Sums and products in finite fields: an integral geometric viewpoint

We prove that if $A\subset {\mathbb{F}}_q$ is such that $$\mathrm{\mid }A\mathrm{\mid }>q^{1\mathrm{/}2+\frac{1}{2d}},$$ then $${\mathbb{F}}_q^{\ast }\subset d{A}^2=A^2+\mathrm{.}\mathrm{.}\mathrm{.}+A^{2}d\text{times},$$ where $$A^2=\{a\cdot a^{\mathrm{\prime }}:a,a^{\mathrm{\prime }}\in A\},$$ and where ${\Bbb F}_q^{*}$denotesthemultiplicativegroupofthefinitefield${\Bbb F}_q$. In particular, we cover ${\mathbb{F}}_q^{\ast }$ by $A^2+A^2$ if $\mathrm{\mid }A\mathrm{\mid }>q^{3\mathrm{/}4}$. Furthermore, we prove that if $$\mathrm{\mid }A\mathrm{\mid }\ge C_{size}^{\frac{1}{d}}{q}^{1\mathrm{/}2+\frac{1}{2(2d-1)}},$$ then $$\mathrm{\mid }d{A}^2\mathrm{\mid }\ge q\cdot \frac{C_{size}^2}{C_{size}^2+1}\mathrm{.}$$ Thus $d{A}^2$ contains a positive proportion of the elements of ${\mathbb{F}}_q$ under a considerably weaker size assumption.We use the geometry of ${\Bbb F}_q^d$, averages over hyper-planes and orthogonality properties of character sums. In particular, we see that using operators that are smoothing on $L^2$ in the Euclidean setting leads to non-trivial arithmetic consequences in the context of finite fields.