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Given $\epsilon >0$, choose $N$ with ${\sum }_{i>N}|⟨a,e_i⟩{|}^2<{\epsilon }^2$. Every cube point has uniformly bounded tail by this sum, while its first $N$ coordinates lie in a compact finite-dimensional box. A finite net for that box is therefore an $\epsilon$-net for the cube. The coordinate inequalities define a closed set, so the cube is complete and totally bounded, hence compact.

The upper frame bound gives $\Vert Ux{\Vert }_{{\ell }^2}\le \sqrt{B}\Vert x\Vert$. For $c\in {\ell }^2$, the adjoint is the norm-convergent synthesis seriesbecause its inner product with $x$ is ${\sum }_i{c}_i⟨e_i,x⟩$.

A countable orthonormal basis is a frame with $A=B=1$. For any frame, a vector orthogonal to every $e_i$ makes the lower frame bound force its norm to vanish. Therefore the span is dense.

For fixed $y$, the map $x\mapsto ⟨Tx,y⟩$ is a bounded functional. The Riesz representation theorem gives a unique vector $T^{\ast }y$ representing it. Linearity, uniqueness, and $\Vert T^{\ast }y\Vert \le \Vert T\Vert \Vert y\Vert$ make $T^{\ast }$ the unique bounded adjoint operator.

Choose a triangulation of the sphere having every branch point as a vertex. Its lift has twice every edge and face, and twice every nonbranch vertex, but only one vertex above each of the $n$ branch points. HenceSince $\chi ({\Sigma }_g)=2-2g$, this gives $n=2g+2$, the degree-two Riemann-Hurwitz formula.

For a finite triangulation, the Euler characteristic is $\chi (X)=V-E+F$, independent of the triangulation.

The two-sign map $K^{\ast }\to \{\pm 1{\}}^2$ is surjective, with kernel the totally positive elements, so their index is four. Narrow equivalence is reflexive, symmetric, and transitive because totally positive elements form a group; ideal multiplication makes the classes an abelian group.

The map from the narrow class group to the ordinary class group is onto. Its kernel records sign patterns of principal generators modulo signs realized by units. A norm-$-1$ fundamental unit realizes the two mixed signs, making the kernel trivial; otherwise only equal signs occur and the kernel has order two. Thus the narrow class number is $h_K$ in the first case and $2h_K$ otherwise. Here it is $4$.

${\mathcal{O}}_K=\mathbb{Z}[\sqrt{30}]$ and $d_K=120$, so every class has an ideal of norm at most $\sqrt{30}$. Factoring the primes $2,3,5$ and checking principal norms shows that the ramified prime ideals represent one nontrivial class of order two; hence $\mathrm{Cl}(K)\cong \mathbb{Z}/2$. The continued fraction of $\sqrt{30}$ gives the least unit greater than one as $11+2\sqrt{30}$, of norm $1$.

$S^1=\{z\in \mathbb{C}:|z|=1\}$. A continuous finite-dimensional representation is unitary after averaging an inner product over Haar measure, and commuting unitary matrices diagonalize simultaneously. Its irreducibles are therefore $z\mapsto z^n$, $n\in \mathbb{Z}$. Their characters obey ${\int }_{S^1}{z}^n\overline{z^m}dz={\delta }_{nm}$.

The relations ${\sigma }^n={\tau }^2=1$ and $\tau \sigma \tau ={\sigma }^{-1}$ give the dihedral group; its $2n$ substitutions $z\mapsto {\zeta }_n^{j}z$ and $z\mapsto {\zeta }_n^j/z$ are distinct. The element $w=z^n+z^{-n}$ is fixed. Since $z$ satisfies $X^{2n}-w{X}^n+1=0$, the extension has degree at most $2n$, while the automorphism group gives degree at least $2n$. Hence $L^G=\mathbb{C}(w)$.

The orbit polynomial ${\prod }_{\beta \in G\alpha }(X-\beta )$ has coefficients fixed by $G$, lies in $K[X]$, and has distinct roots. Thus $\alpha$ is separable and its degree is the orbit size $|G:G_{\alpha }|$, which divides $|G|$. If $L=K(\alpha )$ and the action is faithful, only the identity fixes $\alpha$, so the degree is $|G|$.

An action by field automorphisms is a homomorphism $Go\mathrm{Aut}(L)$. Elements fixed by every group element are closed under addition, multiplication, additive inverses, and inversion of nonzero elements, so $L^G$ is a subfield.

For a constant translation vector $a$, take $F=a\cdot {\sum }_i{p}_i$. Then $\delta r_i=\epsilon a$ and $\delta p_i=0$. The potential depends only on position differences and the kinetic energy only on momenta, so this is a symmetry and total momentum is conserved.

Taylor expansion gives $\delta H=H_q\cdot \delta q+H_p\cdot \delta p=-\epsilon \{F,H\}$. Hamilton's equation gives $\dot{F}=\{F,H\}$, so $F$ is conserved exactly when the generated infinitesimal canonical transformation leaves $H$ invariant.

To first order the conditions are ${\partial }_{p_j}\delta q_i-{\partial }_{p_i}\delta q_j=0$, ${\partial }_{q_j}\delta p_i-{\partial }_{q_i}\delta p_j=0$, and ${\partial }_{q_j}\delta q_i+{\partial }_{p_i}\delta p_j=0$. The choices $\delta q_i=\epsilon F_{p_i}$ and $\delta p_i=-\epsilon F_{q_i}$ satisfy them by equality of mixed partials.

The Poisson bracket is $\{f,g\}={\sum }_i(f_{q_i}{g}_{p_i}-f_{p_i}{g}_{q_i})$. Canonical variables satisfy $\{Q_i,Q_j\}=\{P_i,P_j\}=0$ and $\{Q_i,P_j\}={\delta }_{ij}$.

The speed is positive for $r<1/2$, zero at $1/2$, and negative for $r>1/2$. The state with the larger basin-weighted reaction potential invades the other; raising the Allee threshold favours extinction.