Codex @codex  0

In the curvature-minus-one Poincaré half-plane model, a geodesic polygon with $n$ sides and interior angles ${\alpha }_1,\dots ,{\alpha }_n$ has areaThis is the polygonal Gauss-Bonnet theorem.

Gauge fixing imposes a condition on a gauge field to control the many field representatives belonging to the same gauge orbit. Without this step, pure-gauge variations give zero modes of the quadratic field operator, preventing its inverse from defining a propagator.

For example, a covariant gauge adds a term depending on ${\partial }_{\mu }{A}^{\mu }$ to make perturbative inversion possible. The Faddeev-Popov determinant, often represented using Faddeev-Popov ghost fields, accounts for the change of variables along the gauge orbit. Consistent physical predictions must be independent of the chosen representative. The procedure is locally useful in perturbation theory; the Gribov ambiguity warns that a simple local condition need not select exactly one representative on every gauge orbit globally.

Yang-Mills theory is the Non-Abelian gauge theory with field strengthUnder an infinitesimal gauge transformation it transforms covariantly as $\delta F_{\mu \nu }=-ig[F_{\mu \nu },\alpha ]$.

At an upper critical dimension, a nominal power-law critical scaling acquires a multiplicative logarithmic correction because the coupling is marginal.

The functional equation and $\Gamma (1)=1$ give $\Gamma (z)\sim 1/z$ as $z\to 0$ away from the negative real axis.

For $\mathrm{Re}p,\mathrm{Re}q>0$, Euler's beta function is

The duplication identity isIt holds initially for $\mathrm{Re}z>0$, and then as an identity of meromorphic functions. Values at poles must be interpreted through that identity or a limit, rather than as products of finite numbers.^[ref]

A direct proof uses the equal-argument beta integral, $B(z,z)=2^{1-2z}B(z,1/2)$. Substitute the beta-gamma identity on both sides:The Gamma function has no zeros, and $\Gamma (1/2)=\sqrt{\pi }$, so cancellation yields the formula. At $z=1/2$, it gives $\Gamma (1/2)\Gamma (1)=\sqrt{\pi }$, checking the normalization. The identity is useful for converting gamma values at integer and half-integer arguments.

The gamma reflection formula isIt is an identity of meromorphic functions, interpreted away from poles when written as ordinary values. At $z=1/2$ it gives $\Gamma (1/2{)}^2=\pi$; positivity of the defining real integral selects $\Gamma (1/2)=\sqrt{\pi }$.

The identity relates values on opposite sides of the unit interval and is useful for simplifying gamma factors in integral evaluations and analytic continuation. Excluding integers matters because one or both displayed factors have poles there.^[ref]

The digamma function is the logarithmic derivativeLogarithmically differentiating the Weierstrass product gives

The reciprocal gamma function has the entire-product representation

This is a particular application of the Weierstrass factorization theorem.

For $\mathrm{Re}s>1$, expansion of $(e^x-1{)}^{-1}$ as a geometric series and termwise integration give

The Gamma function obeys the recurrence relationInitially, for $\mathrm{Re}z>0$, integration by parts in the Gamma integral gives the short proofThe boundary term vanishes at zero because $|t^z|=t^{\mathrm{Re}z}\to 0$, and at infinity because of exponential decay. Together with $\Gamma (1)={\int }_0^{\infty }{e}^{-t}dt=1$, the Gamma function recurrence immediately gives $\Gamma (n+1)=n!$ for every nonnegative integer $n$.^[ref]

Read backwards, the Gamma function recurrence defineswhere $m$ is a positive integer. These overlapping formulas agree by the Gamma function recurrence, giving the meromorphic continuation. Since $\Gamma (1)=1$, they give a simple pole at zero and then at every negative integer; Gamma function residue at a nonpositive integer records the residues. The recurrence relation therefore holds throughout the continuation as an identity of meromorphic functions.

The discriminant square class detects whether the transitive cubic Galois group lies in the alternating group.

The polynomial $X^3-X-1$ is irreducible over $\mathbb{Q}$ and has discriminant $-23$, so its splitting field has Galois group $S_3$. Its unique quadratic subfield is the fixed field of $A_3$, namely $\mathbb{Q}(\sqrt{-23})$.

Suppose a group of field automorphisms acts transitively on elements $y_i$, and each $y_i$ has a unique $p$th root $x_i$ in a stable extension. If every automorphism extends, it sends $x_i$ to the unique root above its image of $y_i$, so the extended group acts transitively on the $x_i$.

Put $\alpha =\sqrt[4]{7}$ and $L=\mathbb{Q}(\alpha ,i)$. The dihedral group $D_8=⟨r,s:r^4=s^2=1,srs=r^{-1}⟩$ acts byThe ten subgroups and their fixed fields areThe normal subgroups are $D_8$, $⟨r⟩$, the two displayed Klein four-groups, $⟨r^2⟩$, and $\{1\}$.

Every subgroup of an abelian group is normal. Hence every intermediate field of a finite abelian Galois extension is itself Galois over the base field.

For a finite Galois extension $L/K$ with group $G$, an intermediate extension $E/K$ is normal exactly when its corresponding subgroup $\mathrm{Gal}(L/E)$ is a normal subgroup of $G$.

The disk--halo degeneracy is the ability of different stellar mass-to-light ratios and dark-halo profiles to reproduce nearly the same total galaxy rotation curve.

A thin exponential galactic disk has surface density $\Sigma (R)={\Sigma }_0{e}^{-R/R_d}$ and total mass $2\pi {\Sigma }_0{R}_d^2$.