In an abelian group, the additive energy between two sets is $$E(A,B)=|\{(a\_1,a\_2,b\_1,b\_2)
\in A\times A\times B\times B:a\_1+b\_1=a\_2+b\_2\}|$$ which is ranges from $|A||B|$ to $(|A||B|)^{3/2}$. What I want to know is, given that this is large, should $A$ look like $B$? 


To keep things simple, let's say all sets in question consist of integers. From the symmetry bound $E(A,B)^2\leq E(A,A)E(B,B)$ we see that the "best" additive partner for $A$ is $A$ itself, or any translate of $A$. From Plunnecke-Ruzsa, if we have small numbers $m,n$ then when $E(A,A)$ is (very) large, $E(mA,nA)$ is large, so that $dA+t$ is a good partner for $A$ as long as $d$ is a rational of small height.


If $r(x)$ is the number of solutions to $x=a+b$ with $a\in A$ and $b\in B$ then the fact that $$E(A,B)=\sum\_x r(x)^2$$ tells us that if $E(A,B)\geq K|A||B|$ then there is an $x$ so that $|A\cap x-B|\geq K$. So $B$ contains a subset which looks like (i.e. is a translate of) a subset of $A$. Can we do better?



What pairs of sets have additive energy?
If $\lvert A\rvert \approx \lvert B\rvert$ and $E(A,B)\geq K^{-1}\lvert A\rvert\lvert B\rvert^2$, then the Balog-Szemeredi-Gowers theorem yields large subsets $A'\subset A$ and $B'\subset B$ such that $\lvert A'-B'\rvert\ll K^{O(1)}\lvert A\rvert$. This is essentially the best result possible; from this one can deduce using covering lemmas that $A$ and $B$ are both sets of small doubling, each contained in $K^{O(1)}$-many translates of the other, which is essentially best possible.


The case where $\lvert B\rvert$ is much smaller than $\lvert A\rvert$ is more difficult. Tao and Vu have an asymmetric version of the Balog-Szemeredi-Gowers theorem (their Theorem 2.35) to handle this case: 



> 
> Suppose $E(A,B)\geq K^{-1}\lvert A\rvert\lvert B\rvert^2$, and $L^{-1}\lvert A\rvert \leq \lvert B\rvert\leq \lvert A\rvert$. Then, for any $\epsilon>0$, there are sets $H$ and $X$ such that 
> 
> 
> 1. $\lvert H-H\rvert \ll L^\epsilon K^{O(1)}\lvert H\rvert$,
> 2. $\lvert X+H\rvert=\lvert X\rvert\lvert H\rvert \ll L^\epsilon K^{O(1)} \lvert A\rvert$,
> 3. $\lvert A\cap (X+H)\rvert \gg L^{-\epsilon}K^{-O(1)}\lvert A\rvert$, and
> 4. there is some $x$ such that $\lvert (B-x)\cap H\rvert \gg L^{-\epsilon}K^{-O(1)}\lvert B\rvert$,
> 
> 
> where all the implicit constants depend on $\epsilon$.
> 
> 
> 


That is, up to polynomial losses in $K$, the set $B$ can be approximated by a set $H$ with small doubling constant, and $A$ is roughly a disjoint set of translates of $H$. This is a generalisation of your observation, which says (adjusting notation slightly to fit),



> 
> If $E(A,B)\geq K^{-1}\lvert A\rvert\lvert B\rvert^2$ then there is an $x$ such that $\lvert (B-x)\cap A\rvert\geq K^{-1}\lvert B\rvert$,
> 
> 
> 


so that not only does $A$ contain a translate of $B$, but also that $A$ is essentially a union of translates of $B$.


The dependence of the constants on $\epsilon$ is a little delicate, but can be worked out from the proof in Tao and Vu. I believe that one can take the constants in the $\ll,\gg$ to be independent of $\epsilon$, and replace


$$ K^{O\_\epsilon(1)}\quad\textrm{ by }\quad K^{O(\exp(\epsilon^{-1}))},$$
where the constant is now absolute, provided
$$ \epsilon \gg \log\log L/\log L;$$
this is only a back of the envelope calculation though, and for any particular application one can probably do better by unpacking the proof in Tao and Vu.


