The monomials $x^i{y}^j{z}^k$ with nonnegative integers $i+j+k=m$ form a basis for the vector space of homogeneous polynomials of degree $m$. For each $i=0,\dots ,m$, there are $m-i+1$ choices of $(j,k)$. Hence

The divisor $D={\sum }_i{H}_i$ has degree $md$. The hypothesis gives $\mathrm{deg}(K-D)=2g-2-md<0$. A divisor of negative degree has no nonzero section: if $h\in L(K-D)\setminus \{0\}$, then $(h)+K-D$ would be effective, but its degree would be negative because the divisor of a meromorphic function has degree zero. Thus $\ell (K-D)=0$. The Riemann-Roch theorem now gives

A meromorphic function on a compact Riemann surface has finitely many poles, say $p_j$ of orders $r_j$. For each $p_j$, choose a projective line through it that does not contain $C$. Its hyperplane divisor contains $p_j$ with multiplicity at least one. Include $r_j$ copies of this divisor. The sum then dominates the pole divisor of $f$.

Add further arbitrary hyperplane divisors until their number $m$ satisfies $m\ge d$ and $md>2g-2$. Adding effective divisors cannot destroy the pole bound, so for $D=H_1+\cdots +H_m$ we haveFor a constant function, including zero, no initial poles need to be covered; simply add sufficiently many lines.

Take $\lambda =e^{\pi i/3}$ and the chosen $s=e^{\pi i/6}$. Then $\lambda s=i$, so $i/(\lambda s)=1$, and $-1/\lambda ={\lambda }^2$. In affine coordinates the chosen lift becomesSince ${\lambda }^2$ is a primitive cube root of unity, $1+{\lambda }^2+{\lambda }^4=0$. Three iterations fix $x$ and $y$, whereas one does not. ThusThe other square-root choice replaces this map by $(x,y)\mapsto (1+{\lambda }^{2}x,-y)$. Its cube is $(x,y)\mapsto (x,-y)$, so that lift has order six. This sign dependence is why a lift was specified explicitly.

Both fix the point at infinity. In the complex torus model their linear multipliers have order three or six, so the lattice is stable under a primitive cube root $\zeta =e^{2\pi i/3}$ and hence under $e^{\pi i/3}=1+\zeta$. Choose a shortest nonzero period $\omega$. The triangular lattice $\mathbb{Z}+\zeta \mathbb{Z}$ has covering radius $1/\sqrt{3}<1$: its equilateral triangles of side one each have circumradius $1/\sqrt{3}$. Approximating any period divided by $\omega$ by this lattice leaves a strictly shorter period, which must be zero. ThereforeThus the cubic has the hexagonal lattice shape, independently of which of the two lifts is chosen.

At $\lambda =-1$ and $s=i$, the chosen map isThe square is not the identity on the cubic, since a generic point has $y\ne 0$. Thus $\boxed{\mathrm{ord}(\sigma )=4}$.

Choose the cubic's point $[0:1:0]$ as the origin in its complex torus description. Since $\sigma$ fixes it, its torus form is $z+\Lambda \mapsto az+\Lambda$, with $a\Lambda =\Lambda$. Exact order four implies $a=\pm i$; $a^2=-1$. Therefore $\Lambda$ is stable under multiplication by $i$.

To identify its shape without an unproved uniqueness assertion, choose a shortest nonzero period $\omega$. For any ${\lambda }_0\in \Lambda$, approximate ${\lambda }_0/\omega$ by a Gaussian integer $m+ni$ within distance $1/\sqrt{2}<1$. The difference ${\lambda }_0-(m+ni)\omega$ belongs to $\Lambda$ and is shorter than $\omega$, so it must vanish. HenceThe lattice is determined up to a nonzero complex scaling: it is square.

On the parameter, the two transformations are $S(\lambda )=1/\lambda$ and $T(\lambda )=1-\lambda$. Their compositions yieldEvery step lifts to an invertible projective transformation of the corresponding cubic curves. Therefore all six parameter values give isomorphic curves. Some values coincide for special $\lambda$; this does not affect the conclusion.

Compose the preceding transformations, applying ${\sigma }_{\lambda }$ first and then ${\tau }_{1/\lambda }$. With the square-root convention already specified,This invertible projective transformation fixes $[0:1:0]$ and carries $C_{\lambda }$ to $C_{1-1/\lambda }$. It is important to retain the sign of the chosen lift on the $y$ coordinate when computing its order later.

An appropriate projective transformation isIt fixes $[0:1:0]$ and has an invertible representing matrix. For $X=z-x$, $Y=iy$, $Z=z$,Hence its restriction is an isomorphism $C_{\lambda }\to C_{1-\lambda }$.

Choose $s\in \mathbb{C}$ with $s^2=\lambda$. Since $\lambda \ne 0$, the invertible diagonal projective transformationfixes $[0:1:0]$. Writing its image as $[X:Y:Z]$, direct substitution givesThus it maps $C_{\lambda }$ onto $C_{1/\lambda }$. Either choice of square root works. We choose $s=i$ at $\lambda =-1$ and $s=e^{\pi i/6}$ at $\lambda =e^{\pi i/3}$ for the later order calculations.

Injectivity makes the images of any three distinct points distinct. By the ordered-frame result, there is a projective transformation $L$ agreeing with the given map on three chosen points. Composing by $L^{-1}$ reduces to an injective map fixing those points and preserving every cross-ratio. Change coordinates so that the three points are $\infty ,0,1$.

For any other point with affine coordinate $z$, the convention in the source givesPreservation therefore forces the image of $z$ to have the same coordinate $z$. The normalized map fixes every point and is the identity. Undoing the normalization proves the original map is exactly $L$, hence projective. No prior continuity or surjectivity assumption is needed.

Represent each point by a nonzero column $v_i$. The source's cross-ratio isRescaling representatives cancels between numerator and denominator. If the projective transformation is represented by $A\in {\mathrm{GL}}_2(\mathbb{C})$, then $\det (A{v}_i,A{v}_j)=\det (A)\det (v_i,v_j)$. Two factors of $\det (A)$ occur in both numerator and denominator and cancel. Hence the cross-ratio is invariant under every projective transformation.

Choose representatives $v_1,\dots ,v_{n+1}$ of the first $n+1$ points. The general-position condition says that these vectors form a basis of ${\mathbb{C}}^{n+1}$. Write a representative of the last point as $v_{n+2}={\sum }_i{c}_i{v}_i$. Every $c_i$ is nonzero; otherwise the last point and the other $n$ basis points would lie in a hyperplane. Rescaling $v_i$ by $c_i$ gives representatives with $v_{n+2}={\sum }_i{v}_i$. Normalize representatives $w_i$ of the second frame in the same way.

There is a unique invertible linear map $L$ with $L(v_i)=w_i$ for $i\le n+1$. It also maps $v_{n+2}$ to $w_{n+2}$, so its projectivization has the required effect. For uniqueness, the quotient of two candidate projective transformations fixes the normalized frame. A representing matrix must be diagonal in the $v_i$ basis, with entries $a_i$. Fixing $[{\sum }_i{v}_i]$ forces all $a_i$ to be equal. Such a scalar represents the identity projective transformation. Thus the ordered projective frame determines the transformation uniquely.

Using the two first fundamental forms calculated below, chooseThe surface of revolution is then a catenoid, with metric ${\cosh }^{2}s(d{s}^2+d{v}^2)$. On the helicoid, take $u=\sinh s/a$, assuming $a\ne 0$ so that its parametrization is regular. Substitution givesThus the coordinate mapis a local isometry on sufficiently small patches. The constant $a$ only changes the radial parametrization of the given helicoid, whose pitch is fixed at one. For $a=0$ the displayed helicoid parametrization is not a regular surface.

Two regular coordinate patches are locally isometric under a coordinate correspondence exactly when it preserves their first fundamental forms. More explicitly, a diffeomorphism $\Phi :U\to \tilde{U}$ of parameter domains induces a local isometry if and only ifIn common coordinates this means equality of the three coefficient functions $E,F,G$. The criterion follows because the first fundamental form gives the inner product of every pair of tangent vectors. Equality therefore preserves lengths of all curves; conversely a smooth isometry preserves tangent-vector lengths and their inner products by polarization.

For a regular parametrization $\mathbf{r}(u,v)$, the first fundamental form is the induced Riemannian metricIt measures squared lengths of tangent vectors and therefore lengths of curves. In the plane put $x=\rho \cos \theta$, $y=\rho \sin \theta$. The derivatives are orthogonal, of lengths one and $\rho$, soThese polar coordinates are regular for $\rho >0$.