Parametrize a nonconstant hyperbolic geodesic by unit speed. Its energy is $({\dot{x}}^2+{\dot{y}}^2)/(2y^2)$, and the Euler-Lagrange equation in the cyclic coordinate $x$ gives $\dot{x}/y^2=c$. Unit speed gives ${\dot{x}}^2+{\dot{y}}^2=y^2$. If $c=0$ the curve is vertical. Otherwise, on either side of a possible highest point,The two branches join smoothly at the top. Conversely these curves with their unit-speed parametrizations satisfy the energy equations. Thus the unparametrized complete geodesics are vertical half-lines and upper semicircles centered on the real axis.

Let the tractrix point have positive height $r(\theta )$. Its tangent segment to the axis has length one and makes angle $\theta$, so $r=\sin \theta$. Since its slope is $\tan \theta$, $h^{\prime }(\theta )={\cos }^2\theta /\sin \theta$ for its axial coordinate. Rotating $X(\theta ,v)=(h(\theta ),\sin \theta \cos v,\sin \theta \sin v)$ gives orthogonal coordinate vectors with squared lengths ${\cot }^2\theta$ and ${\sin }^2\theta$. Hence the stated first fundamental form follows.

Now put $x=v$, $y=\csc \theta$. Direct substitution givesOn the specified intervals this coordinate map is a diffeomorphism onto the stated rectangle, so is an isometry there and a local isometry on the tractoid.

Set $b=\csc \beta$, $a=\csc \alpha$. In the unwrapped coordinate, connect $(x_0,b)$ to $(x_0+2\pi ,b)$ along the semicircle centered at $(x_0+\pi ,0)$ of radius $\sqrt{b^2+{\pi }^2}$. Its height lies between $b$ and that radius. The hypothesis $a^2-b^2>{\pi }^2$ puts the whole arc in the closed height strip $b\le y\le a$. Projecting the angular coordinate modulo $2\pi$ therefore gives a geodesic segment in the closed tractoid annulus which winds once and returns to its starting point. Taking $x_0$ to represent any chosen angular position proves the assertion for every point of its $\theta =\beta$ boundary. The returned tangent need not equal the starting tangent, so this proves the claimed returning segment without asserting a periodically smooth closed geodesic.

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.