitrary $m \times j$-matrix. More precisely: ... $\endgroup$ – darij grinberg May 6 '18 at 20:48
  • $\begingroup$ ... Any RREM with $n+m$ columns and $k$ rows can be written as a block matrix $\begin{pmatrix} A & B \\ 0_{\left(k-j\right) \times n} & D \end{pmatrix}$, where $A$ is a RREM with $n$ columns and $j$ rows, where $D$ is a RREM with $m$ columns and $k-j$ rows, and where $B$ is a matrix whose columns above the pivots of $D$ are zero but whose entries are otherwise unconstrained (which leaves $q^{j\left(m-k+j\right)}$ options for $D$). $\endgroup$ – darij grinberg May 6 '18 at 20:49
  • $\begingroup$ Obviously, this is a form of "cell decomposition" of the Grassmannian that holds over any field; I guess it can be defined invariantly using dimensions of intersections with another (fixed, $n$-dimensional) vector subspace, but don't quote me on that. I don't know what motives are (and from what I keep hearing, I am not sure if anyone does). $\endgroup$ – darij grinberg May 6 '18 at 20:52
  • $\begingroup$ To some extent related: mathoverflow.net/questions/208560/… $\endgroup$ – Alexander Chervov May 11 '18 at 19:10

Assume $V$ is a vector space of dimension $m+n$, $M \subset V$ is a subspace of dimension $m$, and $N = V/M$. Let $p:V \to N$ be the projection. Consider the Grassmannian $X = Gr(k,V)$ and its stratification by the dimension of intersection with $M$, i.e., set $$ X_j = \{ U \in Gr(k,V) | \dim(U \cap M) = k - j \}. $$ If $U \in X_k$ then the projection of $U$ to $N$ has dimension $j$. Thus, we have a natural map $$ \pi \colon X_j \to Gr(k-j,M) \times Gr(j,N),\qquad U \mapsto (U \cap M,p(U)). $$ Finally, for any $U_M \in Gr(k-j,M)$ and $U_N \in Gr(j,N)$ we have $$ \pi^{-1}(U_M,U_N) \cong \{U' \in Gr(j,p^{-1}(U_N)/U_M) | U' \cap (U/U_M) = 0\}. $$ This is an open Schubert cell, hence is isomorphic to $\mathbb{A}^{j(m-k-j)}$. Moreover, the map $\pi$ is locally trivial, hence $$ [X_j] = [Gr(k-j,m)][Gr(j,n)][\mathbb{A}^{j(m-k-j)}] $$ in the Grothendieck ring of varieties. Summing up over $j$, we obtain $$ [Gr(k,m+n)] = \sum_j [Gr(k-j,m)][Gr(j,n)][\mathbb{A}^{j(m-k-j)}], $$ a motivic version of the formula..

  • 2
    $\begingroup$ Very nice answer! $\endgroup$ – Libli May 8 '18 at 20:05

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.