Is there a lift of the q-Vandermonde identity to some geometric (motivic) identity for Grassmannians over $F_q$? 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.
 A: 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..
