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Embed the ring of integers into ${\mathbb{R}}^2$ by its two real embeddings. Its image is a full Euclidean lattice of some covolume $d>0$. Fix $C>d$ and consider the symmetric rectangle $|{\sigma }_1(\alpha )|\le T$, $|{\sigma }_2(\alpha )|\le C/T$. Its area is $4C>4d$, so the Minkowski convex body theorem gives a nonzero algebraic integer ${\alpha }_T$ in it. Its algebraic norm is a nonzero integer with $|N({\alpha }_T)|\le C$.

As $T\to \infty$, infinitely many distinct ${\alpha }_T$ must occur: a finite collection of nonzero elements has a positive minimum of $|{\sigma }_2(\alpha )|$, whereas here that coordinate tends to zero. The principal ideals $({\alpha }_T)$ have norm $|N({\alpha }_T)|\le C$. There are only finitely many ideals of bounded norm: they are subgroups of bounded index in a rank-two free abelian group, and integer Hermite normal forms give finitely many possibilities at each such index. Thus infinitely many of these elements generate one ideal. Ratios of two generators of the same ideal are units, because inclusion in both directions supplies inverse algebraic integers. Among infinitely many distinct generators choose two whose ratio is not $\pm 1$. This gives a non-trivial unit without assuming Dirichlet's theorem.

Write $\Gamma =S{L}_2(\mathbb{Z})$ and $G=G{L}_2(\mathbb{Q}{)}^{+}$. The double-coset Hecke algebra consists of $\Gamma$-bi-invariant complex functions on $G$ supported on finitely many double cosets, with convolutionEach double coset has finitely many orbits under left multiplication by $\Gamma$, by rational conjugation of finite-index modular subgroups. Thus the sum is finite and independent of representatives. The characteristic functions of the double cosets form its basis, and its right action on invariant modular forms is $f\ast h={\sum }_{\Gamma x}h(x)f{|}_{k}x$, using the determinant-normalized slash operator.

For a positive integer $n$, let ${\mathcal{M}}_n$ be all integral two-by-two matrices of positive determinant $n$. It is $\Gamma$-bi-invariant. Define $T(n)$ to be its indicator function, equivalently the sum of the distinct double cosets it contains, each with coefficient one. This is the convention consistent with the requested formula. For composite $n$, it need not be the single double coset of $\mathrm{diag}(1,n)$: for instance $2I\in {\mathcal{M}}_4$ cannot lie in that double coset, since multiplying by unimodular integral matrices preserves the greatest common divisor of the entries.

Prove all the needed subgroup facts directly. For a finite-index subgroup $L\subseteq {\mathbb{Z}}^2$, take $a$ to be the least positive first coordinate appearing in $L$ and $d$ the least positive second coordinate on its intersection with the second axis. Euclidean division then shows that the first-coordinate projection is $a\mathbb{Z}$ and $L\cap (\{0\}\times \mathbb{Z})=\{0\}\times d\mathbb{Z}$. Positivity follows, for example, because the finite quotient group kills a nonzero multiple of each coordinate vector. Choose $(a,b)\in L$ and reduce $b$ modulo $d$ to $0\le b<d$. Every vector of $L$ has first coordinate a multiple of $a$, and subtracting that multiple of $(a,b)$ leaves a multiple of $(0,d)$. Therefore these two vectors form a basis of $L$. The parameters $a,d,b$ are unique. Reducing the first coordinate modulo $a$ and then the second modulo $d$ gives precisely $ad$ quotient representatives, so $[{\mathbb{Z}}^2:L]=ad$.

Apply this row Hermite normal form in rank two to the row lattice of an integral matrix $A$ with $\det A=n$. Its row lattice contains $n{\mathbb{Z}}^2$, since $\mathrm{adj}(A)A=nI$, so it has finite index. Its two rows and the displayed two rows are bases of the same row lattice. The two inverse change-of-basis matrices have integer entries, so their determinants are integers whose product is one. Thus the change-of-basis matrix has determinant $\pm 1$; since both orientations are positive, its determinant is one. Thus each orbit under left multiplication by $\Gamma$ has exactly one of the determinant-n matrix representatives for Hecke operatorsThere are ${\sum }_{d\mid n}d$ such representatives, proving finiteness as well as the formula. The subgroup argument is a proof of the relevant Hermite normal form, not an invocation of an unproved lattice classification.

Consequently the normalized Hecke operator isIt preserves $M_k(\Gamma )$: right multiplication by $\Gamma$ permutes the left-multiplication orbits in ${\mathcal{M}}_n$, giving invariance, and cusp holomorphy under rational slash operators gives the holomorphy of each term at every cusp.

For a triangular representative the determinant-normalized slash operator is $f{|}_{k}A=n^{k/2}{d}^{-k}f((az+b)/d)$. Summing the Fourier expansion of a modular form over $b$ kills every index not divisible by $d$, by finite exponential orthogonality. ThusThe Fourier coefficients of a composite-index Hecke operator are thereforeIn particular $a_1(T_{n}f)=a_n(f)$. If $T_{n}f=\lambda f$, comparison of the $q$ coefficients gives

Finally suppose $a_0(f)\ne 0$ and $f$ is a simultaneous eigenfunction. Constant coefficients force ${\lambda }_n={\sigma }_{k-1}(n)$, so $a_n(f)=a_1(f){\sigma }_{k-1}(n)$ for all $n\ge 1$. Weight two cannot occur, by vanishing of weight-two level-one modular forms. For even $k\ge 4$, use the normalized Eisenstein series with its Fourier expansion of a normalized Eisenstein serieswhere $B_k$ is the Bernoulli number. Then $h=f-a_0(f)E_k$ is a cusp form with coefficients $(a_1(f)-a_0(f)c_k){\sigma }_{k-1}(n)$. The Fourier coefficient bound for a cusp form bounds these by $O(n^{k/2})$, but at arbitrarily large primes $p$ their magnitude is $|a_1-a_0{c}_k|(1+p^{k-1})$. Since $k-1>k/2$, the coefficient factor must vanish. All coefficients of $h$ then vanish, including its constant coefficient, so $h=0$ by its cusp expansion and the identity theorem. This proves the noncuspidal level-one Hecke eigenform characterization:

A unimodular matrix is a square integer matrix whose determinant is $1$ or $-1$. The decisive property is that its inverse matrix also has integer entries: multiplication by $U$ gives a bijection of the integer lattice ${\mathbb{Z}}^n$ onto itself. Thus an invertible linear map can preserve not only real coordinates but the requirement that every coordinate is an integer. These matrices form the integral general linear group.^[ref]

For an original two-dimensional example, takeIts determinant is one, and it maps $(a,b)$ to $(a+b,b)$. Conversely, the integer point $(x,y)$ comes from $(x-y,y)$. For instance, $(3,2)$ has new coordinates $(1,2)$ relative to the columns $(1,0)$ and $(1,1)$. No integer point disappears, although the basis vectors and their fundamental cell change shape.

This is useful whenever coordinates must remain integral. A unimodular change of lattice basis can give a shorter or more convenient description without changing the objects represented. Smith normal form and Hermite normal form use controlled integer operations to simplify integer matrices; ordinary Gaussian elimination may divide by numbers that destroy integrality. The geometric freedom is substantial: a sheared basis may look completely different while describing exactly the same infinite set of integer points.

A determinant of $-1$ is allowed too. Swapping two basis columns reverses orientation while preserving the integer lattice. The condition is therefore preservation of the complete integer structure, rather than a requirement to preserve orientation.