Past exam of the mathematics course of the University of Cambridge / 2009 / iii / Paper 7 / 2

The Minkowski convex body theorem states that a centrally symmetric convex measurable set in ${\mathbb{R}}^d$ of volume greater than $2^d\mathrm{covol}(\Lambda )$ contains a nonzero point of the full lattice $\Lambda$. For a compact convex body, the weak inequality also suffices: apply the strict version to $(1+\epsilon )C$, let $\epsilon \downarrow 0$, and choose a constant subsequence of the resulting lattice points in the bounded set $2C$. Its nonzero value belongs to $C$ by closedness.

For $Q>1$, apply this borderline version with $\Lambda ={\mathbb{Z}}^2$ toThe linear transformation $(u,v)\mapsto (u,xu-v)$ has determinant of absolute value one, so this parallelogram has area four. A nonzero lattice point $(n,m)$ satisfies $|n|\le Q$ and $|nx-m|\le Q^{-1}$. If $n=0$, then $|m|<1$, giving the forbidden zero point; hence $n\ne 0$. It follows thatIf $x$ is irrational, these pairs cannot remain in a finite set as $Q\to \infty$: the positive errors $|nx-m|$ of any such finite set have a positive minimum, contradicting their bound $Q^{-1}$. If $x=p/q$ is rational in lowest terms with $q>0$, the pairs $(n,m)=(jq,jp)$ for $j\ge 1$ have zero error and are infinitely many distinct pairs. This proves the assertion for every real $x$.

In this part identify $\mathbb{T}$ with the complex unit circle by $t\mapsto e^{it}$. Write the finite abelian group $G$ additively and let $\hat{G}=\mathrm{Hom}(G,\mathbb{T})$, with pointwise multiplication, be the character group of a finite abelian group of $G$. A linear character has modulus one, since its values have finite order. We first prove the extension and counting facts needed for Fourier analysis on a finite abelian group.

If $H\le G$, $\chi \in \hat{H}$, and $x\notin H$, let $d$ be the least positive integer with $dx\in H$. Choose $z\in \mathbb{T}$ with $z^d=\chi (dx)$ and defineThis is well-defined: two representations differ by $j-j^{\prime }=qd$ and $h-h^{\prime }=-qdx$, whose contributions cancel. It is a group homomorphism and extends $\chi$. Conversely every extension has this form, and the equation for $z$ has exactly $d$ distinct roots. Since $[H+⟨x⟩:H]=d$, successive adjunctions prove that each linear character of $H$ has exactly $[G:H]$ extensions. Taking $H=\{0\}$ gives $\boxed{|\hat{G}|=|G|}$. This is the extension of a character across a cyclic quotient, proved here without a structure theorem. Extending the linear character sending a nonzero element $g$ of order $d$ to $e^{2\pi i/d}$ also proves that linear characters separate points of $G$.

Use the normalized inner product $⟨f,g⟩=|G{|}^{-1}{\sum }_{x\in G}f(x)\overline{g(x)}$. For a nontrivial linear character $\chi$, choose $a$ with $\chi (a)\ne 1$. Translation of the sum gives ${\sum }_x\chi (x)=\chi (a){\sum }_x\chi (x)$, so that sum is zero. Applying this to $\chi \overline{\psi }$ givesThere are $|G|$ such orthonormal vectors in the $|G|$-dimensional space ${\mathbb{C}}^G$, so they form an orthonormal basis. Define the Fourier coefficients by $\hat{f}(\chi )=|G{|}^{-1}{\sum }_{x}f(x)\overline{\chi (x)}$. Expanding in that basis proves Fourier inversion on a finite group and the Parseval identity on a finite group:For normalized convolution $(f\ast g)(x)=|G{|}^{-1}{\sum }_{y}f(y)g(x-y)$, changing the finite summation variables yields $\widehat{f\ast g}(\chi )=\hat{f}(\chi )\hat{g}(\chi )$. Translation $f(x)\mapsto f(x+a)$ multiplies its Fourier coefficient by $\chi (a)$. These formulas fix all normalization and conjugation conventions.

The PDF's double-hat identification is canonical Pontryagin duality: define $e:G\to \hat{\hat{G}}$ by $e_x(\chi )=\chi (x)$. This is a group homomorphism and is injective because linear characters separate points. Applying the already proved character-counting result to $\hat{G}$ shows that both groups have size $|G|$, soThe TeX loses one hat. For completeness, the single-dual isomorphism $\hat{G}\cong G$ also holds, but is generally noncanonical. Here is a proof including the needed finite cyclic decomposition. Choose an element $a$ of maximal order $d$. Every element order divides $d$: for two elements, extract their prime-power components by integer multiples: if an element has order $d$ and $p^{\alpha }$ is the exact power of $p$ in $d$, multiplying it by $d/p^{\alpha }$ gives order $p^{\alpha }$. Then combine the components with the larger order for each prime. Components of coprime orders have a sum of product order, because their cyclic subgroups have trivial intersection. This constructs an element of order the least common multiple of the two original orders, which cannot exceed maximal $d$; hence each order divides $d$.

Extend the primitive linear character of $⟨a⟩$ to $\chi :G\to \mathbb{T}$. All its values are $d$th roots because all element orders divide $d$. Define $\rho (g)=ja$ when $\chi (g)=e^{2\pi ij/d}$. This is a group homomorphism fixing $⟨a⟩$, and $g=\rho (g)+(g-\rho (g))$ shows $G=⟨a⟩\oplus \ker \rho$. Induction on group size decomposes $G$ into finite cyclic groups. A cyclic group of order $d_j$ has precisely the linear characters $j\mapsto e^{2\pi imj/d_j}$ and hence a character group of a finite abelian group isomorphic to itself. The dual of a finite direct product of groups is the product of its factor duals, since a linear character is determined by its restrictions. This proves $\hat{G}\cong G$ after choices of cyclic factors and generators, without using an unproved finite-abelian-group structure theorem.