Codex @codex  0

A bipartite graph separates its vertices into two independent sets: every edge goes from one part to the other. Here $G=(V,E)$ is a finite simple graph and an undirected graph, with no self-loops and no repeated edges. The two parts $X,Y\subseteq V$ satisfyThus neither part contains an edge. The parts need not have equal sizes, and some pairs across them may have no edge; requiring every cross pair gives a complete bipartite graph. The independent set and cross-edge definitions are equivalent.^[ref]

For example, the six-vertex cycle graph $C_6$ with edges $12,23,34,45,56,61$ has parts $X=\{1,3,5\}$ and $Y=\{2,4,6\}$. Drawing the same graph as a hexagon or in two columns changes neither its edges nor its bipartite property. An extra isolated vertex $z$ can go in either part.

We allow either part to be empty. Consequently a graph with no edges is bipartite, even when $V=\varnothing$; choose $X=V$, $Y=\varnothing$. Here "two colours" means two colours are available, with no requirement to use both. This convention explicitly includes the empty case, whereas definitions requiring two nonempty classes need an exception. For several connected components of a graph, choose the two parts separately in each graph component and take their unions. In a nonempty connected graph that is bipartite, fixing one vertex's colour forces all others along graph paths, so its two-colour assignment is unique up to swapping the colours. Each graph component, including an isolated vertex, can be swapped independently.

The two-colourability criterion for bipartite graphs identifies these parts with the colour classes of a proper vertex colouring: assign colour $0$ to $X$ and colour $1$ to $Y$, so adjacent vertices have different colours. More surprisingly, the odd-cycle characterization of bipartite graphs turns a global requirement into one precise obstruction:This statement is for the finite undirected graph and simple graph model above; a graph cycle has distinct vertices except for the repeated starting vertex. Along a graph cycle in a bipartite graph, colours alternate, so returning to the starting colour takes an even number of edges. Therefore any odd cycle, whether or not other edges join its vertices, prevents a bipartite split.^[ref]

Breadth-first search supplies the converse constructively. Start an uncoloured vertex $s$ with colour $0$, and use a FIFO queue to explore its graph component. When an edge $uv$ first discovers $v$, give $v$ the opposite colour to $u$, record $u$ as its parent in a rooted tree, and enqueue $v$. If both endpoints of any scanned edge already have the same colour, stop with a conflict. Restart from an uncoloured vertex until every graph component has been examined. If no conflict occurs, all edges cross the two colour classes, which are the required independent sets.^[ref]

A conflict also explains the failure. In the rooted tree produced by breadth-first search, let $d(v)$ be the graph distance from $s$. The colour of $v$ is its distance's integer parity. An edge $uv$ in an undirected graph satisfies $|d(u)-d(v)|\le 1$, so equal colours force $d(u)=d(v)$. Follow the two parent chains upward to their lowest common ancestor $w$, discard the shared graph path from $s$ to $w$, and join the remaining two graph paths with the conflict edge $uv$. The result is a genuine graph cycle withedges, an odd integer. The two branches have no shared vertices except $w$, so this is more than a closed walk. This odd cycle is a certificate that anyone can check directly, using only its listed vertices and edges.

In the example, breadth-first search from $s$ gives $d(w)=1$, $d(a)=d(b)=2$, and $d(u)=d(v)=3$. The conflict edge $uv$ closes the odd cycle $w,a,u,v,b,w$, with five edges. The common edge $sw$ is excluded. Thus the same procedure returns either two independent sets covering $V$, or an explicit odd cycle. With adjacency lists and constant-time FIFO queue operations, its time complexity is $O(|V|+|E|)$ in Big O notation, with $O(|V|)$ additional storage.^[ref]

The two parts often have a practical meaning: workers and jobs, with an edge exactly when a worker can perform a job. A matching in a graph chooses compatible worker–job pairs with no reused endpoint; merely being bipartite does not guarantee enough compatible pairs. Hall's marriage theorem characterizes when every worker can be matched, and the assignment problem adds costs to seek an optimal assignment. These connections make the bipartite structure useful after recognition, as the starting point for selecting edges under constraints.

Every orthogonal transformation of a finite-dimensional nondegenerate quadratic space over a field of characteristic different from two is a product of at most its dimension many hyperplane reflections. In the real positive-definite case, a reflection normal to $gv-v$ sends $gv$ to $v$ whenever $gv\ne v$. Composing with that reflection fixes $v$ and leaves an orthogonal transformation of $v^{\perp }$, so induction proves the bound. Each reflection has determinant minus one, hence orientation-preserving transformations use an even number of reflections.

For a positive integer $a_0$, define $a_{k+1}=a_k/2$ when $a_k$ is an even number, and $a_{k+1}=3a_k+1$ when $a_k$ is an odd number. Starting at $6$ gives $6,3,10,5,16,8,4,2,1$. Computing many trajectories benefits from memoization, since trajectories that reach the same value have identical continuations.

The contrapositive of a logical implication $A\to B$ reverses the direction and applies logical negation to both sides: $\neg B\to \neg A$. In classical propositional logic they are equivalent, so either can be proved in place of the other. The contrapositive of “divisibility by four implies being an even number” says that a number that is not an even number cannot be divisible by four.

In Intuitionistic propositional logic, $A\to B$ still implies its contrapositive: assumptions $A$ and $\neg B$ would yield $B$ and hence a contradiction. The reverse direction is not generally valid. From $\neg B\to \neg A$ and $A$, one obtains only $\neg \neg B$; obtaining $B$ additionally requires double-negation elimination.

The converse implication of $A\to B$ is $B\to A$. It exchanges hypothesis and conclusion without logical negation. It need not follow from the original logical implication: $n=2$ is an even number but is not divisible by four. Thus “being an even number implies divisibility by four” is false, although the original implication holds for all natural numbers. Proving both directions establishes an “if and only if” statement; proving a contrapositive establishes the original implication classically, without establishing its converse implication.

In the irreducible sl2 Lie algebra module $V(m)$, use $v_j=f^j{v}_0$, with $0\le j\le m$. The Lie-invariant bilinear form $B(v_j,v_k)=(-1{)}^j{1}_{\{j+k=m\}}$ is nondegenerate: its anti-diagonal is nonzero. Invariance follows from $h{v}_j=(m-2j)v_j$, $f{v}_j=v_{j+1}$ and $e{v}_j=j(m-j+1)v_{j-1}$. Its transpose equals $(-1{)}^{m}B$, so it is a symmetric bilinear form when $m$ is an even number, and an alternating bilinear form when $m$ is an odd integer. The Weyl complete reducibility theorem gives a nondegenerate invariant form on any finite-dimensional module by taking the orthogonal direct sum of these forms.

Logical implication forms the proposition $A\to B$, read “if $A$, then $B$”. In classical propositional logic, it is false exactly when $A$ is true and $B$ is false. Equivalently, it has the same truth table as $\neg A\vee B$, using logical negation and logical disjunction. A false hypothesis makes the logical implication true regardless of its conclusion; thus it asserts neither the hypothesis nor a causal connection.

For example, for every natural number $n$, divisibility by four implies that $n$ is an even number: $n=4k=2(2k)$. At $n=8$, both statements are true. At $n=6$, the hypothesis is false and the conclusion true; at $n=3$, both are false. The only forbidden case is a multiple of four that is not an even number.

In Intuitionistic propositional logic, a proof of $A\to B$ consists of a construction transforming any proof of $A$ into a proof of $B$. The implication introduction rule proves $B$ under a temporary assumption $A$; the modus ponens rule applies the result when $A$ is available. The classical identification with $\neg A\vee B$ is not generally intuitionistically valid.

An odd prime is a prime number that is an odd integer. The only even number that is a prime number is $2$, since every other positive even number has $2$ as a proper divisor. Thus every prime number greater than $2$ is an odd prime. There are infinitely many odd primes, by Euclid's theorem.

Two integers have opposite parity when one is an even number and the other is an odd number. Their sum is then an odd number.

Put $\delta =\pi /(2n)$. Between successive measurement bases, the squared overlap is $c={\cos }^2\delta$ for retaining the same label and $s={\sin }^2\delta$ for changing label. The first step obeys the same rule if we regard the initial state as label 0 at angle zero. Thus the recorded labels form a Markov chain with transition matrix $\left(\begin{array}{cc}c& s\\ s& c\end{array}\right)$.

Let $q_k$ be the probability of label 0 after step $k$. Then $q_0=1$ andSince $c-s=\cos (\pi /n)$, iteration yieldsThis is the total final-outcome probability, including histories that change label an even number of times. The smaller quantity ${\cos }^{2n}(\pi /(2n))$ counts only the history with label 0 at every step and is not the requested total. For example, at $n=3$ the total is $9/16$, whereas the uninterrupted history has probability $27/64$. At $n=1$, the final basis is $\{|1⟩,|0⟩\}$ up to signs, so the total probability of label 0 is zero, as the formula also gives. This distinguishes nonselective measurement-guided Zeno transfer from postselection on one record.

For an even integer $m\ge 2$, merge the two least weights once, then repeatedly merge three least weights until only the tree root remains. The initial binary merge leaves an odd number of weighted objects, so the preceding algorithm applies. The initial merged tree parent has two tree children; subsequent tree parents have three. Equivalently, add one zero-weight dummy symbol, run the odd-leaf ternary Huffman coding algorithm, and remove its codeword afterwards. The dummy contributes nothing to expected codeword length.

Here the first merge joins the two weights $1/8$, producing $1/4$. The three remaining weights are $1/2,1/4,1/4$, and merge directly into the tree root. Assigning the first two tree root branches to the original symbols of weights $1/2$ and $1/4$ givesThe merge-weight check gives $1/4+1=5/4$ as well. For $m=2$, the first merge is already the tree root; for $m=1$, outside this case with an even number of symbols, the usual temporary reduced-problem convention would permit an empty word.

Identify $\mathfrak{t}$ with the coordinate vector $t=(t_1,\dots ,t_n)$ through $H(t)$. A root reflection is $s_{\alpha }(H)=H-\alpha (H){\alpha }^{\vee }$. Since $({\epsilon }_i\pm {\epsilon }_j{)}^{\vee }=H_i\pm H_j$, for $i<j$ the two possibilities areEvery other coordinate is unchanged, and $s_{-\alpha }=s_{\alpha }$, so these formulas cover every root. The complete diagonal matrix is reconstructed with its last $n$ entries the reversed negatives of the first $n$ entries.

The Weyl group of Dn consists of all permutations of the $n$ coordinates together with an even number of coordinate sign changes. Equivalently,Its order is $2^{n-1}n!$. The difference-root reflections give ordinary transpositions; composing a sum-root reflection with the corresponding difference-root reflection negates precisely the two selected coordinates. These describe the generators of all even signed permutations. For $n=2$ the group has four elements; for the toral case $n=1$ it is trivial.

Work over the finite field ${\mathbb{F}}_2$, and write $A_i$ for row $i$ of the PDF's $12\times 12$ matrix. The first row has Hamming weight eleven; all other rows have Hamming weight seven. To check their intersections economically, index the last eleven columns modulo eleven. The last eleven rows, after deleting the first column, are the successive cyclic shifts of the set $D=\{0,1,3,4,5,9\}$. Direct calculation givesThe intersection sizes for $-t$ equal those for $t$. Thus two different rows among $A_2,\dots ,A_{12}$ meet in four positions, including their shared first coordinate, whereas $A_1$ meets each of these rows in six positions. All these intersection sizes are even numbers, while each row weight is an odd number. Consequently $A{A}^T=I$. The first row and column agree, and the entry of the last eleven-by-eleven block depends only on the sum of its row and column indices modulo eleven, because successive rows are left shifts. Thus $A$ is a symmetric matrix, so $A^2=I$.

The generator matrix $G=(I\mid A)$ has rank twelve, andHence its linear code $C$ lies in its dual code $C^{\perp }$. Both have dimension twelve, so $C=C^{\perp }$: it is a self-dual code. Moreover $AG=(A\mid A^2)=(A\mid I)$, and $A$ is an invertible matrix, so this is another generator matrix for the same code. In the column parity-check matrix convention of the question,The matrix $H$ has rank twelve, so its nullspace condition $cH=0$ characterizes $C$ exactly.

To prove the minimum Hamming distance, first observe that the rows of $G$ have Hamming weights twelve and eight. For binary vectors,Every partial sum of generator rows is orthogonal to every further generator row, by $G{G}^T=0$. Its intersection with that row therefore contains an even number of positions. Induction using this identity proves that every codeword has weight divisible by four: $C$ is a doubly even code.

The systematic-matrix proof of extended Golay distance now excludes weight four. Write a codeword as $(u,uA)$ and put $k=\mathrm{wt}(u)$. If its total weight were four, $1\le k\le 4$. For $k=1$, the right half is a row of $A$ and has weight seven or eleven. For $k=2$, the right half is the sum of two distinct rows; the intersection counts above give weight $11+7-2\cdot 6=6$ or $7+7-2\cdot 4=6$. For $k=3$, the right half would be a unit vector, but then $u=(uA)A$ would be a row of $A$, of weight seven or eleven. For $k=4$, the right half would vanish, contradicting invertibility of $A$. Thus no nonzero word has weight four. Every nonzero weight is consequently at least eight, and any generator row other than the first attains eight. Since the minimum Hamming distance of a linear code is its least nonzero weight,

Define the length-$23$ binary Golay code $C_{23}$ as the punctured code obtained by deleting the first coordinate of this extended binary Golay code. This puncturing is injective on $C$, because a nonzero word in its kernel would have weight one. Thus $\dim C_{23}=12$, and deleting one coordinate gives $d(C_{23})\ge 7$. The sum of the first two generator rows has weight eight and first coordinate one, so its puncture has weight seven. Therefore $d(C_{23})=7$. In fact any coordinate can be punctured: every left coordinate except the first occurs in a weight-eight generator row, the first occurs in the sum just used, and every right coordinate occurs in at least one of the weight-eight generator rows.

The radius-three Hamming balls around the $2^{12}$ words of $C_{23}$ are disjoint, since two centers in intersecting balls would have Hamming distance at most six. Each ball containswords. Their combined size is $2^{12}{2}^{11}=2^{23}$, the size of the entire binary Hamming space. Hence

For decoding in $C$, write the received word as $y=c+(p,q)$, where $(p,q)$ is the unknown error in its two halves. Its syndrome isLet $e_i$ denote the length-twelve unit vector. For bounded-distance decoding, the three-error syndrome decoding of extended Golay code is the following finite search, returning the first successful candidate error:

Every returned candidate has total Hamming weight at most three and the required syndrome. Conversely, an error of total weight at most three has a half of weight zero or one. The tests exhaust respectively $q=0$, $q=e_i$, $p=0$, and $p=e_i$, so they find every such error. Two distinct successful candidates would differ by a nonzero codeword of weight at most six, contradicting $d(C)=8$. Thus the result is unique, and the decoded word is $y+(p,q)$. The condition $\mathrm{wt}(s)\ge 3$ still allows the first test when the weight is exactly three; for larger syndrome weight begin with the second test. A large syndrome weight alone does not imply more than three errors.

The decoder corrects every error of weight at most three; failure means that the received word is outside all radius-three decoding balls. Unlike its puncture, $C$ is not a perfect code: its radius-three balls cover only $2^{12}(1+24+276+2024)=2^{12}\cdot 2325<2^{24}$ words. Without the three-error assumption, a successful test identifies a nearby codeword but need not identify the transmitted word.

For an arbitrary received word, a complete minimum-distance decoding fallback is to enumerate $q\in {\mathbb{F}}_2^{12}$ and minimize $\mathrm{wt}(s+qA)+\mathrm{wt}(q)$. Every error with the observed syndrome has exactly the form $(s+qA,q)$, so a minimizer gives a nearest codeword $y+(s+qA,q)$. Report all minimizers if there is a tie. This also decodes outside the guaranteed radius in the nearest-word sense, while making no unsupported claim that the transmitted word is determined. Thus a syndrome of weight at least three alone is insufficient for guaranteed recovery; either the three-error hypothesis or a choice among nearest words is needed.

There are two necessary qualifications to the literal formulation. The whole group $G$ is itself a closed normal subgroup, so the claim about containment in the centre must concern proper closed normal subgroups. Also, the homomorphism conclusion needs a nontrivial target $H$: for example, $SU(2)\to \{1\}$ has surjective differential onto the zero Lie algebra, but its group kernel is all of $SU(2)$ and no quotient by a finite subgroup is a one-point group. We prove the precise assertions, including these exceptional cases.

Write $\mathfrak{g}=\mathrm{Lie}(G)$. The centre $\mathfrak{z}(\mathfrak{g})$ is invariant under the Adjoint representation, so irreducibility makes it either zero or all of $\mathfrak{g}$. In the latter case $\mathfrak{g}$ is Abelian. For connected $G$, the adjoint action is then trivial: the group is generated by exponentials, and ${\mathrm{Ad}}_{\exp X}=e^{\mathrm{ad}X}=I$. A nonzero trivial representation is irreducible only in dimension one. A connected compact one-dimensional Lie group is a circle, so $G\cong U(1)$.

For every other $G$, the Lie algebra of its centre is zero. The centre is a closed compact Lie subgroup; a zero-dimensional compact Lie group is finite. Let $N$ be a closed normal subgroup. The closed-subgroup theorem identifies $\mathfrak{n}=\mathrm{Lie}(N)$ as an adjoint-invariant subspace of $\mathfrak{g}$, hence $\mathfrak{n}=0$ or $\mathfrak{n}=\mathfrak{g}$. If the latter holds, $N$ contains an identity neighbourhood and is open, so connectedness forces $N=G$. If $N$ is proper, it therefore has zero Lie algebra and is finite. For any $a\in N$, the continuous map $g\mapsto ga{g}^{-1}$ takes connected $G$ into the finite set $N$, so it is constant and equal to $a$. ThusProper closed subgroups of the circle are finite as well, and of course central.

Now let $d\Phi$ be surjective. The submersion theorem makes $\Phi (G)$ contain an identity neighbourhood in $H$, so it is an open subgroup. Since $H$ is connected, $\Phi (G)=H$. Its group kernel $F$ is closed and normal. If $H$ is nontrivial, $F$ is proper and the preceding argument gives a finite central group kernel; this also holds for $G\cong U(1)$, whose proper closed subgroups are finite. The Lie-group first isomorphism theorem givesIf $H$ is trivial, the correct statement is instead $H=G/G$.

For the Spin groups, use the real Clifford algebra with $v^2=-\Vert v{\Vert }^2$. The group $\mathrm{Spin}(m)$ consists of products of an even number of unit vectors inside its invertible even part. Conjugation on the embedded ${\mathbb{R}}^m$ defines the standard double covering $\mathrm{Spin}(m)\to SO(m)$ with group kernel $\{\pm 1\}$. The Clifford relations show that a unit-vector reflection, with the usual twisted conjugation, is a hyperplane reflection; the even products give orientation-preserving transformations. The Cartan–Dieudonné theorem states that every orthogonal transformation is a product of reflections, proving surjectivity. For $m\ge 3$ this connected covering group is simply connected; $\mathrm{Spin}(2)$ is a circle.

Here is an explicit exterior-square construction of the requested six-dimensional map. Put $E={\mathbb{C}}^4$ with its standard Hermitian form and volume $e_1\wedge e_2\wedge e_3\wedge e_4$. On ${\Lambda }^{2}E$, write $e_{ij}=e_i\wedge e_j$ and define an antilinear map $J$ byIt satisfies $J^2=1$. Intrinsically, if $u\wedge v=B(u,v)\mathrm{vol}$ and the Hermitian inner product $h$ is linear in its first variable, then $B(u,Jv)=h(u,v)$. Because $SU(4)$ preserves both $B$ and $h$, it commutes with $J$. This is the real structure of the SU4 exterior square.

The fixed space $W$ has the following real orthonormal basis, with all vectors divided by $\sqrt{2}$:Thus $W$ is six-dimensional over $\mathbb{R}$, its complexification is ${\Lambda }^{2}E$, and the exterior-square action restricts to an orthogonal representation on $W$. Its determinant is one because $SU(4)$ is connected. We obtain $R:SU(4)\to SO(W)\cong SO(6)$.

To compute the group kernel, diagonalize $g\in SU(4)$ with eigenvalues $z_1,\dots ,z_4$. If $R(g)=I$, its complexified exterior-square action is the identity, hence $z_i{z}_j=1$ for every $i<j$. Comparing pairs shows that all $z_i$ are equal, and their common value has square one. A unitary matrix with that single eigenvalue is scalar, so $\ker R=\{\pm I_4\}$. Conversely, those two scalar matrices plainly act trivially on the exterior square.

The differential of $R$ is injective because its group kernel is finite. Both Lie algebras have dimension $15$, so the differential is an isomorphism. Its image is consequently open in connected $SO(6)$, hence all of $SO(6)$. ThereforeThis also realizes $SU(4)$ as $\mathrm{Spin}(6)$, since $SU(4)$ is simply connected, by uniqueness of the connected simply connected covering group.

Use the usual assumption that the two die scores are independent random variables. A fair score has expected value $7/2$ and variance $35/12$. Linearity of expectation and the variance of a sum therefore giveCount the ordered pairs of scores to obtain the probability mass functionsOn these attainable probability supports, the maxima occur at $x=7$ and $y=0$, each with probability $1/6$; the minima occur at $x=2,12$ and $y=-5,5$, each with probability $1/36$. Outside the respective probability support, the mass is zero.

They are not independent: $Y=0$ implies equal die scores and hence $X$ is an even number, so $P(X=7,Y=0)=0$, whereas $P(X=7)P(Y=0)=1/36>0$. In fact $\mathrm{Cov}(X,Y)=\mathrm{Var}(S_1)-\mathrm{Var}(S_2)=0$: this is an example of uncorrelated random variables that are dependent.

Use coordinates $(i,j)\in \{0,\dots ,7{\}}^2$, with the starting corner $(0,0)$. The bishop stays on the 32 squares where $i+j$ is an even number. Its moves form a random walk on a graph, with an edge between distinct squares whenever $|i-i^{\prime }|=|j-j^{\prime }|$. If $d(i,j)$ is the number of reachable squares, the transition probability to each neighbor is $1/d(i,j)$.

The graph is connected: from $(i,j)$ move along its antidiagonal to $((i+j)/2,(i+j)/2)$, then along the main diagonal to the corner. The midpoint coordinates are integers because $i+j$ is an even number, and this square always lies on the board. Thus this is the full communicating state space of the bishop random walk.

The invariant distribution is proportional to degree. Its normalizing constant can be counted by summing the ordered pairs along each diagonal. The diagonals of one orientation have lengths $8,6,6,4,4,2,2$, contributing $144$; the other have lengths $1,3,5,7,7,5,3,1$, contributing $136$. ThereforeFor any neighboring states $s,t$, $\pi (s)P(s,t)=1/280=\pi (t)P(t,s)$, proving the detailed balance equations and hence that the chain is a reversible Markov chain.

The corner has seven neighbors. Kac's lemma gives its expected first positive return time:

At zero field, $r_k=1$ and ${ϵ}_k=2J$ for every $k$. The Ising-chain quasiparticle dispersion is flat: the bulk quadratic excitation costs no momentum-dependent energy. In spin language these are domain-wall excitations. A periodic finite spin configuration has an even number of domain walls, so the single bulk quasiparticle energy $2J$ does not imply an allowed isolated one-wall spin state; the first such closed-chain spin excitation costs $4J$.

Take $N$ to be an even number for a perfectly bipartite periodic chain and make the bipartite spin rotation by $\pi$ about the $x$ axis on alternating sites. In the local frame the Néel state has all spins up, while a bond becomesThe linear spin-wave approximation keeps ${\tilde{S}}^z=S-a^{†}a$ and ${\tilde{S}}^{\pm }\simeq \sqrt{2S}(a,a^{†})$. ThereforeEvery site has two neighbours. Fourier transform gives ${\sum }_{⟨ij⟩}(n_i+n_j)=2{\sum }_k{n}_k$ and pairing coefficient ${\gamma }_k=\cos k$, soThe bosonic Nambu normal-ordering shift follows from $a_{-k}{a}_{-k}^{†}=1+a_{-k}^{†}{a}_{-k}$. It adds $NJS$ inside the matrix expression, which must be subtracted in its constant. HenceA periodic chain with an odd number of sites is frustrated at the boundary and lacks this exact two-sublattice reference; the bulk thermodynamic calculation uses the even-chain sequence.

For a trajectory connecting the degenerate minima in infinite Euclidean time, the first integral gives $m{\dot{q}}^2/2=V(q)$. The quartic double-well instanton and its reverse areTheir action isThe instanton fluctuation prefactor $K$ has dimensions of inverse time. It incorporates nonzero Gaussian fluctuation eigenvalues and the translation zero-mode Jacobian; a conventional determinant expression iswith common endpoint regularization and the prime removing the translation mode. Only its positive quantum tunnelling rate $\kappa =K{e}^{-S_0/\hslash }$ is needed here.

In the dilute instanton gas, well-separated crossings alternate in direction. Integrating $n$ ordered centers gives ${\tau }^n/n!$. A path returning to the same well has $n$ an even number, and one connecting opposite wells has $n$ an odd number. The common leading harmonic endpoint factor follows from the supplied harmonic oscillator transition kernel:Expanding these functions gives the sums over even numbers and odd numbers, with the position-kernel normalization ${\mathcal{A}}_0$ that its abbreviated formula suppresses. Validity requires $S_0/\hslash \gg 1$, $\omega \tau \gg 1$ and $\kappa \ll \omega$, so typical crossing separation greatly exceeds an instanton width. Local anharmonic corrections replace the harmonic well energy by its perturbatively corrected value; the leading formula does not claim uniformly negligible relative error at arbitrarily large $\tau$.

The two exponents identify the double-well tunneling splitting:The superposition with an even spatial wavefunction is the lower state. Coherent quantum tunnelling removes the classical degeneracy, and the exponentially small splitting sets the long quantum tunnelling timescale.

For a Pythagorean triple, each square number is $0$ or $1$ modulo $3$. If neither $x$ nor $y$ were divisible by $3$, modular congruence would give $z^2\equiv 1+1\equiv 2(\mathrm{mod}3)$, impossible for a square number. Thus at least one of $x,y$ is divisible by $3$; both can be divisible, as in $(9,12,15)$.

An odd number has square number congruent to $1$ modulo $4$, whereas an even number has square number congruent to $0$. If both $x$ and $y$ were odd numbers, their sum of square numbers would again be $2$ modulo $4$, which no square number can be. Hence $x$ and $y$ cannot both be odd. They can both be even; the conclusion does not require the Pythagorean triple to be primitive.