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Use the usual strictly increasing spline knot sequence, and initially assume $k\ge 2$. The explicit formula for the order-$k$ divided difference givesEach summand is a truncated power function in $t$, with its junction at $t_j$. It is a piecewise polynomial function of degree at most $k-1$ and has $k-2$ continuous derivatives. Finite summation preserves these properties.

For $t\ge t_{i+k}$ all summands vanish. For $t\le t_i$, all spline knot data come from the polynomial $u\mapsto (u-t{)}^{k-1}$. An order-$k$ divided difference annihilates a polynomial of smaller degree, so $M_i(t)=0$ there too. Thus the support is contained in $[t_i,t_{i+k}]$.

On the last open spline knot interval only the $j=i+k$ term survives; it is a nonzero multiple of $(t_{i+k}-t{)}^{k-1}$. On the first open spline knot interval, compare the truncated spline knot data with the untruncated polynomial data: only the $j=i$ value differs. Since the untruncated divided difference is zero, the result is a nonzero multiple of $(t_i-t{)}^{k-1}$. Consequently both outer pieces are nonzero, giving exact closed support. The Cox-de Boor recurrence used below also proves positivity throughout the open support. ThereforeThis is simple-knot B-spline regularity. For $k=1$ the B-splines are interval indicators; they are piecewise constant and need not be continuous. The notation $C^{k-2}$ should then be read as no continuity requirement, not as a classical negative-order differentiability space.

The divided difference is a finite linear combination of spline knot values, so it commutes with the finite integral. For every spline knot $t_j$ in the defining interval,It follows thatbecause the order-$k$ divided difference of a monic degree-$k$ polynomial is its leading coefficient. This proves the unit-integral normalization of a B-spline.

For the partition of unity, let $N_{i,r}$ denote the partition-normalized order-$r$ B-spline. The Cox-de Boor recurrence isIt follows by substituting the explicit divided difference formula into both sides and collecting each truncated power function. The starting case is $N_{i,1}=1_{[t_i,t_{i+1})}$, with endpoint conventions irrelevant when $r\ge 2$. The two weights are nonnegative wherever their respective lower-order B-splines are nonzero. Induction therefore gives $N_{i,r}\ge 0$ and strict positivity in its open support.

Extend the finite spline knot sequence strictly in both directions, with no finite accumulation. At each point only finitely many B-splines are nonzero. For the full extended family the sum is one for order one. Summing the Cox-de Boor recurrence, the combined coefficient of $N_{j,r-1}$ isInduction gives a full partition of unity at every order. On $t_k<t<t_{n+1}$, all extended order-$k$ functions with index $i\le 0$ or $i\ge n+1$ vanish by their supports. HenceThe same argument gives the useful global subpartition of unity for B-splines, $0\le {\sum }_{i=1}^n{N}_i(t)\le 1$, including outside the basic spline knot interval.

A regular perturbation admits an asymptotic expansion about the limiting problem on the relevant scale without losing its solutions or conditions. A singular perturbation changes the limiting structure, so some solutions require a different scale: a differential equation may lose an order, or a polynomial may lose its degree.

The two finite roots approach $s=\pm \sqrt{2}$. Set $x=s+ϵa+O({ϵ}^2)$. Substitution into the polynomial gives $s^2-2=0$ at leading order and $s^3+2sa=0$ at the next order. Since $s^2=2$, $a=-1$ for either sign. Each branch is a regular perturbation of a simple algebraic root, withThe limiting quadratic has only two roots, whereas the original equation has three. The missing root is a divergent algebraic root in a singular perturbation. The dominant balance for algebraic roots sets $x={ϵ}^{-1}X$, giving $X^3+X^2=2{ϵ}^2$. Its nonzero leading root is $X=-1$. Put $X=-1+a{ϵ}^2+O({ϵ}^4)$; the left side is $a{ϵ}^2+O({ϵ}^4)$, so $a=2$. Hence the second nonzero term occurs at order $ϵ$, not order one:For small nonzero real $ϵ$ these three branches are real and distinct; the divergent branch changes sign with $ϵ$. The two bounded branches are regular even though the full root problem is singular.

We give a quantitative construction for the whole open set, including the case of infinitely many component intervals. Pointwise divergence at a single point is insufficient here. Write $z=e^{2\pi it}$ and use normalized measure on $\mathbb{T}$.

First let $K$ be a finite union of closed arcs with $|K|\le 3\epsilon$, where $\epsilon$ is small. Choose a smooth function nonnegative function $u$, equal to one on $K$, with $u\le 1$ and $\eta :=\int u\le 6\epsilon$. This follows by making the transition regions around the finitely many arc endpoints sufficiently short. Define the analytic Herglotz integralIts real part is $(\eta +P[u](z))/(2\eta )>0$, where $P[u]$ is the disk Poisson integral, and $G(0)=1$. Smoothness of $u$ makes $G$ continuous and smooth function up to the circle; on $K$ its boundary real part is at least $1/(2\eta )$. The branch $H=\log G$ therefore has $H(0)=0$ andThe complex logarithm has no branch problem because $G$ lies in the right half-plane. Its smooth function boundary values give absolutely summable Taylor coefficients. Choose an analytic polynomial $A$, with zero constant coefficient, uniformly within $1/4$ of $H$ on the closed disk. Put $P(t)=\mathrm{Im}A(e^{2\pi it})/\pi$. Then $\Vert P{\Vert }_{\infty }\le 1/2+1/(4\pi )<1$.

If $D=\mathrm{deg}A$ and $m>D$, the modulated trigonometric polynomial $Q(t)=e^{2\pi imt}P(t)$ has spectrum in $[m-D,m+D]$. A sharp Fourier partial sum at $m$ selects exactly the negative-frequency half of $P$, since $A(0)=0$:Consequently, for every $t\in K$,The carrier frequency $m$ can be arbitrarily large. This compact-set Fourier amplification lemma uses a bounded imaginary part to obtain a large one-sided Fourier multiplier; no assertion about pointwise divergence is used.

Now let $E$ be any open set of measure $\delta$. DefineIf this is zero take $f=0$. Otherwise choose a decreasing sequence ${\delta }_j\downarrow 0$, with ${\delta }_1=\delta$, so small for $j\ge 2$ thatFor $j=1$ this inequality also holds: $B(\delta )>4$ and $B(\delta )/2\ge B(\delta )/4+1$. Let $L_j$ be an increasing compact exhaustion of $E$.

Using compact batching of a small open set, we recursively choose finite closed-arc unions $K_j\subset E$, of measure at most $3{\delta }_j$, such that $L_j$ is covered by the interiors of $K_1,\dots ,K_j$ andHere is the measure-theoretic justification. At step $j$, the uncovered portion has measure at most ${\delta }_j$. Its intersection with $L_j$ after removing the previously covered interiors is compact, and can be covered by finitely many closed arcs lying in $E$ with total length at most ${\delta }_j$ plus an arbitrarily small excess. By inner regularity of Lebesgue measure, a further compact subset of the uncovered open portion captures all but ${\delta }_{j+1}$ of its measure; cover it similarly. The combined finite arc union has length at most $3{\delta }_j$. This both covers the outstanding part of $L_j$ and ensures the next remainder estimate. In particular $E\subset {\bigcup }_j{K}_j$.

Apply the finite-arc construction to $K_j$ to get $P_j$ of degree $D_j$. Choose carriers $m_j$ recursively so that the bands $[m_j-D_j,m_j+D_j]$ are positive and strictly ordered with no overlap. SetThe frequency-separated Fourier block series and the Weierstrass M-test give a continuous complex-valued function with $\Vert f{\Vert }_{\infty }\le {\sum }_j{2}^{-j}=1$. At the cutoff $m_j$, all earlier bands have been included in full and every later band is absent. Thus on $K_j$,Every point of $E$ lies in some $K_j$, soValues of $\psi$ outside $0<\delta \le 1$ may be defined arbitrarily. The construction handles an arbitrary open set, not just one point or a finite collection of intervals. It also prevents cancellation explicitly: earlier bands contribute a bounded complete sum, and later bands contribute nothing at the chosen cutoff.

Use the uniform probability measure on the Boolean hypercube. The influence of a variable is the flip probabilityThe polynomial defining $f$ is interpreted over ${\mathbb{F}}_2$. Flipping the first coordinate adds $x_2+\cdots +x_n$ modulo two. This nonempty parity is balanced for $n\ge 2$, so

An explicit Tribes function works for every $n\ge 2$. Choose the largest integer $w\ge 1$ with $w{2}^w\le n$, let $m=\lfloor 2^w\log 2\rfloor$, and take the OR of $m$ disjoint ANDs of $w$ coordinates. Ignore the unused coordinates. Its zero probability is $(1-2^{-w}{)}^m$, between $1/4$ and $3/4$, and a used coordinate has influenceHere $n<2^{w+1}(w+1)$ by maximality. For the balance claim, $w=1$ gives probability $1/2$; for $w\ge 2$, $m\ge 2^w\log 2-1$ gives zero probability at most $e^{-\log 2+2^{-w}}<3/4$, while $(1-1/q{)}^q\ge 1/4$ for $q=2^w\ge 2$ gives a lower bound $(1/4{)}^{\log 2}>1/4$. Only the example was requested, but these checks fix its parameters unambiguously.

For the lower bound we prove the required consequence of Beckner's inequality with constants. Write the Fourier-Walsh transform as $f={\sum }_S\hat{f}(S){\chi }_S$, where ${\chi }_S(x)=(-1{)}^{{\sum }_{i\in S}{x}_i}$ and $\hat{f}(S)=\mathbb{E}(f{\chi }_S)$. The discrete derivative of a Boolean function is $D_{i}f=(f-f(\cdot +e_i))/2$. Since $f$ takes values zero and one, $D_{i}f$ takes values $0,\pm 1/2$. Consequently Parseval's identity givesThe factor four is required for the printed $0/1$ convention. Also $V:=\mathrm{Var}(f)={\sum }_{S\ne \varnothing }|\hat{f}(S){|}^2\ge 3/16$ under the balance hypothesis.

The true Beckner's inequality states $\Vert T_{\rho }h{\Vert }_q\le \Vert h{\Vert }_p$ for $1<p\le q<\infty$, $0\le \rho \le \sqrt{(p-1)/(q-1)}$, with $T_{\rho }h={\sum }_S{\rho }^{|S|}\hat{h}(S){\chi }_S$. Apply it with $p=3/2$, $q=2$, $\rho =1/\sqrt{2}$ to $h_i={\chi }_{\{i\}}{D}_{i}f$, a function independent of coordinate $i$. Its nonzero values still have magnitude $1/2$. We obtainLet $a={\max }_i{I}_i(f)>0$. For an integer $\ell \ge 1$, the low-degree Fourier weight satisfiesFor the high-degree part the total influence identity givesTogether these imply $4V\le na(2^{\ell -1}{a}^{1/3}+1/(\ell +1))$.

Suppose, seeking a contradiction, that $a<\log n/(100n)$, with natural logarithms. Then $a<1/(100e)$ and $\ell =\lfloor \log (1/a)/(6\log 2)\rfloor \ge 1$. Hence $2^{\ell -1}{a}^{1/3}\le a^{1/6}/2$. Put $y=\log n$. The first term on the right is bounded bythe middle expression is maximized at $y=7$. Moreover $\log (1/a)\ge y/2$, since $y\le 100e^{y/2}$. ThusWe would have $4V<1/10$, contradicting $4V\ge 3/4$. ThereforeThis derives the stated Kahn-Kalai-Linial theorem bound from the actual noise operator on the Boolean hypercube inequality, rather than using the desired conclusion as an input.

For two-node Gaussian quadrature with linear weight, the monic node polynomial $q(x)=x^2-sx+p$ must be orthogonal to $1$ and $x$ for weight $x$. Indeed, exactness through degree three makes the quadrature of $q$ and $xq$ vanish at its nodes. The moment equations areso $s=6/5$ and $p=3/10$. Thus the Gaussian quadrature nodes, labelled in increasing order, areIntegrating the linear Lagrange interpolation cardinal functions gives the weightsThey are positive, sum to $1/2$, and reproduce the first moment $1/3$. Any cubic polynomial has remainder of degree at most one after division by $q$; orthogonality annihilates the other part both in the integral and at the nodes. This proves degree-three exactness, not merely the two moment conditions for the weights.

The general modular form, the nonuniform Frankl-Wilson theorem, can be stated as follows. Let $p$ be a prime number, and let $L\subseteq {\mathbb{F}}_p$ have $s$ elements. Suppose $\mathcal{F}\subseteq \mathcal{P}([n])$ has $|A|\mathrm{mod}p\notin L$ for every member, while $|A\cap B|\mathrm{mod}p\in L$ for any distinct members. ThenThe uniform Frankl-Wilson theorem sharpens this to $|\mathcal{F}|\le \left(\genfrac{}{}{0ex}{}{n}{s}\right)$ when all members have size $k$ and $0\le s\le \min \{k,n-k\}$. Both forms require a prime modulus and exclusion of the self-intersection residue.

For the general proof, work over the finite field ${\mathbb{F}}_p$. Associate to each member $A$ the intersection polynomialAt the characteristic vectors of sets $1_B$, this is zero if $B\ne A$, and nonzero if $B=A$. Thus these polynomials, regarded as functions on the Boolean hypercube, are linearly independent: evaluating any relation at $1_A$ isolates its coefficient. Apply multilinear reduction on the Boolean cube, replacing every positive power of a variable by that variable. Values on the Boolean hypercube remain unchanged, and the resulting multilinear polynomials have degree at most $s$. Their ambient space has a monomial basis consisting of $x_S={\prod }_{i\in S}{x}_i$ for $|S|\le s$, with dimension ${\sum }_{j=0}^{\min (s,n)}\left(\genfrac{}{}{0ex}{}{n}{j}\right)$. This proves the general bound, including $s=0$.

For completeness, obtain the uniform sharpening without dividing by factorials in a finite field. For $s\ge 1$ put $a=k\mathrm{mod}p$ and $g(x)={\sum }_i{x}_i-a$. The functions $x_{S}g$ with $|S|<s$ are linearly independent. Indeed, a relation gives $gh=0$ with $\mathrm{deg}h<s$, so $h$ is supported only on weights in $H=\{j\in [0,n]:j\equiv a(\mathrm{mod}p)\}$. With boundary weights $-1,n+1$, this set has a gap of at least $s+1$: either consecutive allowed weights differ by $p\ge s+1$, or its terminal gap does, since $n\ge k+s\ge a+s$. The alternating sum of $h$ over any interval of $s$ free coordinates vanishes by its degree. Across that gap only one endpoint level can contribute, forcing $h$ to vanish there. Delete that level and repeat across the enlarged gap until $H$ is empty. Hence $h=0$, and independence of the square-free monomials gives the claim. Adjoin these functions to the $f_A$. Evaluation at each family vector eliminates the coefficients of $f_A$, since $g$ vanishes there; the claim eliminates all remaining coefficients. Counting dimensions yieldsThe case $s=0$ permits at most one member directly. The gap argument is an instance of the modular layer vanishing lemma.

For the final application, enumerate the distinct sets as $A_1,\dots ,A_m$ and let $v_i=1_{A_i}\in {\mathbb{R}}^n$ be their characteristic vectors of sets. If $m\le 1$, the desired bound is immediate for $n\ge 1$. Otherwise $|A_i|\ge k$, because $A_i$ intersects another member in $k$ points. At most one member can have size $k$: two distinct $k$-sets cannot have intersection size $k$. The Gram matrix of these vectors isFor real coefficients $c_i$,All terms are nonnegative, with $k>0$. If the expression vanishes, every coefficient corresponding to a set of size greater than $k$ is zero. At most one coefficient remains, and the first term forces it to be zero as well. Thus the characteristic vectors of sets are linearly independent in ${\mathbb{R}}^n$, proving the constant-intersection family boundThis proof explicitly covers the possible member of size exactly $k$; assuming all diagonal corrections were strictly positive would miss that case. Positivity of $k$ is essential: when $k=0$, the empty set and all $n$ singleton sets give $n+1$ members.

A linear map $\alpha$ is a nilpotent endomorphism if ${\alpha }^q=0$ for some positive integer $q$, and a semisimple endomorphism if it is diagonalizable over $\mathbb{C}$. Decompose $V$ into its generalized eigenspaces $V_{\lambda }=\ker (\alpha -\lambda I{)}^d$, where $d=\dim V$. Define ${\alpha }_s$ on $V_{\lambda }$ to be $\lambda I$, and set ${\alpha }_n=\alpha -{\alpha }_s$. Then ${\alpha }_s$ is diagonalizable, ${\alpha }_n$ is nilpotent, and both preserve these vector subspaces and commute. ThusFor uniqueness, suppose $\alpha =S+N$ with $S$ semisimple, $N$ nilpotent and $SN=NS$. Both commute with $\alpha$, so preserve each $V_{\lambda }$. On an eigenspace of $S$ of eigenvalue $\mu$ inside $V_{\lambda }$, the map $\alpha$ is $\mu I+N$ and has only the eigenvalue $\mu$. Since the same subspace lies in $V_{\lambda }$, $\mu =\lambda$. Hence $S=\lambda I$ on $V_{\lambda }$, proving $S={\alpha }_s$ and $N={\alpha }_n$. This is the additive Jordan–Chevalley decomposition.

We need a polynomial consequence of this decomposition. Hermite interpolation supplies a polynomial $p$ with $p(\alpha )={\alpha }_s$, by prescribing $p(t)\equiv \lambda (\mathrm{mod}(t-\lambda {)}^d)$ for each eigenvalue. For any endomorphism $T$, its semisimple part can likewise be expressed as a polynomial in $T$ with zero constant term: if zero is an eigenvalue, its interpolation condition already forces this; if not, add the independent condition $p(0)=0$.

On $\mathrm{End}(V)$, the maps $\mathrm{ad}{\alpha }_s$ and $\mathrm{ad}{\alpha }_n$ commute. The first is diagonalizable, with eigenvalues $\lambda -\mu$ on $\mathrm{Hom}(V_{\mu },V_{\lambda })$. The second is nilpotent, sincewhich vanishes for $k\ge 2q-1$ when $N^q=0$. Uniqueness therefore proves adjoint compatibility of additive Jordan decomposition: $\mathrm{ad}{\alpha }_s=(\mathrm{ad}\alpha {)}_s$.

Now assume $\alpha \in M$. The condition $[\alpha ,W]\subseteq U\subseteq W$ implies that both $W$ and $U$ are invariant under $\mathrm{ad}\alpha$, and every polynomial in $\mathrm{ad}\alpha$ with zero constant term maps $W$ into $U$. Define $\beta$ to be multiplication by $\overline{\lambda }$ on $V_{\lambda }$. It commutes with $\alpha$. On $\mathrm{Hom}(V_{\mu },V_{\lambda })$, $\mathrm{ad}\beta$ acts by $\overline{\lambda -\mu }$. Polynomial interpolation on the finite set of differences gives $\mathrm{ad}\beta =q(\mathrm{ad}{\alpha }_s)$ with $q(0)=0$. The preceding paragraph then expresses $\mathrm{ad}\beta$ as a polynomial in $\mathrm{ad}\alpha$ with zero constant term. Consequently $[\beta ,W]\subseteq U$, so $\beta \in M$.

The assumed trace orthogonality nilpotence lemma now follows directly. On $V_{\lambda }$, $\alpha =\lambda I+{\alpha }_n$ and $\beta =\overline{\lambda }I$, while the matrix trace of the nilpotent restriction of ${\alpha }_n$ is zero. HenceEvery summand is nonnegative, so every eigenvalue of $\alpha$ is zero. Its Jordan–Chevalley decomposition therefore has ${\alpha }_s=0$, and $\alpha$ is nilpotent. Notice that neither $U$ nor $W$ was required to be a Lie subalgebra.

Choose a positive system of a root system and the corresponding triangular decomposition of a Lie algebra $L={\mathfrak{n}}^{-}\oplus H\oplus {\mathfrak{n}}^{+}$. A primitive element of a Lie algebra representation of weight $\omega \in H^{\ast }$ is a nonzero vector $v$ with $hv=\omega (h)v$ for every $h\in H$ and ${\mathfrak{n}}^{+}v=0$. Thus it is a highest-weight vector; the choice of positive roots is part of this definition. A highest-weight representation is generated by such a vector.

Let $\mathfrak{b}=H\oplus {\mathfrak{n}}^{+}$ be the corresponding Borel subalgebra. Define its one-dimensional Lie algebra representation ${\mathbb{C}}_{\omega }$ by the scalar $\omega (h)$ on $H$ and zero on ${\mathfrak{n}}^{+}$. This respects the Lie bracket, since $[\mathfrak{b},\mathfrak{b}]\subseteq {\mathfrak{n}}^{+}$. The induced Verma moduleis nonzero: the Poincaré-Birkhoff-Witt theorem identifies its underlying vector space with $U({\mathfrak{n}}^{-})$, and $v=1\otimes 1$ is a primitive element of weight $\omega$. Its weight spaces are finite dimensional, its top weight space is the line $\mathbb{C}v$, and all other weights are $\omega -\sum n_i{\alpha }_i$ with $n_i\ge 0$ and at least one positive coefficient.

A proper submodule cannot contain $v$, because $v$ generates $M(\omega )$. More strongly it has no component of weight $\omega$: any finite sum of distinct weight vectors can be projected onto its individual components by a polynomial in a generic element of $H$. Therefore the sum $N$ of all proper submodules still misses the top weight line and is proper. It contains every proper submodule, so the irreducible quotient of a Verma moduleis irreducible and retains the nonzero primitive element $v+N$. This constructs the requested representation for every $\omega \in H^{\ast }$. It is not asserted to be finite dimensional for arbitrary $\omega$.

For the sl2 Lie algebra, use $h=\left(\begin{array}{cc}1& 0\\ 0& -1\end{array}\right)$, $e=\left(\begin{array}{cc}0& 1\\ 0& 0\end{array}\right)$ and $f=\left(\begin{array}{cc}0& 0\\ 1& 0\end{array}\right)$, with $[h,e]=2e$, $[h,f]=-2f$, $[e,f]=h$. The classification of finite-dimensional sl2 representations gives one irreducible $V_n$ for every integer $n\ge 0$. In a basis $v_0,\dots ,v_n$ its action iswhere $v_{-1}=v_{n+1}=0$. These formulas satisfy the three Lie brackets. Any nonzero invariant subspace contains a weight vector by polynomial projection using $h$; repeated application of $e$ gives $v_0$, and applications of $f$ then give the entire basis. Thus $V_n$ is irreducible.

Conversely, in any finite-dimensional irreducible module, start with an eigenvector of $h$ and apply $e$ until reaching a nonzero vector $v$ with $ev=0$. This process terminates because $e$ increases the eigenvalue by two and only finitely many eigenvalues occur. Write $hv=av$. The sl2 highest-weight lowering formula givesLet $f^{n}v\ne 0$ and $f^{n+1}v=0$, which again follows from the finite set of weights. Applying $e$ to the latter identity yields $(n+1)(a-n)f^{n}v=0$, so $a=n$. The resulting $n+1$ distinct weight vectors span an invariant subspace, hence the entire irreducible module. Therefore $\dim V_n=n+1$ and its primitive weight is ${\omega }_n(h)=n$, with $n=0,1,2,\dots$; its weights are $n,n-2,\dots ,-n$.

Construct the Euclidean lattices in an orthogonal coordinate system whose squared axis lengths areThus an integer coordinate vector $z$ has squared norm ${\sum }_i{a}_i{z}_i^2$. These four numbers are linearly independent over $\mathbb{Q}$: the four sign changes of $\sqrt{2},\sqrt{3}$ in their field extension isolate the four coefficients of a rational relation. The independence will distinguish the two flat tori; equality of their length spectra will hold for every positive choice of axis weights.

LetHere and below these integer coordinates are interpreted in the weighted orthogonal axes. Direct multiplication gives $S^T=-S$, $S^2=-3I$, andIn particular both $L_{\pm }$ are full-rank Euclidean lattices. Denote their four generator columns by $v_i^{\pm }$. Modulo $3$ the two matrices have the same columns, and the nonzero words of the given ternary tetracode $T$ are exactlyThese are the reductions of $\pm v_i^{\pm }$, respectively. The first and fourth displayed words are the specified generators, and the middle two arise from their sum and difference, up to sign. Thus reduction modulo $3$ maps either Euclidean lattice onto precisely $T$.

To construct a length-preserving correspondence, find the kernels $M_{\pm }=L_{\pm }\cap 3{\mathbb{Z}}^4$ of this reduction. The inverse matrix is $A_{\pm }^{-1}=A_{\pm }^T/12$. Hence $3w\in L_{\pm }$ exactly when $A_{\pm }^{T}w\equiv 0(\mathrm{mod}4)$. These congruences are equivalent toFor clarity, the first $+$ congruence is $-(w_1+w_2+w_3+w_4)=0(\mathrm{mod}4)$; subtracting the others gives evenness of each required pair sum. For the $-$ matrix the first congruence is $2w_1-(w_1+w_2+w_3+w_4)=0(\mathrm{mod}4)$; the remaining congruences again give equality of the four parities. Substitution proves the converse in each case.

Let $R_i$ be the orthogonal reflection changing only coordinate $i$. If the coordinates of $w$ are even, this changes their sum by a multiple of $4$; if all are odd, it changes the sum by $2$ modulo $4$. ThereforeEach Euclidean lattice is the disjoint union of its kernel and the eight cosets $\pm v_i^{\pm }+M_{\pm }$, since $T$ has nine words. Define a map $\Phi :L_{+}\to L_{-}$ by using $R_1$ on $M_{+}$ and $R_i$ on each of $v_i^{+}+M_{+}$ and $-v_i^{+}+M_{+}$. Each piece maps bijectively onto the corresponding $-$ coset. Thus $\Phi$ is a bijection satisfyingIt also commutes with $z\mapsto -z$. The closed geodesics of a flat torus have lengths $\Vert z\Vert$ for nonzero translation vectors $z$ in its Euclidean lattice. Consequently the two length spectra agree, including the multiplicities of translation classes and with either consistent orientation convention. If only primitive closed geodesics are counted, the same conclusion follows by removing iterates in increasing order of length: the total multiplicity at $\ell$ is the sum of primitive multiplicities at $\ell /k$, and there are only finitely many such contributions below any fixed length. The correspondence need not itself preserve primitiveness to give this conclusion.

It remains to prove the two flat tori are not isometric. An isometry of flat tori lifts to an affine Euclidean isometry; its linear part of an affine map must carry $L_{+}$ onto $L_{-}$. In integer coordinates write this part as $B$. For every $z\in L_{+}$, both $z$ and $Bz$ have integer coordinates and have equal squared norms. Rational independence of the $a_i$ impliesApply this to $z=A_{+}n$ for all $n\in {\mathbb{Z}}^4$. Two squared linear forms agreeing on all integer points agree as polynomials; their difference factors, so one form is the other or its negative. Since $A_{+}$ is invertible, $B$ is consequently a diagonal sign change $D=\mathrm{diag}({\epsilon }_1,\dots ,{\epsilon }_4)$.

Modulo $3$, such a $D$ must preserve $T$. Each nonzero word of $T$ has exactly three nonzero coordinates, and for each choice of zero coordinate there are just two such words, differing by an overall sign. Preserving these words forces the three signs on each support to agree. The four overlapping supports then force ${\epsilon }_1={\epsilon }_2={\epsilon }_3={\epsilon }_4$. Thus only $D=I$ and $D=-I$ are possible. Neither changes $L_{+}$.

Finally $v_1^{+}\notin L_{-}$: it has the same ternary word as $v_1^{-}$, but their difference is $6e_1=3(2,0,0,0)$, which fails the congruence defining $M_{-}$. Hence $L_{+}\ne L_{-}$ and no such linear isometry exists. This completes the tetracode length-isospectral lattice construction.

Use the unique curvature $-1$ hyperbolic metric on a closed Riemann surface of genus $g\ge 2$, supplied by the uniformization theorem. This is the implicit setting for the partition and spectral theorems; a genus-zero or genus-one surface does not have this hyperbolic metric.

A partition is a pants decomposition by disjoint essential simple closed geodesics, whose complement is a union of pairs of pants. A pair of pants is a sphere with three disks removed, furnished here with geodesic boundary. Its Euler characteristic is $-1$, so there must be $2g-2$ pants. Every pant has three boundaries and every cutting curve occurs twice, giving $3g-3$ cutting curves. The Bers pants decomposition theorem asserts that a constant $B_g$ depending only on $g$ bounds the lengths of all cuffs in some such pants decomposition of every closed genus-$g$ hyperbolic surface.

To see what geometry the cuffs determine, cut each pair of pants along its three perpendicular seams into two congruent right-angled hyperbolic hexagons. The three alternate sides are ${\ell }_1/2,{\ell }_2/2,{\ell }_3/2$, where the ${\ell }_i>0$ are its boundary lengths. The right-angled hyperbolic hexagon identity determines the seam $s_i$ opposite the half-cuff ${\ell }_i/2$ byWe use the standard existence and uniqueness theorem for a right-angled hyperbolic hexagon with prescribed positive alternating side lengths. Thus the three boundary lengths determine the pair of pants up to isometry.

Fix a topological pants decomposition, label its cuffs, and choose reference seam endpoints and orientations. Gluing two boundaries of the same length requires a translation along that boundary; its signed distance is a twist ${\tau }_i$. The gluing graph, the $3g-3$ positive cuff lengths, and the $3g-3$ twists therefore determine the unmarked hyperbolic surface up to isometry. A full boundary translation has period ${\ell }_i$, so for an unmarked gluing one may take $0\le {\tau }_i<{\ell }_i$, with endpoints identified. For a marked surface the whole real twist records the number of Dehn twists, and the Fenchel–Nielsen coordinates areThe topological gluing data are necessary when the decomposition is not fixed. Different markings or decomposition graphs can describe the same unmarked surface.

Teichmüller space ${\mathcal{T}}_g$ consists of pairs $(X,f)$, where $X$ is a closed genus-$g$ Riemann surface and $f:{\Sigma }_g\to X$ is an orientation-preserving marking, modulo the equivalence $(X,f)\sim (Y,h)$ when a conformal orientation-preserving map $a:X\to Y$ has $a\circ f$ homotopic to $h$. The Fenchel–Nielsen coordinates identify it with ${\mathbb{R}}_{>0}^{3g-3}\times {\mathbb{R}}^{3g-3}$, of real dimension $6g-6$. The moduli space of Riemann surfaces forgets the marking by quotienting by the mapping class group.

The Wolpert generic spectral rigidity theorem says that the locus of closed genus-$g$ hyperbolic surfaces possessing an isospectral but nonisometric partner lies in a locally real-analytic exceptional locus of lower dimension in Teichmüller space. Thus a generic surface is determined, up to isometry, by its unmarked length spectrum, equivalently by its Laplacian spectrum. Here an isometry may reverse orientation; the length spectrum cannot distinguish the two orientations of the same metric. The equivalence of the two spectra is the compact hyperbolic Selberg trace formula.

The simplifying result is the Buser finite length spectrum theorem: for every $g\ge 2$ and $\epsilon >0$ there is $L=L(g,\epsilon )$ such that two closed genus-$g$ hyperbolic surfaces with hyperbolic systoles at least $\epsilon$ have the same entire unmarked length spectrum if their length multisets up to $L$ agree. Multiplicities are included. We will count primitive unoriented closed geodesics; counting all iterates is equivalent by successively removing shorter iterates. The theorem is uniform on the thick part of moduli space, not merely a cutoff chosen separately for a particular pair.

Here is a proof using compactness and polynomial trace equations. We state the subsidiary facts and explain their role. First, Mumford's compactness theorem makes the genus-$g$ thick moduli space compact. It can also be seen from the Bers pants decomposition theorem: cuffs in a Bers decomposition lie in $[\epsilon ,B_g]$, there are finitely many pants decomposition graphs, and unmarked twists may be reduced modulo their cuff lengths. These data lie in finitely many compact boxes of Fenchel–Nielsen coordinates and cover the thick moduli space.

Second, these finite compact boxes give a compact family $K$ of marked hyperbolic structures on a fixed smooth surface ${\Sigma }_g$ representing every member of the thick moduli space. We use the standard smooth dependence of the glued metrics and their hyperbolic holonomy representations on Fenchel–Nielsen coordinates. Choose smooth representatives over finitely many parameter neighborhoods and conjugate the hyperbolic holonomy representation using a fixed lifted Riemannian orthonormal frame. Passing to finite compact subboxes gives a compact set of representations, together with a compact family of representative smooth metrics. Changing the markings between boxes causes no problem: all are markings from the same fixed ${\Sigma }_g$, and the union is finite.

Third, uniformization theorem identifies each marked hyperbolic surface with ${\mathbb{H}}^2/\rho ({\pi }_1{\Sigma }_g)$ for a faithful discrete hyperbolic holonomy representation into ${\mathrm{PSL}}_2(\mathbb{R})$. We use the standard lifting fact that a closed orientable hyperbolic holonomy representation admits a lift to ${\mathrm{SL}}_2(\mathbb{R})$, and lifts may be chosen continuously on small parameter neighborhoods. Choose lifts on finitely many such neighborhoods, allowing the finitely many possible sign choices. A representation is described by the matrices of $2g$ standard fundamental group generators. The polynomial encoding of hyperbolic geodesic lengths uses, for each fixed group word $w$, its squared matrix traceis a polynomial in their entries: products are polynomial, and the inverse of a determinant-one matrix is its polynomial adjugate. For a nontrivial hyperbolic word, the hyperbolic translation length obeysThus equality of the positive lengths is exactly equality of these trace polynomials, independent of lift signs.

Fourth, the length spectrum of a closed hyperbolic surface is locally finite with finite multiplicities. We need the uniform finiteness of short geodesic classes across the compact marked family. More uniformly, for each finite $b$, only finitely many unoriented primitive conjugacy classes can have length at most $b$ in any metric of $K$. To justify uniformity, compactness of the smooth metrics gives a common bilipschitz equivalence comparison with one fixed metric $h_0$ on ${\Sigma }_g$. If ${\ell }_h(\alpha )\le b$, the minimizing geodesic has $h_0$ length at most $Cb$, so ${\ell }_{h_0}(\alpha )\le Cb$. Local finiteness for $h_0$ then gives a finite list. Local finiteness itself follows from proper discontinuity of the cocompact group of the hyperbolic holonomy representation: conjugate a geodesic axis to meet a fixed compact fundamental set, and a bounded-length axis gives a group element moving that compact set a bounded distance, of which there are only finitely many.

Finally, the Hilbert basis theorem says that a polynomial ring in finitely many real variables is Noetherian. Consequently a decreasing sequence of sets cut out by polynomial equations eventually stabilizes. Indeed, their vanishing ideals form an increasing chain and stabilize. Finite unions of such sets are still algebraic: one can take all products of one defining polynomial from each component. This finiteness fact is the engine that turns arbitrarily many spectral comparisons into finitely many.

Enumerate the primitive unoriented conjugacy classes as ${\alpha }_1,{\alpha }_2,\dots$. For each $n$ letand let $F_n$ be the finite set of primitive classes that can have length at most $b_n$ anywhere in $K$. Define $Z_n$ in the finite-dimensional space of pairs of generator matrices as follows: it consists of pairs $(\rho ,\sigma )$ for which there exists an injectionand also an injection satisfying the analogous equations with $\rho$ and $\sigma$ interchanged. There are only finitely many injections. Each choice imposes finitely many polynomial equations; taking their finite unions and then intersecting the two directions proves that $Z_n$ is a real affine algebraic set. On determinant-one representations these equations are exactly the desired matching of lengths. Defining the polynomials on the ambient matrix-entry space is harmless; the argument will only apply them to representations in $K$.

Put $W_n=Z_1\cap \cdots \cap Z_n$. The sets $W_n$ form a decreasing chain of affine algebraic sets, hence for some $N$,For representations in $K$, membership in this intersection is equivalent to equality of the full length spectra. One direction follows by matching equal lengths, including their multiplicities. For the other, fix a length $\ell$. All primitive classes of length $\ell$ on the first surface occur in some finite initial segment. Its injective matching shows that the second surface has at least that multiplicity at $\ell$; the reverse injection gives the reverse inequality. Local finiteness makes both multiplicities finite. Doing this for every $\ell$ proves multiset equality.