Minkowski's theorem

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https://en.wikipedia.org/wiki/Minkowski%27s_theorem

 

A set in ℝ² satisfying the hypotheses of Minkowski's theorem.

In mathematics, Minkowski's theorem is the statement that every convex set in ${\displaystyle {\mathbb{R}}^n}$ which is symmetric with respect to the origin and which has volume greater than ${\displaystyle 2^n}$ contains a non-zero integer point (meaning a point in ${\displaystyle {\mathbb{Z}}^n}$ that is not the origin). The theorem was proved by Hermann Minkowski in 1889 and became the foundation of the branch of number theory called the geometry of numbers. It can be extended from the integers to any lattice ${\displaystyle L}$ and to any symmetric convex set with volume greater than ${\displaystyle 2^{n}d(L)}$, where ${\displaystyle d(L)}$ denotes the covolume of the lattice (the absolute value of the determinant of any of its bases).

Formulation

Suppose that L is a lattice of determinant d(L) in the n-dimensional real vector space ${\displaystyle {\mathbb{R}}^n}$ and S is a convex subset of ${\displaystyle {\mathbb{R}}^n}$ that is symmetric with respect to the origin, meaning that if x is in S then −x is also in S. Minkowski's theorem states that if the volume of S is strictly greater than 2ⁿ d(L), then S must contain at least one lattice point other than the origin. (Since the set S is symmetric, it would then contain at least three lattice points: the origin 0 and a pair of points ± x, where x ∈ L \ 0.)

Example

The simplest example of a lattice is the integer lattice ${\displaystyle {\mathbb{Z}}^n}$ of all points with integer coefficients; its determinant is 1. For n = 2, the theorem claims that a convex figure in the Euclidean plane symmetric about the origin and with area greater than 4 encloses at least one lattice point in addition to the origin. The area bound is sharp: if S is the interior of the square with vertices (±1, ±1) then S is symmetric and convex, and has area 4, but the only lattice point it contains is the origin. This example, showing that the bound of the theorem is sharp, generalizes to hypercubes in every dimension n.

Proof

The following argument proves Minkowski's theorem for the specific case of ${\displaystyle L={\mathbb{Z}}^2.}$

Proof of the ${\mathbb{Z}}^2$ case: Consider the map

${\displaystyle f:S\to {\mathbb{R}}^2/2L,(x,y)\mapsto (xmod2,ymod2)}$

Intuitively, this map cuts the plane into 2 by 2 squares, then stacks the squares on top of each other. Clearly f (S) has area less than or equal to 4, because this set lies within a 2 by 2 square. Assume for a contradiction that f could be injective, which means the pieces of S cut out by the squares stack up in a non-overlapping way. Because f is locally area-preserving, this non-overlapping property would make it area-preserving for all of S, so the area of f (S) would be the same as that of S, which is greater than 4. That is not the case, so the assumption must be false: f is not injective, meaning that there exist at least two distinct points p₁, p₂ in S that are mapped by f to the same point: f (p₁) = f (p₂).

Because of the way f was defined, the only way that f (p₁) can equal f (p₂) is for p₂ to equal p₁ + (2i, 2j) for some integers i and j, not both zero. That is, the coordinates of the two points differ by two even integers. Since S is symmetric about the origin, −p₁ is also a point in S. Since S is convex, the line segment between −p₁ and p₂ lies entirely in S, and in particular the midpoint of that segment lies in S. In other words,

${\displaystyle \frac{1}{2}\left(-p_1+p_2\right)=\frac{1}{2}\left(-p_1+p_1+(2i,2j)\right)=(i,j)}$

is a point in S. But this point (i, j) is an integer point, and is not the origin since i and j are not both zero. Therefore, S contains a nonzero integer point.

Remarks:

• The argument above proves the theorem that any set of volume $>det(L)$ contains two distinct points that differ by a lattice vector. This is a special case of Blichfeldt's theorem.
• The argument above highlights that the term $2^{n}det(L)$ is the covolume of the lattice $2L$.
• To obtain a proof for general lattices, it suffices to prove Minkowski's theorem only for ${\mathbb{Z}}^n$; this is because every full-rank lattice can be written as $B{\mathbb{Z}}^n$ for some linear transformation $B$, and the properties of being convex and symmetric about the origin are preserved by linear transformations, while the covolume of $B{\mathbb{Z}}^n$ is $|det(B)|$ and volume of a body scales by exactly $\frac{1}{det(B)}$ under an application of $B^{-1}$.

Applications

Bounding the shortest vector

Minkowski's theorem gives an upper bound for the length of the shortest nonzero vector. This result has applications in lattice cryptography and number theory.

Theorem (Minkowski's bound on the shortest vector): Let $L$ be a lattice. Then there is a $x\in L\setminus \{0\}$ with $\Vert x{\Vert }_{\mathrm{\infty }}\le {\left|det(L)\right|}^{1/n}$. In particular, by the standard comparison between $l_2$ and $l_{\mathrm{\infty }}$ norms, $\Vert x{\Vert }_2\le \sqrt{n}{\left|det(L)\right|}^{1/n}$.

Proof

Let $l=min\{\Vert x{\Vert }_{\mathrm{\infty }}:x\in L\setminus \{0\}\}$, and set $C=\{y:\Vert y{\Vert }_{\mathrm{\infty }}<l\}$. Then $\text{vol}(C)=(2l{)}^n$. If $(2l{)}^n>2^n|d(L)|$, then $C$ contains a non-zero lattice point, which is a contradiction. Thus $l\le |d(L){|}^{1/n}$. Q.E.D.

Remarks:

• The constant in the $L^2$ bound can be improved, for instance by taking the open ball of radius $<l$ as $C$ in the above argument. The optimal constant is known as the Hermite constant.
• The bound given by the theorem can be very loose, as can be seen by considering the lattice generated by $(1,0),(0,n)$. But it cannot be further improved in the sense that there exists a global constant ${\displaystyle c}$ such that there exists an ${\displaystyle n}$-dimensional lattice ${\displaystyle L}$ satisfying ${\displaystyle \Vert x{\Vert }_2\ge c\sqrt{n}\cdot {\left|det(L)\right|}^{1/n}}$for all ${\displaystyle x\in L\setminus \{0\}}$. Furthermore, such lattice can be self-dual. 
• Even though Minkowski's theorem guarantees a short lattice vector within a certain magnitude bound, finding this vector is in general a hard computational problem. Finding the vector within a factor guaranteed by Minkowski's bound is referred to as Minkowski's Vector Problem (MVP), and it is known that approximation SVP reduces to it using transference properties of the dual lattice. The computational problem is also sometimes referred to as HermiteSVP.
• The LLL-basis reduction algorithm can be seen as a weak but efficiently algorithmic version of Minkowski's bound on the shortest vector. This is because a $\delta$-LLL reduced basis $b_1,\dots ,b_n$ for $L$ has the property that $\Vert b_1\Vert \le {\left(\frac{1}{\delta -.25}\right)}^{\frac{n-1}{4}}det(L{)}^{1/n}$; see these lecture notes of Micciancio for more on this. As explained in, proofs of bounds on the Hermite constant contain some of the key ideas in the LLL-reduction algorithm.

Applications to number theory

Primes that are sums of two squares

The difficult implication in Fermat's theorem on sums of two squares can be proven using Minkowski's bound on the shortest vector.

Theorem: Every prime with $p\equiv 1\mathrm{mod}4$ can be written as a sum of two squares.

Proof

Since $4|p-1$ and $a$ is a quadratic residue modulo a prime $p$ if and only if $a^{\frac{p-1}{2}}=1(\text{mod}p)$ (Euler's Criterion) there is a square root of $-1$ in $\mathbb{Z}/p\mathbb{Z}$; choose one and call one representative in $\mathbb{Z}$ for it $j$. Consider the lattice $L$ defined by the vectors $(1,j),(0,p)$, and let $B$ denote the associated matrix. The determinant of this lattice is $p$, whence Minkowski's bound tells us that there is a nonzero $x=(x_1,x_2)\in {\mathbb{Z}}^2$ with $0<\Vert Bx{\Vert }_2^2<2p$. We have $\Vert Bx{\Vert }^2=\Vert (x_1,j{x}_1+p{x}_2){\Vert }^2=x_1^2+(j{x}_1+p{x}_2{)}^2$ and we define the integers $a=x_1,b=(j{x}_1+p{x}_2)$. Minkowski's bound tells us that $0<a^2+b^2<2p$, and simple modular arithmetic shows that $a^2+b^2=x_1^2+(j{x}_1+p{x}_2{)}^2=0\mathrm{mod}p$, and thus we conclude that $a^2+b^2=p$. Q.E.D.

Reciprocal polynomial

From Wikipedia, the free encyclopedia
https://en.wikipedia.org/wiki/Reciprocal_polynomial

In algebra, given a polynomial

${\displaystyle p(x)=a_0+a_{1}x+a_2{x}^2+\cdots +a_n{x}^n,}$

with coefficients from an arbitrary field, its reciprocal polynomial or reflected polynomial, denoted by p^∗ or p^Ris the polynomial

${\displaystyle p^{\ast }(x)=a_n+a_{n-1}x+\cdots +a_0{x}^n=x^{n}p(x^{-1}).}$

That is, the coefficients of p^∗ are the coefficients of p in reverse order. Reciprocal polynomials arise naturally in linear algebra as the characteristic polynomial of the inverse of a matrix.

In the special case where the field is the complex numbers, when

${\displaystyle p(z)=a_0+a_{1}z+a_2{z}^2+\cdots +a_n{z}^n,}$

the conjugate reciprocal polynomial, denoted p^†, is defined by,

${\displaystyle p^{†}(z)=\overline{a_n}+\overline{a_{n-1}}z+\cdots +\overline{a_0}{z}^n=z^n\overline{p({\overline{z}}^{-1})},}$

where ${\displaystyle \overline{a_i}}$ denotes the complex conjugate of ${\displaystyle a_i}$, and is also called the reciprocal polynomial when no confusion can arise.

A polynomial p is called self-reciprocal or palindromic if p(x) = p^∗(x). The coefficients of a self-reciprocal polynomial satisfy a_i = a_n−i for all i.

Properties

Reciprocal polynomials have several connections with their original polynomials, including:

1. deg p = deg p^∗ if ${\displaystyle a_0}$ is not 0.
2. p(x) = xⁿp^∗(x⁻¹).
3. α is a root of a polynomial p if and only if α⁻¹ is a root of p^∗.
4. If p(x) ≠ x then p is irreducible if and only if p^∗ is irreducible.
5. p is primitive if and only if p^∗ is primitive.

Other properties of reciprocal polynomials may be obtained, for instance:

• A self-reciprocal polynomial of odd degree is divisible by x+1, hence is not irreducible if its degree is > 1.

Palindromic and antipalindromic polynomials

A self-reciprocal polynomial is also called palindromic because its coefficients, when the polynomial is written in the order of ascending or descending powers, form a palindrome. That is, if

${\displaystyle P(x)=\sum _{i=0}^n{a}_i{x}^i}$

is a polynomial of degree n, then P is palindromic if a_i = a_n−i for i = 0, 1, ..., n.

Similarly, a polynomial P of degree n is called antipalindromic if a_i = −a_n−i for i = 0, 1, ..., n. That is, a polynomial P is antipalindromic if P(x) = –P^∗(x).

Examples

From the properties of the binomial coefficients, it follows that the polynomials P(x) = (x + 1)ⁿ are palindromic for all positive integers n, while the polynomials Q(x) = (x – 1)ⁿ are palindromic when n is even and antipalindromic when n is odd.

Other examples of palindromic polynomials include cyclotomic polynomials and Eulerian polynomials.

Properties

• If a is a root of a polynomial that is either palindromic or antipalindromic, then 1/a is also a root and has the same multiplicity.
• The converse is true: If a polynomial is such that a is a root then if 1/a is also a root of the same multiplicity, then the polynomial is either palindromic or antipalindromic.
• For any polynomial q, the polynomial q + q^∗ is palindromic and the polynomial q − q^∗ is antipalindromic.
• It follows that any polynomial q can be written as the sum of a palindromic and an antipalindromic polynomial, since q = (q + q^∗)/2 + (q − q^∗)/2.
• The product of two palindromic or antipalindromic polynomials is palindromic.
• The product of a palindromic polynomial and an antipalindromic polynomial is antipalindromic.
• A palindromic polynomial of odd degree is a multiple of x + 1 (it has –1 as a root) and its quotient by x + 1 is also palindromic.
• An antipalindromic polynomial over a field k with odd characteristic is a multiple of x – 1 (it has 1 as a root) and its quotient by x – 1 is palindromic.
• An antipalindromic polynomial of even degree is a multiple of x² – 1 (it has −1 and 1 as roots) and its quotient by x² – 1 is palindromic.
• If p(x) is a palindromic polynomial of even degree 2d, then there is a polynomial q of degree d such that p(x) = x^dq(x + 1/x).
• If p(x) is a monic antipalindromic polynomial of even degree 2d over a field k of odd characteristic, then it can be written uniquely as p(x) = x^d(Q(x) − Q(1/x)), where Q is a monic polynomial of degree d with no constant term.
• If an antipalindromic polynomial P has even degree 2n over a field k of odd characteristic, then its "middle" coefficient (of power n) is 0 since a_n = −a_2n – n.

Real coefficients

A polynomial with real coefficients all of whose complex roots lie on the unit circle in the complex plane (that is, all the roots have modulus 1) is either palindromic or antipalindromic.

Conjugate reciprocal polynomials

A polynomial is conjugate reciprocal if ${\displaystyle p(x)\equiv p^{†}(x)}$ and self-inversive if ${\displaystyle p(x)=\omega p^{†}(x)}$ for a scale factor ω on the unit circle.

If p(z) is the minimal polynomial of z₀ with |z₀| = 1, z₀ ≠ 1, and p(z) has real coefficients, then p(z) is self-reciprocal. This follows because

${\displaystyle z_0^n\overline{p(1/\overline{z_0})}=z_0^n\overline{p(z_0)}=z_0^n\overline{0}=0.}$

So z₀ is a root of the polynomial ${\displaystyle z^n\overline{p({\overline{z}}^{-1})}}$ which has degree n. But, the minimal polynomial is unique, hence

${\displaystyle cp(z)=z^n\overline{p({\overline{z}}^{-1})}}$

for some constant c, i.e. ${\displaystyle c{a}_i=\overline{a_{n-i}}=a_{n-i}}$. Sum from i = 0 to n and note that 1 is not a root of p. We conclude that c = 1.

A consequence is that the cyclotomic polynomials Φ_n are self-reciprocal for n > 1. This is used in the special number field sieve to allow numbers of the form x¹¹ ± 1, x¹³ ± 1, x¹⁵ ± 1 and x²¹ ± 1 to be factored taking advantage of the algebraic factors by using polynomials of degree 5, 6, 4 and 6 respectively – note that φ (Euler's totient function) of the exponents are 10, 12, 8 and 12.

Per Cohn's theorem, a self-inversive polynomial has as many roots in the unit disk ${\displaystyle \{z\in \mathbb{C}:|z|<1\}}$ as the reciprocal polynomial of its derivative.

Application in coding theory

The reciprocal polynomial finds a use in the theory of cyclic error correcting codes. Suppose xⁿ − 1 can be factored into the product of two polynomials, say xⁿ − 1 = g(x)p(x). When g(x) generates a cyclic code C, then the reciprocal polynomial p^∗ generates C^⊥, the orthogonal complement of C. Also, C is self-orthogonal (that is, C ⊆ C^⊥), if and only if p^∗ divides g(x).