For $|\varphi |<1$, the bilateral sum ${\sum }_{k\in \mathbb{Z}}{\varphi }^{|k|}{z}^k$ equals $(1-{\varphi }^2)/((1-\varphi z)(1-\varphi z^{-1}))$ on the unit circle. Multiplying by ${\sigma }^2/(\pi (1-{\varphi }^2))$ gives the one-sided autoregressive spectrum and reads off its covariance coefficients.

For an autologistic binary-image model on a finite undirected graph, let $x_v\in \{0,1\}$ and give the configuration prior mass proportional toSuppose $Y_v\mid x$ are independent Gaussian distribution observations with expected value $x_v$ and unit variance. Their local log-likelihood ratio for $x_v=1$ versus $0$ isWriting $z_v={\sum }_{w\sim v}{x}_w$, the full posterior conditional probability is thereforeThus the local evidence changes the one-site field while retaining the same neighbor interaction. For a noise variance ${\sigma }^2>0$, replace $y_v-1/2$ by $(y_v-1/2)/{\sigma }^2$.

The graphical conditional independence survives this update: vertices in any independent color class are conditionally independent given the other vertices and the observations. On a nearest-neighbor square grid, the two checkerboard colors give such classes, so one color can be updated in parallel in a Gibbs sampler. An arbitrary graph need not be bipartite and may require more colors. Updating every vertex simultaneously from its old neighbors is not the same as this exact conditional block update.

For a nonzero finite-dimensional complex vector space $V$, a unital matrix algebra acting irreducibly on $V$ is the full endomorphism algebra. Its Jacobson radical annihilates $V$: otherwise the radical times $V$ is all of $V$, contradicting nilpotence. Faithfulness kills the radical, and the Artin–Wedderburn theorem then leaves one simple matrix factor acting on its standard column module. This is a matrix-algebra theorem, distinct from the orbit-counting Burnside lemma and the finite-group Burnside's theorem.

For a projective complete intersection curve cut out by a homogeneous regular sequence of positive degrees $a,b$ in ${\mathbb{P}}_k^3$, the Koszul resolution gives Hilbert polynomial $abm+ab(4-a-b)/2$. Hence its arithmetic genus is $1+ab(a+b-4)/2$. The same resolution shows $H^0(C,{\mathcal{O}}_C)=k$, and the dimension of a scheme is one, so this genus equals ${\dim }_k{H}^1(C,{\mathcal{O}}_C)$. Smoothness and irreducibility of the curve are not required. For degrees five and seven, the value is $141$.

Uniform convergence in probability of empirical criteria on a compact set proves consistency for minimizers in that set; it does not confine a minimizer over a larger domain. A bounded identifiable regression function can approach its true value again at infinity. For median regression, $a(\theta )={\theta }^2/(1+{\theta }^4)$ has unique zero at zero, but a small positive fitted sample median $c$ has a second solution ${\theta }^2=(1+\sqrt{1-4c^2})/(2c)$. This solution diverges as $c\downarrow 0$, although uniform convergence on any fixed compact interval holds. Constraining the estimator, or proving suitable global separation and localization, repairs the argument.

If ${\varphi }_B=Z_{\varphi }^{1/2}{\varphi }_R$, differentiating a renormalized $n$-point function at fixed bare parameters gives $(\mu {\partial }_{\mu }+\beta {\partial }_g+n{\gamma }_{\varphi })G_{n,R}=0$. With this convention the field scaling dimension at a fixed point is its engineering dimension plus ${\gamma }_{\varphi }$. The critical exponent $\eta$ for scalar two-point decay is $2{\gamma }_{\varphi }$.

With Adjoint representation convention ${\mathrm{ad}}_X(Y)=[X,Y]$, the matrix exponential identity is $e^{-X}Y{e}^X=e^{-{\mathrm{ad}}_X}Y$. Inverse conjugation reverses product order, so it is a right action or antirepresentation, whereas ${\mathrm{Ad}}_g(Y)=gY{g}^{-1}$ is an ordinary left group representation. The first-order terms $Y\mp [X,Y]$ detect a sign mismatch immediately. Negating every adjoint generator without reversing the Lie bracket does not give a new Lie-algebra homomorphism.

For bounded operators $A,B$ on a Hilbert space, moving products across the inner product givesThe order reversal and minus sign matter. If $A,B$ are self-adjoint, their commutator is skew-adjoint, and $i[A,B]$ is self-adjoint.

For unbounded densely defined operators, the same manipulation gives a pairing identity only on vectors for which all the displayed compositions are defined. It does not by itself prove equality of the full adjoint operator domains. For example, a formally self-adjoint Dolbeault Laplacian and mutually formal-adjoint Lefschetz operators give the formal identity $[{\Delta }_{\bar{\partial }},L{]}^{\ast }=[\Lambda ,{\Delta }_{\bar{\partial }}]$ on suitable smooth test functions; operator realizations additionally require compatible domains or boundary conditions.

Let $T_m$ consist of a rooted path of $m\ge 1$ edges followed by two leaves at its far end. Its only vertex with two children has depth $m$. An adjacency-preserving rooted-tree embedding must preserve this depth, so $T_m$ embeds into $T_n$ only when $m=n$. This gives an infinite bad sequence, while rooted-tree homeomorphic embeddings can stretch the stems.

The signed instantaneous contribution of apsidal precession to azimuthal motion during isotropic stellar mass loss is $\dot{\varpi }/\dot{\theta }=-(\dot{\mu }/\mu n)(1-e^2{)}^{3/2}\sin f/[e(1+e\cos f{)}^2]$. Its maximum absolute magnitude is small when $n|\mu /\dot{\mu }|\gg F(e)$, where $F(e)\sim 1/e$ near a circular orbit and $F(e)\to 3\sqrt{3}/4$ as $e\to 1$. The low-orbital eccentricity coordinate singularity can be avoided by evolving the eccentricity vector.

The dual of an abelian variety represents algebraically trivial line bundles rigidified at the origin. Its group law is tensor product. A group homomorphism $f:A\to B$ induces $\hat{f}:\hat{B}\to \hat{A}$ by pullback of these line bundles.

A smooth projective curve $C\subset A$ generates $A$ by its point differences if and only if it meets every prime Weil divisor. If a prime Weil divisor avoids $C$, its restriction to $C$ has degree zero; constancy of line bundle degree in a family shows that every translate either contains $C$ or misses it. This forces invariance under $C-C$ and puts the generated subgroup inside the line bundle translation stabilizer. If this subgroup were all of $A$, the divisor would be algebraically trivial, contradicting its positive intersection with an ample line bundle. Conversely a proper closed generated subgroup admits a disjoint divisor by the pole divisor avoiding a fiber construction, after translating the fiber to contain $C$.

For odd $p$, remove the finite torsion subgroup ${\mu }_{p-1}$ from the Galois group of $\mathbb{Q}({\mu }_{p^{\infty }})/\mathbb{Q}$ by taking its fixed field. For $p=2$, take the fixed field of $\{\pm 1\}$ instead. The remaining Galois group is ${\mathbb{Z}}_p$, using the p-adic logarithm on $1+p{\mathbb{Z}}_p$ or $1+4{\mathbb{Z}}_2$. For a number field $K$, the cyclotomic Zp-extension is the compositum with this rational tower.

At least one finite prime ramifies in a Zp-extension; otherwise the entire tower would lie in a finite Hilbert class field. Ramification can occur only over $p$: local class field theory makes inertia away from $p$ an image of a unit group with finite maximal pro-$p$ quotient, whereas ${\mathbb{Z}}_p$ has no nontrivial finite subgroup. Nonzero inertia is open in ${\mathbb{Z}}_p$, so it becomes total after passing to a sufficiently high finite layer.

A structural-zero component in a count mixture produces a count identically equal to zero. If its mixture weight is $\pi$ and the ordinary component also assigns mass $p_0$ to zero, the observed zero probability is $\pi +(1-\pi )p_0$. Thus an observed zero does not reveal the latent component membership. A shared component for a repeated count profile makes the whole profile zero and induces dependence between its components.

Buy one maturity-$(T-1)$ bond and short $1+f$ maturity-$T$ bonds. Reinvest the first bond's unit maturity payoff in $1/P_{T-1}(T)$ maturity-$T$ bonds. This self-financing portfolio pays $r_T-f$ at $T$. The replication uses only the adjacent maturities and does not require completeness of the whole bond market.

For a real scalar coupled through $-g\bar{\psi }\varphi \psi$, the scalar-fermion Feynman vertex is $-ig$. Two-fermion scattering has scalar exchange in the $t$ and $u$ channels with a relative minus from exchanging identical external fermions. Fermion-antifermion scattering has $t$ exchange and $s$ annihilation, again with a relative fermionic minus for a fixed state ordering. Fermion-scalar scattering has the two orderings along one open fermion line, using internal momenta $p_1+p_2$ and $p_1-p_4$; these have the same sign. The Mandelstam variables specify the propagator denominators.

For one four-component Dirac field with scalar interaction $-y\bar{\psi }\psi \varphi$ and the mostly-plus propagator convention, the fermion loop numerator is $4[M^2-k\cdot (k-p)]$. Reduce it to a bubble and tadpole: $-4y^2[(2M^2+p^2/2)J-I]$. Dimensional regularization gives pole parts $J=2/(\epsilon 16{\pi }^2)$ and $I=-2M^2/(\epsilon 16{\pi }^2)$. The local pole then requires both wavefunction renormalization and a scalar mass counterterm.

In signature $(-,+)$, choose real gamma matrices ${\gamma }^0=\left(\begin{array}{cc}0& -1\\ 1& 0\end{array}\right)$, ${\gamma }^1=\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)$ and ${\gamma }_{\ast }=-{\gamma }^0{\gamma }^1$. The Dirac equation then gives $({\partial }_{\tau }+{\partial }_{\sigma }){\psi }_{+}=0$ and $({\partial }_{\tau }-{\partial }_{\sigma }){\psi }_{-}=0$. Positive chirality is right-moving and negative chirality is left-moving in this orientation. Reversing the definition of ${\gamma }_{\ast }$ exchanges the labels without changing the physics.

Contracting an antighost with the cubic FP ghost term of the BRST operator produces the bilinear generators. They give $[L_m,b_n]=(m-n)b_{m+n}$ and $[L_m,c_n]=-(2m+n)c_{m+n}$. Their zero-mode additive constant depends on the vacuum and normal ordering convention; it must agree with the corresponding $c_0$ term in the charge.