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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.

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  • $\begingroup$ 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
    Commented May 6, 2018 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
    Commented May 6, 2018 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
    Commented May 6, 2018 at 20:52
  • $\begingroup$ To some extent related: mathoverflow.net/questions/208560/… $\endgroup$
    – Alexander Chervov
    Commented May 11, 2018 at 19:10

1 Answer 1

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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..

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    $\begingroup$ Very nice answer! $\endgroup$
    – Libli
    Commented May 8, 2018 at 20:05

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