How to obtain a lower bound on the volume of a progression which contains translates of another progression?

Question

I am trying to solve a problem in Tao's Additive Combinatorics(Problem 3.2.2).

For any additive set $A$ and a natural number $d$(rank of progression), for vectors ($v_1$,$\cdots$,$v_d$)$\in A^d$, $a\in A$ and collection of natural numbers $(N_1,\cdots,N_d)$ a progression $P$ of rank $d$ is the collection $\{a+n_1v_1+\cdots+n_dv_d: 0\leq n_j\leq N_j \}$. The volume of progression $P$ is defined as $vol(P)=\prod_{j=1}^{d}(N_j+1)$.

Suppose $P=\{a+n_1v_1+\cdots+n_dv_d: 0\leq n_j\leq N_j \}$ is a progression of rank $d$ and $S$ is another progression which contains the translates $P+e_1,\cdots, P+e_K$ where $e_i\in A$. Given that $v_1,\cdots,v_d,e_1,\cdots,e_K$ are linearly independent over $Z$. Prove that rank of $S$ is $\geq d+K-1$ and $vol(S)\geq 2^{K-1}vol(P)$.

My working so far:

It is easy to construct a progression $S$ with rank $r+K-1$ and volume $2^{K-1}vol(P)$. I also showed that the rank of $S$ is $\geq d+K-1$ by showing that $v_1,\cdots,v_d,e_1,\cdots,e_K$ are linearly dependent over $Z$ if rank of $S$ is $<d+K-1$. However, I am unable to show that $vol(S)\geq 2^{K-1}vol(P)$. The trivial lower bounds on $vol(S)$ are $2^{d+K-1}$ (obtained from the fact that $rank(S)\geq r+K-1$) and $Kvol(P)$ (obtained from the fact that $vol(S)\geq |S|\geq |P|K \geq Kvol(P)$).

Can someone help me in proving the fact $vol(S)\geq 2^{K-1}vol(P)$? Can someone prove $vol(S)\geq 2^{K-1}vol(P)$ when $A=\mathbb{Z}^n$ or $A=\mathbb{R}^n$ and $n>d+K$?

Answer 1

This is not an answer, but is too long for the usual comment format. The main difficulty is that the structures of $P$ and $S$ might be quite different. If $S$ were pure like $P$, then we would have $vol(S)=|S|$ and we would be done. It might be helpful to try to vizualize what happens in a reasonably complex example.

Here is an example with $d=1,K=2$, $P=[0,2]v$.

Let $F$ be the abelian group defined by generators $s_1,s_2,s_3,s_4$ and the unique relation $10s_1+24s_2+19s_3+8s_4=0$.

Let $S$ be the progression $S=[0,13]s_1+[0,31]s_2+[0,24]s_3+[0,9]s_4$, let $e_1=s_1+2s_2+4s_3+8s_4, e_2=3s_1+7s_2+6s_3+s_4,v=5s_1+12s_2+9s_3+s_4$, and let $P$ be the progression $[0,2]v$. Then the following shows that $e_1+P$ and $e_2+P$ are subprogressions of $S$ :

$$ \begin{array}{rcl} e_1 &=& s_1+2s_2+4s_3+8s_4 \\ e_1+v &=& 6s_1+14s_2+13s_3+9s_4 \\ e_1+2v &=& s_1+2s_2+3s_3+2s_4 \\ e_2 &=& 3s_1+7s_2+6s_3+s_4 \\ e_2+v &=& 8s_1+19s_2+15s_3+2s_4 \\ e_2+2v &=& 13s_1+31s_2+24s_3+3s_4 \\ \end{array} $$

The structure of $P$ (on the left) and its image in $S$ (on the right) seem quite unrelated, except for the neat sequence $s_4,2s_4,3s_4$ in the last three lines. Hope this helps all the same.