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If $Z_1,\dots ,Z_{\nu }$ are independent standard normal random variables and $\lambda ={\sum }_j{\mu }_j^2$, then ${\sum }_j(Z_j+{\mu }_j{)}^2$ has a noncentral chi-squared distribution with $\nu$ degrees of freedom and noncentrality $\lambda$. For $\lambda =0$ it reduces to the chi-squared distribution. Squaring a shifted signed normal Wald statistic gives the case $\nu =1$.

For an anti-Hermitian $SU(N)$ connection one-form, normalize the Chern-Simons three-form by $1/(8{\pi }^2)$ and use $A^g=g^{-1}Ag+g^{-1}dg$. Its change is $-d\mathrm{Tr}(dg{g}^{-1}\wedge A)/(8{\pi }^2)-\mathrm{Tr}(g^{-1}dg{)}^3/(24{\pi }^2)$. On a closed three-manifold the first term integrates to zero, while the latter is an integer with the fundamental matrix trace. Hence the integral is defined modulo integers; integer-level exponentials are invariant under large Yang-Mills gauge transformations.

For a rank-$r$ complex vector bundle, let $h=c_1(S^{\ast })$ for the tautological line $S$ on its projective bundle. The Leray-Hirsch theorem gives the free basis $1,h,\dots ,h^{r-1}$ over base cohomology. The uniquely determined coefficients in the displayed monic relation define the integral Chern classes. The line normalization uses the Euler class of the canonically oriented underlying real two-plane bundle, and uniqueness makes this definition intrinsic and natural under pullback. No choice of a trivializing cover or classifying map remains in the definition.

The Total Chern class of a direct sum of complex vector bundles is the product of their total Chern classes. In each degree, $c_k(E\oplus F)={\sum }_{i+j=k}{c}_i(E)⌣c_j(F)$. The degree-zero class is one, classes beyond the bundle rank vanish, and a trivial bundle has total class one.

On a compact oriented $2r$-manifold without boundary, an integral product of Chern classes of total complex degree $r$ pairs with the fundamental class to give a Chern number. A unitary connection supplies Chern-Weil theory representatives, but the integer is independent of that connection. The Second Chern number on a four-manifold is one important example; the integral of the First Chern class over a closed surface is another.

For one particle species in chemical equilibrium under $2\varphi \leftrightarrow 3\varphi$, chemical balance gives $2{\mu }_{\varphi }=3{\mu }_{\varphi }$, hence ${\mu }_{\varphi }=0$. Elastic $2\varphi \leftrightarrow 2\varphi$ imposes no such constraint. In an expanding universe the number-changing reaction must remain fast compared with the Hubble parameter; after chemical freeze-out a nonzero chemical potential can develop even if elastic collisions maintain kinetic equilibrium.

A chemical system is out of local chemical equilibrium when reaction adjustment cannot keep pace with transport, imposed changes or external driving. Comparing the chemical relaxation time with transport and irradiation times identifies plausible departures. Disequilibrium chemistry in an exoplanet atmosphere combines this principle with mixing, horizontal advection and atmospheric photochemistry.

Cantelli's inequality uses a mean and variance to control deviations in one specified direction. For a real random variable $X$ with expected value $\mu$, finite variance ${\sigma }^2$, and $a>0$,Applying the same statement to $-X$ gives the lower tail probability bound $\mathbb{P}(X-\mu \le -a)\le {\sigma }^2/({\sigma }^2+a^2)$. No independence or particular probability distribution is required. If ${\sigma }^2=0$, the random variable equals $\mu$ almost surely and both positive-threshold tail probabilities are zero.

For variance one and a threshold two above the mean, the upper-tail probability is at most $1/5$. Applying the two-sided Chebyshev's inequality to contain that upper event would give the weaker bound $1/4$. For the standard normal distribution, the actual upper tail probability at two is about $0.0227501$, much smaller than either bound. Yet the two-point probability distribution in equality in Cantelli's inequality attains $1/5$, showing why the smaller Gaussian value cannot be promised from these two moments alone.

A short explanation is to centre $Y=X-\mu$ and choose a nonnegative shift $b$. On $Y\ge a$, the square $(Y+b{)}^2$ is at least $(a+b{)}^2$, so Markov's inequality givesThe best shift is $b={\sigma }^2/a$, giving the displayed bound. Choosing the shift before squaring exploits the direction of the deviation; squaring the unshifted random variable treats both directions together. This shifted-square proof is also the strategy requested in the University of Utah's 2005 qualifying problem on Cantelli's inequality. ^[ref].

The one-sided improvement should be used for a one-sided question. Adding upper and lower bounds by the union bound gives $2{\sigma }^2/({\sigma }^2+a^2)$ for $|X-\mu |\ge a$. When $a\ge \sigma >0$, that is at least the Chebyshev's inequality bound ${\sigma }^2/a^2$; when $a<\sigma$, both two-sided expressions exceed one and must be capped. The separate one-sided bounds generally cannot be attained simultaneously by the same probability distribution.

An operator $({\bar{L}}_i\varphi e_{Rj})({\varphi }^{†}\varphi )/{\Lambda }^2$ contains three factors of $v+h$. Its contribution to the mass matrix is proportional to $v^3$, while its single-Higgs coupling is three times the corresponding mass-over-v contribution. With the normalization $-\sqrt{2}{\lambda }^{\prime }/{\Lambda }^2$, the mass-basis coupling is $Y_h=M_{\mathrm{diag}}/v+v^2{\hat{\lambda }}^{\prime }/{\Lambda }^2$. Nonzero off-diagonal coefficients permit charged-lepton flavour-violating Higgs decays.

For ${\psi }^C={\eta }_{C}C{\bar{\psi }}^T$ and ${\gamma }^{0T}={\gamma }^0=-C{\gamma }^0{C}^{-1}$, the Dirac adjoint transforms as ${\bar{\psi }}^C=-{\eta }_C^{\ast }{\psi }^T{C}^{†}$. Anticommuting the components of the fermion bilinear gives $(\bar{\psi }\psi {)}^C=\bar{\psi }(C^{†}C{)}^T\psi$. Invariance of the scalar bilinear therefore fixes $C^{†}C=I$.

If $C{\gamma }^{\mu T}{C}^{-1}=-{\gamma }^{\mu }$, transposition shows that $C^T{C}^{-1}$ commutes with every gamma matrix. Irreducibility of the complex Dirac representation and the Schur lemma make it scalar. Transposing $C^T=zC$ again yields $z^2=1$, so $C^T=\pm C$.

The unitary charge-conjugation operator acts on field operators and commutes with numerical chiral projectors, so $\hat{C}(P_L\psi ){\hat{C}}^{-1}=P_L{\psi }^c$. This component is left-chiral. The algebraic conjugate of an already projected spinor instead satisfies $(P_L\psi {)}^c=P_R{\psi }^c$, because its adjoint carries the opposite projector. Keeping the two operations separate avoids a false chirality conclusion.

For a self-adjoint element $h$, the exponentials $\exp (ith)$ are unitary. A continuous algebra character therefore satisfies $|\exp (it\varphi (h))|\le 1$ for every real $t$, forcing $\varphi (h)$ to be real. Decomposing an arbitrary element into real and imaginary self-adjoint parts proves the displayed identity. Thus every algebra character is a star-preserving scalar homomorphism; commutativity of the whole algebra is not needed for this fact.

If forcing has the $\kappa$-chain condition for a regular uncountable $\kappa$, every forced club subset of $\kappa$ contains a ground-model club below the forcing condition. Bound the possible witnesses above each ordinal using antichains, then take closure points of the bound function.

The Koszul complex on $f_1,\dots ,f_r$ has term $K_i={\bigwedge }^i{R}^r$ and differential $\partial (e_{j_1}\wedge \cdots \wedge e_{j_i})={\sum }_a(-1{)}^{a-1}{f}_{j_a}{e}_{j_1}\wedge \cdots \wedge \widehat{e_{j_a}}\wedge \cdots \wedge e_{j_i}$. Its degree-zero homology is $R/(f_1,\dots ,f_r)$. For a regular sequence it is exact in positive degrees, hence a Koszul resolution of that quotient. The variables of a polynomial ring form a regular sequence, so this resolves the coefficient field in length equal to the number of variables.

A double complex has modules $C_{p,q}$ and differentials lowering either index, each squaring to zero and anticommuting with the other. The total chain complex has term ${⨁}_{p+q=i}{C}_{p,q}$ and differential the sum of the two differentials. For the tensor product of two chain complexes, a sign $(-1{)}^p$ on the second differential ensures anticommutation. If the double complex is in the first quadrant and one direction has homology only in degree zero, its total homology is computed by the surviving degree-zero complex. Applying this twice to two free resolutions proves the balanced calculation of the Tor functor.

The chain complex with $R$ in a single degree $n$ and zero differential. A map from it into a complex chooses a cycle in degree $n$.

For $n\ge 1$, the chain complex with copies of $R$ in degrees $n,n-1$ and identity differential. It is acyclic; a map from it chooses an arbitrary degree-$n$ element and its boundary.

The tensor product of homologically graded chain complexes has differential $D(x_p\otimes y)=d{x}_p\otimes y+(-1{)}^p{x}_p\otimes dy$. The sign makes the two mixed terms in $D^2$ cancel. Total degree is the sum of the two degrees.

A graded Hom functor construction consists of maps shifting degree by $j$. One homological convention is $d_{M}f=f{d}_C+(-1{)}^{j-1}{d}_{C^{\prime }}f$. Its degree-zero cycles are chain maps, and its degree-zero boundaries are null-homotopic chain maps, so its zeroth homology is the module of chain homotopy classes. This convention differs by degree-dependent signs from the alternative convention $d^{\prime }f-(-1{)}^{j}fd$.