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Let the Bravais lattice have primitive vectors $a_1,a_2,a_3$, and putThe reciprocal lattice has the basisThe scalar triple product and cross product identities giveConsequently every $q={\sum }_j{n}_j{b}_j$ satisfies $q\cdot l\in 2\pi \mathbb{Z}$ for every lattice point $l={\sum }_i{m}_i{a}_i$, and conversely these three conditions force the coefficients of $q$ in the reciprocal basis to be integers. This proves that the displayed vectors generate ${\Lambda }^{\ast }$.

In the Born approximation, suppose that the crystal potential is the sum of translates of one atomic potential $V_0$ over the finite set $S$ of lattice sites. Its Fourier transform at the momentum transfer $Q=k-k^{\prime }$ factors asThus the single-atom scattering amplitude is multiplied by the crystal lattice structure factor $\Delta (Q)$. Writingseparates the sum into three finite geometric series. For $l_i=-L_i/2,\dots ,L_i/2$ this givesWhen the $L_i$ are large, a factor has a sharp maximum when ${\alpha }_i=2\pi n_i$. All three factors are therefore simultaneously large precisely when $Q\in {\Lambda }^{\ast }$. These are the reciprocal-lattice peaks of a finite crystal.

For the stated Body-centered cubic lattice basis,the reciprocal-basis formula givesEquivalently,The shortest nonzero reciprocal vectors have squared Euclidean norm $2(2\pi /a{)}^2$, so

In elastic scattering, $|k|=|k^{\prime }|=k$. If $q=k-k^{\prime }\in {\Lambda }^{\ast }$ and $\theta$ is the angle between $k$ and $k^{\prime }$, Euclidean geometry givesThe first possible diffraction peak therefore occurs atas $ka\to \infty$, by the small-angle approximation. This is the elastic Bragg scattering condition for the shortest reciprocal-lattice vector.

The $[SU(3){]}^{2}U(1)$ condition gives $u+d=2q$, which is even because all hypercharges are integers. The sum $u+d$ and difference $u-d$ have the same parity, so $u-d$ is even. ThereforeSolving the sum and difference equations gives

Consider the $N+1$ fractional partsin $[0,1)$. Divide that interval into $N$ intervals of length $1/N$. By the pigeonhole principle, two fractional parts, say those indexed by $i<j$, lie in the same interval. Thus, for some integer $p$,Set $q=j-i$. Then $1\le q\le N$, and division by $q$ gives the Dirichlet approximation theorem

Choose a number field $L$ containing ${\alpha }_1,\dots ,{\alpha }_k$. At every place $v$ of $L$, putThe triangle inequality giveswhere $c_v=1$ at every non-Archimedean place because the coefficients are integers, while at an Archimedean place one may takeRaise these inequalities to the local weights and multiply over all places. The definition of the Absolute multiplicative Weil height and the product formula then giveThis is the height bound for a polynomial evaluation.

Let $f$ be the monic cubic minimal polynomial of the algebraic integer $\alpha$, let ${\alpha }_1=\alpha ,{\alpha }_2,{\alpha }_3$ be its conjugates, and let ${\beta }_1=\beta ,{\beta }_2$ be the two conjugates of a nonrational $\beta \in {\mathcal{O}}_K$. Because the degrees three and two are coprime, the fields are linearly disjoint, andis a nonzero integer.

AssumeFor large $H(\beta )$ this makes ${\beta }_1$ bounded. Since $\beta$ is a quadratic algebraic integer, the height-Mahler measure formula givesand therefore $|{\beta }_2|\asymp H(\beta {)}^2$. The two remaining factors with $j=1$ are bounded, while the three factors with $j=2$ are $O(|{\beta }_2|)$. Consequently

We now use the standard effective norm form consequence of Part (b). Forwhere $1,\omega$ is an integral basis of $K$, the effective norm-form height estimate supplies effective constants $A,C>0$, depending only on $\alpha$ and $K$, such thatfor every nonzero $z\in V\cap {\mathcal{O}}_L$. Its proof factors $(z)$, balances a generator using the Dirichlet unit theorem, and applies the Baker lower bound for a homogeneous linear form in logarithms to the linear relations defining $V$; the coprime degrees $3$ and $2$ exclude a unit-family degeneracy.

Apply this estimate to $z=\alpha -\beta$. Since $\beta =\alpha -z$, the height inequalities imply $H(\beta )\le 2H(\alpha )H(z)$. HenceChoose the effective value $\epsilon =1/(2A)$. The last inequality bounds $H(\beta )$ effectively. The rational integers $\beta \in \mathbb{Z}$ are already covered by the stronger degree-three Liouville approximation theorem, and the Northcott theorem leaves only finitely many remaining quadratic integers of bounded height. Taking the minimum ofover this effective finite set gives an effective $c>0$ and provesfor every $\beta \in {\mathcal{O}}_K$.

If a periodic unimodular function $a$ and its inverse have absolutely summable Fourier coefficients, replacing ${\hat{\phi }}_0$ by $\hat{\phi }=a{\hat{\phi }}_0$ preserves the orthonormal integer-translate span and integrability of an integrable scaling function. In physical space this is an absolutely summable combination of integer function translations. The new MRA low-pass filter is $m(t)=a(2t)m_0(t)/a(t)$. A rough phase can therefore preserve the multiresolution analysis while changing MRA low-pass filter regularity.

A positive integer is an integer greater than zero.

A prime factor of a nonzero integer is a prime number that divides it.

Positive integers are equivalent when they have the same finite set of prime factors. Equivalently, each divides a positive power of the other. This equivalence relation is induced by the radical of an integer; its equivalence classes correspond to finite prime supports. The empty support gives the singleton $\{1\}$, and every nonempty support gives an infinite class by varying exponents.

A Pythagorean triple consists of three positive integers satisfying $x^2+y^2=z^2$. It is primitive when their greatest common divisor is one. Reducing their square numbers modulo $3$ shows that at least one leg is divisible by $3$; reducing modulo $4$ shows that both legs cannot be odd.

The radical of a positive integer is the product of its distinct prime factors, with $\mathrm{rad}(1)=1$. It is the squarefree integer recording which primes occur, independently of their exponents. Equality of radicals is equivalent to mutual divisibility by sufficiently high positive powers.

A rational matrix is a matrix whose entries are rational numbers. Multiplying by a common denominator produces an integer matrix and preserves its homogeneous solution set.

A reduced fraction $a/b$ has coprime integers $a,b$ and positive denominator $b$. Distinct reduced fractions with denominators at most $Q$ differ by at least $Q^{-2}$, since their difference has a nonzero integer numerator over a denominator at most $Q^2$.

The residue class $[a{]}_m$ consists of all integers congruent to $a$ modulo $m$.

A semistandard Young tableau fills a Young diagram with positive integers, permitting repeated entries, so that entries weakly increase along rows and strictly increase down columns. Its content records the number of occurrences of each entry. The Kostka number $K_{\lambda \mu }$ counts such fillings of shape $\lambda$ and content $\mu$.

For every matrix $A$ over a principal ideal domain, there are invertible matrices $U,V$ such thatThe nonzero diagonal entries are unique up to multiplication by units. Over the integers, $U$ and $V$ are unimodular and the $d_i$ may be chosen positive; their products $d_1\cdots d_k$ are the greatest common divisors of the $k$-rowed minors.

A squarefree divisor of an integer $n$ is a divisor $d\mid n$ that is a squarefree integer. It is specified by choosing a subset of the distinct prime factors of $n$.

A square number is an integer of the form $n^2$ for some integer $n$. Consecutive nonnegative squares differ by $(n+1{)}^2-n^2=2n+1$.

For a finite set $S$ of primes, an integer is $S$-smooth when every prime dividing it belongs to $S$. Bounding the largest prime factor is equivalent to restricting the integer to be $S$-smooth for a suitable finite $S$.

For coefficients $b_n$ supported on an interval of $H$ consecutive integers, set $B=\sum b_n$ and $B_p(a)={\sum }_{n\equiv a(\mathrm{mod}p)}{b}_n$. ThenThe orthogonality of roots of unity gives $p{\sum }_a|B_p(a)-B/p{|}^2={\sum }_{r=1}^{p-1}|{\sum }_n{b}_{n}e(rn/p){|}^2$. The distinct fractions $r/p$ have circular spacing at least $Q^{-2}$; now apply the exponential-sum large sieve.