Introduction to Decoupling

Fourier analysis and diophantine equations

Sample problem: Let $H{L}_{s,k}(N)$ be the number of integer solutions of $n_1^k+\cdots n_s^k=n_{s+1}^k+\cdots n_{2s}^k,\text{with}1\le m\le N$

For fixed $s,k$, what are the asymptotics of $H{L}_{s,k}(N)$ as$N\to \infty$ ?

Proposition

Let $h(x)=h_{k,N}(x):={\sum }_{n=1}^{N}e(n^{k}x).$ ${\int }_0^1|h_{k,N}(x){|}^{2s}dx=H{L}_{s,k}(N)$, the number of integer solutions of $n_1^k+\cdots n_s^k=n_{s+1}^k+\cdots n_{2s}^k,\text{with}1\le m\le N$

Proof If $m\in \mathbb{Z}$, then ${\int }_0^{1}e(mx)dx=\{\begin{array}{ll}1& \text{if }m=0\\ 0& \text{if }m\ne 0\end{array}$ Let $h(x):={\sum }_{n=1}^{N}e(n^{k}x)$. $|h|{}^{2s}=h^s{\bar{h}}^s={\sum }_{n_1,\dots ,n_{2s}=1}^{N}e((n_1^k+\cdots +n_s^k-n_{s+1}^k-\cdots -n_{2s}^k)x).$ ${\int }_0^1|h|{}^{2s}={\sum }_{n_1,\dots ,n_{2s}=1}^N{\int }_0^{1}e((n_1^k+\cdots +n_s^k-n_{s+1}^k-\cdots -n_{2s}^k)x)dx,$

Vinogradov system

In the 1930s, Vinogradov studied the number of solutions of the following system of equations: $n_1^j+\cdots n_s^j=n_{s+1}^j+\cdots n_{2s}^j,\text{with}1\le j\le k,$ $J_{s,k}(N)$ = the number of integer solutions of above system of equations with $1\le n_i\le N.$

(Sharp estimate for Vinogradov system)

For every $ϵ>0$, there is a constant $C(ϵ,k)$ so that $J_{s,k}(N)\le C(ϵ,k)N^{ϵ}\left(N^s+N^{2s-\frac{k(k+1)}{2}}\right).$ This upper bound is sharp up to the factor $C(ϵ,k)N^{ϵ}$.

Equivalently, we can write $J_{s,k}(N)$ as an integral $J_{s,k}(N)={\int }_{[0,1{]}^k}|f(x{)}^{2s}|dx$ where $f(x)$ has a nice Fourier series.

Sketch of Decoupling idea: Consider $k=2,s=3$, for $x=(x_1,x_2)\in {\mathbb{R}}^2$, define $f(x)={\sum }_{n=1}^{N}e(\frac{n}{N}{x}_1+\frac{n^2}{N^2}{x}_2)$, the frequency $(\frac{n}{N},\frac{n^2}{N^2})$ lie on a parabola. We write $Q_s(x)$ for a square of side length $S$ centered at $x$. To get sharp estimate, we need to bound ${\int }_{Q_{N^2}(0)}|f(x){|}^{6}dx$. $f(0)=N$ ${\int }_{Q_N}|f(x)|{}^{2}dx={\int }_{Q_N}\left({\sum }_{n=1}^{N}e\left(\frac{n}{N}{x}_1+\frac{n^2}{N^2}{x}_2\right)\right)\left({\sum }_{m=1}^{N}e\left(-\frac{m}{N}{x}_1-\frac{m^2}{N^2}{x}_2\right)\right)dx$ $={\sum }_{n=1}^N{\sum }_{m=1}^N{\int }_{Q_N}e\left(\frac{n}{N}{x}_1+\frac{n^2}{N^2}{x}_2-\frac{m}{N}{x}_1-\frac{m^2}{N^2}{x}_2\right)$ and because $f(x)$ is $N$-periodic in $x_1$ variable, we get ${\int }_{Q_N}e\left(\frac{n}{N}{x}_1+\frac{n^2}{N^2}{x}_2-\frac{m}{N}{x}_1-\frac{m^2}{N^2}{x}_2\right)dx=\{\begin{array}{ll}|Q_N|& \text{if }m=n\\ 0& \text{if }m\ne n\end{array},$ so ${\int }_{Q_N}|f(x)|{}^{2}dx=N\cdot |Q_N|$ Let $S=\{x:|f(x)|\ge 10\sqrt{N}\}$, we get $N\cdot |Q_N|={\int }_{Q_N}|f(x)|{}^{2}dx\ge {\int }_S|f(x){|}^2\ge |S|100N,$ so $|S|\le \frac{1}{100}|Q_N|$.

So because of orthogonality, $|f(x)|\le 10\sqrt{N}$ for most points $x$. $|f(x)|$ is roughly constant on each unit square.

Let $U_{\lambda }(f):=\{x:|f(x)|>\lambda \}$. $|U|=$ measure of $U$.

We claim $|U_{\frac{N}{10}}(f)\cap Q_{N^2}|\le C_{ϵ}{N}^{1+ϵ}$

Tool 1: Orthogonality

The functions $\{e(\frac{n}{N}{x}_1+\frac{n^2}{N^2}{x}_2){\}}_{n=1}^N$ are orthogonal on each $Q_N$. ${\int }_{Q_N}|f(x)|{}^{2}dx=N\cdot |Q_N|=N\cdot N^2=N^3\ge {\int }_{|U_{\frac{N}{10}}(f)\cap Q_N|}(\frac{N}{10}{)}^2=|U_{\frac{N}{10}}(f)\cap Q_N|\frac{N^2}{100}$ So $|U_{\frac{N}{10}}(f)\cap Q_N|\le CN$ for any $x$, so $|U_{\frac{N}{10}}(f)\cap Q_{N^2}(0)|\le C{N}^3$.

Tool 2: Pieces of the sum.

If $I\subset \{1,...,N\},$ let $f_I={\sum }_{n\in I}e(\frac{n}{N}{x}_1+\frac{n^2}{N^2}{x}_2)$. We can partition $\{1,...,N\}$ into intervals $I$ of length $L$ and then $f(x)={\sum }_{I\text{of length}L}{f}_I(x).$

Lemma: If $x\in U_{\frac{N}{10}}(f)$, then $x\in U_{\frac{L}{20}}(f_I)$ for most I. Let $K=\{I:|f_I(x)|>\frac{L}{20}\}$ $|f(x)|=|{\sum }_{I}f(x)|\le {\sum }_I|f_I(x)|={\sum }_{I\in K}|f_I(x)|+{\sum }_{I\notin K}|f_I|\le |K|\cdot L+\frac{L}{20}(\frac{N}{L}-|K|)$ for $x\in U_{\frac{N}{10}}(f)$, $\frac{N}{10}\le \frac{19}{20}|K|L+\frac{N}{20},⟹|K|>\frac{N}{19L}.$

Tool 3: The shape of $f_I$.

The set of frequencies $\{\left(\frac{n}{N},\frac{n^2}{N^2}\right){\}}_{n\in I}$ lie in a small box. Theorem. (Shannon-Nyquist) If $\hat{g}$ supported in $[-\frac{1}{2},\frac{1}{2}]$, then $g$ can be recovered from $g(n)$ for $n\in \mathbb{Z}$. If $\hat{g}$ supported in $[-\frac{1}{2},\frac{1}{2}]$, then $g$ is roughly constant on each interval of length 1. $|f_I|$ roughly constant on each rectangular tile of a tiling that is “dual to I” So $U_{\frac{L}{20}}(f_I)$ is a union of these dual rectangles.

Section 4 of survey Larry Guth

Tool 4: Transversality.

For different I, the rectangles are oriented in different directions.

This is where Kakeya Problem and Harmonic Analysis intersects,