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Every identity-preserving biholomorphism of $\mathbb{C}/\Lambda$ lifts to multiplication by a nonzero complex number $\zeta$ satisfying $\zeta \Lambda =\Lambda$. Relative to a $\mathbb{Z}$-basis of $\Lambda$, multiplication by $\zeta$ is represented by a unimodular matrix $A\in {\mathrm{GL}}_2(\mathbb{Z})$. Its real determinant is $|\zeta {|}^2$, while complex multiplication preserves orientation, so $\det A=1$ and $|\zeta |=1$. The Cayley-Hamilton theorem givesConsequently $2\mathrm{Re}\zeta =\mathrm{tr}(A)\in \{-2,-1,0,1,2\}$, and $\zeta$ is a root of unity of order $1$, $2$, $3$, $4$, or $6$.

On the torus ${\mathbb{R}}^d/{\mathbb{Z}}^d$, an integer matrix $A$ defines an endomorphism. The coordinate circles give a basis of the first homology group, and the induced map on that group is $A$. The Künneth theorem identifies higher integral homology with the exterior powers of ${\mathbb{Z}}^d$, so the degree-$q$ map is ${\bigwedge }^{q}A$. In top degree it is multiplication by $\det A$. The map is a homeomorphism precisely when $A$ is a unimodular matrix; a merely nonsingular integer matrix need not have an integer inverse.

The adjugate matrix is the transpose of the cofactor matrix: $\mathrm{adj}(A{)}_{ij}=(-1{)}^{i+j}\det A_{\hat{j},\hat{i}}$. Cofactor expansion gives the adjugate identityFor an invertible matrix this yields $A^{-1}=\mathrm{adj}(A)/\det A$. Integer entries of $A$ imply integer cofactors, so $\det A=\pm 1$ makes its inverse integral. Conversely, if $A$ and $A^{-1}$ both have integer entries, their determinants are integers and $(\det A)(\det A^{-1})=1$. The only integer units are $\pm 1$. Hence an invertible integer matrix has an integer inverse exactly when it is a unimodular matrix, with determinant $\pm 1$. Invertibility is implicit whenever $A^{-1}$ is used.

Use integer coefficients. The mapping-torus homology from Mayer–Vietoris works for every continuous map $f:X\to X$, without assuming that the mapping torus is a fiber bundle. Project to the time circle and choose two open arcs, one containing the seam and one away from it. The inverse image $U$ of the seam arc is a mapping cylinder neighborhood with a deformation retraction to its target copy of $X$; the inverse image $V$ of the other arc is a product with a deformation retraction to $X$. Their intersection has two components, each with a deformation retraction to $X$. Label the upper component first. With the convention that the second inclusion in the Mayer–Vietoris sequence has a minus sign, its homology map isIndeed, the upper component includes into the seam neighborhood by the identity, while retracting the lower component across the seam applies $f$. Put $u=a+b$ and $v=b$, and replace target coordinates $(p,q)$ by $(p+q,-q)$. The map becomes $((f_{\ast }-\mathrm{id})v,u)$. Negating $v$ gives a direct sum of $\mathrm{id}-f_{\ast }$ and an isomorphism. Canceling the latter summands in the long exact sequence yieldsIn particular, there is a short exact sequenceNo splitting of this short exact sequence is assumed at this stage.

For the torus, take the coordinate loops $c_1(t)=[(t,0)]$ and $c_2(t)=[(0,t)]$ as the basis of $H_1$. Their lifts to ${\mathbb{R}}^2$ have endpoint displacements $e_1,e_2$. Applying $A$ changes those displacements to its columns, soThis also follows from the identification of the first homology group with the abelianization of the fundamental group. The Künneth theorem gives $H_2(X)=\mathbb{Z}$, with generator the product orientation class. If $\alpha ,\beta$ are the dual degree-one cohomology classes, then $\alpha ⌣\beta$ evaluates to one on this generator, andThus the degree-two homology map is multiplication by $\det A$, as in homology of an integer torus endomorphism. The claim that the torus map is a homeomorphism requires $A$ to be a unimodular matrix, equivalently an integer inverse or $\det A=\pm 1$; nonsingularity over $\mathbb{R}$ alone would only give a covering map of finite degree. The particular matrix here is unimodular.

For this matrix, $\det A=1$ andIts kernel is zero and its Smith normal form is $\mathrm{diag}(1,2)$, so its cokernel is $\mathbb{Z}/2$. The maps $\mathrm{id}-f_{\ast }$ on $H_0$ and $H_2$ are zero. Substituting into the short exact sequences, and splitting the degree-one extension because its quotient $\mathbb{Z}$ is free, gives

We use left multiplication by $\Gamma =S{L}_2(\mathbb{Z})$. For an integral matrix $A$ of determinant $n>0$, the Bezout identity gives a determinant-one row operation sending its first column to $(a,0{)}^t$, where $a$ is the positive gcd of that column. The resulting matrix is $\left(\begin{array}{cc}a& b\\ 0& d\end{array}\right)$ with $d=n/a>0$. Adding a multiple of the second row to the first makes $0\le b<d$.

These representatives are unique. Left multiplication by a unimodular matrix preserves the gcd of the first column, so two representatives in one orbit have the same $a$ and hence $d$. A matrix taking $(a,0{)}^t$ to itself has the form $\left(\begin{array}{cc}1& t\\ 0& 1\end{array}\right)$, so it changes $b$ by $td$; the prescribed range makes $b$ unique. Thus the determinant-n matrix representatives for Hecke operators are exactly ${\Pi }_n$. In the printed set, $0\le b<d$ already forces $d>0$, and $ad=n>0$ forces $a>0$.

For positive determinant extend the slash operator for modular forms byThe determinant factor makes this a right action: $(f{|}_{k}A){|}_{k}B=f{|}_k(AB)$. Right multiplication by $\delta \in \Gamma$ permutes the left cosets represented by ${\Pi }_n$, since it preserves the set of determinant-$n$ integral matrices. Hence $T_{n}f$ satisfies the modular transformation law. Its summands are holomorphic on the upper half-plane. For an upper-triangular representative,Substitute the Fourier expansion of a modular form. The sum over $b$ is zero unless the original index is divisible by $d$, in which case it equals $d$. Writing that index as $dr$ givesOnly nonnegative powers occur, so $T_{n}f$ is holomorphic at infinity, and all level-one cusps are equivalent to infinity. We have proved $T_{n}f\in M_k(\Gamma )$ and the Fourier coefficients of a composite-index Hecke operator formulaHere $\gcd (n,0)=n$, giving $a_0(T_{n}f)={\sum }_{a\mid n}{a}^{k-1}{a}_0(f)$. In particular cusp forms stay cusp forms. For nonzero level-one cusp forms the weight is an even integer at least twelve, so the powers $a^{k-1}$ are integers; the zero cusp space causes no exception. Thus every $T_n$ preserves integral Fourier coefficients, and closure under addition, multiplication and integer scalars provesThis is the integral Hecke algebra of level-one cusp forms.

We will also need commutativity. The coefficient formula directly givesFor the first identity, the divisors of $u$ and $v$ in the iterated coefficient sum combine uniquely into a divisor of $uv$. For the second, write a coefficient index as $p^t{r}_0$ with $p\nmid r_0$; the formula isApplying $T_p$ splits this into the two sums with shifts $t+1$ and $t-1$; their overlapping terms give exactly the stated recurrence. Since $T_1$ is the identity, induction expresses each $T_{p^r}$ as a polynomial in $T_p$, and coprime multiplicativity expresses every $T_n$ as a product of these polynomials. Distinct prime operators commute by the coprime identity. Hence all Hecke operators commute, and so do all elements of $\mathbb{T}$.

Let $\Lambda =S_k(\Gamma ,\mathbb{Z})$ and $m=\dim S_k(\Gamma )$. In the permitted basis $F_j={\Delta }^j{E}_4^{a_j}{E}_6^{b_j}$, the modular discriminant begins $q+O(q^2)$ and the normalized Eisenstein series begin $1+O(q)$, soThe matrix of $a_1,\dots ,a_m$ on this integral basis is unitriangular and has determinant one. Integer elimination therefore gives an integral basis $f_1,\dots ,f_m$ with $a_i(f_j)={\delta }_{ij}$ for $1\le i,j\le m$. Thus $a_1,\dots ,a_m$ are a $\mathbb{Z}$-basis of the dual module ${\mathrm{Hom}}_{\mathbb{Z}}(\Lambda ,\mathbb{Z})$.

The map $\alpha$ is well-defined by lattice preservation and is $\mathbb{Z}$-linear. Because $\alpha (T_j)=a_j$, it is surjective. For injectivity suppose $\alpha (T)=0$. For every $f\in \Lambda$ and every $n\ge 1$, commutativity givesThe constant term is also zero because $Tf$ is a cusp form. The identity theorem applied to its Fourier expansion of a modular form gives $Tf=0$. The integral basis spans the complex cusp space, so $T=0$ as an endomorphism. This proves the perfect integral Hecke pairing andIf $m=0$, both modules are zero and the asserted basis is empty.

Let $\pi :\mathbb{C}\to \mathbb{C}/\Lambda$ be the universal covering map. Since the complex plane is a simply connected domain, the lifting criterion for a covering space gives a holomorphic lift between one-dimensional complex tori $F:\mathbb{C}\to \mathbb{C}$ such thatFor every $\lambda \in \Lambda$, the difference $F(z+\lambda )-F(z)$ belongs to the discrete period lattice $\Lambda$. It depends continuously on $z$, so it is constant. Differentiating shows that the derivative $F^{\prime }$ is $\Lambda$-periodic. It is bounded on the compact closure of a fundamental parallelogram, and periodicity makes it bounded on all of $\mathbb{C}$. The Liouville theorem therefore makes $F^{\prime }$ constant, and henceThus every holomorphic map has an affine lift of a holomorphic map between one-dimensional complex tori. If “map of complex tori” means an identity-preserving map, choose $F(0)=0$; then $b=0$, so $F$ is the required linear map. Without that convention the statement must say affine, since a nonzero translation in a group of the torus lifts to $z\mapsto z+b$.

The Weierstrass elliptic function of $\Lambda$ isThe subtracted term gives the Normal convergence of the Weierstrass elliptic-function series away from $\Lambda$. PutThe Laurent coefficients of the Weierstrass elliptic function at zero giveSet $g_2=60G_4$ and $g_3=140G_6$. Direct substitution shows that the principal part and constant term ofvanish at zero. Since $H$ is an elliptic function, translation gives the same cancellation at every point of the period lattice. Every apparent isolated singularity is therefore a removable singularity, so $H$ is an entire function. It is periodic and hence bounded on the translates of a compact fundamental parallelogram. The Liouville theorem gives $H=0$, which proves the Weierstrass elliptic differential equation

Finally suppose that $f$ is a biholomorphic group homomorphism. Its identity-preserving lift has the form $F(z)=\zeta z$ by the first part. Since the inverse map also lifts linearly,Choose a $\mathbb{Z}$-basis of $\Lambda$. Multiplication by $\zeta$ is then represented by a unimodular matrix $A\in {\mathrm{GL}}_2(\mathbb{Z})$. As a real-linear transformation of $\mathbb{C}$, it has determinant $|\zeta {|}^2>0$; hence $\det A=1$ and $|\zeta |=1$. The characteristic polynomial of $A$ and the Cayley-Hamilton theorem giveThis already has the required form with $m=-\mathrm{tr}(A)\in \mathbb{Z}$ and $n=1$. Moreover $2\mathrm{Re}\zeta =\mathrm{tr}(A)$, so $|\zeta |=1$ forces $\mathrm{tr}(A)\in \{-2,-1,0,1,2\}$. Thus $\zeta$ is a root of unity, of order $1$, $2$, $3$, $4$, or $6$, completing the description of an automorphism of a one-dimensional complex torus.

The quadratic form factors as a sum of two squares:The change of variableshas determinant $-1$, so its matrix is a unimodular matrix and it maps ${\mathbb{Z}}^2$ bijectively to itself. The original equation is therefore equivalent toThe sum of two squares theorem also decides this systematically: $365=5\cdot 73$, and every prime factor congruent to $3$ modulo $4$ occurs to an even power, vacuously here. Thus $365$ is a sum of two integer squares. Explicitly,Inverting the change of variables gives, for example, $(x,y)=(32,-15)$. ConsequentlyThis is the unimodular reduction of a discriminant-minus-four form to two squares.

Put $Q=G/H$ and write $M^H$ for the invariant submodule. The five-term exact sequence in group cohomology associated with the Lyndon–Hochschild–Serre spectral sequence isThe inflation map in group cohomology composes a cocycle on $Q$ with the quotient homomorphism $G\to Q$. The restriction map in group cohomology restricts a cocycle from $G$ to $H$. The quotient action on the middle term is, for $q=gH$ and a one-cocycle $f$,This is independent of the lift and of the representative at the level of cohomology. Finally, the transgression in group cohomology extends a $Q$-invariant class on $H$ to a one-cochain on $G$; its coboundary is $H$-basic and descends to the two-cocycle on $Q$ representing $d_2[f]$. Changing the extension changes that cocycle by a group coboundary.

For the application, choose free generators $x_1,\dots ,x_n$ of $F$ and normal generators $r_1,\dots ,r_n$ of $R$. Since a finite nonabelian simple group $K$ is a perfect group, its abelianization is zero. The five-term sequence for $1\to R\to F\to K\to 1$ with trivial coefficients containsThe first term is zero because $K$ is finite, and the last term is zero because a free group has cohomological dimension one. It remains to prove that restriction is surjective.

The invariant submodule of homomorphisms $R\to \mathbb{Z}$ is exactlyThe images of the relators $r_1,\dots ,r_n$ generate $R/[F,R]$, so such a homomorphism $\lambda$ is determined by the integer vector $b=(\lambda (r_1),\dots ,\lambda (r_n){)}^T$. Let $A$ be the relator exponent-sum matrix. This square integer matrix presents $K^{\mathrm{ab}}$, which is zero, so $A$ is a unimodular matrix. There is therefore an integer vector $v$ satisfying $Av=b$. Define $\psi :F\to \mathbb{Z}$ by assigning to $x_i$ the $i$th entry of $v$. The definition of $A$ gives $\psi (r_j)=\lambda (r_j)$ for every $j$. Since the relator images generate $R/[F,R]$, the restriction of $\psi$ to $R$ equals $\lambda$. Restriction is surjective, exactness now gives

For a finite group presentation, the relator exponent-sum matrix records in row $i$ the total exponent of each generator in relator $r_i$. It presents the abelianization of the group. For a balanced presentation of a perfect group, it is a unimodular matrix.