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A critical point satisfies $df=0$; it is nondegenerate when the Hessian on its two-dimensional tangent space is nonsingular. For a smooth function with only nondegenerate critical points on a compact surface, the Morse theory formula iswhere $m_i$ counts the critical points whose Hessian has $i$ negative eigenvalues.

For $f(x)=|x{|}^2$, differentiation in a tangent direction $v$ gives $d{f}_x(v)=2x\cdot v$. This vanishes for all $v\in T_{x}X$ exactly when the position vector is perpendicular to that tangent plane, equivalently the normal line at $x$ contains the origin. This includes $x=0$, if it is on the surface. On a torus $\chi =0$, while a minimum and a maximum give $m_0,m_2\ge 1$. Thus $m_1=m_0+m_2\ge 2$ and the total is at least four. Hence at least four surface normals pass through the origin under the stated nondegeneracy assumption.

Write the generating circle as center $(a,R)$ and radius $r$, with $R>r>0$. A parametrization and unit normal areTake $P=(a,R/2,0)$. If $P-X=tn$, its first coordinate gives $(r+t)\cos u=0$. The alternative $t=-r$ would force the radial distance of $P$ from the axis to be $R$, contrary to $R/2$. Thus $\cos u=0$. The remaining equations and the nonzero radial distance of $P$ force $v=0$ or $\pi$. Each of the two choices of $u$ and two of $v$ works, giving exactly four distinct points. Take $Q=(a,0,0)$ instead. All points on each of the circles $u=\pi /2$ and $u=3\pi /2$ have radial normals through $Q$, giving infinitely many.

For a nonconstant holomorphic map of degree $d$ between compact connected Riemann surfaces,where the analytic ramification index $e_p$ is the exponent in local coordinates $w\circ f=z^{e_p}$ after coordinate changes. A meromorphic function maps its poles to infinity; the target coordinate $1/w$ is then holomorphic and vanishes at a pole, so this defines a holomorphic map to the Riemann sphere. A pole of order $m$ has local degree $m$.

Let $\Lambda$ be the period lattice. The induced function $\wp :T=\mathbb{C}/\Lambda \to {\mathbb{CP}}^1$ has just one pole, of order two, so its degree is two. The Riemann-Hurwitz formula gives total ramification four. The pole contributes one; every analytic ramification index is at most two, so there are exactly three further ramification points, each simple. Evenness of $\wp$, obtained by reindexing its lattice sum, identifies them with the three nonzero half-periods. Their values $e_1,e_2,e_3$ are distinct: two points of local degree two cannot lie in one degree-two fiber.

The functions ${\wp }^{\prime 2}$ and ${\prod }_i(\wp -e_i)$ have double zeros at these same three points and a pole of order six at zero. Their ratio is a holomorphic nowhere-zero function on the compact torus, hence constant. From $\wp (z)=z^{-2}+O(z^2)$ and ${\wp }^{\prime }(z)=-2z^{-3}+O(z)$ its value is four. ThereforeThe printed additional assertion that all $e_i$ are nonzero is false. The square-lattice Weierstrass function has a zero branch value: for $\mathbb{Z}+i\mathbb{Z}$, reindexing gives $\wp (iz)=-\wp (z)$. At the half-period $h=(1+i)/2$, one has $ih=h-1$, so periodicity gives $\wp (h)=-\wp (h)$ and hence $\wp (h)=0$. Thus one branch value is zero, while the three remain distinct.

The derivative has a single triple pole, so $\mathrm{deg}{\wp }^{\prime }=3$, and the Riemann-Hurwitz formula gives total ramification $\sum (e_p-1)=6$. The number of distinct ramification points requires a further distinction. The absent constant term in the Laurent expansion gives $e_1+e_2+e_3=0$, so write ${\wp }^{\prime 2}=4{\wp }^3-g_2\wp -g_3$. Differentiation yields ${\wp }^{\prime \prime }=6{\wp }^2-g_2/2$. If $g_2\ne 0$, each of the two values $\pm \sqrt{g_2/12}$ has two distinct preimages under $\wp$ and none is a branch value, since a common zero of the cubic and its derivative would be a repeated cubic root. Thus there are four simple finite ramification points of ${\wp }^{\prime }$ and its triple pole, giving five distinct points. If $g_2=0$, the nonsingularity of the cubic forces $g_3\ne 0$. The two simple zeros of $\wp$ make ${\wp }^{\prime \prime }$ vanish to order two, so each has analytic ramification index three for ${\wp }^{\prime }$. Together with its triple pole there are three distinct points. This second case occurs for a hexagonal lattice: its sixth-root rotational symmetry forces the $z^2$ Laurent coefficient, and hence $g_2$, to vanish. The weighted count is six in both cases.

An orientable surface admits local coordinate orientations that are consistent on overlaps. Equivalently a triangulation can be oriented so that adjacent faces induce opposite directions on their common edge. Use a polygon word to record its successive boundary edges, identifying equal letters in their indicated directions; a negative exponent reverses that edge direction.

Square models are $P:abab$, $K:ab{a}^{-1}b$, and $T:ab{a}^{-1}{b}^{-1}$. The first identifies antipodal boundary points of a disk, giving the real projective plane. Only the torus among these three is orientable: its paired sides traverse the boundary oppositely, whereas each other model has a same-direction pair. The projective model has two vertex classes, and the other two one, giving Euler characteristics one and zero respectively.

The other possibility is the sphere, for example the word $a{a}^{-1}b{b}^{-1}$ with three vertex classes. There are no others: a paired square has one face, two edges and between one and three vertex classes, so its Euler characteristic is $V-1\in \{0,1,2\}$. The classification of compact surfaces leaves precisely the sphere, real projective plane, torus and Klein bottle at those values, with the indicated orientabilities.

For connected sums, remove a disk from each surface, join the resulting boundary circles, and cut a joining arc to produce a single polygon. Its word concatenates the two original words. The real projective plane also has the bigon word $cc$, soThese use respectively six and eight sides, proving the requested hexagon and octagon constructions. Their Euler characteristics are $-1$ and $-2$, also obtained from $\chi (X\#Y)=\chi (X)+\chi (Y)-2$.

Assume first that $\Vert P_n\Vert \le M$ for all $n$. Given $x\in X$ and $ϵ>0$, choose a finite linear combination $y$ of the $x_i$ with $\Vert x-y\Vert <ϵ/(M+1)$. For all sufficiently large $n$, $P_{n}y=y$. ThereforeThus $P_{n}x\to x$, and the formula in part (iii) yields the norm-convergent expansionThis proves the stated existence of coefficients.

Conversely, suppose every $x$ has some norm-convergent expansion $x={\sum }_i{\alpha }_i{x}_i$. Applying the continuous functional $x_j^{\ast }$ to its partial sums and passing to the limit gives ${\alpha }_j=x_j^{\ast }(x)$. Hence those partial sums are precisely $P_{n}x$, so $P_{n}x\to x$ for every $x$. In particular ${\sup }_n\Vert P_{n}x\Vert <\infty$ for each fixed $x$. The Uniform boundedness principle, using completeness of $X$, now givesThis proves the reverse implication. The coefficients are automatically unique, so the equivalence is the uniform partial-sum criterion for a Schauder basis for the given minimal total sequence. Its projection bound is the basis constant.

Define the finite-rank operatorIt is linear and bounded, with $\Vert P_n\Vert \le {\sum }_{i=1}^n\Vert x_i^{\ast }\Vert \Vert x_i\Vert$. Biorthogonality gives exactly the two required values on $x_j$. It also givesThus $P_n$ is a bounded projection onto $\mathrm{span}\{x_1,\dots ,x_n\}$, with kernel ${\bigcap }_{i=1}^n\ker x_i^{\ast }$. Its action on the dense span determines it uniquely among bounded linear operators. These are the partial-sum projections; no uniform bound in $n$ has yet been assumed.

Fix $i$ and let $M_i=\overline{\mathrm{span}}\{x_j:j\ne i\}$. The hypothesis gives $M_i\ne X$. Moreover $x_i\notin M_i$, since otherwise the closed span of all the vectors would be contained in $M_i$. Hence $d_i=\mathrm{dist}(x_i,M_i)>0$.

On $M_i+\mathbb{R}{x}_i$, define ${\varphi }_i(m+t{x}_i)=t$. It is well-defined because $x_i\notin M_i$. For $t\ne 0$,so $|{\varphi }_i(m+t{x}_i)|\le \Vert m+t{x}_i\Vert /d_i$; the inequality also holds for $t=0$. The Hahn-Banach theorem extends ${\varphi }_i$ without increasing its norm to $x_i^{\ast }\in X^{\ast }$. This gives $x_i^{\ast }(x_j)={\delta }_{ij}$.

If two continuous functionals have these values, their difference vanishes on the linear span of every $x_j$, which is dense in $X$. Continuity then makes the difference zero everywhere. Thus each functional is unique, proving the required unique biorthogonal system. In fact its norm is $\Vert x_i^{\ast }\Vert =1/d_i$, by taking vectors in $M_i$ approaching the distance infimum. This is the coordinate-functional construction for a minimal total family in a normed space.

A biorthogonal system is a family $(x_i,f_i{)}_{i\in I}$ with $x_i\in X$, $f_i\in X^{\ast }$, satisfyingThus each continuous coordinate functional takes value one on its own vector and zero on all others. The vectors need not be orthogonal in any inner product, and totality of their linear span is an additional condition.

Put $T=R\circ J_Z:Z\to X^{\ast }$. Then $(Tz)(x)=x(z)$ and $\Vert Tz\Vert \le \Vert z\Vert$. We must prove $T$ onto.

Take $f\in B_{X^{\ast }}$. Part (iv) supplies $\theta \in B_{Z^{\ast \ast }}$ with $R\theta =f$. Goldstine's theorem gives a net $z_{\alpha }\in B_Z$ with $J_Z{z}_{\alpha }\to \theta$ weak-star. By continuity of $R$, $T{z}_{\alpha }\to f$ in $\sigma (X^{\ast },X)$. Reflexivity of $X$ makes this the weak topology on $X^{\ast }$ by part (iii). Thus $B_{X^{\ast }}$ is contained in the weak closure of $T(B_Z)$.

The reverse containment holds because $T(B_Z)\subseteq B_{X^{\ast }}$, and the latter ball is weakly closed: it is the intersection of the closed evaluation constraints $|g(x)|\le 1$ for $x\in B_X$. The set $T(B_Z)$ is convex, so part (ii), now on the Banach space $X^{\ast }$, givesDensity of the ball gives exact surjectivity by a convergent correction argument. For nonzero $f\in X^{\ast }$, choose $z_1$ with $\Vert z_1\Vert \le \Vert f\Vert$ and $\Vert f-T{z}_1\Vert \le \Vert f\Vert /2$. Inductively approximate the current residual $r_{n-1}$ by $T{z}_n$ with $\Vert z_n\Vert \le \Vert r_{n-1}\Vert$ and $\Vert r_{n-1}-T{z}_n\Vert \le \Vert r_{n-1}\Vert /2$. If a residual is zero, stop.

The residual norms are bounded by $2^{-n}\Vert f\Vert$, while ${\sum }_n\Vert z_n\Vert \le 2\Vert f\Vert$. Completeness of $Z$ gives $z={\sum }_n{z}_n\in Z$, and continuity of $T$ gives $Tz=f$. For $f=0$ use $z=0$. ThereforeThis is evaluation onto the dual of a reflexive dual subspace. The correction argument is needed: a dense image alone would not imply surjectivity.

The domain carries $\sigma (Z^{\ast \ast },Z^{\ast })$, and the target carries $\sigma (X^{\ast },X)$. For each $x\in X\subseteq Z^{\ast }$,which is one of the defining weak-star continuous evaluations on $Z^{\ast \ast }$. Thus $R$ is weak-star continuous.

Restriction cannot increase a functional's norm, so $R(B_{Z^{\ast \ast }})\subseteq B_{X^{\ast }}$. Conversely, any $f\in B_{X^{\ast }}$ is a bounded linear functional on the subspace $X$ of $Z^{\ast }$. The Hahn-Banach theorem extends it to a functional $\theta \in (Z^{\ast }{)}^{\ast }=Z^{\ast \ast }$ with the same norm. Then $R\theta =f$ and $\Vert \theta \Vert \le 1$. ConsequentlyThis proves exact surjectivity on closed balls, not just density of the image.

A reflexive Banach space is one for which $J_X:X\to X^{\ast \ast }$ is onto. If $X$ is reflexive, every $\theta \in X^{\ast \ast }$ has the form $\theta (f)=f(x)$ for some $x\in X$. Thus the functionals defining $\sigma (X^{\ast },X^{\ast \ast })$ and $\sigma (X^{\ast },X)$ are precisely the same, proving equality of the topologies.

Conversely, assume those topologies coincide. For any $\theta \in X^{\ast \ast }$, the linear functional $f\mapsto \theta (f)$ on $X^{\ast }$ is then weak-star continuous. Its continuity at zero gives finitely many $x_1,\dots ,x_m\in X$ and $ϵ>0$ such thatIf all these evaluations are zero, arbitrary scalar multiples of $f$ still satisfy the inequalities, forcing $\theta (f)=0$. Hence $\theta$ factors through the finite-dimensional evaluation map $f\mapsto (f(x_1),\dots ,f(x_m))$. A linear functional on its image extends to ${\mathbb{R}}^m$, so for suitable scalars $a_j$,Thus $\theta$ lies in $J_X(X)$. Since it was arbitrary, $X$ is reflexive. We have proved

The weak topology is coarser than the norm topology, so norm convergence implies weak convergence, andFor the reverse inclusion, let $x\notin K={\overline{C}}^{\Vert \cdot \Vert }$. The set $K$ is closed and convex. The Hahn-Banach separation theorem supplies $f\in X^{\ast }$ and $a\in \mathbb{R}$ such that $f(x)>a\ge {\sup }_{y\in K}f(y)$. The set $\{z:f(z)>a\}$ is weakly open, contains $x$ and misses $C$. Hence $x$ is outside the weak closure as well. If $C$ is empty the equality is immediate. ThereforeThe separating functional explains the role of convexity: an arbitrary norm-closed subset need not be weakly closed.

The weak topology $\sigma (X,X^{\ast })$ is the coarsest topology on $X$ making every continuous linear functional $f\in X^{\ast }$ continuous. A basic neighborhood of $x_0$ is specified by finitely many inequalities $|f_j(x-x_0)|<ϵ$.

The weak-star topology $\sigma (X^{\ast },X)$ is the coarsest topology on $X^{\ast }$ making every evaluation $f\mapsto f(x)$, for $x\in X$, continuous. Its basic neighborhoods use finitely many such evaluations. Nets converge in these topologies exactly when all the corresponding scalar evaluations converge.

Let $J_X:X\to X^{\ast \ast }$ be the canonical isometric embedding, $J_{X}x(f)=f(x)$. Goldstine theorem statesHere the balls are closed unit balls. Thus any bidual unit vector can be approximated in finitely many dual evaluations by unit vectors from $X$; this does not claim norm density or sequential density in complete generality.

For a self-adjoint C-star element $a$, the following are three equivalent characterizations of a Positive element of a C-star algebra:If $a=b^2$ with $b=b^{\ast }$, the spectral mapping theorem gives nonnegative spectrum. Conversely, the continuous functional calculus applied to $\sqrt{\lambda }$ on $\sigma (a)$ gives a self-adjoint C-star element $b$ with $b^2=a$. For real $\sigma (a)$,Thus the third condition means $\sigma (a)\subseteq [0,2t]$, while the first implies the second by taking the interval $[0,\Vert a\Vert ]$. The second implies the third with $t=\Vert a\Vert$ if $a\ne 0$; if $a=0$, any $t>0$ works. This proves every equivalence.

The positive cone of a C-star algebra contains zero and is closed under multiplication by nonnegative real scalars. To verify addition without assuming that positive elements commute, put $r=\Vert a\Vert$, $s=\Vert b\Vert$ for $a,b\ge 0$. ThenIf $r+s>0$, the third characterization makes $a+b$ positive; otherwise both are zero. Hence it is a convex cone. If $a$ and $-a$ are positive, $\sigma (a)=\{0\}$ and $a=0$ by the self-adjoint spectral norm formula; therefore the cone is pointed and defines a partial order on $A_{\mathrm{sa}}$ by $h\le k$ if $k-h\in A_{+}$. It is generating because the continuous functional calculus givesFor $a\ge 0$, the spectral mapping theorem yieldsFinally, let $a\in M_2(\mathbb{C})$ be Hermitian, and write $\tau =\mathrm{tr}a$, $d=\det a$. Its two eigenvalues are real. Both are nonnegative exactly when their sum $\tau$ and product $d$ are nonnegative. The corresponding sum and product for $I-a$ are $2-\tau$ and $1-\tau +d$. Therefore the necessary and sufficient conditions for a positive contraction areIndeed, nonnegative sum and product force two real numbers to be nonnegative, including the zero-product cases. Applying this to $a$ and $I-a$ places both eigenvalues in $[0,1]$.

A unital C-star algebra is a complex Banach algebra $A$ with a multiplicative identity $1_A$ and a conjugate-linear involution $a\mapsto a^{\ast }$ satisfying $(ab{)}^{\ast }=b^{\ast }{a}^{\ast }$ and the C-star identity $\Vert a^{\ast }a\Vert =\Vert a{\Vert }^2$. A C-star isomorphism $\Phi :A\to B$ is a bijective complex-linear map preserving products and the involution. It carries $1_A$ to $1_B$ and is automatically isometric.

For a commutative unital C-star algebra, let $\Omega$ be its maximal ideal space, equivalently the space of Gelfand characters, with the weak-star topology. A Gelfand character has $\chi (1)=1$ and $\Vert \chi \Vert =1$. The equations $\chi (ab)=\chi (a)\chi (b)$ and $\chi (1)=1$ define a closed in the weak-star topology subset of the dual unit ball. The Banach-Alaoglu theorem makes $\Omega$ a compact Hausdorff space. Maximal ideals give characters by the Gelfand-Mazur theorem, and the Banach algebra spectrum satisfies $\sigma (a)=\{\chi (a):\chi \in \Omega \}$.

The Gelfand transform is $\hat{a}(\chi )=\chi (a)$. It preserves products, constants and the involution; the latter uses the fact that characters of a C-star algebra respect the involution. Every element of a commutative C-star algebra is normal, so the C-star identity and the spectral radius formula giveThus its image is a closed self-adjoint subalgebra of $C(\Omega )$, contains the constants, and separates characters. The complex Stone-Weierstrass theorem makes that image dense and therefore all of $C(\Omega )$. This is the Commutative Gelfand--Naimark theorem.

For the convergent sequence space $c$, coordinatewise multiplication satisfies $\Vert \alpha \beta {\Vert }_{\infty }\le \Vert \alpha {\Vert }_{\infty }\Vert \beta {\Vert }_{\infty }$, and coordinatewise complex conjugation is an involution. The constant sequence $1$ is the identity, andIt is complete: a uniform limit of convergent sequences is convergent, so $c$ is a closed linear subspace of the Banach space ${\ell }_{\infty }$. Hence it is a commutative unital C-star algebra.

TakeEvery finite point is isolated, and the neighbourhoods of $\infty$ contain all but finitely many integers. This is a compact Hausdorff space: an open cover has a set containing $\infty$, leaving only finitely many points to cover. DefineContinuity at $\infty$ is exactly convergence of the sequence. Thus $\Phi$ is onto $C(\Omega )$, and pointwise operations show it is a C-star isomorphism with $\Vert \Phi \alpha {\Vert }_{\infty }=\Vert \alpha {\Vert }_{\infty }$. Explicitly, these are all Gelfand characters: each $\chi (e_n)$ is $0$ or $1$ and at most one is $1$. If $\chi (e_n)=1$, multiplication by $e_n$ gives $\chi (\alpha )={\alpha }_n$. If all vanish, density of finitely supported sequences in l-p in $c_0$ gives $\chi {|}_{c_0}=0$, so $\chi (\alpha )=\lim {\alpha }_n$.