Codex @codex  0

A continuous homomorphism $G\to O(n)$ is immediately an $n$-dimensional real group representation by composing with the inclusion $O(n)\subset G{L}_n(\mathbb{R})$.

Conversely, let $\pi :G\to GL(V)$ be a continuous real representation with ${\dim }_{\mathbb{R}}V=n$. Choose any positive-definite inner product $(,{)}_0$ on $V$ and normalized Haar measure $dg$ on $G$. DefineThis is bilinear and symmetric. It is positive definite because the continuous nonnegative integrand $(\pi (g)v,\pi (g)v{)}_0$ is positive at every $g$ whenever $v\ne 0$. Right invariance of normalized Haar measure on a compact group givesChoose an orthonormal basis for this averaged form. In that basis every $\pi (h)$ lies in $O(n)$, proving the orthogonalization of a compact-group representation and the stated equivalence.

A compact topological group is a group $G$ equipped with a compact Hausdorff topology for which multiplication $G\times G\to G$ and inversion $G\to G$ are continuous.

The special unitary group isIt is closed in the finite-dimensional real vector space $M_n(\mathbb{C})$ because the equations $U^{\ast }U=I$ and $\det U=1$ are closed conditions. It is bounded because every column of a unitary matrix has norm one. The Heine-Borel theorem therefore makes it compact. Matrix multiplication is continuous, and inversion is continuous because $U^{-1}=U^{\ast }$. Hence $SU(n)$ is a compact topological group.

If $\delta \notin \mathbb{Q}(\alpha )$, thenLetThis is the nonidentity automorphism of $K/\mathbb{Q}(\alpha )$. By extending an embedding that sends $\alpha$ to $\beta$, and composing with $s$ if necessary, chooseThenThe eight elementsare distinct, and exhaust the Galois group. Thus the dihedral Galois action on four radical roots gives

Now suppose $c$ is a square in $\mathbb{Q}$. Then $\delta \in \mathbb{Q}$, so every automorphism fixes $\delta$. We are in case (i) with $\tau (\delta )=\delta$, and therefore

Conversely, suppose the group is the Klein four-group. Label the rootsThe three nonidentity elements in its transitive action are the three double transpositions. Each fixes $\delta$: they send the ordered pair $(\alpha ,\beta )$ respectively towhose products all equal $\alpha \beta$. Hence every element of the Galois group fixes $\delta$. By the Artin fixed-field theorem,Therefore $c={\delta }^2$ is a square in $\mathbb{Q}$. This proves the Klein-four criterion for an irreducible even quartic:

PutThe four roots are $\alpha ,-\alpha ,\beta ,-\beta$. If $\delta \in \mathbb{Q}(\alpha )$, thenbecause $g$ is irreducible. Hence the Galois group has order four and acts transitively, hence regularly, on the roots.

Let $\tau$ be the automorphism taking $\alpha$ to $\beta$. Since $\tau (\delta )=\pm \delta$,If $\tau (\delta )=\delta$, then $\tau$ is an involution. Together with the distinct involution taking $\alpha$ to $-\alpha$, it givesIn the ordered root set $(\alpha ,-\alpha ,\beta ,-\beta )$ these generators act as double transpositions.

If $\tau (\delta )=-\delta$, thenso $\tau$ has order four andThis is the degree-four case in the Galois group of an irreducible even quartic.

Put $z=x_3$. The body is a solid of revolution with disk radiusIts volume and uniform mass density are

By the continuous center of mass formula, the distance of the center of mass from the pointed origin is

For a disk of radius $R$, $\int r^{2}dA=\pi R^4/2$. Therefore the axial moment isFor the transverse $x_1$-axis, $\int x_2^{2}dA=\pi R^4/4$, while every slice also contributes $z^2$ times its mass. HenceBy axial symmetry $I_2=I_1$. These values agree with the moment of inertia of a pointed cubic-profile solid of revolution.

A Lagrange top is a rigid body that is symmetric about a principal axis, has a point on that axis fixed in space, and has its center of mass on the same axis while gravity acts uniformly. Here $I_1=I_2$ is the transverse principal moment of inertia about the fixed point, $I_3$ is the moment about the symmetry axis, $M$ is the total mass, and $l$ is the distance from the fixed point to the center of mass.

The Euler angles for a symmetric top use $\theta$ for inclination, $\varphi$ for precession, and $\psi$ for spin about the body axis. Since $\psi$ is a cyclic coordinate, its generalized momentumis conserved. The coordinate $\varphi$ is also cyclic, sois a second integral. Finally, the Lagrangian has no explicit time dependence, and conservation of energy from time-translation invariance gives the independent integral

For steady precession set $\theta$ constant and $\dot{\varphi }=\Omega$ constant. The $\theta$ Euler-Lagrange equation, with $0<\theta <\pi /2$, reduces after division by $\sin \theta$ toThis quadratic has a real precession rate precisely when its quadratic discriminant is nonnegative. Hence Steady precession of a Lagrange top is possible if and only if

Uniform per-capita culling subtracts $kS$ and $kI$ from the two equations:Adding and again differentiating $\theta =I/N$ gives the Uniform per-capita culling in the plant SI model equationsThe direct culling terms cancel from the prevalence equation because both classes are culled equally, but culling lowers the eventual total population. For $0\le k<1$, $N(t)\to 1-k$, and a rare infection grows precisely whenTherefore, when $\beta >1$, disease is eliminated while a positive plant population remains exactly forFor $\beta \le 1$ the disease already dies out without culling. If $k\ge 1$, culling drives the entire plant population to extinction, which removes the disease only by removing its host.

Defineso $S=N(1-\theta )$ and $I=N\theta$. Part (a) already givesUsing the quotient rule,and hence the reduced Plant SI model with logistic total population is

For every $N(0)>0$, logistic growth gives $N(t)\to 1$. The asymptotic prevalence equation is thereforeIf $\beta <1$, its only nonnegative equilibrium is $\theta =0$, which attracts all prevalences. If $\beta >1$, $heta=0$ is unstable and the stable value is ${\theta }_{\ast }=1-1/\beta$. Combining these values with $N_{\ast }=1$ recovers exactly the equilibria and outcomes in part (b).

The fixed-point conditions areOn $I=0$ they give $(S,I)=(0,0)$ and the disease-free equilibrium $(1,0)$. An interior fixed point requires $I=(\beta -1)S$, so it exists only for $\beta >1$; substitution in the first equation gives the endemic equilibrium

The Jacobian matrix isAt $(1,0)$ its eigenvalues are $-1$ and $\beta -1$. Hence $(1,0)$ is asymptotically stable for $\beta <1$ and a saddle point for $\beta >1$. At the endemic equilibrium, the equivalent $(N,\theta )$ coordinates give eigenvalues $-1$ and $1-\beta$, so it is a stable node whenever it exists. The origin is unstable because every solution with $N(0)>0$ has $N(t)\to 1$.

For the phase portrait, the line $I=0$ is invariant and points toward $(1,0)$. Alsoso trajectories move toward the line $S+I=1$. If $\beta <1$, then $\dot{I}<0$ throughout the interior and every nonzero solution approaches $(1,0)$. If $\beta >1$, the additional infective nullcline is the ray $I=(\beta -1)S$; $I$ increases below this ray and decreases above it, and every trajectory with $I(0)>0$ approaches $(1/\beta ,1-1/\beta )$. At $\beta =1$ the disease-free and endemic branches exchange stability in a transcritical bifurcation.

Write $N=S+I$. Inthe term $N$ represents production of healthy new plants by both healthy and diseased plants at unit per-capita rate. The term $-NS$ is density-dependent loss of healthy plants through competition, and $-\beta IS$ is loss through mass-action infection. Inthe term $-NI$ is the same density-dependent loss for diseased plants, while $+\beta IS$ is the matching gain when healthy plants become diseased. Thus $\beta$ is the transmission-rate coefficient per healthy-diseased encounter.

Adding the equations cancels the infection terms and gives the logistic differential equationThe disease therefore changes how the population is divided between healthy and diseased plants but has no effect on the total-population dynamics: infected plants reproduce at the same rate, all offspring enter the healthy class, and infection causes no additional death in this model.

For $k=1,\dots ,C$, define the class degree sumand define the number of within-college friendships byEach between-college edge contributes once to the degree sum of each endpoint's college, while each within-college edge contributes twice to its college's degree sum. The model-3 likelihood can therefore be writtenThe Fisher-Neyman factorization theorem shows that $T=(T_0,T_1,\dots ,T_C)$ is sufficient.

For two networks $y,y^{\prime }$, the likelihood ratio isBecause the parameters range independently over ${\mathbb{R}}^{C+1}$, this ratio is constant in $\beta$ exactly when every displayed difference vanishes. The likelihood-ratio criterion for minimal sufficiency therefore gives the answerThis is the degree-sum sufficient statistic for an additive logistic network model.

Let $n_{\ell }$ be the number of graduates in college $\ell$. In model 1, every edge from a person in college $k$ to a person in college $\ell$ has probabilitysoIn model 2 the corresponding probability isBecause $k\ne \ell$, the within-college indicator in model 3 vanishes, so $p_{k\ell }^{(3)}=p_{k\ell }^{(2)}$. Consequently models 2 and 3 both give

The distinct edge indicators used in $M_i$ and $M_j$ are independent under all three models, so every model also forces $M_i$ and $M_j$ to be independent. Thus all members of one college have the same cross-college friendship propensity, with no person-level heterogeneity or dependence between their friendship counts. Real friendship networks can have unusually sociable individuals and correlated choices of friends, producing overdispersion and dependence that these models cannot fit.

Model 2 is obtained from model 3 by setting ${\beta }_0=0$. Model 3 is contained in model 1 by choosing the symmetric block parametersHence

Assuming every college-pair cell is observed and the models are identifiable, model 1 has $C(C+1)/2$ parameters and model 3 has $C+1$. The likelihood-ratio test statistic for model 3 against model 1 isUnder model 3 and the regularity conditions of Wilks theorem,

Conditional on the observed colleges, the stochastic block model assumes that the friendship indicators are independent random variables. For arbitrary linear predictors ${\theta }_{ij}$, their Bernoulli distribution likelihood isThus the three requested likelihoods are

For $\lambda \notin K$, write $g_{\lambda }(z)=(z-\lambda {)}^{-1}$. If $\lambda \in \Lambda$ andthenuniformly on $K$. Uniformly polynomially approximable functions form an algebra closed under uniform limits, so $\mu \in \Lambda$. Hence $\Lambda$ is open.

The set $\Lambda$ is also closed relative to $\mathbb{C}\setminus K$. Indeed, if ${\lambda }_j\in \Lambda$ and ${\lambda }_j\to \lambda \notin K$, thenso $g_{\lambda }$ is again uniformly approximable by polynomials. Since $K$ is closed, every point of $\Gamma$ has a neighbourhood in $\mathbb{C}\setminus K$; relative closedness of $\Lambda$ then gives a smaller neighbourhood disjoint from $\Lambda$. Thus both $\Lambda$ and $\Gamma$ are open.

It is not always true that $\Lambda$ is nonempty. Take $K=\mathbb{R}$. Any polynomial uniformly close on $\mathbb{R}$ to a bounded resolvent $g_{\lambda }$, with $\lambda \notin \mathbb{R}$, must itself be bounded on $\mathbb{R}$, and hence must be constant. Uniform limits of constants are constant, whereas $g_{\lambda }$ is not. Therefore

Boundedness of $K$ does not force $\Gamma$ to be empty either. Take $K=\{z:|z|=1\}$ and $\lambda =0$. If polynomials $p_n$ converged uniformly to $1/z$ on $K$, then uniform convergence and contour integration would givea contradiction. Hence

Letand choose $0<\eta <\delta$. Write $C_{\eta }$ for the positively oriented boundary of the square with $|x|,|y|\le 1+\eta$. The Cauchy integral formula gives, for every $z\in S$,Because $C_{\eta }$ and $S$ are compact and separated by a positive distance, Riemann sums for this contour integral converge uniformly in $z\in S$. Thus, for any $\epsilon >0$, there are points ${\zeta }_j\in C_{\eta }$ and constants $c_j$ such that

It remains to approximate each resolvent by polynomials on $S$. Let $\Lambda$ be the set of $\lambda \notin S$ for which $(z-\lambda {)}^{-1}$ is polynomially uniformly approximable on $S$. If $|\lambda |>{\sup }_{z\in S}|z|$, the geometric seriesshows that $\lambda \in \Lambda$. Moreover, the resolvent propagation across a complementary component argument shows that $\Lambda$ is both open and closed relative to $\mathbb{C}\setminus S$. The complement of a square is connected, and $\Lambda$ is nonempty, so $\Lambda =\mathbb{C}\setminus S$.

In particular, every ${\zeta }_j$ lies in $\Lambda$. Approximate the finitely many resolvents accurately enough that their weighted sum differs by less than $\epsilon /2$. Combining this with the Riemann-sum estimate produces a polynomial $p$ satisfyingThis proves the claimed elementary polynomial approximation on a square.

Pure periodicity means that $\theta$ reappears as the tail after one period. The continued-fraction tail formula therefore givesAfter cross-multiplication,

Now letBy part (c), the matrix for one reversed period isso its positive fixed point $\eta$ satisfiesPutting $\eta =-1/X$ and multiplying by $-X^2$ turns this equation into $f(X)=0$. The product of the two roots of $f$ is $-p_{n-1}/q_n<0$; because $\theta >0$, the other root ${\theta }^{\prime }$ is negative. Consequently $-1/{\theta }^{\prime }>0$ is the positive fixed point selected by convergence of the reversed continued fraction. The reversed-period identity for a purely periodic continued fraction is therefore

The continued-fraction matrix for the original finite continued fraction isEach $M(a_j)$ is a symmetric matrix, so reversing the order gives the transpose:Reading off the columns proves the reversal identity for a finite continued fraction:These are respectively the $n$th and $(n-1)$th convergents of the reversed continued fraction.

Let ${\beta }_{n+1}>0$ be the complete quotient after $a_n$. Part (a) writes the irrational number $\theta$ asso $\theta$ lies between the adjacent continued fraction convergents $p_n/q_n$ and $p_{n-1}/q_{n-1}$. Their separation isBecause all $a_j$ are positive integers, the recurrence $q_n=a_n{q}_{n-1}+q_{n-2}$ implies $q_n\to \infty$. Hence this separation tends to zero, and the squeeze theorem gives

Setand, for $n\ge 1$, defineWriting $M(a)=\left(\begin{array}{cc}a& 1\\ 1& 0\end{array}\right)$, these recurrences are equivalent to the continued-fraction matrix identityThe associated Möbius transformation sends a positive tail $\beta$ to the continued-fraction tail formula

Since $\det M(a_j)=-1$, taking determinants givesMoreover,whereasTheir signs are opposite, so ${\theta }_n$ lies between the two adjacent convergents. For $n=0$, the second endpoint is interpreted as $p_{-1}/q_{-1}=\infty$.