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A covering-surface formulation of the Ahlfors second fundamental theorem is as follows. For a regular finite bordered Riemann surface $X$ mapped holomorphically to a fixed metric bordered Riemann surface $Y$, let $S=A_f(X)/A(Y)$ be the average sheet number and $L$ the pulled-back length of the relative boundary, namely the boundary mapped into the interior of $Y$. With the modern sign convention for the Euler characteristic,where $h$ depends on $Y$ and its metric, not on $X$ or the map. The associated area comparison is $|S_D-S|\le h_{D}L$ for each fixed regular target region $D$. These formulations and the sign convention are explained in www.math.purdue.edu/~eremenko/dvi/ahl.pdf .

The island version of the Ahlfors second fundamental theorem for a disk source, which is the useful form here, says that for $q\ge 3$ fixed Jordan domains $D_j$ with disjoint closures on the Riemann sphere,Here $n_j$ counts simply connected islands of a meromorphic function over $D_j$, once each without weighting by mapping degree, andAn island of a meromorphic function is a relatively compact inverse-image component mapped properly onto its target domain; components reaching the source boundary are not counted. The simply connected island estimate is recorded, for spherical disks, in Lemma 2 of pdfs.semanticscholar.org/d247/562607e237af7c8fc6e683b77be6b94829fa.pdf . Regular target regions give the same estimate with a region-dependent constant. General Jordan domains can be enclosed in slightly larger regular Jordan domains still having disjoint closures. A simple island over an enlarged domain restricts to one over the original domain, so it suffices to prove the conclusion for regular domains.

Suppose a nonconstant meromorphic function on the plane has no simple islands over five such domains. On every counted simply connected island its proper map has integer degree at least two. The area formula therefore givesTogether with area comparison and the Ahlfors second fundamental theorem, this yieldsThe area comparison can also be seen directly. The two-form $[1_D/A(D)-1/(4\pi )]dA$ on the target sphere has zero integral and a bounded one-form primitive. One construction solves its Poisson equation: the gradient of the spherical Green function has an integrable $1/\mathrm{distance}$ singularity, so convolution against the bounded density gives a bounded primitive. Pulling back and applying Stokes theorem, or smooth approximation at the domain boundary, bounds the integral by a constant times $L(R)$.

We now derive the needed length-area exhaustion of the complex plane. The Cauchy-Schwarz inequality on a circle givesIf $L(R)/S(R)\ge \epsilon >0$ for all sufficiently large $R$, then $S(R)>0$ andIntegration makes $1/S(R)$ negative, which is impossible. Hence there are radii $R_k\to \infty$ with $L(R_k)/S(R_k)\to 0$. This contradicts $S/2\le CL$ and proves the Ahlfors five islands theorem: one of the five target domains has a bounded simply connected inverse-image component mapped biholomorphically onto it.

Four domains do not suffice. Take the Weierstrass elliptic function $\wp$ for the square lattice $\mathbb{Z}+i\mathbb{Z}$. The equationshows the three distinct finite branch values $e_1,e_2,e_3$; the fourth is infinity, because all poles are double. Equivalently, the degree-two map from the elliptic torus to the Riemann sphere has these four branch values, with all their preimages of local degree two. Choose four small disjoint Jordan domains, each containing one of these four totally ramified Weierstrass values. A degree-one island of a meromorphic function over any one would contain a preimage of its central value of local degree one, contradicting total ramification. Thus $\wp$ provides four domains with no simple island.

The printed proximity integral is missing a factor $1/(2\pi )$. Denote that literal integral by $m_{\mathrm{print}}$ and write $\overline{m}=m_{\mathrm{print}}/(2\pi )$. The correct Annular spherical Jensen formula, with the printed definitions of $N$ and $T$, isThe following derivation proves this identity and also isolates the normalization issue.

For the chordal metric put $u_a(z)=-\log k(f(z),a)$ and let $M_a(s)=(2\pi {)}^{-1}{\int }_0^{2\pi }{u}_a(s{e}^{i\theta })d\theta$. In a finite target coordinate,For $a=\infty$ use $u_{\infty }=\frac{1}{2}\log (1+|f{|}^2)$. Differentiating away from the preimages of $a$ gives half the squared spherical derivative of a meromorphic function. At an $a$-point of local degree $d$, $u_a=-d\log |z-z_0|+O(1)$; the second distributional derivatives give the Laplacian of $\log |z-z_0|$ is $2\pi {\delta }_{z_0}$. Thus, including poles by changing spherical coordinates,Because $a$ is off the image of $|z|=1$, $u_a$ is smooth near that circle and $I(a)=M_a^{\prime }(1)$ is finite and independent of $r$.

Apply Green's second identity to the harmonic weight $w(z)=\log (|z|/r)$ on $r<|z|<1$. On the outer boundary $w=\log (1/r)$ and ${\partial }_{n}w=1$; on the inner boundary $w=0$ and ${\partial }_{n}w=-1/r$. HenceThe distributional formula makes the left side $2\pi T(r)-2\pi N(r;a)$, and $M_a(r)-M_a(1)=\overline{m}(r;a)$, proving the corrected Annular spherical Jensen formula. Initially take an inner circle avoiding $a$-points. Logarithmic singularities have continuous angular means, and a point on the inner circle has zero logarithmic counting weight, so continuity extends the identity to every $0<r\le 1$.

There is no way to absorb the missing factor into a different constant $I(a)$. For the holomorphic function $f(z)=z$ and $a=\infty$, direct integration givesThe value of $(T-m_{\mathrm{print}})/\log (1/r)$ tends to $1/2$ as $r\downarrow 0$ and to $\pi$ as $r\uparrow 1$. Therefore the literal printed identity has no constant $I(\infty )$, whereas the corrected identity has $I(\infty )=1/2$.

For the singularity criterion, define the nondecreasing normalized spherical areaTonelli's theorem applied to the logarithmic weight gives $T(e^{-l})={\int }_0^{l}s(t)dt$. Averages of a nondecreasing function converge to its supremum: for fixed $t_0<l$, $(1-t_0/l)s(t_0)\le T(e^{-l})/l\le s(l)$. Thus the limit, not just the lower limit, exists in $[0,\infty ]$ and equalsIf $f$ has a removable singularity or pole at zero, it extends as a holomorphic map to the Riemann sphere; the pulled-back spherical density is smooth and bounded on the closed disk, hence its area is finite. Conversely an essential singularity gives, by the meromorphic version of the Great Picard theorem, infinitely many preimages of every spherical target with at most two exceptions in each punctured neighborhood. The area formula givesThe formula follows by applying change of variables on inverse-function neighborhoods away from the discrete critical points, then summing with multiplicity; the countable critical values have area zero. This proves the finite spherical area criterion for an isolated singularity and the requested equivalence between a finite lower limit and a removable singularity or pole.

Use the Nevanlinna proximity function $m(r,f)=(2\pi {)}^{-1}\int {\log }^{+}|f(r{e}^{i\theta })|d\theta$ and the Nevanlinna integrated counting function $N(r,f)$ for poles, counting multiplicity. If the pole order at zero is ${\nu }_0$, thenBoundedness of the Nevanlinna characteristic refers to $r\uparrow 1$, equivalently to a uniform bound on $r_0\le r<1$ for any fixed $r_0>0$. This avoids the harmless origin normalization when ${\nu }_0>0$.

The Nevanlinna first main theorem states, for every fixed finite value $a$ and every nonconstant meromorphic function on the disk,For $a=\infty$, this is the definition of $T$. The error is bounded independently of $r$. Translation changes $T$ by $O(1)$, while Jensen's formula gives $T(r,h)-T(r,1/h)=\log |c|$ when $h(z)=c{z}^{\nu }(1+O(z))$ near zero; these observations explain the first theorem.

The Nevanlinna second main theorem in the unit disc states that for $q\ge 3$ distinct values $a_j$ on the Riemann sphere,as $r\uparrow 1$ outside an exceptional set $E$ satisfying ${\int }_{E}dr/(1-r)<\infty$. The truncated Nevanlinna counting function $\overline{N}$ counts each distinct preimage once. The boundary-distance error cannot simply be suppressed: on the disk it need not be $o(T)$. The disk error and exceptional-set estimates follow from the logarithmic-derivative estimates in mathweb.tifr.res.in/Documents/Publications/Lectures/tifr17.pdf .

Suppose first that $T(r,f)$ is bounded. Since $m(r,f)\ge 0$, its poles satisfy the Blaschke condition, counting their orders: let $r\uparrow 1$ in their integrated count and use $1-|b|\le -\log |b|$. Let $B$ be their Blaschke product, including the origin factor if necessary. Then $g=Bf$ extends to a holomorphic function on the whole disk, and $|B|\le 1$ givesHere the last bound follows from bounded $T$ and the uniform lower bound $N(r,f)\ge {\nu }_0\log r_0$ on $r\ge r_0$.

The nonnegative subharmonic function ${\log }^{+}|g|$ has a finite harmonic majorant. Indeed its Poisson integrals $H_R$ on expanding disks dominate it by the maximum principle for subharmonic functions, increase with $R$, and have $H_R(0)\le C$. The Harnack inequality for harmonic functions bounds them on every compact subdisk, so their increasing limit is a finite harmonic function $H\ge {\log }^{+}|g|$. Since the disk is simply connected, $H$ has a harmonic conjugate, giving a holomorphic function $A$ with $\mathrm{Re}A=H$. ConsequentlyThis proves the constructive half of the quotient characterization of bounded characteristic.

Conversely, if $f=u/v$ with $u,v$ bounded holomorphic functions and $v\not\equiv 0$, each pole of $f$ is a zero of $v$ of at least the same order. The elementary inequality ${\log }^{+}|u/v|\le {\log }^{+}|u|+{\log }^{+}|1/v|$ and the Nevanlinna first main theorem giveAny cancellation at an origin zero contributes only a bounded multiple of $\log r$ on $r\ge r_0$, which is included in this $O(1)$. The zero function is immediate; constant nonzero $v$ is handled directly. Thus bounded characteristic is exactly a quotient of two bounded analytic functions.

An unbounded example with bounded characteristic isIt is a ratio of the bounded holomorphic functions $1$ and $1-z$. More explicitly, Jensen's formula gives the mean of $\log |1-r{e}^{i\theta }|$ as zero, so $m(r,1/(1-z))=m(r,1-z)\le \log 2$, although $g(r)\to \infty$.

Finally, set $Y=(1+|g{|}^2{)}^{p/2}\ge 1$. Pointwise ${\log }^{+}|g|\le p^{-1}\log Y$. Jensen's inequality for the concave logarithm givesThere is no pole term because $g$ is holomorphic. Hence the stated integral bound implies bounded characteristic. This argument in fact works for every $p>0$.

The modular group is $S{L}_2(\mathbb{Z})$ acting on the complex upper half-plane byIts central element $-I$ acts trivially on points, so the effective group is $PS{L}_2(\mathbb{Z})$. It is generated by $S:\tau \mapsto -1/\tau$ and $T:\tau \mapsto \tau +1$: the Euclidean algorithm reduces the bottom row of a matrix using these operations until it is a translation. The relations $S^2=(ST{)}^3=1$ hold in the projective group.

A standard fundamental domain of the modular group isTo see that each orbit meets it, use $\mathrm{Im}(\gamma \tau )=\mathrm{Im}\tau /|c\tau +d{|}^2$. Choose a coprime integer pair $(c,d)$ minimizing the nonzero lattice modulus $|c\tau +d|$; a modular matrix with that bottom row maximizes the imaginary part. Translate to the central strip. If the result had modulus below one, inversion would further increase its imaginary part, a contradiction. Interior representatives are unique: for a point in the interior, $|c\tau +d|>1$ when $c\ne 0$. For $|c|\ge 2$ use $\mathrm{Im}\tau >\sqrt{3}/2$; for $|c|=1$ use the strip and the strict unit-circle bound. Thus such a matrix lowers the imaginary part, which would contradict the same fact applied to its inverse if both representatives were in the interior. Matrices with $c=0$ are translations, and cannot move one interior strip point to another.

The vertical sides are identified by $T$, and the circular halves by $S$. The exceptional points with elliptic stabilizers of the modular group are represented by $i$ and $\rho =e^{2\pi i/3}$, of projective stabilizer orders two and three. The rational boundary points together with infinity form one modular cusp orbit. In the hyperbolic metric the region has area ${\int }_{-1/2}^{1/2}(1-x^2{)}^{-1/2}dx=\pi /3$.

A useful subgroup is the level-two principal congruence subgroup,Reduction modulo two is onto $S{L}_2({\mathbb{F}}_2)$, which has six elements and permutes the three nonzero vectors. Hence its kernel has index six, and the quotient is $S_3$. In the projective action, $\Gamma (2)$ is torsion-free. Indeed finite-order nonidentity modular elements have trace zero or $\pm 1$; matrices in $\Gamma (2)$ have even trace, and trace zero is impossible because $a,d$ are odd while $b,c$ are even: $ad-bc=1$ with $d=-a$ would imply $bc=-a^2-1\equiv 2(\mathrm{mod}4)$. The subgroup has three modular cusp classes, represented by $\infty ,0,1$, each of width two. Parity of a primitive pair distinguishes these classes, and elementary level-two row operations reduce each class to its representative. Six translates of $\mathcal{F}$ supply a fundamental region. Its compactified level-two modular curve is obtained by adjoining these three modular cusp points.

Invariant holomorphic functions arise naturally from theta constants. In the radian convention of Question 4, put ${\Theta }_j(\tau )={\theta }_j(0,\tau )$, and defineHere ${\Theta }_2={\sum }_{n\in \mathbb{Z}}{q}^{(n+1/2{)}^2}$. The theta-constant inversion and translation laws, obtained by Gaussian Poisson summation and by reindexing, interchange ${\Theta }_2,{\Theta }_4$ under $S$ with the common factor $\sqrt{-i\tau }$, leave ${\Theta }_3$ fixed up to that factor, and under $T$ interchange ${\Theta }_3,{\Theta }_4$ while multiplying ${\Theta }_2$ by $e^{\pi i/4}$. Together with the Jacobi abstruse identity ${\Theta }_3^4={\Theta }_2^4+{\Theta }_4^4$, this givesThese transformations generate the six permutations of $0,1,\infty$. In particular $T^2$ and $S{T}^2{S}^{-1}$ fix $\lambda$; with $-I$, these generate $\Gamma (2)$. Thus the modular lambda function is invariant under this subgroup, but not under the whole modular group. Its even theta constants do not vanish on $\mathbb{H}$, so it is holomorphic there and avoids $0,1,\infty$. Near infinity, $\lambda =16q+O(q^2)$, with $q=e^{\pi i\tau }$, which is the width-two modular cusp coordinate. The two transformation laws give modular cusp values one at zero and infinity at one, with the same simple local orders. Thus $\lambda$ has exactly one simple pole on the compactified level-two quotient and no pole in the half-plane. A meromorphic map with one simple pole has degree one, so it identifies that quotient with the sphere and the open quotient with $\mathbb{C}\setminus \{0,1\}$. This proves the coordinate assertion and describes all three modular cusps.

Symmetrizing this subgroup invariant produces a full modular-invariant function:Substitution shows that this rational expression is unchanged by both $\lambda \mapsto 1-\lambda$ and $\lambda \mapsto \lambda /(\lambda -1)$, hence by $S$ and $T$. It is the Klein j-invariant. Its denominator does not vanish on $\mathbb{H}$, so it is holomorphic there; at the modular cusp it has $j\sim q^{-2}$, a simple pole in $Q=e^{2\pi i\tau }$. It therefore remains nonconstant without being a bounded holomorphic function on the compactified quotient. In fact a holomorphic invariant extending holomorphically to the compactified modular curve would be constant by the maximum principle.

This also connects invariance to elliptic curves: changing an oriented basis of the lattice $\mathbb{Z}+\tau \mathbb{Z}$ gives precisely the modular action after a rescaling. On the compact full modular quotient, $j$ has just the one simple modular cusp pole, so its map to the sphere has degree one. It therefore classifies the resulting homothety of complex lattices classes, while the modular lambda function retains a level-two labeling of the branch points in the Legendre model $y^2=x(x-1)(x-\lambda )$. Finally, a modular form of nonzero weight obeys a transformation law with a factor $(c\tau +d{)}^k$, not plain invariance; ratios of equal-weight forms instead give weight-zero modular functions. The distinction between forms, invariant holomorphic functions on the open half-plane, and their modular cusp behavior is essential to this construction.

For fixed $\mathrm{Im}\tau >0$, Gaussian decay of $|q{|}^{n^2}$ gives locally uniform convergence of the theta function series and all its $z$-derivatives. It is therefore entire. Shifting the series argument and reindexing givesSince ${\theta }_4(z)={\theta }_3(z+\pi /2)$, the corresponding second multiplier for ${\theta }_4$ has the opposite sign:Thus $h={\theta }_3/{\theta }_4$ obeys $h(z+\pi )=h(z)$ and $h(z+\pi \tau )=-h(z)$. These give candidate period lattices; to show they are exact, first locate every zero.

At $z_{\ast }=(\pi +\pi \tau )/2$,because the terms with indices $n$ and $-n-1$ cancel. The absolute convergence justifies that pairing. Quasi-periodicity propagates this zero to every $z_{\ast }+m\pi +n\pi \tau$.

For exhaustiveness and simplicity, let $v={\theta }_3^{\prime }/{\theta }_3$. It satisfies $v(z+\pi )=v(z)$ and $v(z+\pi \tau )=v(z)-2i$. Take a fundamental parallelogram for ${\Lambda }_0=\pi \mathbb{Z}+\pi \tau \mathbb{Z}$, translated so its boundary has no zero. The two sloping edges cancel by the first identity; the bottom and reversed top giveThe argument principle therefore counts one zero with multiplicity in every cell. Since a known zero class already exists, all zeros are simple and exactlyThe denominator's simple zeros are instead $\pi \tau /2+{\Lambda }_0$. These two cosets are disjoint: $\pi /2$ cannot lie in ${\Lambda }_0$ because $\mathrm{Im}\tau >0$. Thus there are no cancelled zeros or poles in the quotient.

Any period of $h$ or $h^2$ must translate its zero set onto itself, so must belong to ${\Lambda }_0$. A translation $m\pi +n\pi \tau$ multiplies $h$ by $(-1{)}^n$; the quotient is nonconstant, so an odd $n$ cannot be a period of $h$. Squaring removes precisely that sign. This proves the period lattice of theta3 over theta4 and its squared counterpart:The second lattice is for the squared function; no additional translation can preserve its zero set.

Take a real modulus $0<k<1$ and complementary modulus $k^{\prime }=\sqrt{1-k^2}$. The Jacobi elliptic sine is constructed by inverting a Schwarz-Christoffel mapping. The main reason for the Schwarz-Christoffel derivative is local angle behavior: if a prevertex $a_j$ corresponds to a polygon angle $\pi {\alpha }_j$, straightening that corner gives $U(w)-U(a_j)\sim C_j(w-a_j{)}^{{\alpha }_j}$. Hence $U^{\prime }$ has exponent ${\alpha }_j-1$. Along each straight boundary side its argument is constant. Dividing $U^{\prime }$ by ${\prod }_j(w-a_j{)}^{{\alpha }_j-1}$ removes those angle changes; Schwarz reflection extends the quotient across the real boundary and the corner singularities. With all vertices finite, its behavior at infinity is also regular, because the sum of the derivative exponents is $-2$. It is therefore a nonzero constant on the sphere. This yieldsThis argument explains the exponents, the branch choices and the role of boundary reflection, rather than just writing down the formula.

For a rectangle, four angles are $\pi /2$, so choose prevertices $-1/k,-1,1,1/k$ and normalize the inverse map asUse the branch positive on $(-1,1)$. SetThe two middle corner values are $U(\pm 1)=\pm K$. The right side length isas the substitution $t=(1-k^{\prime 2}{s}^2{)}^{-1/2}$ shows. The selected branch makes the increment along that side $i{K}^{\prime }$. Thus the other corner values are $K+i{K}^{\prime }$ and $-K+i{K}^{\prime }$. The two boundary intervals through infinity have combined horizontal length $2K$; the substitution $t=1/(ks)$ gives length $K$ from $1/k$ to infinity. In particular $U(\infty )=i{K}^{\prime }$. As in the square mapping, the boundary goes once around a rectangle, and the argument principle gives a conformal bijectionIts inverse $s(z)=\mathrm{sn}(z,k)$ maps that rectangle onto the complex upper half-plane, and near zero has $s(0)=0$, $s^{\prime }(0)=1$. Differentiating the inverse relation givesThese identities continue meromorphically.

All four rectangle edges have real images, so Schwarz reflection extends $s$ across them. Reflection in the bottom edge gives $s(\bar{z})=\overline{s(z)}$. Reflection in the right edge, combined with the first reflection, gives $s(2K-z)=s(z)$. The even inverse-integral derivative gives $s(-z)=-s(z)$, and henceReflection in the top edge and then the bottom gives $s(z+2i{K}^{\prime })=s(z)$. The resulting period lattice of the Jacobi elliptic sine isTo see the poles, use $U^{\prime }(w)\sim -1/(k{w}^2)$ near the boundary point infinity. Then $U(w)-i{K}^{\prime }\sim 1/(kw)$, so the inverse has a simple pole at $i{K}^{\prime }$ with residue $1/k$. The real half-period shift gives another pole at $2K+i{K}^{\prime }$ with opposite residue. Reflection tiles the plane with copies of the rectangle, so these are exactly two simple poles per displayed fundamental cell. They also show there is no additional period: a period must permute these two pole classes; exchanging them would shift by $2K$ modulo the displayed lattice, which reverses the function's sign and residues rather than preserving it.

For any finite $w$, integrate $s^{\prime }(z)/(s(z)-w)$ around a fundamental parallelogram avoiding its zeros and poles. The opposite edges cancel by periodicity, so the number of zeros of $s-w$ equals the number of poles, namely two. Therefore every value has two inverse images modulo periods, counted with multiplicity. The value infinity likewise has the two simple poles as preimages. Usually the finite-value preimages are distinct; the involution $z\mapsto 2K-z$ interchanges them. At $w=\pm 1,\pm 1/k$, a critical point supplies one double inverse image instead. For example $s(K)=1$, $s^{\prime }(K)=0$, $s^{\prime \prime }(K)=-(1-k^2)\ne 0$. Thus the printed “exactly two” requires the standard multiplicity convention; it would be false if interpreted as two distinct solutions at these branch values of a holomorphic map.

Now choose $z_0=K$ and put $f(z)=s(z+K)$. Since $s(2K-z)=s(z)$, $f$ is even, elliptic for $\Lambda$, and has degree two. The Weierstrass elliptic function for the same lattice is even and has one double pole per cell, hence also degree two. Its generic fibers are precisely $\{z,-z\}$. Consequently an even $f$ is constant on those fibers and descends to a meromorphic function $R$ of $\wp$. At the four fixed classes of $z\mapsto -z$, the even local Laurent expansion makes this descent meromorphic in the squared local coordinate. A meromorphic function on the Riemann sphere is rational, so $f=R(\wp )$. Generic map degrees multiply:Thus $R$ has degree one and the degree-two even elliptic function conclusion isThe determinant condition follows because $f$ is nonconstant. This proves the existence requested, with the explicit allowable shift $z_0=K$.

For genuine degrees three and two, the resultant is the Sylvester matrix determinantEquivalently, if $r_1,r_2,r_3$ are the roots of $p$, counted with multiplicity, then $R(p,q)=a^2{\prod }_{i}q(r_i)$. Thus $R(p,q)\ne 0$ means that the polynomials have no common root, or equivalently are coprime over $\mathbb{C}$. One can also see this from the matrix: its singularity gives a nonzero relation $Ap+Bq=0$ with $\mathrm{deg}A<2$, $\mathrm{deg}B<3$. If $p,q$ were coprime, $p\mid B$ would force $B=0$, then $A=0$, a contradiction. Conversely, a common factor supplies such a relation after dividing $p,q$ by that factor. When a leading coefficient is zero, use the resultant for the actual polynomial degrees rather than assuming the displayed fixed-degree criterion unchanged.

An algebraic addition theorem for a meromorphic function $f$ means a nonzero polynomial $P(U,V,W)$ withidentically where the values are finite. A local relation extends meromorphically to the other regular arguments.

For a polynomial $f$ of positive degree $d$, let $z_i$ be the $d$ roots of $f(z)-U$ and $w_j$ the $d$ roots of $f(w)-V$. FormIts coefficients are separately symmetric polynomials in the two sets of roots. The Fundamental theorem of symmetric polynomials and the fixed nonzero leading coefficient of $f$ show they are polynomials in $U,V$. The expression is monic of degree $d^2$ in $W$, so it is nonzero. If $U=f(z)$, $V=f(w)$, one factor vanishes at $W=f(z+w)$. This proves the addition relation, including at exceptional multiple-root values by polynomial identity. A constant polynomial instead has the relation $W-f(0)=0$.

For the elliptic case, let $E=\mathbb{C}/\Lambda$. A nonconstant elliptic function $f$ defines a finite morphism $E\to {\mathbb{P}}^1$, so the function field of $E$ is a finite algebraic extension of $\mathbb{C}(f)$. The elliptic function-field decomposition expresses it as $\mathbb{C}(\wp ,{\wp }^{\prime })$, withThe Weierstrass addition formula makes translation algebraic:Differentiating and using the cubic differential equation also makes ${\wp }^{\prime }(z+w)$ rational in the four separate values. Therefore $f(z+w)$ lies in a finite algebraic extension of $\mathbb{C}(f(z),f(w))$. Its algebraic equation, after clearing denominators, is the required polynomial relation. This sketches why every elliptic function, not only $\wp$, has an algebraic addition theorem. Constants are already covered.

For the final assertion, consider $f(z)=e^{e^z}$, an entire function. If $c$ is a period, then $e^{e^z(e^c-1)}=1$ for every $z$. The continuous exponent must be a constant integer multiple of $2\pi i$; its derivative forces $e^c=1$. Hence its periods are exactly $2\pi i\mathbb{Z}$, making it a simply periodic function.

Suppose it had an algebraic addition theorem $P$. There are only finitely many values $V_0$ for which $P(U,V_0,W)$ is identically zero: a nonzero coefficient polynomial in $V$ already bounds this exceptional set. Choose a positive irrational $\alpha$ with $e^{\alpha }$ outside this set, and set $c=\log \alpha$. Specializing $w=c$ gives a nonzero polynomial $Q(U,W)=P(U,e^{\alpha },W)$ and, with $t=e^z$,All exponents with nonzero coefficients are distinct, by irrationality of $\alpha$. Divide by the exponential with largest real exponent and let $t\to +\infty$; its nonzero coefficient would have to tend to zero. This contradiction proves that simple periodicity does not imply an algebraic addition theorem. It is the simply periodic entire function without an algebraic addition theorem example.

A skew Young diagram is connected when any two of its cells can be joined by a sequence of cells sharing edges; touching only at a corner does not connect them. A nonempty border strip, or rim hook, is a connected skew diagram containing no $2\times 2$ square. Its height is its number of occupied rows minus one.

For a weak composition $\alpha =({\alpha }_1,\dots ,{\alpha }_{\ell })$, a border-strip tableau of shape $\lambda /\mu$ and type $\alpha$ is a chainwhose $i$th difference is a border strip of size ${\alpha }_i$. Filling that strip with $i$ gives the equivalent weakly increasing row-and-column filling. A zero part corresponds to an empty difference and has height zero. Put $\mathrm{ht}(T)={\sum }_i\mathrm{ht}({\lambda }^{(i)}/{\lambda }^{(i-1)})$. For positive $r$, $p_r={\sum }_j{x}_j^r$; write $p_{\alpha }={\prod }_{{\alpha }_i>0}{p}_{{\alpha }_i}$. Thus zero type parts are ignored: one does not multiply by the finite-variable value $p_0=N$.

We first prove the one-strip multiplication formula using alternants. Choose $N$ sufficiently large and let ${\beta }_i={\mu }_i+N-i$ be the decreasing beta numbers of a partition. Expanding the determinant by permutations showsIndeed, in each determinant monomial every variable occurs in exactly one row factor; multiplying that variable by its $r$th power is equivalent to raising that row exponent.

A term vanishes if the new exponent equals another exponent. Otherwise suppose the changed exponent moves from position $i$ to position $j\le i$ when restored to decreasing order. Its determinant sign is $(-1{)}^{i-j}$. The resulting partition $\lambda$ hasand all other rows unchanged. The new diagram contains $\mu$, and its size increases by $r$. Between successive added rows, the overlap is exactly the single column ${\mu }_{t-1}+1$. Thus the added cells form a connected strip without a $2\times 2$ square, of height $i-j$. Conversely every added border strip has consecutive occupied rows and exactly this single-column overlap, forcing these formulas and the corresponding exponent shift. The correspondence is therefore bijective. Divide by the unchanged Vandermonde determinant and use the bialternant formula to get the power-sum border-strip multiplication identity

Apply this formula successively to each positive part of $\alpha$. Each sequence of choices is exactly a border-strip tableau, and its product of signs is $(-1{)}^{\mathrm{ht}(T)}$. ThereforeCommutativity of the power sums also shows that the resulting coefficient does not depend on the order of the positive parts of the type.

Take $\mu$ empty and restrict to $N\ge \ell (\lambda )$ variables. Multiply by $a_{\delta }$, with $\delta =(N-1,\dots ,0)$:In an alternant $a_{\nu +\delta }$, the exponents are distinct and their decreasing arrangement is $\nu +\delta$. Consequently the monomial $x^{\lambda +\delta }$ occurs only for $\nu =\lambda$, and in that determinant its coefficient is one. This proves the Frobenius alternant character formula

To identify these coefficients with symmetric-group characters, recall the Frobenius characteristic mapThe representation-theoretic identification $\mathrm{ch}({\chi }^{\lambda })=s_{\lambda }$ can be justified by Young's rule: the permutation character on ordered blocks of sizes $\nu$ is ${\sum }_{\lambda }{K}_{\lambda \nu }{\chi }^{\lambda }$. Its characteristic is $h_{\nu }$, because its fixed block decompositions allocate whole cycles to blocks, and the exponential generating series for those allocations is ${\prod }_{j}H(t_j)$. On the symmetric-function side, Hall duality and Schur orthonormality give $h_{\nu }={\sum }_{\lambda }{K}_{\lambda \nu }{s}_{\lambda }$. Inverting the unitriangular Kostka matrix establishes the identification. Thus the coefficient of $s_{\lambda }$ in $p_{\rho }$ is the irreducible character value on cycle type $\rho$. We have supplied the border-strip computation rather than assuming the character rule.

Grouping border-strip chains by their first removal of a cycle length $r$ now yields the Murnaghan–Nakayama ruleOne may choose the $r$-cycle to be last in a strip-building chain because the power sums commute. The empty-shape value on the empty type is one, providing the recursion's initial condition.

For the staircase $\delta =(m-1,m-2,\dots ,1)$, use $m$ beta numbers including its padded zero row. Its beta set isRemoving a rim hook of length $r$ lowers one beta number by $r$ into an unoccupied nonnegative position, reversing the exponent-shift description above. For a positive even $r$, this move either hits another occupied even position or goes below zero. Therefore there are no removable even-size border strips.

Under the Hall inner product of symmetric functions, the adjoint of multiplication by $p_r$ is $r\partial /\partial p_r$: this follows on the power-sum basis from $z_{\nu }/z_{\nu \setminus r}=r{m}_r(\nu )$. Taking adjoints of the one-strip multiplication rule expresses this derivative on $s_{\delta }$ as the signed sum over removable size-$r$ strips. That sum is empty for even $r$, soThe rational symmetric-function algebra is freely generated by the power sums, and its characteristic is zero. Vanishing of each of these derivatives therefore proves the staircase Schur functions use only odd power sums statement:

Finally, the cycle type of an $n$-cycle has one strip of size $n$. A straight diagram is a border strip exactly when its second row has length at most one: otherwise it contains the upper-left $2\times 2$ square; conversely the first row with a single-column tail is connected and contains none. These diagrams are exactly the hook partitions $(n-s,1^s)$, of height $s$. There is one strip filling in this case and none otherwise, giving

In $N$ variables, pad $\lambda$ to length $N$ and set $\delta =(N-1,N-2,\dots ,0)$. Define the alternantThe last equality is the Vandermonde determinant with the indicated descending-exponent convention.

We use the Jacobi–Trudi identity proved above and give the determinant step explicitly. LetPut $b_i={\lambda }_i+N-i\ge 0$. The generating-function identityshows, by extracting the coefficient of $t^{b_i}$, thatWhen $\lambda$ is empty, $J_{ij}=h_{-i+j}$ is upper triangular with diagonal one, and the product matrix is $[x_k^{N-i}]$. Thus $\det B=a_{\delta }$. Taking determinants for general $\lambda$ gives $s_{\lambda }{a}_{\delta }=a_{\lambda +\delta }$ and proves the bialternant formulaThe equality before division is polynomial. Consequently the quotient extends at coincident variables as the Schur polynomial.