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Take the circular cylinderIts first fundamental form is $R^{2}d{\theta }^2+d{z}^2$. The universal covering parametrization ${\mathbb{R}}^2\to S$ is therefore a local isometry from the Euclidean plane, after scaling the first coordinate. By geodesic preservation by a local isometry, every geodesic on the cylinder lifts to a straight line defined for all $t\in \mathbb{R}$. Thus the cylinder is geodesically complete. It is noncompact because $z$ is unbounded, and it is not a plane because one of its principal curvatures is $1/R$.

A twice differentiable curve $\gamma :I\to S$ is a geodesic when its velocity is parallel along the curve:For a surface embedded in ${\mathbb{R}}^3$, the ambient acceleration criterion for a surface geodesic gives the equivalent conditionThe parameter $t$ is then an affine parameter, and $|\dot{\gamma }|$ is constant.

Subtracting the global weighted mean from $h$ leaves its derivative unchanged, so it suffices to assumeWrite $w=e^{-\Phi }$, choose $R\ge R_0$, and putThe weighted mean of $h$ on $[-R,R]$ isBy the Cauchy-Schwarz inequality,Part (c), applied on $[-R,R]$, therefore gives for $E={\int }_{\mathbb{R}}|h^{\prime }{|}^{2}w$Part (a) gives independently

Choose ${ϵ}_0>0$ so small that ${ϵ}_0{K}_1\le 1/2$, and then choose $R$ large enough that $q/Z\le {ϵ}_0{\lambda }_1/2$. Multiply the last inequality by $ϵ={ϵ}_0{\lambda }_R$ and add it to the local inequality. This yieldswhere $c_R=\frac{{\lambda }_R}{2}\min \{1,{ϵ}_0{\lambda }_1\}>0$. Hence, with ${\lambda }_0=c_R/(1+ϵ)$,

LetThen $q^{\prime }=h^{\prime }$ and $q$ has zero weighted mean. Part (b) givesTherefore, with $Z_R={\int }_{-R}^R{e}^{-\Phi }$,Thus the compact-interval Weighted Poincare inequality holds with

LetThe fundamental theorem of calculus and the Cauchy-Schwarz inequality giveThus the one-half Hölder seminorm is at most $m_R^{-1/2}{E}_R(h{)}^{1/2}$.

Writing $Z_R={\int }_{-R}^{R}w>0$ and using the zero weighted mean, for each $x$ we haveSince $|x-y|\le 2R$, the preceding estimate yieldsConsequently the required inequality holds, for example, with

SetThe ground-state transform for a weighted Dirichlet energy, obtained by expanding the square and using integration by parts, isThe boundary term vanishes because $h$, and hence $g$, has compact support.

Since $V(x)\to \ell >0$, choose $R_0$ so that $V\ge \ell /2$ when $|x|\ge R_0$, and setFor every $R\ge R_0$, discard the nonnegative $|g^{\prime }{|}^2$ term and the nonnegative contribution from $R_0\le |x|\le R$ to obtainBecause $|g{|}^2=|h{|}^2{e}^{-\Phi }$, this is exactly the required estimate.

Fix $z_0\in S^1$, and let ${\delta }_{z_0}\in C(S^1{)}^{\ast }$ be the point evaluation functional ${\delta }_{z_0}(f)=f(z_0)$. Define orbit averages in the dual space byEach ${\mu }_n$ has norm at most one and satisfies ${\mu }_n(1_{S^1})=1$. The space $C(S^1)$ is separable, so part (a) gives a subsequence converging pointwise to some $\mu \in C(S^1{)}^{\ast }$. In particular,For every $f\in C(S^1)$, the Cesàro average telescopes at the endpoints:whose absolute value is at most $2\Vert f{\Vert }_{\infty }/n$. Passing to the convergent subsequence provesThus $\mu$ is an invariant functional of a composition operator.

It need not be unique when $T$ is not the identity operator. For the nonidentity homeomorphism $\tau (z)=\overline{z}$, both $1$ and $-1$ are fixed. Hence the two distinct point evaluationshave value one on $1_{S^1}$ and are invariant under $T$. Therefore nonidentity does not imply uniqueness.

The fixed-point space $I=\ker (U-I)$ is closed. If $y\in I$ and $w=Ux-x\in W$, then unitarity givesso $I\perp W$.

Conversely, $z\in W^{\perp }$ exactly whenfor every $x\in H$. Thus $U^{\ast }z=z$, equivalently $Uz=z$, and $z\in I$. Hence $W^{\perp }=I$, so the double orthogonal complement identity givesThe orthogonal decomposition by a closed subspace now yields

Define the Cesàro averageFor $y\in I$, $A_{n}y=y$. For a generator $w=Uz-z$ of $W$, the sum telescopes:Moreover $\Vert A_n\Vert \le 1$, so approximation extends this convergence from $W$ to $\overline{W}$. Writing $x=Px+(I-P)x$ according to the orthogonal decomposition gives the Von Neumann mean ergodic theorem

Choose a countable dense subset $\{x_1,x_2,\dots \}$ of the normed vector space $X$. For each fixed $j$, the scalar sequence $(f_n(x_j))$ is bounded becauseRepeated use of the Bolzano-Weierstrass theorem followed by a diagonal subsequence argument produces a subsequence $(f_n{)}_{n\in \Lambda }$ for which $f_n(x_j)$ converges for every $j$.

Let $D$ be the linear span of the $x_j$. On every $x\in D$, defineFinite linear combinations show that $f:D\to \mathbb{R}$ is linear, and passage to the limit in $|f_n(x)|\le \Vert x\Vert$ gives $|f(x)|\le \Vert x\Vert$. Thus $f$ is a bounded linear functional on the dense linear subspace $D$ and extends uniquely to an element of the dual space $X^{\ast }$ with $\Vert f\Vert \le 1$.

For arbitrary $x\in X$ and $y\in D$,First choose $y$ close to $x$, then let $n\in \Lambda$ tend to infinity. This provesThis is the Sequential Banach-Alaoglu theorem for a separable predual.

The given homeomorphisms and homotopy invariance of homology givebecause $n\ge 2$, while $M\simeq S^{n-1}$ is path connected and $L$ is the disjoint union of two copies of $S^{n-1}$. HenceThe degree-one and degree-zero part of the Mayer–Vietoris sequence is thereforeTaking the rank of an abelian group throughout this short exact sequence givesso $\mathrm{rank}{H}_0(N)=2$. Since the zeroth homology group is freely generated by the path components,

Represent a class in $H_k(K)$ by a chain cycle $z\in Z_k(K)$. Since every simplex of $K=M\cup N$ belongs to at least one of the two simplicial subcomplexes, split the chain in the simplicial chain complex asBecause $\partial z=0$, we haveThe left side lies in $C_{k-1}(M)$ and the right side in $C_{k-1}(N)$, so this common chain is supported in $L=M\cap N$. It is a cycle because ${\partial }^2=0$. The connecting homomorphism is thereforeChanging the decomposition or the representative changes $\partial m$ only by a boundary in $L$, which is why the construction descends to homology classes.

Let $i:L\hookrightarrow M$, $j:L\hookrightarrow N$, $r:M\hookrightarrow K$, and $s:N\hookrightarrow K$ be the inclusions. The Mayer–Vietoris sequence is the long exact sequence in homologyThe same sequence continues through degree zero, or uniformly in all degrees when written with reduced homology.

Now let $n=m=p$ be prime and put $\alpha =\sqrt[p]{p}$ and $A=\mathbb{Z}[\alpha ]$. The Eisenstein criterion applied at $p$ gives $[\mathbb{Q}(\alpha ):\mathbb{Q}]=p$, while the discriminant of elements of a number field of its power basis iswhere $f(X)=X^p-p$. By the discriminant-index formula for an integral lattice, every prime dividing $[{\mathcal{O}}_L:A]$ must therefore be $p$.

Suppose $A\ne {\mathcal{O}}_L$. The finite abelian group ${\mathcal{O}}_L/A$ is then a nontrivial finite p-group, so it contains an element of order $p$. Consequently there is an $x\in {\mathcal{O}}_L\setminus A$ such thatwith some $c_k$ not divisible by $p$.

The polynomial $X^p-p$ is Eisenstein, so total ramification from an Eisenstein polynomial gives a unique prime ideal $P$ above $p$ with normalized discrete valuationLet $k_0$ be the least index for which $p\nmid c_{k_0}$. The term $c_{k_0}{\alpha }^{k_0}$ has valuation $k_0<p$; every earlier term has valuation at least $p$, and every later term with coefficient prime to $p$ has a distinct, larger valuation. The non-Archimedean valuation therefore givesThis contradicts $x\in {\mathcal{O}}_L$, since an algebraic integer has nonnegative valuation at every prime ideal. Thus $A={\mathcal{O}}_L$, and

Write $\alpha =\sqrt[n]{m}$ and $\zeta =e^{2\pi i/n}$. The $n$ field embeddings of $L$ into $\mathbb{C}$ send $\alpha$ to ${\zeta }^j\alpha$, so the field trace isFor the trace pairing $T(x,y)={\mathrm{Tr}}_{L/\mathbb{Q}}(xy)$, the basis ${\beta }_i={\alpha }^i$ therefore has dual basis

Every element of $\mathbb{Z}[\alpha ]$ is an algebraic integer. If $x\in {\mathcal{O}}_L$ and $z\in \mathbb{Z}[\alpha ]$, then $xz$ is an algebraic integer and its trace is an integer. Thus $x$ belongs to the trace-dual lattice of $\mathbb{Z}[\alpha ]$, which is generated over $\mathbb{Z}$ by the displayed dual basis. Since $1/n=m/(nm)$, every such linear combination lies in $(nm{)}^{-1}\mathbb{Z}[\alpha ]$. Hence

Put $\alpha =\sqrt[n]{m}$. It is an algebraic integer because it satisfies the monic polynomial $X^n-m$. Choose a prime number $q$ dividing the square-free integer $m$, and choose a prime ideal $\mathfrak{p}$ of ${\mathcal{O}}_L$ above $q$. Fromand prime ideal factorization, the $\mathfrak{p}$-adic exponents satisfybecause $q$ occurs to exponent one in $m$. Hence $n$ divides the ramification index of a prime ideal $e_{\mathfrak{p}}$. If $d=[L:\mathbb{Q}]$, then $e_{\mathfrak{p}}\le d$, while the equation ${\alpha }^n=m$ gives $d\le n$. Thereforeso

Writing $\rho (A)X=A_{1}X{A}_1^{-1}$, we haveand $\rho (I)=I$, so this conjugation representation is indeed a group representation.

The underlying three-dimensional module isUsing $M_3(\mathbb{C})=\mathrm{End}(W)\cong W\otimes W^{\ast }$, the self-duality of finite-dimensional SU2 representations, and $V_1\otimes V_1\cong V_2\oplus V_0$, we obtainTherefore the irreducible decomposition is