Can one divide algebraic manifolds ? Make sense: $Gr(2,n)/ Gr(2,n+m) = P^{n-1}/P^{n+m-1} P^{n-2}/P^{n+m-2}$

Question

Let's start from a little bit far.

Basic probability theory - chain rule reads:

$$ P(AB) = P(A)P(B|A)$$

Example: consider n+m balls, where n - white balls, m - black balls, consider A - first chosen ball is white, B - second chosen ball is also white. The formula above gives:

$$ \frac{ \binom{ n}{2} } {\binom{ n + m}{2}} = \frac{n} {n+m} \frac{n - 1} {n+m - 1 } $$

q-Example: consider q-analogs of binomial coefficients, obviosly, similar fact is true for them:

$$ \frac{ \binom{ n}{2}_{\!q} } {\binom{ n + m}{2}_{\!q} } = \frac{ [n]_{q} } {[n+m]_{q} } \frac{[n - 1]_{q} } {[n+m - 1]_{q} } $$

Now, number of points of the Grassmanian over the finite field $F_q$ is exactly given by the q-binomial coefficient and $[n]_q$ is the one for $P^{n-1}(F_q)$. Optimistically enumaration relation is manifstation of some deeper relation on the geometric level. (See for example: Is there a lift of the q-Vandermonde identity to some geometric (motivic) identity for Grassmannians over Fq?)

So we come up with:

Question 1: Can one make sense of: $$Gr(2,n)/ Gr(2,n+m) = P^{n-1}/P^{n+m-1} P^{n-2}/P^{n+m-2} $$

Motivation. Combinatorics of binomial coefficients is very very related to probability theory, on the other hand they might be thought as $F_{q=1}$ of the Grassmanian geometry, since q-binomial coefficients behave pretty much similar to ordinary, it would be tempting to have a lift of probability ideas to generic $q$, and if so, we should have the relation above as a kind of q-chain rule.

It might be easier to analyse more simple question:

Question 2: (Weak version of 1). Is there any geometric relation between: $$ Gr(2,n) P^{n+m-1} P^{n+m-2} = Gr(2,n+m) P^{n-1} P^{n-2} $$

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The idea to divide manyfolds might sound crazy, however, it might have some ground in the past. There are ideas of the so-called "fractional motives". Which might be something like motives of such results of the division. What I vaguely remember from Yuri Manin's talks about ten years ago, is that one the main example comes from considerations by physicts Doron Gepner who discovered that CFT on certain CY can be factorized to product of other CFT (minimal models) which does not have direct geometric interpetation but if the that would exist - it should be a manyfold with some fractional dimension. Manin mentioned some similar constructions for some Frobenius manyfolds...

Answer 1

Let me begin with a sketch of an answer to your question two, namely giving an interpretation of the cross-multiplied equality $$\text{Gr}(2,n)\mathbb{P}^{n+m-1}\mathbb{P}^{n+m-2}=\text{Gr}(2,n+m)\mathbb{P}^{n-1}\mathbb{P}^{n-2}.$$ I claim this equality holds in the Grothendieck ring of stacks. This is the Grothendieck ring of varieties (the free Abelian group on isomorphism classes of algebraic varieties, subject to the cut-and-paste relation $$[X]=[U]+[X\setminus U]$$ where $U$ is Zariski-open in $X$, with multiplication given by the Cartesian product), and with the classes $[\mathbb{A}^1], [\mathbb{A}^n]-1$ inverted. Indeed, let $$Y=\text{Fl}_{1,2}(U)\times \text{Fl}_{1,2}(W),$$ where $\text{Fl}_{1,2}(V)$ denotes the space of flags of signature $(1,2)$ in a vector space $V$, and where $\dim(U)=n, \dim(W)=n+m$. Now $$[\text{Fl}_{1,2}(k^r)]=[\mathbb{P}^{r-1}][\mathbb{P}^{r-2}],$$ because forgetting the subspace of dimension $2$ witnesses $\text{Fl}_{1,2}(k^r)$ as a Zariski-locally trivial $\mathbb{P}^{r-2}$ bundle over $\mathbb{P}^{r-1}$. On the other hand, $$[\text{Fl}_{1,2}(k^r)]=[\mathbb{P}^1][\text{Gr}(2,r)],$$ because forgetting the subspace of rank $1$ witness $\text{Fl}_{1,2}(k^r)$ as a Zariski-locally trivial $\mathbb{P}^1$-bundle over $\text{Gr}(2,r)$. So we have $$[Y]=[\mathbb{P}^1][\text{Gr}(2,n)][\mathbb{P}^{n+m-1}][\mathbb{P}^{n+m-2}]=[\mathbb{P}^{n-1}][\mathbb{P}^{n-2}][\text{Gr}(2,n+m)][\mathbb{P}^1].$$ This is your desired equality, with both sides multiplied by $\mathbb{P}^1$. But in the Grothendieck ring of stacks, $[\mathbb{P}^1]=1+[\mathbb{A}^1]$ is invertible, because $[\mathbb{P}^1]([\mathbb{A}^1]-1)=[\mathbb{A}^2]-1$, which we inverted by fiat. So the desired equality holds in this ring.

In fact, it is an easy exercies to show that $\text{Gr}(a,b)$ and $\mathbb{P}^c$ are invertible for any $a,b,c$, so one may divide out to obtain your equality in Question 1.

That said, I'm dubious that there is any real value in thinking about this in terms of "fractional motives" -- as you can see, the underlying geometry is just a linearization of the combinatorial reasoning in your example.