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A root system spanning a real inner product space is irreducible when it cannot be partitioned into two nonempty mutually orthogonal root systems. Its Dynkin diagram is connected. This condition concerns the roots themselves, rather than an arbitrary ambient space larger than their span.

For commuting diagonalizable operators $A_1,\dots ,A_r$ on a finite-dimensional vector space, the joint spectrum is the set of tuples $(a_1,\dots ,a_r)$ for which there is a common nonzero eigenvector $v$ with $A_{i}v=a_{i}v$. Commuting self-adjoint operators on a finite-dimensional inner product space admit a simultaneous orthonormal eigenbasis.

A matrix element $A_{ij}$ is the entry in row $i$ and column $j$ of a matrix $A$.

For a linear operator $T$ on an inner product space with an orthonormal basis $(e_i)$, its matrix elements are $T_{ij}=⟨e_i,T{e}_j⟩$, using the convention that the inner product is linear in its second argument. For a finite-dimensional space, the expansion $T{e}_j={\sum }_i{T}_{ij}{e}_i$ explains why these numbers represent the linear operator in that basis. In quantum mechanics, the expression $⟨\varphi |T|\psi ⟩$ is also called a matrix element between the two states; it need not be diagonal or an expectation value.

An operator $F$ on an inner product space is monotone whenfor every $v,w$. The gradient of every differentiable convex function is monotone.

An orthogonal change of coordinates relates two orthonormal bases of a real inner product space. With the new basis vectors as columns of an orthogonal matrix $Q$, old and new coordinate columns obey $x=Qy$. Then $x^{T}x=y^{T}y$, so Euclidean norm, Euclidean distance and angles are preserved. A quadratic form $x^{T}Ax$ becomes $y^T(Q^{T}AQ)y$. For a symmetric matrix, orthogonal diagonalization chooses $Q$ so that the transformed quadratic form is a sum of eigenvalues times squared coordinates. This explains the principal-axis descriptions of a quadric surface.

A finite-dimensional linear subspace $W$ of an inner product space has a unique closest vector to any given vector $x$. This is its orthogonal projection $P_{W}x$. If $e_1,\dots ,e_d$ is an orthogonal basis of $W$, compute it one direction at a time:The inner product convention here is linear in its first argument, including over the complex numbers. For an orthonormal basis, each denominator is one. When $W$ consists only of the zero vector, the empty sum gives $P_{W}x=0$. The finite-dimensional formula is described in Francis Narcowich, Inner Products and Norms: minimization problems.

Take $W=\mathrm{span}\{(1,2{)}^T\}$ and $x=(3,1{)}^T$, using the dot product. With $e=(1,2{)}^T$,The orthogonal projection residual has $r\cdot e=2-2=0$. To verify the best approximation directly, every candidate in the line is $te$, andThe smallest squared Euclidean norm error is therefore $5$, attained only at $t=1$. The diagram connects the perpendicular residual to this one-variable quadratic polynomial.

The general mechanism is the Pythagorean theorem in an inner-product space. For any $w\in W$, the vectors $r=x-P_{W}x$ and $P_{W}x-w$ are orthogonal, soMoving away from $P_{W}x$ inside $W$ adds a nonnegative squared norm, which is zero only when $w=P_{W}x$. This proves both optimality and uniqueness.

The useful feature is separation of preparation from evaluation. Construct an orthonormal basis once with the Gram-Schmidt process, then obtain each new best approximation using inner products instead of a fresh optimization problem. When $W$ is the column space of a matrix, this gives linear least squares; when $W$ is spanned by orthogonal polynomials, it gives a least-squares polynomial in an orthogonal-polynomial basis. The ambient inner product space need not be a Hilbert space for this finite-dimensional construction. In an infinite-dimensional target linear subspace, existence requires additional hypotheses, supplied by orthogonal decomposition by a closed subspace.

Two vectors are orthogonal when their inner product is zero.

For real Euclidean vectors, the inner product is the dot product, so $(3,4)$ and $(4,-3)$ are orthogonal: $3(4)+4(-3)=0$. For nonzero Euclidean vectors, this means their angle is $90^{\circ }$. The zero vector is orthogonal to every vector, although it has no direction or defined angle.^[ref]

In an abstract inner product space, orthogonality always refers to its chosen inner product. For example, the weighted inner product $⟨u,v{⟩}_w=4u_1{v}_1+u_2{v}_2$ makes $(1,1)$ and $(1,-4)$ orthogonal, since $4-4=0$. Their ordinary dot product is $-3$, so their arrows are not perpendicular in the usual Cartesian coordinate system. The algebraic definition also applies to functions, where arrows need not be available.

The parallelogram law for a quadratic form $Q$ isConversely, under the usual divisibility assumptions, a function satisfying this identity and the appropriate homogeneity comes from a symmetric bilinear form by polarization.

For an inner product space, taking $Q(v)=\Vert v{\Vert }^2$ gives the displayed squared-norm identity by expanding and cancelling the mixed inner products. It is the geometric ingredient that makes minimizing sequences converge in the Hilbert projection theorem.

Use an inner product linear in its first argument. The adjoint operator ${\alpha }^{\ast }$ is the unique linear operator such thatIn a finite-dimensional inner product space, existence follows by choosing an orthonormal basis: the matrix of the adjoint is the conjugate transpose of the matrix of $\alpha$. Uniqueness follows from nondegeneracy of the inner product.

If $\alpha (W)\subseteq W$, then for $w\in W$ and $v\in W^{\perp }$,Thus ${\alpha }^{\ast }v\in W^{\perp }$. Conversely, if ${\alpha }^{\ast }(W^{\perp })\subseteq W^{\perp }$, the same equality gives $⟨\alpha w,v⟩=0$ for every $v\in W^{\perp }$, so $\alpha w\in (W^{\perp }{)}^{\perp }=W$. This proves the adjoint criterion for an invariant orthogonal complement.

The equality $(W^{\perp }{)}^{\perp }=W$ uses the finite-dimensional setting appropriate here. In a general Hilbert space it holds for closed subspaces, and an adjoint is assured for bounded operators. Without closedness the converse only implies invariance of the closure. For example in ${\ell }^2$, take $W$ to be the finite-support sequences and $\alpha v=⟨v,e_1⟩(2^{-j}{)}_{j\ge 1}$. Then $W^{\perp }=\{0\}$ is automatically adjoint-invariant, but $\alpha e_1\notin W$. Thus the finite-dimensional or closed-subspace hypothesis cannot be omitted from an infinite-dimensional reading.

For a real inner product space $V={\mathbb{R}}^5$, defineThis map takes values in the Special orthogonal Lie algebra: its matrices are skew-symmetric. It sends $e_i\wedge e_j$ to $E_{ij}-E_{ji}$, so it is an isomorphism between the two ten-dimensional spaces. For $g\in SO(5)$, invariance of the inner product givesThis proves the exterior square realization of the orthogonal adjoint representation, including equivariance, rather than merely matching dimensions.

After complexification, part (a) is therefore the weight decomposition of the Adjoint representation. Its zero weight space is the complex centralizer of $\mathfrak{t}$. The two zero-weight wedges are nonzero multiples of $e_1\wedge e_2$ and $e_3\wedge e_4$, which map under $\Phi$ to the two infinitesimal rotation generators of $T$. HenceIf a larger torus contained $T$, its Lie algebra would commute with $\mathfrak{t}$ and therefore lie in $\mathfrak{t}$. Equality of Lie algebras and connectedness of both tori force equality of the groups. Thus $T$ is a maximal torus.

Let ${\epsilon }_j$ denote the real coordinate functional $x_j$. In the real weight convention, with the common factor $i$ removed from the differentiated torus weights, the root system isThese are the roots of a root system of type B2 root system. In the imaginary-valued compact Lie-algebra convention each displayed functional is multiplied by $i$. Every root space is one-dimensional, as part (a) explicitly shows.

Write $r=x-u^{\ast }$. If $u^{\ast }$ is a best approximation in the finite-dimensional linear subspace, then for every $v$ in it and every real $t$,A nonzero real linear coefficient would make this negative for sufficiently small $t$ of the appropriate sign. Hence $\mathrm{Re}(r,v)=0$. In a complex inner product space, apply the argument to $iv$ too; the imaginary part also vanishes. Thus $(x-u^{\ast },v)=0$ for every $v$ in the subspace.

Conversely, if that orthogonality holds, every other candidate has the form $u^{\ast }+v$, and the Pythagorean theorem in the inner product norm givesEquality requires $v=0$. ThereforeThe next part constructs this orthogonal projection explicitly. No completeness of the whole inner product space is required for projection onto a finite-dimensional subspace.

The finite-dimensional inner product space has the orthogonal decomposition $U=W\oplus W^{\perp }$. Define $\tau (w+z)=w$ for $w\in W$, $z\in W^{\perp }$. This is a linear map with image $W$, satisfies ${\tau }^2=\tau$, and is self-adjoint by taking inner products between decomposed vectors. Hence it is an orthogonal projection.

Any other orthogonal projection with image $W$ has kernel $W^{\perp }$ by part (a); it is identity on $W$ by idempotence and zero on its kernel. Its values are therefore forced to agree with $\tau$, proving uniqueness.

If ${\tau }_2{\tau }_1=0$, every $w_1\in W_1$ has $w_1={\tau }_1{w}_1$, so ${\tau }_2{w}_1=0$ and $w_1\in W_2^{\perp }$. Conversely, if $W_1\subset W_2^{\perp }$, every image ${\tau }_{1}u$ is killed by ${\tau }_2$. Therefore

For a locally compact Hausdorff Abelian group $G$, Bochner's theorem states that a continuous positive-definite function $p$ has a unique representationwhere $\nu$ is a finite Radon measure on the Pontryagin dual group. Conversely every such measure gives a continuous positive-definite function. Here positive definiteness means ${\sum }_{i,j}{c}_i\overline{c_j}p(x_i{x}_j^{-1})\ge 0$ for all finite choices. The Abelian and continuity hypotheses are part of the theorem; the representation is not claimed for arbitrary non-Abelian groups or discontinuous functions.

First prove existence without presupposing the representation. The elementary consequences of positive definiteness give $p(x^{-1})=\overline{p(x)}$, $p(e)\ge 0$ and $|p(x)|\le p(e)$. If $p(e)=0$, then $p=0$ and take $\nu =0$. Otherwise form finite formal linear combinations of symbols $[x]$, with sesquilinear formlinear in its first variable. Positive definiteness makes this form positive semidefinite. The Cauchy-Schwarz inequality for such a form shows that null vectors are orthogonal to everything, so quotienting by them gives an inner product space. Complete it to a Hilbert space $H$. Since $G$ is Abelian, translation $U_a[x]=[ax]$ preserves the form and extends to a unitary operator. Its inverse is $U_{a^{-1}}$, so $(U_a)$ is a unitary representation. With $v=[e]$,The translated copies of $v$ span a dense subspace. MoreoverFinite sums and density, together with $\Vert U_a\Vert =1$, give strong continuity on all of $H$. This constructs the cyclic unitary representation of a positive-definite function directly.

For $f\in L^1(G)$ define the integrated unitary representation $\pi (f)=\int f(x)U_{x}dx$, with the integral applied to each vector. Its norm is at most $\Vert f{\Vert }_1$. Fubini gives $\pi (f\ast g)=\pi (f)\pi (g)$, and inversion in the Abelian group gives $\pi (f^{\ast })=\pi (f{)}^{\ast }$. Thusis a commutative C-star algebra. It is nondegenerate: choose nonnegative compactly supported functions $u_V$ of Haar integral one with supports shrinking to the identity element. Strong continuity impliesfor every $w\in H$. These form an approximate identity in the representation. Neighbourhoods may form a net, since no metrizability is being assumed here.

The Commutative Gelfand--Naimark theorem identifies $\mathcal{A}$ with $C_0(X)$, where $X$ is its character space. The positive vector functional $\omega (B)=⟨Bv,v⟩$ becomes a bounded positive functional on $C_0(X)$, so the Riesz-Markov-Kakutani representation theorem supplies a finite positive Radon measure $\rho$ withThese general representation theorems are applied to the explicitly constructed algebra; they are not an assumption of Bochner's conclusion.

For each $\theta \in X$, the functional $f\mapsto \theta (\pi (f))$ is a nonzero algebra character of $L^1(G)$, since $\pi (L^1(G))$ is dense in $\mathcal{A}$. By Question 3 it has the form $\int f(x){\chi }_{\theta }(x)dx$. The resulting map $\iota :X\to \hat{G}$ is continuous: pointwise convergence of $\theta$ on $\mathcal{A}$ gives convergence on all $\pi (f)$, and the topological character identification in Question 3 gives compact-open convergence of ${\chi }_{\theta }$. Push $\rho$ forward under $\iota$ to obtain $\nu$. This finite measure is Radon: compact subsets approximating the measure of any inverse-image Borel set have compact images, giving inner regularity; finiteness and complements then give outer regularity.

To recover $p(x)$, note that $U_x\pi (u_V)=\pi (T_x{u}_V)\in \mathcal{A}$. HenceThe left side tends to $p(x)$. On the right, $|{\Phi }_{\chi }(u_V)|\le 1$ and ${\Phi }_{\chi }(u_V)\to 1$ uniformly on compact subsets of $\hat{G}$. For completeness, the evaluation $(\chi ,x)\mapsto \chi (x)$ is jointly continuous: near $x_0$ restrict $x$ to a compact neighbourhood, control the character uniformly there, and use continuity of the fixed character. A finite cover of a compact set of characters then makes those characters equicontinuous at $e$, proving the stated uniform convergence. Continuity of $\iota$ gives uniform convergence on compact subsets of $X$, and the finite Radon measure $\rho$ makes the remaining tails arbitrarily small. Thus passage to the limit is valid even for a net, andTaking $x=e$ also gives the required total mass $p(e)$.

Conversely, a finite positive Radon measure givesIts transform is continuous. Indeed, choose a compact set of characters with arbitrarily small measure outside it; joint continuity of evaluation gives uniform continuity in $x$ near the specified point on that compact set, while the outside contribution is bounded by twice its measure. This proves continuity without using an unjustified dominated-convergence assertion for arbitrary nets.

Finally suppose two finite Radon measures have the same transform. Multiply by $f\in L^1(G)$ and integrate over $G$. Fubini shows that the measures agree on every function $\chi \mapsto {\Phi }_{\chi }(f)$. Question 3 showed that these functions are uniformly dense in $C_0(\hat{G})$. Boundedness of both measure functionals extends equality to all of $C_0(\hat{G})$; uniqueness in the Riesz representation theorem gives equality of the measures. This proves existence, uniqueness and the converse, completing Bochner's theorem. In particular $p(e)=1$ corresponds exactly to a probability measure on the dual group.

A topological spectrum can be described sequentially by based spaces $E_j$, $j\ge 0$, and structure mapsIterating these maps gives compatible maps $S^{k-j}\wedge E_j\to E_k$. Equivalently, an indexed prespectrum on a universe for spectra $U$ has spaces $E(V)$ and unital associative structure maps $S^{W\ominus V}\wedge E(V)\to E(W)$ for finite-dimensional inclusions $V\subset W\subset U$. Here $S^A$ denotes the one-point compactification of the real inner product space $A$.

A map of topological spectra $u:D\to E$ is a family of based maps $u_j:D_j\to E_j$ satisfyingThe indexed version requires the same compatibility for every inclusion of indexing subspaces. These are point-set maps, before passing to spectrum homotopy or the stable homotopy category.

An Omega-spectrum is a topological spectrum whose adjoint structure mapsare weak homotopy equivalences. Consequently ${\pi }_{k}E={\pi }_{k+j}{E}_j$ whenever $k+j\ge 0$, using the structure identifications. In the genuine indexed point-set convention, the word spectrum already includes the stronger condition that $E(V)\to {\Omega }^{W\ominus V}E(W)$ is a homeomorphism; a general system without that condition is called an indexed prespectrum. Its associated spectrum is obtained by spectrification. The sequential and indexed conventions present the same stable theory.

A cell spectrum is built from the zero topological spectrum by successive stable cell attachments. Concretely, attach a free spectrum on the inclusion $S^{m-1}\hookrightarrow D^m$ at some level $j$ by a pushout along its boundary map; its quotient is the stable cell sphere ${\Sigma }^{m-j}\mathbb{S}$. Iterating such attachments and taking the colimit gives a cell spectrum, with successive quotients wedges of integer suspensions of spectra of the sphere spectrum. Negative cell degrees are allowed by choosing $j>m$. In the genuine indexed model apply spectrification to this cellular prespectrum. The compatibility of the structure maps is part of both the objects and their maps.

A reduced crystallographic abstract root system in a finite-dimensional real inner product space $E$ is a finite set $R$ such that: it spans $E$ and does not contain zero; the only scalar multiples of $\alpha \in R$ belonging to $R$ are $\pm \alpha$; every root reflectionpermutes $R$; and each Cartan integer $2(\beta ,\alpha )/(\alpha ,\alpha )$ is integral. These are the crystallographic and reduced conventions appropriate to roots of a complex semisimple Lie algebra. Without crystallographic integrality, the numerical restrictions in the following subparts would not hold.

Use the nonnegative Laplacian convention $\tilde{\Delta }=-{\sum }_{j=1}^{d+1}{\partial }_j^2$ and $\Delta =-\mathrm{div}\mathrm{grad}$ on the unit sphere. In polar coordinates the Euclidean Riemannian metric is $d{r}^2+r^2{g}_{S^d}$, its volume form is $r^{d}drd{V}_{S^d}$, and therefore the Laplacian in polar coordinates isConsequently the required restriction identity isFor the opposite sign convention all eigenvalues below change sign, and the restriction identity changes accordingly.

Let ${\mathcal{P}}_n$ be the vector space of degree-$n$ homogeneous polynomials in $d+1$ variables, and let ${\mathcal{H}}_n$ be its subspace of harmonic polynomials. Write $H(r\omega )=r^{n}Y(\omega )$ for $H\in {\mathcal{H}}_n$. Substitution in the Laplacian identity givesThus each restricted harmonic polynomial is a spherical harmonic with eigenvalue ${\lambda }_n=n(n+d-1)$.

To count these spherical harmonics, give polynomials the Fischer inner product $⟨x^{\alpha },x^{\beta }{⟩}_F=\alpha !{\delta }_{\alpha \beta }$. Multiplication by $x_j$ is adjoint to ${\partial }_j$, so multiplication by $r^2$ is adjoint to the ordinary signed Laplacian ${\sum }_j{\partial }_j^2$. In these finite-dimensional inner product spaces, the orthogonal complement of $r^2{\mathcal{P}}_{n-2}$ is exactly ${\mathcal{H}}_n$. Hence the Fischer decomposition isMultiplication by $r^2$ is injective, while restriction of a homogeneous polynomial to the sphere is injective: homogeneity recovers its value at every nonzero point. Counting monomials now gives the eigenvalue multiplicitywhere the second term is zero for $n<2$.

It remains to establish that these are all the eigenvalues and their full multiplicities. Iterating the Fischer decomposition expresses every restricted polynomial as a sum of the restricted harmonic polynomials. Restricted polynomials separate points and contain the constants, so the Stone-Weierstrass theorem makes their span dense in continuous functions on the sphere, hence dense in its L2 space. The Laplace-Beltrami operator is self-adjoint, and the distinct numbers ${\lambda }_n$ strictly increase for $d\ge 1$. Its different eigenspaces are therefore orthogonal. Any additional eigenfunction, or any additional vector in one of these eigenspaces orthogonal to its known spherical harmonics, would be orthogonal to a dense subspace and hence zero. The standard compact elliptic spectral theorem then shows that the spectrum of the round sphere is precisely ${\lambda }_n=n(n+d-1)$ with multiplicity $m_n$. For $S^1$ this gives multiplicity one at zero and two at $n^2$ for $n\ge 1$. If the degenerate case $d=0$ is included, $S^0$ has two points and its Laplacian is zero on a two-dimensional function space.

The Hecke operators organize modular forms simultaneously by their transformation law and their Fourier coefficients. Work with $\Gamma =S{L}_2(\mathbb{Z})$, $M_k$ its space of weight-$k$ modular forms, and $S_k$ its cusp form subspace, with even $k\ge 4$. Odd weights give zero spaces. The normalization below makes the first coefficient of $T_{n}f$ equal to the $n$th coefficient of $f$, which is the key link between operators and arithmetic.

For $A=\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)$ with positive determinant, the determinant-normalized slash operator for modular forms isThe automorphy factor identity proves $(f{|}_{k}A){|}_{k}B=f{|}_k(AB)$. Positive scalar matrices act trivially. Let ${\mathcal{A}}_n$ be the integral matrices of determinant $n>0$. Left multiplication by $\Gamma$ reduces their first column, by Bezout identity, to $(a,0)$ with $a>0$; the determinant then gives $d=n/a>0$. A row shear makes $0\le b<d$. These are unique determinant-n matrix representatives for Hecke operators, so defineRight multiplication by any element of $\Gamma$ permutes these left cosets. The slash operator for modular forms identity therefore proves that $T_{n}f$ has the same modular transformation law as $f$. Each summand is holomorphic on the upper half-plane. Its cusp behavior follows explicitly from the coefficient calculation below, so $T_n$ preserves both $M_k$ and $S_k$.

Insert $f={\sum }_{r\ge 0}{a}_r{q}^r$ in the sum. The finite sum over $b$ is zero unless $d\mid r$, in which case it is $d$. Writing $r=d\ell$ leaves the term $a_{d\ell }{q}^{a\ell }$ with multiplier $n^{k-1}{d}^{1-k}=a^{k-1}$. Thus the Fourier coefficients of a composite-index Hecke operator areFor $m=0$ use $\gcd (0,n)=n$. In particular the constant coefficient becomes ${\sigma }_{k-1}(n)a_0$, so cusp forms remain cusp forms, andfor a prime number $p$. The absence of negative powers proves cusp holomorphy. Notice that no primality assumption is needed in the composite-index formula.

The Hecke multiplication relations can be derived directly, rather than assumed from the coset description. On Fourier series define auxiliary maps $U_p(\sum a_m{q}^m)=\sum a_{pm}{q}^m$ and $V_{p}f(q)=f(q^p)$. They need not separately preserve $M_k$, but $U_p{V}_p=1$. With $c=p^{k-1}$, the coefficient formula givesMultiplying on the left by $U_p+c{V}_p$, the term $j=0$ in the $U_p$ sum contributes $U_p^{r+1}$; the terms with $j\ge 1$ become $c^j{V}_p^{j-1}{U}_p^{r-j}$. Together with the $c{V}_p$ sum this is exactly $T_{p^{r+1}}+c{T}_{p^{r-1}}$. Hence the Hecke prime-power recurrence from q-shift operators isFor coprime $m,n$, substituting the coefficient formula twice gives a sum over $a\mid \gcd (m,r)$ and $b\mid \gcd (n,r)$ with coefficient $(ab{)}^{k-1}{a}_{mnr/(ab{)}^2}(f)$. The divisors $a,b$ correspond uniquely to a divisor $ab\mid \gcd (mn,r)$, so $T_m{T}_n=T_{mn}$. The prime-power recurrence, by induction on the smaller exponent, givesCombining the coprime and prime-power formulas provesIn particular the Hecke operators commute, and the modular Hecke algebra is generated by the prime operators.

The analytic reason for simultaneous diagonalization is the Petersson inner productwhich is positive definite on $S_k$ and convergent because of exponential cusp decay. Here is the adjoint argument. For a matrix $A$, set $H_A=\Gamma \cap A^{-1}\Gamma A$. The left cosets in $\Gamma A\Gamma$ are represented by $A\gamma$ with $\gamma \in H_A\backslash \Gamma$. Unfolding this finite coset sum in the Petersson inner product givesUnder $z=A\tau$, hyperbolic area is unchanged and the determinant factors in the slash operator for modular forms giveThe integration group becomes $A{H}_A{A}^{-1}=\Gamma \cap A\Gamma A^{-1}$. Folding this finite cover back to $\Gamma \backslash \mathbb{H}$ identifies the adjoint with the correspondence for $A^{-1}$. For $A\in {\mathcal{A}}_n$, the matrix $\mathrm{adj}(A)=n{A}^{-1}$ is again integral with determinant $n$; scalar matrices act trivially, and adjugation permutes the double cosets making up ${\mathcal{A}}_n$. The real normalizing factor $n^{k/2-1}$ is unchanged. Thus this proves that Hecke operators are self-adjoint for the Petersson inner product:The valence formula for the modular group makes $S_k$ a finite-dimensional vector space. Commuting self-adjoint operators on a finite-dimensional inner product space have a common orthogonal eigenbasis: diagonalize one operator, use commutativity to preserve each of its eigenspaces, and repeat within those spaces. Therefore $S_k$ has a basis of Hecke eigenforms, with real Hecke operator eigenvalues.

If $f$ is a common Hecke eigenform, say $T_{n}f={\lambda }_{n}f$, its first Fourier coefficient cannot vanish: $a_1=0$ would imply $a_n=a_1(T_{n}f)={\lambda }_n{a}_1=0$ for every $n$. Normalize $a_1=1$. Then the Hecke operator eigenvalue ${\lambda }_n$ is precisely the Fourier coefficient $a_n$. The Hecke multiplication relations becomeConsequently the coefficients are multiplicative at coprime indices, and their local generating function satisfiesThis gives the Euler product of a Hecke eigenformin a half-plane of absolute convergence. The Fourier coefficient bound for a cusp form already supplies such a half-plane; Question 4 improves it to $\mathrm{Re}s>(k+1)/2$.

The L-function of a cusp form also has an analytic description independent of the Hecke eigenform condition. Termwise Mellin transform in an initial right half-plane givesAt infinity the cusp form decays exponentially. Modular inversion gives $f(i/y)=(iy{)}^{k}f(iy)$, so at zero it is $O(y^{-k}{e}^{-2\pi /y})$. The integral therefore converges locally uniformly for every $s$, continuing $\Lambda$ to an entire function. Changing $y$ to $1/y$ provesFor a normalized Hecke eigenform, the Euler product of a Hecke eigenform and this functional equation describe the same completed L-function of a cusp form.

There is also an integral arithmetic structure. One standard consequence of the level-one dimension formula and the valence formula for the modular group is the integral triangular basis of level-one modular forms: if $k=12m+s$ with $s\in \{0,4,6,8,10,14\}$, choose $F_s=1,E_4,E_6,E_4^2,E_4{E}_6,E_4^2{E}_6$ respectively. Thenform a basis of $M_k$, and the terms $1\le j\le m$ form a basis of $S_k$. Their first nonzero terms are $q^j$ with coefficient one and all coefficients are integers. Successive elimination of coefficients of $q,q^2,\dots ,q^m$ shows that these cusp basis elements generate exactly the lattice of rational cusp forms with integral Fourier coefficients. The composite-index coefficient formula preserves this lattice, so each $T_n$ has an integral matrix on it. Its characteristic polynomial is monic with integral coefficients. This proves that Hecke eigenvalues are algebraic integers; self-adjointness also makes every root of that characteristic polynomial real. The field generated by the eigenvalues is a finite totally real number field: a finite set of operators spans the finite-dimensional rational modular Hecke algebra, and every conjugate of each eigenvalue remains a root of its real-rooted characteristic polynomial.

Two basic examples connect the operator theory to familiar modular forms. The normalized Eisenstein series $E_k$ has coefficients proportional to ${\sigma }_{k-1}(n)$. At a prime number its coefficient formula useswhere the second term is zero unless $p\mid m$. This follows by writing $m=p^{r}u$ with $p\nmid u$ and summing a finite geometric series. It proves $T_p{E}_k=(1+p^{k-1})E_k$, and the Hecke multiplication relations give $T_n{E}_k={\sigma }_{k-1}(n)E_k$. Meanwhile $S_{12}=\mathbb{C}\Delta$ implies $T_n\Delta =\tau (n)\Delta$, because $a_1(\Delta )=1$. Thus the Ramanujan tau function is multiplicative and satisfies $\tau (p^{r+1})=\tau (p)\tau (p^r)-p^{11}\tau (p^{r-1})$. For example, $\tau (2)=-24$ gives $\tau (4)=(-24{)}^2-2^{11}=-1472$. This illustrates how Hecke operators turn analytic Fourier coefficients into a coherent arithmetic system.

Write $r=x-u^{\ast }$. If $u^{\ast }$ is a best approximation, then for every $v\in {\mathcal{U}}_n$ and real $t$, the competitor $u^{\ast }+tv$ satisfiesThe linear coefficient must be zero, since otherwise choosing sufficiently small $t$ of the appropriate sign makes this expression negative. Hence $\mathrm{Re}(r,v)=0$. In a real inner product space this is the desired orthogonality; in a complex inner product space, apply the same argument to $iv$ to show that the imaginary part also vanishes. Thus $(r,v)=0$ for all $v\in {\mathcal{U}}_n$.

Conversely, suppose this orthogonality holds. Every competitor $u\in {\mathcal{U}}_n$ has $u-u^{\ast }\in {\mathcal{U}}_n$. The Pythagorean identity then givesEquality forces $u=u^{\ast }$. We have proved the characterization, as well as uniqueness:This is the orthogonal projection onto a finite-dimensional subspace. Completeness of the ambient space is not needed. Existence follows, for example, by solving the invertible Gram matrix system in a basis of the finite-dimensional linear subspace.