The Mycielski construction replaces a graph $G$ with a graph $\mu (G)$ having one shadow vertex for each original vertex and one further vertex adjacent to every shadow. It satisfiesand preserves triangle-freeness. Iterating it from the five-cycle gives triangle-free graphs of arbitrarily large chromatic number.

Here $G$ is a finite simple graph with at least one vertex; it need not have an edge. The Mycielskian separates two ways of needing many colours: a large clique forces distinct colours immediately, but a triangle-free graph can impose incompatible choices through its whole pattern of graph adjacency. The Mycielski construction increases that global constraint while preserving the absence of triangles in a graph.

Write the vertex set of $G$ as $\{v_1,\dots ,v_n\}$, with $n\ge 1$. Keep these vertices, add distinct Mycielski shadow vertices $u_1,\dots ,u_n$, and add a Mycielski apex $w$. The vertex set and complete list of edges areThere are no other edges: the Mycielski shadow vertices form an independent set, $w$ has no original vertex as a neighbour, and $u_i$ is not adjacent to $v_i$. Each Mycielski shadow vertex copies its original vertex's graph neighbourhood within the original vertex set, not its position in the drawing. This is the construction in the inspected research paper.^[ref]

For $n$ original vertices and $m$ original edges, direct counting gives $2n+1$ vertices and $3m+n$ edges. The original vertices induce a copy of $G$, so no original graph adjacency is lost. A drawing crossing is not an extra vertex.

Start with the complete graph $K_2$, whose only edge is $v_1{v}_2$, abbreviating $\{v_1,v_2\}$. Its Mycielskian has exactly the edges $v_1{v}_2,u_1{v}_2,u_2{v}_1,w{u}_1,w{u}_2$. Reading them as $v_1,v_2,u_1,w,u_2,v_1$ gives the cycle graph $C_5$. A proper vertex colouring gives $v_1,u_1$ colour 1, $v_2,u_2$ colour 2, and $w$ colour 3. Two colours cannot close an odd cycle, so its chromatic number is exactly three.

This is the least number of colors needed to color all points of Euclidean space so that points at Euclidean distance one receive different colors. Any finite unit-distance graph embedded in the space gives a lower bound. If $p$ is prime, the characteristic vectors of all $(2p-1)$-subsets of $[4p-1]$, divided by $\sqrt{2p}$, give such a graph: its edges are exactly pairs intersecting in $p-1$ points. The prime-power modular intersection bound bounds an independent set by $\left(\genfrac{}{}{0ex}{}{4p-1}{p-1}\right)$, and hence gives $\chi ({\mathbb{R}}^{4p-1})\ge \left(\genfrac{}{}{0ex}{}{4p-1}{2p-1}\right)/\left(\genfrac{}{}{0ex}{}{4p-1}{p-1}\right)$.

Lateral chromatic aberration, also called transverse chromatic aberration, is colour-dependent image scale: an object away from the optical axis can have different image heights at different wavelengths. Its changing transverse magnification produces displaced coloured edges. The on-axis parallel bundle in the chromatic aberration diagram illustrates axial chromatic aberration, and does not measure lateral chromatic aberration. ^[ref].

Axial chromatic aberration, also called longitudinal chromatic aberration, is colour-dependent focus position along the optical axis. For incident light rays parallel to the optical axis, it appears as a changing focal length. The chromatic aberration diagram isolates this effect using an ideal thin lens at each colour.

For $A_i=r_i{e}^{i{\varphi }_i}$ with all $r_i>0$, the translation-invariant total phase is $\Phi ={\varphi }_1+{\varphi }_2+{\varphi }_3$. With $\alpha >0$, the displayed equation makes $\Phi =0$ stable and $\Phi =\pi$ unstable. A spatial translation can remove two phases, so the stable locked states have all amplitudes positive real in a suitable origin. This is a statement modulo translation, not that every translated representative is real; the zero solution and rolls belong to boundary strata.

Assume $\alpha >0$, as in the stable phase-locked branch. At a positive equilibrium set $x_i=r_i^2$ and $p=r_1{r}_2{r}_3$. The equations imply $a{x}_i+b{x}_{i+1}+c{x}_{i+2}-\alpha p/x_i=\mu$. Suppose $b>a>c$ and relabel cyclically so $x_1$ is maximal. If $x_2\ge x_3$, subtracting equations one and two makes the cubic side positive unless all are equal, whereas $\alpha p(1/x_1-1/x_2)$ is nonpositive. If $x_2\le x_3$, subtracting equations two and three gives the opposite sign contradiction. Interchanging the orientation handles $c>a>b$. Thus all positive equilibria have equal amplitudes. With $\alpha \ne 0$, a state with exactly two nonzero amplitudes cannot be an equilibrium because it immediately forces the third.

Assume $\alpha >0$. For equal positive amplitudes $r$, let $s=a+b+c$ and use $\mu +\alpha r-s{r}^2=0$. The radial Jacobian eigenvalue is $\alpha r-2s{r}^2$. The conjugate pair has real part $(b+c-2a)r^2-2\alpha r$ and imaginary parts $\pm \sqrt{3}(b-c)r^2$. When $b+c>2a>0$ and $b\ne c$, the upper hexagon branch loses amplitude stability through this pair at the displayed radius. The associated parameter is ${\mu }_H=2{\alpha }^2(4a+b+c)/(b+c-2a{)}^2$. Nonlinear terms decide the direction and stability of the resulting periodic branch.

A Chi-bar-squared distribution is a mixture distribution ${\sum }_{j=0}^m{w}_j{\chi }_j^2$, where $w_j\ge 0$ and ${\sum }_j{w}_j=1$, and ${\chi }_0^2$ denotes a point mass at zero. It arises when a Gaussian likelihood is optimized under inequality constraints: projection onto a constraint cone selects different dimensions according to which inequalities are active. For example, if $W,Z$ are independent standard Gaussian random variables, then $Z^2+(-W{)}_{+}^2$ has distribution $\frac{1}{2}{\chi }_1^2+\frac{1}{2}{\chi }_2^2$. On $W\ge 0$ it is $Z^2$; on $W<0$ it is $Z^2+W^2$. Symmetry makes the sign of $W$ independent of $W^2$, proving the stated weights and component laws.

For $d$ independent centered normal random variables of common variance $v>0$, $|\mathbf{X}{|}^2/v$ has a chi-squared distribution with $d$ degrees of freedom. Its density gives the displayed moments by a gamma integral. In three dimensions, $\mathbb{E}|\mathbf{X}{|}^{2n}=(2n+1)!!v^n$. Radial odd moments need not vanish, although any odd Cartesian product does.

Stanford's archive records a technical report by Herman Chernoff, dated November 1951, on asymptotic efficiency of hypothesis tests based on sums of observations.^[ref] This is historical source metadata, rather than a claim to have inspected the report's full proof. The modern theorem cited above connects that efficiency question to the log-likelihood ratio accumulated across repeated observations.

Let $P$ be the Bernoulli distribution with success probability $0.2$ and $Q$ the Bernoulli distribution with success probability $0.8$. For $0\le s\le 1$,The two summands have product $0.16$. Their arithmetic mean is at least their geometric mean, with equality at $s=1/2$. ThusThe same argument gives $C=-\ln (2\sqrt{p(1-p)})$ for complementary parameters $0<p<1$. At $p=1/2$ the laws agree and $C=0$.

With equal prior probabilities, count the successes $K$ among $n$ observations. The likelihood ratio is $Q^{\otimes n}/P^{\otimes n}=4^{2K-n}$, so choose $Q$ when $K>n/2$ and $P$ when $K<n/2$. If $n$ is even, either decision or a fair tie randomization gives the same average error at $K=n/2$. Under $P$, $K$ has the binomial distribution with parameters $n,0.2$. Symmetry therefore gives the exact optimal average errorThe last term applies only for even $n$; it represents half the tie probability. Direct finite sums give $p_1^{\ast }=0.2$, $p_5^{\ast }=0.05792$ and $p_{20}^{\ast }\simeq 0.00157912$. Twenty observations reduce the optimal average error to about 0.158%. The finite-sample Chernoff error bound at $n=20$ is $\frac{1}{2}(0.8{)}^{20}\simeq 0.00576461$, a guarantee rather than the exact error.

Reducing the exact error from $p_{20}^{\ast }$ to $p_{50}^{\ast }\simeq 1.29363\times 10^{-6}$ illustrates the value of repeated evidence. The example assumes a single fixed coin law throughout the experiment. If repeated sensor readings share an unmodelled common error, increasing their number need not supply this much independent information.

For equal prior probabilities and independent and identically distributed observations on a finite alphabet, minimum sum of errors in a simple hypothesis test givesFor nonnegative $a,b$ and $0<s<1$, $\min (a,b)\le a^s{b}^{1-s}$. Applying this termwise and using the independence of random variables to factor the sum yieldsOptimizing $s$, with an infimum if needed, proves $p_n^{\ast }\le \frac{1}{2}{e}^{-nC(P,Q)}$. This short argument gives a bound for every $n$. The theorem identifying the optimal error exponent also needs the converse: no decision rule can achieve a strictly larger limiting exponent. A log-likelihood ratio is a sum of independent contributions, so large deviation principle supplies the relevant converse idea.

For probability mass functions $P,Q$, their Chernoff affinity at $0<s<1$ is the sum $Z(s)={\sum }_{x}P(x{)}^{s}Q(x{)}^{1-s}$. It weights pointwise agreement between the two laws. The weighted arithmetic-geometric mean inequality gives $P(x{)}^{s}Q(x{)}^{1-s}\le sP(x)+(1-s)Q(x)$, hence $Z(s)\le 1$. With positive common support, $Z(0)=Z(1)=1$. Smaller affinity means a stronger separation at that weighting.

At $s=1/2$, $Z(1/2)={\sum }_x\sqrt{P(x)Q(x)}$ is the Bhattacharyya coefficient. The quantity $-\ln Z(1/2)$ is the Bhattacharyya distance. Allowing all weights $s$ gives Chernoff information at least this large. Symmetry under exchanging $P,Q$ sends $s$ to $1-s$, so Chernoff information is symmetric even though an optimizing weight need not be $1/2$.

The boundary four-form is $\Theta =\mathrm{Tr}[\delta A(FA+AF-\frac{1}{2}{A}^3)]$. To derive the identity, put $B=dA$ and $a=\delta A$. The variations of the three polynomial terms in $Q_5$ are respectively $3\mathrm{Tr}(a{B}^2)+d\mathrm{Tr}[a(BA+AB)]$, $3\mathrm{Tr}[a(A^{2}B+B{A}^2)]+\frac{3}{2}d\mathrm{Tr}(a{A}^3)$, and $3\mathrm{Tr}(a{A}^4)$. Their sum gives the identity because $F=B+A^2$. For variations in a Lie algebra with basis $T_{\alpha }$, the bulk field equations are $\mathrm{Tr}(T_{\alpha }{F}^2)=0$, not necessarily the unprojected matrix equation $F^2=0$. Conjugation of both the Lie algebra and the gauge curvature proves covariance of these equations under Yang-Mills gauge transformations.

For $F=dA$ on a five-dimensional oriented spacetime, the action $S=-\frac{1}{2}\int F\wedge \ast F+c\int A\wedge F\wedge F$ has bulk variation $\int \delta A\wedge (-d\ast F+3cF\wedge F)$. Two integrations by parts make the three variations of the Chern-Simons 5-form contribute equally. Under $A\mapsto A+d\lambda$, the extra action is $c\int d(\lambda F\wedge F)$, a boundary term by the Generalized Stokes theorem. Its bulk field equation is gauge invariant. The Chern-Simons term is metric-independent and has no bulk contribution to the stress-energy tensor when the differential-form field is held fixed in metric variation.

For an oriented three-dimensional slice $\Sigma$, use the Skew-Hermitian trace conventionSince the differential of the integrand is $\mathrm{tr}(F\wedge F)$, a space-time slab oriented by $dt\wedge {\mathrm{vol}}_{\Sigma }$ has instanton charge $k=N_{\mathrm{CS}}(t_{+})-N_{\mathrm{CS}}(t_{-})$ when the side boundary contributes no flux. For a pure gauge $A=h^{-1}dh$, this reduces to the winding integral $(24{\pi }^2{)}^{-1}{\int }_{\Sigma }\mathrm{tr}(h^{-1}dh{)}^3$. Gauge transformations of nonzero winding can shift it by an integer.

The complex orientation of complex cobordism defines Chern classes of actual complex bundles by the splitting principle and elementary symmetric expressions in the first Chern classes of line summands. For an actual rank-$n$ complex bundle, its top Chern class is its Euler class in a generalized cohomology theory. Chern classes extend stably by the Whitney formula; this stable extension must not be confused with the Euler class of a merely stably complex real bundle.

Here the $x_j$ are the formal Chern roots of a vector bundle. The degree-four term is $(c_1^2+c_2)/12$. Applied to the tangent bundle of a compact complex surface, Riemann–Roch for the structure sheaf gives the Noether formula.

For a vector bundle $E$ of rank $r$ on a smooth algebraic variety, algebraic Chern classes have $c_0(E)=1$, vanish for $i>r$, commute with pullback, and obey the Whitney sum formula for Chern classes for every short exact sequence of vector bundles. A line bundle $\mathcal{O}(D)$ has total class $1+[D]$. The projective-bundle splitting construction makes pullback injective and supplies a filtration by line bundles, so these axioms determine all classes as elementary symmetric functions of their first classes. In particular, $c_i(E^{\ast })=(-1{)}^i{c}_i(E)$ and $c_1(L\otimes M)=c_1(L)+c_1(M)$ in the ordinary Chow ring.

CheZ is an enzyme that binds phosphorylated CheY and accelerates removal of its phosphate.

Accelerated dephosphorylation shortens the lifetime of the motor-regulating CheY signal. The output depends on both phosphorylation and removal, so equal kinase activity can produce different steady signalling when CheZ activity changes. ^[ref]