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Cut the square model of the Klein bottle along the midline parallel to the pair of sides whose identification reverses orientation. Each half becomes a Möbius band, and the two new boundary circles are identified. If $a$ and $b$ are the core loops of these two bands, each boundary travels twice around its core. The Seifert-van Kampen theorem therefore gives

Writing $x=a^{-1}b$ and $y=a$ converts this into the standard Klein-bottle presentationThe orientation homomorphism sends both $a$ and $b$ to the nonzero element of $C_2$. Its kernel is the index-two subgroupBy the classification of connected covering spaces, $H$ determines the orientation double cover of the Klein bottleChoosing torus generators $\alpha ,\beta$ along the two translation directions, we may takeThese commute: the standard relation says that $a$ conjugates $a^{-1}b=x$ to $x^{-1}$, so $a^2$ centralizes $x$.

The space $Y$ is the mapping cylinder of $p$ with its unused end omitted. Sliding $(x,t)$ to $(x,0)$ gives a deformation retraction of $Y$ onto the bottom quotient $T^2/\sim \cong K$. Hence

Finally suppose $U\subseteq X$ is open and homeomorphic to $Y$, and identify the bottom copy of $K$ inside $U$. This $K$ is compact and therefore closed in the Hausdorff space $X$. Apply van Kampen, in its groupoid form if intersections are disconnected, toThe intersection $U\setminus K$ deformation retracts to a positive-height torus, and its map to ${\pi }_1(U)\cong {\pi }_1(K)$ has image preciselyThe orientation homomorphismis zero on this intersection, so it is compatible with the trivial homomorphism from ${\pi }_1(X\setminus K)$ to $C_2$. The pushout property in van Kampen extends it to a surjectionThereforeThis is the Klein-bottle mapping-cylinder obstruction to simple connectivity.

Two nonzero fractional ideals $I,J$ of ${\mathcal{O}}_K$ are in the same ideal class whenfor some $\alpha \in K^{\times }$, equivalently when $I{J}^{-1}$ is principal. Multiplication is well defined on classes:The identity is $[{\mathcal{O}}_K]$, the inverse of $[I]$ is $[I^{-1}]$, and commutativity comes from ideal multiplication. Nonzero fractional ideals are invertible because ${\mathcal{O}}_K$ is a Dedekind domain. Thus these classes form the abelian ideal class group $\mathrm{Cl}(K)$.

For finiteness, begin with a nonzero integral ideal $I$. By hypothesis choose $0\ne \alpha \in I$ withSince $(\alpha )\subseteq I$, the idealis integral, represents $[I{]}^{-1}$, and satisfiesHence every class has an integral representative of bounded norm. There are only finitely many integral ideals of bounded norm, so $\mathrm{Cl}(K)$ is finite. This is the bounded-norm ideal representatives argument.

Now take $K=\mathbb{Q}(\sqrt{-33})$. Since $-33\equiv 3(\mathrm{mod}4)$,The imaginary-quadratic Minkowski bound for ideal classes isEvery class is therefore represented by an integral ideal of norm at most seven.

The ramified prime ideals above two and three arewithNeither is principal, since the norm equationhas no integer solution. Their product is also nonprincipal, since $a^2+33b^2=6$ has no solution. Thusare four distinct classes, all of order at most two.

The prime five is inert, so it contributes no ideal of norm five. The prime seven splits asSincecomparison of norms, all equal to $42$, confirms the factorization and givesThe conjugate prime has the inverse class, which is the same because this class has order two. Ideals of norms four and six yield respectively the principal class and $[{\mathfrak{p}}_2{\mathfrak{p}}_3]$. The Minkowski bound now shows that the four displayed classes exhaust the group. ThereforeThis is the Ideal class group of Q of square root of minus thirty-three.

The simple continued fraction algorithm for $\sqrt{29}$ begins with $a_0=5$. Its complete quotients repeat afterThus

The convergents through one period areThe last one gives the negative Pell equation solutionTherefore

The Mackey restriction formula says, for an $H$-representation $V$,Frobenius reciprocity saysApplying both with an irreducible character $\chi$ givesThe identity double coset contributes one, and all terms are nonnegative integers. Therefore Mackey irreducibility criterion says that ${\mathrm{Ind}}_H^G\chi$ is irreducible exactly when every term belonging to a nonidentity double coset is zero.

Now take $G=S{L}_2(k)$. Writeso $B=T⋉U$. For $q\ge 4$, commutators of $T$ with $U$ generate $U$, because conjugation sends $u(b)$ to $u(a^{2}b)$ and one can choose $a^2\ne 1$. Every degree-one character of $B$ is therefore trivial on $U$ and has the formwhere $\theta :k^{\times }\to {\mathbb{C}}^{\times }$ is a multiplicative character.

The Bruhat decomposition of SL2 over a finite field has two double cosets,and $B\cap wB{w}^{-1}=T$. Conjugation by $w$ inverts $T$, soThe only nonidentity Mackey term vanishes exactly when these two one-dimensional characters differ. ConsequentlyThus the trivial character is excluded, as is the unique quadratic character when $q$ is odd. All other degree-one characters of $B$ induce irreducibly; the induced representations have degree $[G:B]=q+1$. This is the irreducible principal series of finite SL2.

In the Erdős-Rényi model $G(n,p)$, the vertex set is $[n]$ and every one of the $\left(\genfrac{}{}{0ex}{}{n}{2}\right)$ possible edges is present independently with probability $p$.

Let $X$ count isolated vertices. If $p\ge (1+\epsilon )\log n/n$, the union bound givesThus $\mathbb{P}(E_n)\to 1$.

Chebyshev inequality states thatNow suppose $p\le (1-\epsilon )\log n/n$. Monotonicity lets us use the largest allowed $p$. Withthe supplied lower exponential estimate showsFor distinct vertices $i,j$,HenceandChebyshev with $t=\mathbb{E}X$ gives $\mathbb{P}(X=0)\to 0$, so $\mathbb{P}(E_n)\to 0$.

For connectivity, the lower-threshold result is immediate because a connected graph has no isolated vertex:For the upper threshold, if a graph is disconnected it has a component with vertex set $S$ of some size $1\le k\le n/2$. In particular, every one of the $k(n-k)$ edges from $S$ to its complement is absent. ThereforeChooseFor $1\le k\le \eta n$, the standard bound $\left(\genfrac{}{}{0ex}{}{n}{k}\right)\le (en/k{)}^k$ givesThe sum of these terms tends to zero. For $\eta n\le k\le n/2$, use $\left(\genfrac{}{}{0ex}{}{n}{k}\right)\le 2^n$ and $k(n-k)\ge \eta n^2/2$:Thus $\mathbb{P}(F_n)\to 1$ above the threshold. Together,This is the connectivity threshold in the Erdős-Rényi model.

The gravitational three-body Lagrangian is

Put $M_{12}=m_1+m_2$ and $M=M_{12}+m_3$. Solving the definitions of the Jacobi coordinates for three particles givesSubstitution into the kinetic energy makes all cross terms cancel:where

The pair separations areHence the gravitational potential energy iswhich is independent of $\mathbf{c}$. Thus $\mathbf{c}$ is an ignorable coordinate andThe center of mass moves uniformly because the isolated system has no external force.

Finally, substituting the inverse coordinate transformation intoagain cancels every cross term and yieldsThis is the angular momentum decomposition in three-body Jacobi coordinates.

This is attenuation bias from classical measurement error, also called regression dilution. The latent predictor is $Z$, but $X_1$ and $X_2$ are noisy proxies. The noisier proxy produces a slope closer to zero.

For a regression with an intercept, the population slope isHere all variables are centred and the noises are independent. Thereforeso regressing $Y$ on $Z$ has slope $\beta =2$. Forwe haveThusWith the stated values, the slopes arematching the simulation.

Under this independent additive measurement-error model, the magnitude of the $Y$-on-$X_1$ slope is generally smaller than the magnitude of the $Y$-on-$Z$ slope whenever ${\sigma }_{x_1}^2>0$. Increasing ${\sigma }_y$ does not change the population slope, because response noise contributes neither to $\mathrm{Cov}(X_1,Y)$ nor to $\mathrm{Var}(X_1)$. It increases residual variance and the sampling variability of the estimate; with one million observations, doubling it should leave the displayed slope close to $1.6$ while increasing its standard error.