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The Gauss map of an oriented surface is its smooth unit normal $N:S\to S^2$. With the convention that the shape operator is $A_p=-d{N}_p$, the second fundamental form is ${\mathrm{II}}_p(v,w)=⟨A_{p}v,w⟩$ and the normal curvature in a nonzero direction $w$ is $k_n(w)=\mathrm{II}(w,w)/⟨w,w⟩$. The two principal curvatures are the real eigenvalues $k_1,k_2$ of this self-adjoint operator; the mean curvature is $H=(k_1+k_2)/2$.

For directions successively spaced around the tangent plane, write their angles to a principal direction as ${\theta }_j={\theta }_0+(j-1)\pi /m$. The Euler formula for normal curvature givesThe geometric series ${\sum }_{j=0}^{m-1}{e}^{2\pi ij/m}=0$ for $m\ge 2$ makes the cosine sum zero. ThereforeHere equal angular spacing is understood in the usual directed sense. If the printed consecutive angles were only unsigned, without requiring the same rotational sense, the assertion would be false: for $m=3$ the directions $0,\pi /3,0$ satisfy those unsigned angle conditions but do not generally give $3H$.

A minimal surface has $H=0$, so at each point $k_2=-k_1$. In an orthonormal principal basis, $A=\mathrm{diag}(k_1,-k_1)$ and $A^2=k_1^{2}I$. Hencewhich proves the identity with $\boxed{\mu (p)=k_1^2=-K(p)\ge 0}$, where $K$ is the Gaussian curvature.

The converse is false. The identity only forces $k_1^2=k_2^2$, which allows either $k_2=-k_1$ or $k_2=k_1$. A sphere of radius $R$ has $dN=I/R$ for its outward normal, so it satisfies the identity with $\mu =R^{-2}$, yet $H=-1/R\ne 0$ under the chosen convention. Thus a conformal Gauss map can also arise from an umbilic nonminimal surface.

The monodromy theorem states that if a holomorphic germ can be analytically continued along every path in a domain, its continuations along paths homotopic with fixed endpoints coincide. In a simply connected domain this gives a single-valued holomorphic function on the whole domain.

Write ${\zeta }_j=e^{2\pi ij/n}$. Remove the radial rays $R_j=\{t{\zeta }_j:t\ge 1\}$ and set $U=\mathbb{C}\setminus {\bigcup }_j{R}_j$. This is an open star-shaped domain containing zero, hence simply connected, and $h=z^n-1$ never vanishes there. Locally its logarithm continues along all paths in $U$, so monodromy gives a holomorphic logarithm $L$ andEquivalently, integrate $h^{\prime }/h$ on the simply connected domain and choose the additive constant to obtain $e^L=h$.

Take $r$ copies of the slit domain, labeled by $j\in \mathbb{Z}/r\mathbb{Z}$, carrying values $e^{(L+2\pi ij)/r}$. Across each cut identify banks so that a positive loop around its simple zero increments the sheet index by one. This cyclic gluing produces the unbranched Riemann surface over $\mathbb{C}\setminus Z$, on which $w^r=z^n-1$ is single valued. Completing each finite puncture gives local coordinate $w$, withThere is one ramification point of index $r$ over each of the $n$ roots of unity.

At infinity a loop around all $n$ zeros acts on sheet labels by $j\mapsto j+n(\mathrm{mod}r)$. With $d=\gcd (n,r)$ this permutation has $d$ cycles, each of length $r/d$. In the compactification there are therefore $d$ points over infinity, each with ramification index $r/d$. Infinity is unbranched precisely when $d=r$, namely when $r\mid n$. Counting branch values in the base sphere,If “branch points” denotes ramification points upstairs instead, the count is $n$ or $n+d$ respectively, with exactly the same stated equivalence.

The spectrum of a bounded operator is $\sigma (T)=\{\lambda \in \mathbb{C}:T-\lambda I\text{ has no bounded everywhere-defined inverse}\}$. Initially define $T$ on finite linear combinations of the orthonormal basis. The bilateral shift operator $S{e}_n=e_{n+1}$ preserves the Hilbert space norm and extends to a unitary operator on $H$; its inverse is $S^{\ast }{e}_n=e_{n-1}$. Thus $T=S+S^{\ast }$ extends uniquely by density, is bounded with norm at most two, and is self-adjoint.

For $v_N=(2N+1{)}^{-1/2}{\sum }_{n=-N}^N{e}_n$, $\Vert v_N\Vert =1$ and the only nonzero coefficients of $(T-2I)v_N$ occur at $-N-1,-N,N,N+1$. HenceIt follows that $\Vert T{v}_N\Vert \to 2$, proving $\boxed{\Vert T\Vert =2}$. Also $\boxed{2\in \sigma (T)}$: a bounded inverse of $T-2I$ would give $1\le \Vert (T-2I{)}^{-1}\Vert \Vert (T-2I)v_N\Vert \to 0$. Thus two belongs to the approximate point spectrum.

An eigenvector $x=\sum x_n{e}_n$ would satisfy $\lambda x_n=x_{n-1}+x_{n+1}$. Self-adjointness and the norm bound force $\lambda \in [-2,2]$. If $|\lambda |<2$, write $\lambda =2\cos \theta$ with $0<\theta <\pi$; the general complex solution of the linear recurrence relation isThe Cesàro mean of $|x_n{|}^2$ over $n=-N,\dots ,N$ tends to $|A{|}^2+|B{|}^2$, since the mixed geometric sum divided by $2N+1$ tends to zero. For an ${\ell }^2$ sequence the same mean tends to zero. Therefore $A=B=0$. At $\lambda =\pm 2$, the general solution is $(A+Bn)(\pm 1{)}^n$, again square summable only when zero. This also covers $\lambda =0$ through $\theta =\pi /2$. Thus $\boxed{T\text{ has no eigenvectors}}$.

Finally the bounded sequence $e_{3j}$ has images $T{e}_{3j}=e_{3j-1}+e_{3j+1}$ with pairwise disjoint supports. Distinct images have distance two, so no subsequence converges. Hence $\boxed{T\text{ is not compact}}$.

For a character $\psi$ of a subgroup $H\le G$, the induced character isIt is the character of the induced representation $\mathbb{C}[G]{\otimes }_{\mathbb{C}[H]}{V}_{\psi }$, and has degree $[G:H]\psi (1)$. The character inner product is $⟨\alpha ,\beta {⟩}_G=|G{|}^{-1}{\sum }_g\alpha (g)\overline{\beta (g)}$. Substituting the displayed formula and putting $g=xh{x}^{-1}$ yieldsThus Frobenius reciprocity isNow assume index two, so $H$ is normal. Choose $g\notin H$ and write ${\psi }^g(h)=\psi (g^{-1}hg)$. Directly from the induction formula,If ${\psi }^g\ne \psi$, this norm is one, so induction is irreducible of degree $2d$: $\psi$ is monogamous. If ${\psi }^g=\psi$, the norm is two. Since a norm is the sum of squared nonnegative integer multiplicities, induction is the sum of two distinct irreducibles, each once. Each is related to $\psi$, so its degree is at least $d$ by restriction and reciprocity; their degrees sum to $2d$, forcing both to have degree $d$. Thus $\psi$ is bigamous.

Every irreducible $\chi$ of $G$ restricts to a nonzero sum of irreducibles. Choose a constituent $\psi$. If $\psi$ is bigamous, $\chi$ is one of its two degree-$d$ partners; since its restriction already contains a degree-$d$ constituent, $\mathrm{Res}\chi =\psi$. If $\psi$ is monogamous, $\chi =\mathrm{Ind}\psi$ and $\mathrm{Res}\chi =\psi +{\psi }^g$, giving precisely two distinct monogamous partners of equal degree.

The complex irreducible character degrees of $A_5$ are $1,3,3,4,5$, with squared sum $60$. Conjugation by the outside coset preserves degrees, so it fixes the unique degree-$1$, degree-$4$ and degree-$5$ characters and either fixes or swaps the two degree-$3$ characters. The two possibilities for $G$ areBoth lists have squared sum $120=|G|$. They occur for $A_5\times C_2$ and $S_5$ respectively; in $S_5$, odd conjugation swaps the two classes of $5$-cycles and hence the two degree-$3$ characters.

The Turán graph $T_r(n)$ is the complete $r$-partite graph whose vertex-class sizes differ by at most one; empty classes are allowed when $r>n$. Its edge count is $t_r(n)=\frac{1}{2}(n^2-{\sum }_{i=1}^r{n}_i^2)$. Turan theorem states that every $n$-vertex graph containing no $K_{r+1}$ has at most $t_r(n)$ edges.

Here is a symmetrization proof, including the equality case. Let $G$ be any $K_{r+1}$-free graph maximizing the edge count. If nonadjacent $u,v$ have unequal degrees, replace the lower-degree vertex by a twin of the higher-degree vertex: this increases edges and cannot create a forbidden clique, since any new clique containing the twin corresponds to an old clique containing its model. Thus all nonadjacent pairs have equal degrees.

If such $u,v$ had distinct neighbourhoods, choose $w\in N(u)\setminus N(v)$ (their equal degrees ensure such a vertex exists), and clone $v$ onto $u$. The new graph has the same maximum number of edges and is still clique-free. Originally $w$ and $v$ were nonadjacent, so $d(w)=d(v)$. After cloning, $w$ loses precisely its edge to $u$, while $v$ keeps its degree. They remain nonadjacent but now have different degrees, contradicting the preceding necessary property of every extremal graph. Hence nonadjacent vertices have identical neighbourhoods.

Nonadjacency together with equality is therefore an equivalence relation, making $G$ complete multipartite. It has at most $r$ classes, otherwise choosing one vertex from each of $r+1$ classes gives a clique. For fixed total size, $\sum n_i^2$ is minimized by $r$ classes balanced as equally as possible: splitting a nontrivial class when fewer than $r$ classes are used adds edges, and moving a vertex from a class of size at least two greater than another also adds edges. Thus every extremal graph is $T_r(n)$, up to isomorphism. This proves both the bound and its equality characterization. In particularFor the bipartite assertion, take a maximum matching in a graph of $a$ edges. No edge joins two unmatched vertices. Let $U$ be its $2a$ endpoints. For each matched pair $uv$, both $u$ and $v$ cannot have neighbours outside $U$: those two neighbours lie in opposite vertex classes and would replace $uv$ by two disjoint edges, increasing the matching. Thus at most $n-a$ edges from this pair lead outside $U$, giving $|F|\le a(n-a)$. There are at most $a^2$ edges within $U$. ConsequentlyIf $e(G)>(k-1)n$, necessarily $a\ge k$.

Finally let $A,B,C$ be the three classes of the triangle-free graph. If $m=e(A,B)=0$, the desired bound is immediate. Otherwise set $k=\lceil m/n\rceil$. The preceding result gives $k$ disjoint $AB$ edges. For each edge $uv$, the $C$-neighbourhoods of its endpoints are disjoint, or there would be a triangle, so $d_C(u)+d_C(v)\le n$. The other $2(n-k)$ vertices of $A\cup B$ each have at most $n$ neighbours in $C$. HenceSince $m\le kn$,Equality is attained by retaining every $AC$ and $BC$ edge and no $AB$ edges.