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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.

Let $V$ be a finite-dimensional real or complex inner product space, and let $A\subseteq \mathrm{End}(V)$ be a unital subalgebra stable under the adjoint operator. Define its commutant of an operator algebra by $A^{\prime }=\{T:Ta=aT\text{ for all }a\in A\}$. The double commutant theorem for star-algebras asserts $\boxed{A^{\prime \prime }=A}$. The inclusion $A\subseteq A^{\prime \prime }$ follows immediately from the definition; we prove the reverse.

Write $N=\dim V$, fix a basis $v_1,\dots ,v_N$, and act diagonally on $V^N$. The linear subspaceis invariant under every diagonal $a$. Since $A$ is closed under adjoints, its orthogonal complement $W^{\perp }$ is invariant too: $⟨aw,z⟩=⟨w,a^{\ast }z⟩=0$ for $w\in W^{\perp }$, $z\in W$. Hence the orthogonal projection $P$ onto $W$ commutes with every diagonal $a$.

Writing $P$ as an $N$ by $N$ block matrix $(P_{ij})$, this says $P_{ij}a=a{P}_{ij}$, so each $P_{ij}\in A^{\prime }$. If $b\in A^{\prime \prime }$, it commutes with every $P_{ij}$, and its diagonal action therefore commutes with $P$ and preserves $W$. Because $1\in A$, $(v_1,\dots ,v_N)\in W$. Therefore there is one $a\in A$ such thatEquality on the basis gives $b=a$, proving the theorem over either scalar field.

The identity hypothesis is necessary. For a possibly nonunital adjoint-stable subalgebra, adjoining the ambient identity does not change its commutant of an operator algebra, so the exact statement is $A^{\prime \prime }=A+\mathbb{F}I$. For example $A=\{0\}$ has $A^{\prime \prime }=\mathbb{F}I$ on a nonzero $V$. Thus the usual exam formulation takes the algebra to contain the identity.

Here a reduced root system is crystallographic, as appropriate to a semisimple Lie algebra. It is a finite spanning set $\Phi \subset E\setminus \{0\}$ in a real inner product space of dimension $r$, invariant under each Weyl reflectionwith $2(\beta ,\alpha )/(\alpha ,\alpha )\in \mathbb{Z}$ for every pair of roots and $\Phi \cap \mathbb{R}\alpha =\{\alpha ,-\alpha \}$. A base of a root system $\Delta =({\alpha }_1,\dots ,{\alpha }_r)$ is a basis such that the coefficients of every root are integers that are either all nonnegative or all nonpositive. These define the positive system of a root system. We use the existing Cartan matrix conventionUsing the alternative denominator $({\alpha }_i,{\alpha }_i)$ transposes all the matrices below.

The classification of rank-two root systems gives $A_1\times A_1$, $A_2$, $B_2=C_2$, and $G_2$. In orthonormal coordinates we may choose the following simple roots and Cartan matrices:

Each complete root system consists of these positive roots and their negatives. To see why the list is exhaustive, the off-diagonal Cartan matrix entries of two distinct simple roots are nonpositive integers, and their product is $4{\cos }^2\theta <4$. Thus the product is $0,1,2$ or $3$, corresponding to simple-root angles $\pi /2,2\pi /3,3\pi /4$ or $5\pi /6$. These determine the four systems and their length ratios.

The Weyl group of a rank-two root system is generated by the two simple Weyl reflections. Their product is a rotation of order $m=2,3,4,6$, respectively, so the groups are the dihedral groups of order $2m=4,6,8,12$. The first is also the Klein four-group and the second the symmetric group $S_3$. The root-orthogonal lines cut the plane into $2m$ Weyl chambers, each of angle $\pi /m$. A chosen fundamental chamber of a root system is $(x,{\alpha }_1)>0$, $(x,{\alpha }_2)>0$; its walls are the two root-orthogonal lines. The Weyl group acts simply transitively on these open chambers.

The crystallographic hypothesis matters: if the integer-pairing condition is dropped, reduced noncrystallographic systems of type $I_2(m)$ also occur. They are not additional Lie-algebra root systems.

The appropriate abstract object is a finite reduced crystallographic root system $R$ in a real inner product space $E$. Its axioms are: $R$ is finite, spans $E$, and does not contain zero; for $\alpha \in R$, $R\cap \mathbb{R}\alpha =\{\alpha ,-\alpha \}$; each root reflectionpermutes $R$; and every Cartan integer $n_{\alpha \beta }=2(\alpha ,\beta )/(\beta ,\beta )$ is an integer. The restriction to a reduced root system and crystallographic integrality distinguishes roots of complex semisimple Lie algebras from more general reflection configurations.

For any two roots of a root system, the Cauchy-Schwarz inequality givesIf the inner product is zero, both Cartan integers vanish. Otherwise their signs agree, and the absolute value of each is a positive integer. Dividing their product by an integer of absolute value at least one provesFor nonproportional roots the product is strictly less than four. In a reduced root system, proportional roots are just $\pm \alpha$ and have Cartan integers $\pm 2$, so the printed bound is intentionally looser than the resulting bound of three.

A fundamental system of a root system is a basis $\Delta$ of $E$ made of roots, such that each root is an integer combination of $\Delta$ with either all coefficients nonnegative or all nonpositive. Its members are the simple roots. Suppose distinct $\alpha ,\beta \in \Delta$ had $n_{\alpha \beta }>0$. Thenhas a positive coefficient of $\alpha$ and a negative coefficient of $\beta$, contradicting the defining sign condition. Thus $n_{\alpha \beta }\le 0$. Distinct simple roots are linearly independent, so their Cartan-integer product is strictly less than four. Combining integrality and the sign condition givesHere nonpositive is the intended sense of the printed convention that includes zero among “negative” numbers; orthogonal simple roots really do give zero.

To form a Dynkin diagram, place a vertex at each simple root. Join two vertices by $n_{\alpha \beta }{n}_{\beta \alpha }$ bonds, hence zero, one, two or three. A multiple bond has an arrow toward the short root. Indeed $n_{\alpha \beta }/n_{\beta \alpha }=(\alpha ,\alpha )/(\beta ,\beta )$ determines the squared length ratio, and the diagram with the Cartan matrix reconstructs the angles and relative lengths. A single bond joins equal-length roots.

The connected finite Dynkin diagrams are the following. The descriptions include bond multiplicities and arrow directions, so distinguish dual diagrams:

There are no other connected finite Dynkin diagrams. Low-rank conventions also identify $B_1=C_1=A_1$ and $D_3=A_3$; $D_2$ is disconnected, so introduces no further connected type. Affine diagrams are outside this finite classification.

Finally suppose the underlying graph contained a cycle on $m\ge 3$ distinct simple roots ${\alpha }_1,\dots ,{\alpha }_m$. Put $u_i={\alpha }_i/\Vert {\alpha }_i\Vert$. Any bonded pair hasand all other distinct pairs have nonpositive inner products. The cycle contributes at least $m$ bonded pairs, soBut the simple roots, and hence these normalized vectors, are linearly independent, making the displayed sum nonzero. Positive definiteness gives a contradiction. Thus the underlying graph of a finite Dynkin diagram has no cycle. This acyclicity of a finite Dynkin diagram argument also excludes cycles with extra chords or multiple bonds; multiple bonds themselves are not treated as two-edge cycles.

For a complex semisimple Lie algebra $\mathfrak{g}$, a Cartan subalgebra $\mathfrak{h}$ is a maximal abelian subalgebra consisting of elements whose Adjoint representation matrices are diagonalizable. Equivalently it is a nilpotent self-normalizing Lie subalgebra. The root-space decomposition iswhere the roots are the nonzero weights of that Adjoint representation.

We use these properties of the Killing form $B$: it is nondegenerate on $\mathfrak{g}$ and on $\mathfrak{h}$, it is invariant, distinct root spaces are orthogonal unless their roots sum to zero, and $B$ pairs ${\mathfrak{g}}_{\alpha }$ nondegenerately with ${\mathfrak{g}}_{-\alpha }$. Define $t_{\alpha }$ by $B(t_{\alpha },h)=\alpha (h)$, and choose $e\in {\mathfrak{g}}_{\alpha }$, $f\in {\mathfrak{g}}_{-\alpha }$ with $B(e,f)=1$. Since weights add, $[e,f]\in \mathfrak{h}$, and invariance givesThe nonisotropic root lemma shows $\alpha (t_{\alpha })\ne 0$. Here is its short proof: if this number were zero, the span of $e,f,t_{\alpha }$ would be a solvable Lie algebra with $t_{\alpha }$ central. Apply the Lie theorem to its action on $\mathfrak{g}$. The commutator $\mathrm{ad}{t}_{\alpha }=[\mathrm{ad}e,\mathrm{ad}f]$ would be strictly upper triangular and hence nilpotent. But $t_{\alpha }\in \mathfrak{h}$ makes it diagonalizable. It would therefore be zero, putting $t_{\alpha }$ in the zero center of a Lie algebra of $\mathfrak{g}$, contrary to its definition. SetThenso their span is the sl2 subalgebra associated with a root.

The abstract reduced crystallographic root system axioms are as follows. In a finite-dimensional real inner product space $E$, the set $R$ is finite, consists of nonzero vectors and spans $E$; for every $\alpha \in R$, one has $R\cap \mathbb{R}\alpha =\{\alpha ,-\alpha \}$; the root reflection $s_{\alpha }(\beta )=\beta -2(\beta ,\alpha )\alpha /(\alpha ,\alpha )$ preserves $R$; and $2(\beta ,\alpha )/(\alpha ,\alpha )$ is an integer for every pair of roots. We verify the positive-definite real form as well as these axioms, rather than assuming the complex Killing form is already positive.

First $R$ is finite and has no zero element by its definition. The roots span ${\mathfrak{h}}^{\ast }$ over $\mathbb{C}$: any $h\in \mathfrak{h}$ annihilated by all roots commutes with the whole root-space decomposition and is central, hence zero. Therefore the $t_{\alpha }$, and also the $H_{\alpha }$, span $\mathfrak{h}$ over $\mathbb{C}$. For every root $\beta$, the classification of finite-dimensional sl2 representations applied to the Adjoint representation of the root subalgebra gives $\beta (H_{\alpha })\in \mathbb{Z}$. Put ${\mathfrak{h}}_{\mathbb{R}}={\mathrm{span}}_{\mathbb{R}}\{H_{\alpha }:\alpha \in R\}$. Every root takes real values on this space, andThe strict inequality follows because the roots span ${\mathfrak{h}}^{\ast }$. This also shows that the complexification of ${\mathfrak{h}}_{\mathbb{R}}$ injects into $\mathfrak{h}$: an equality $h_1+i{h}_2=0$ with real $h_j$ would contradict positivity of $B(h_1,h_1)$ and $B(h_2,h_2)$. Since its complex span is all of $\mathfrak{h}$, it is a real form. Moreover $B(H_{\alpha },H_{\alpha })=4/\alpha (t_{\alpha })>0$, so $t_{\alpha }$ is a real scalar multiple of $H_{\alpha }$. The dual inner product thus makes $E={\mathrm{span}}_{\mathbb{R}}R$ a Euclidean space, as in the Euclidean subspace of a Cartan subalgebra.

To prove reducedness without assuming it, consider the root-subalgebra modulewith absent root spaces understood to be zero. Its $H_{\alpha }$ weights are even, so every nontrivial Irreducible Lie algebra representation in it has even positive highest weight and one-dimensional weight-zero space. The action of $X_{\alpha }$ on its weight-zero space $\mathfrak{h}$ has image exactly $\mathbb{C}{X}_{\alpha }$, of dimension one. Therefore there is precisely one nontrivial irreducible summand, the already embedded adjoint ${\mathfrak{sl}}_2$ module of highest weight $2$. Thus $\dim {\mathfrak{g}}_{\alpha }=1$ and $2\alpha \notin R$. If $\beta =c\alpha$ is any root on the same real line, the integral numbers $\beta (H_{\alpha })=2c$ and $\alpha (H_{\beta })=2/c$ have product $4$. Hence $c$ is one of $\pm \frac{1}{2},\pm 1,\pm 2$; the half and double cases are excluded by applying the preceding argument to the appropriate root. This proves the root-space reducedness lemma and $R\cap \mathbb{R}\alpha =\{\pm \alpha \}$.

For a root $\beta$ not parallel to $\alpha$, the sum of root spaces ${⨁}_{j\in \mathbb{Z}}{\mathfrak{g}}_{\beta +j\alpha }$ is stable under the root ${\mathfrak{sl}}_2$. Its integer $H_{\alpha }$ weights are symmetric under sign in each irreducible summand. Therefore the weight $-\beta (H_{\alpha })$ also occurs, at the root $\beta -\beta (H_{\alpha })\alpha$. The Killing form normalization givesso this root is exactly $s_{\alpha }(\beta )$. For $\beta =\pm \alpha$ the reflection just swaps the two roots. This proves reflection invariance and the Cartan integer condition. All axioms of the reduced crystallographic root system have now been verified.

A normalization is important here. I use the covolume-one theta function of a fractional ideal, for which the requested limit is $1$, and also give the formula for the unnormalized Gaussian theta sum.

Let $n=[K:\mathbb{Q}]$, let $d_v=[K_v:\mathbb{R}]\in \{1,2\}$, and choose one field embedding ${\sigma }_v$ for each Archimedean place. The Minkowski embedding of a number field identifieswith a real inner product space havingIts Euclidean Lebesgue measure is $d\mu ={\prod }_{v\text{ real}}d{x}_v{\prod }_{v\text{ complex}}2d\mathrm{Re}{x}_{v}d\mathrm{Im}{x}_v$. With this convention, the covolume of a fractional ideal lattice $j(\mathfrak{b})$ iswhere $D_K$ is the field discriminant and $N(\mathfrak{b})$ is the positive absolute norm of a fractional ideal. Thus the Euclidean lattice ${\Lambda }_{\mathfrak{b}}=C_{\mathfrak{b}}^{-1/n}j(\mathfrak{b})$ has covolume $1$. DefineThis Gaussian theta sum converges absolutely whenever every $y_v>0$.

The trace dual of a fractional ideal isHere ${\mathfrak{D}}_{K/\mathbb{Q}}^{-1}$ is the inverse different. Since $N({\mathfrak{D}}_{K/\mathbb{Q}})=|D_K|$, we have $C_{{\mathfrak{b}}^{\vee }}=C_{\mathfrak{b}}^{-1}$. There is a subtle distinction between the trace pairing and the positive inner product: the dual lattice of $j(\mathfrak{b})$ is $\overline{j({\mathfrak{b}}^{\vee })}$, where the bar conjugates the complex coordinates and fixes the real ones. Indeed,Consequently ${\Lambda }_{\mathfrak{b}}^{\ast }=\overline{{\Lambda }_{{\mathfrak{b}}^{\vee }}}$. Coordinatewise complex conjugation preserves the weighted squared lengths in the Gaussian theta sum.

Here are the precise analytic formulas used in the proof. For a Schwartz function on $K_{\infty }$, take the Fourier transform to beFor a full Euclidean lattice $\Lambda$ of covolume $C$, the Poisson summation formula for a Euclidean lattice isFor $f_y(x)=\exp (-\pi {\sum }_v{d}_v{y}_v|x_v{|}^2)$, the Gaussian Fourier transform, applied in orthonormal real coordinates, givesMore generally, $\widehat{\exp (-\pi x^{T}Ax)}(z)=(\det A{)}^{-1/2}\exp (-\pi z^T{A}^{-1}z)$ for a real positive-definite matrix $A$ that is symmetric. Each complex coordinate contributes two real coordinates, which explains its exponent $d_v=2$.

Apply the Poisson summation formula for a Euclidean lattice to ${\Lambda }_{\mathfrak{b}}$, whose covolume is $1$, and use the preceding description of its dual lattice. The anisotropic theta functional equation isFor comparison, the unnormalized Gaussian theta sumhas the functional equationand its corresponding limit is $C_{\mathfrak{b}}^{-1}$ rather than $1$. Explicitly, our normalization is $\Theta (y,\mathfrak{b})=\theta (C_{\mathfrak{b}}^{-2/n}y,\mathfrak{b})$.

For the small-parameter asymptotic of a lattice theta sum, $y\to 0$ means that every coordinate tends to zero. In $\Theta (y^{-1},{\mathfrak{b}}^{\vee })$, the zero lattice vector contributes $1$. Every nonzero vector contributes a term tending to zero. Once every $y_v\le 1$, all these terms are bounded by the summable Gaussian theta sum ${\sum }_{\lambda \in {\Lambda }_{{\mathfrak{b}}^{\vee }}}{e}^{-\pi |\lambda {|}^2}$. The dominated convergence theorem therefore gives $\Theta (y^{-1},{\mathfrak{b}}^{\vee })\to 1$. Using the anisotropic theta functional equation,The condition that every coordinate tends to zero matters; $\Vert y\Vert \to 0$ alone would not justify this argument.

Put $e=x-u^{\ast }$. If $u^{\ast }$ minimizes distance, then for every $v\in {\mathcal{U}}_n$ and every real $t$,Both signs of arbitrarily small $t$ force $\mathrm{Re}(e,v)=0$. In a complex inner product space, apply the same argument to $iv$ to obtain the imaginary part too. Conversely, if $e$ is orthogonal to the subspace, the Pythagorean theorem for inner product spaces givesThus orthogonality characterizes the unique best approximation:Finite dimension also guarantees existence by solving the invertible Gram matrix normal equations; completeness of the ambient inner product space is unnecessary.

Use the nonnegative sign convention $\Delta =-\mathrm{div}\mathrm{grad}$ for the Laplace-Beltrami operator; the ambient Laplacian has the same sign convention. Changing both signs changes the signs of the eigenvalues below. Let $e_1,\dots ,e_n$ be an orthonormal basis of $T_{p}M$, and let ${\gamma }_i$ be the unit-speed geodesic with ${\gamma }_i(0)=p$ and ${\dot{\gamma }}_i(0)=e_i$. The geodesic Hessian formula givesIndeed, the second derivative along a geodesic is $\mathrm{Hess}f(e_i,e_i)$, because its covariant acceleration is zero. Taking the trace of the Hessian matrix in this orthonormal basis gives the displayed Laplace-Beltrami operator.

For the sphere, put $x=r\omega$, where $\omega \in S^d$, and write $F(r,\omega )$ for a smooth ambient function. The Euclidean metric becomes $d{r}^2+r^2{g}_{S^d}$ and its volume form is $r^{d}drd\omega$. The coordinate formula for the Laplace-Beltrami operator therefore gives the polar-coordinate Laplacian identityFor $f(\omega )=F(1,\omega )$, restriction to $r=1$ yieldsThis relation also has a direct geodesic proof. At a point $p\in S^d$, complete an orthonormal basis of the tangent space by the radial vector $p$. Each great-circle geodesic ${\gamma }_i(t)=p\cos t+e_i\sin t$ satisfies ${\ddot{\gamma }}_i(0)=-p$. The chain rule gives $(F\circ {\gamma }_i{)}^{\prime \prime }(0)=D^{2}F(e_i,e_i)-{\partial }_{r}F$. Summing and comparing with the ambient trace gives precisely the radial correction above.

Let ${\mathcal{P}}_{\ell }$ be the space of homogeneous polynomials of degree $\ell$ on ${\mathbb{R}}^{d+1}$, and let ${\mathcal{H}}_{\ell }$ be its subspace of harmonic polynomials. For $H\in {\mathcal{H}}_{\ell }$, homogeneity gives $H(r\omega )=r^{\ell }H(\omega )$. Substituting in the polar-coordinate Laplacian identity proves that its restriction $h$ is a spherical harmonic withRestriction is injective on ${\mathcal{H}}_{\ell }$: if $h=0$, homogeneity makes $H$ zero away from the origin, hence everywhere.

To count and exhaust these eigenfunctions, use the harmonic decomposition of homogeneous polynomialsHere is an algebraic proof rather than an assumption about the spectrum. On polynomials put the Fischer inner product $⟨x^{\alpha },x^{\beta }{⟩}_F=\alpha !{\delta }_{\alpha \beta }$. Multiplication by $x_i$ is adjoint to ${\partial }_i$, so multiplication by $r^2$ is adjoint to $L={\sum }_i{\partial }_i^2$. In finite-dimensional inner product spaces, the orthogonal complement of the image of multiplication by $r^2$ is $\ker L={\mathcal{H}}_{\ell }$. This proves the decomposition. Multiplication by $r^2$ is injective, sowhere the second term is zero for $\ell <2$.

Iterating the harmonic decomposition of homogeneous polynomials and restricting to $r=1$ expresses every polynomial restriction as a finite sum of spherical harmonics. Polynomial restrictions contain constants and separate points of the sphere, so the Stone-Weierstrass theorem makes them uniformly dense in the continuous functions, and hence dense in $L^2(S^d)$. The Laplace-Beltrami operator is self-adjoint, and eigenfunctions with different eigenvalues are orthogonal. If a smooth eigenfunction had an eigenvalue not in the list, it would be orthogonal to a dense subspace and would vanish. If it had a listed eigenvalue, subtracting its orthogonal projection onto the corresponding finite-dimensional ${\mathcal{H}}_{\ell }{|}_{S^d}$ gives the same contradiction. Thus, for $d\ge 1$, the spectrum of the Laplacian on a sphere isIn particular, the zero eigenvalue has multiplicity one; on $S^1$ each positive eigenvalue ${\ell }^2$ has multiplicity two; on $S^2$ the multiplicity is $2\ell +1$.

Hecke operators give a commuting family of arithmetic symmetries of modular forms. They preserve the weight and cusp forms, and diagonalizing them turns the Fourier coefficients of a modular form into multiplicative arithmetic data. Odd weight spaces are zero by the action of $-I$; in weight zero the operators act on constants by $T_{n}1={\sigma }_{-1}(n)$. For the substantive theory, fix an even positive weight $k$ and $\Gamma =S{L}_2(\mathbb{Z})$, and use the following normalization consistently.

Extend the slash operator for modular forms to $G{L}_2^{+}(\mathbb{Q})$ byIt is a right action and positive scalar matrices act trivially. Let ${\mathcal{D}}_n$ be the integral matrices of determinant $n>0$. The determinant-n matrix representatives for Hecke operators for $\Gamma \backslash {\mathcal{D}}_n$ areInteger row operations reduce the first column to $(a,0)$, after which a row shear reduces $b$ modulo $d$; uniqueness follows from these same conditions. The Hecke operator isRight multiplication by $\Gamma$ permutes the left cosets, so $T_{n}f$ is again invariant under the weight-$k$ slash operator for modular forms. Each summand is a holomorphic function on the complex upper half-plane, and the coefficient formula below proves holomorphy at a cusp. Thus $T_n$ acts on $M_k(\Gamma )$ and preserves $S_k(\Gamma )$. Geometrically, these operators sum over index-$n$ sublattices, or equivalently the associated finite-degree correspondences between complex tori; the factors above compensate for the weight's response to lattice scaling.

Write $f={\sum }_{r\ge 0}{a}_r{q}^r$. The sum over $b$ is zero unless $d\mid r$, in which case it is $d$. Substituting into the displayed definition proves the Fourier coefficients of a composite-index Hecke operator formulawith $\gcd (n,0)=n$. In particular, $a_0(T_{n}f)={\sigma }_{k-1}(n)a_0(f)$ and $a_1(T_{n}f)=a_n(f)$. At a prime this becomeswhere $a_{r/p}=0$ if $p\nmid r$.

The Hecke multiplication relations areThey can be checked by applying the coefficient formula twice and regrouping the divisors. More explicitly, coprime indices factor independently, so $T_m{T}_n=T_{mn}$ if $(m,n)=1$. At a prime, separating the terms in the divisor sum according to whether one more factor $p$ is available givesIteration gives $T_{p^a}{T}_{p^b}={\sum }_{j=0}^{\min (a,b)}{p}^{j(k-1)}{T}_{p^{a+b-2j}}$; combining the prime factors gives the full relation. Hence all Hecke operators commute, $T_1=1$, and the Hecke algebra of modular forms is generated by the $T_p$ together with scalars.

The Petersson inner product on the finite-dimensional cusp form space isThe integrand is invariant and the integral converges because cusp forms decay exponentially at the cusp. The Hecke operators are self-adjoint for the Petersson inner product at level one. To see the underlying adjoint calculation, unfold the finite correspondence for a determinant-$n$ matrix and change variable $\tau \mapsto \alpha \tau$. The invariant hyperbolic measure and the determinant-normalized slash factors transfer $\alpha$ from one side of the inner product to ${\alpha }^{-1}$. Scalar matrices act trivially, so one can replace ${\alpha }^{-1}$ by the integral adjugate $n{\alpha }^{-1}$, which again has determinant $n$. This reverses the correspondence and permutes the same collection of terms, with the same real prefactor $n^{k/2-1}$. Summing givesCommuting self-adjoint operators on a finite-dimensional inner product space admit a common orthonormal basis of eigenvectors. Thus $S_k(\Gamma )$ has a basis of simultaneous Hecke eigenforms, with real eigenvalues.

A nonzero simultaneous cuspidal Hecke eigenform has $a_1\ne 0$: if $T_{n}f={\lambda }_{n}f$ and $a_1=0$, comparison of the first coefficients gives $a_n={\lambda }_n{a}_1=0$ for every $n\ge 1$, so $f=0$. Normalize $a_1=1$. ThenConsequently the full Fourier series is determined by the eigenvalues, and a simultaneous eigenvalue system has a one-dimensional eigenspace. The Hecke multiplication relations translate intoIn particular the coefficients are multiplicative at coprime indices and are determined by the prime coefficients. This is the elementary multiplicity-one statement for normalized level-one Hecke eigenforms.

The Hecke eigenvalues are algebraic integers. One way to explain the arithmetic input is to use the rational structure supplied by the graded ring $M_{\ast }(\Gamma ,\mathbb{Q})=\mathbb{Q}[E_4,E_6]$. The rational cusp form space has a basis with integral Fourier coefficients, for example from appropriate monomials in $E_4,E_6,\Delta$. Its subgroup of forms whose entire expansions are integral is a full lattice: a finite initial coefficient map is injective by the valence formula for the modular group, so it embeds this subgroup into a finite-rank integer module. The coefficient formula shows that every $T_n$ preserves this lattice, and therefore has a monic integral characteristic polynomial. Its eigenvalues are algebraic integers. Simultaneous diagonalization also makes their number field finite over $\mathbb{Q}$; conjugating the rational coefficient data gives another eigenform, whose eigenvalues are again real by self-adjointness. Thus the coefficient field is totally real.

For a normalized cuspidal Hecke eigenform, define its L-function of a cusp form by $L(f,s)={\sum }_{n\ge 1}{a}_n{n}^{-s}$. The coefficient relations yield the Euler product of a Hecke eigenformIndeed the prime-power recurrence gives ${\sum }_{r\ge 0}{a}_{p^r}{X}^r=(1-a_{p}X+p^{k-1}{X}^2{)}^{-1}$, and multiplicativity assembles these factors. These identities hold analytically in an initial right half-plane, as well as formally. For instance boundedness of $y^{k/2}|f(\tau )|$ on the full complex upper half-plane gives $a_n=O(n^{k/2})$, so $\mathrm{Re}s>k/2+1$ suffices for absolute convergence.

The Mellin transform of a cusp-form L-function relates this arithmetic to the modular transformation:Splitting at one and using $f(i/y)=i^k{y}^{k}f(iy)$ givesExponential cusp decay makes the last integral entire in $s$. This proves the analytic continuation and functional equation of the L-function of a cusp form, complementing its Euler product. The root number is $i^k=(-1{)}^{k/2}$.

Two examples show the scope of the theory. For even $k\ge 4$, the Eisenstein series $E_k$ is an eigenform with $T_n{E}_k={\sigma }_{k-1}(n)E_k$. This follows from its divisor-sum expansion and the same multiplicativity relations; its associated positive-index Dirichlet series is proportional to $\zeta (s)\zeta (s-k+1)$. The constant coefficient gives $M_k(\Gamma )=\mathbb{C}{E}_k\oplus S_k(\Gamma )$, so an eigenbasis of the cusp space together with $E_k$ gives an eigenbasis of the whole modular-form space. The ordinary Petersson inner product above is only used on the cusp space, where it converges. In weight twelve the valence formula for the modular group gives $\dim S_{12}=1$: cancellation of the first coefficients of two independent cusp forms would give a nonzero form with order at infinity at least two, exceeding $12/12$. Thus the normalized modular discriminant is a Hecke eigenform; writing $\Delta ={\sum }_{n\ge 1}\tau (n)q^n$ gives the Ramanujan tau function relations $\tau (mn)=\tau (m)\tau (n)$ for coprime $m,n$ and $\tau (p^{r+1})=\tau (p)\tau (p^r)-p^{11}\tau (p^{r-1})$.

Use the normalizationThis is ridge regression; the penalty is the squared Euclidean norm. The centering removes the need for an intercept. Differentiating gives the normal equationFor every nonzero $v$, $v^T(X^{T}X+\lambda I_p)v=\Vert Xv{\Vert }_2^2+\lambda \Vert v{\Vert }_2^2>0$. The coefficient matrix is therefore a positive-definite matrix, even if the design matrix has deficient matrix rank. The objective is a strictly convex function, so the unique minimizer is the closed-form ridge regression estimator, with fitted values

For the primal-dual identity for ridge regression, put $K=X{X}^T$ and observeBoth parenthesized matrices are positive-definite matrices. Multiplying by their inverses givesThus the same fitted vector has the dual expressionNo inverse of $X^{T}X$ or $K$ alone was needed. This is also the kernel-ridge hat matrix for the linear kernel.

Here a positive-definite kernel means a symmetric real-valued function $k$ whose every finite kernel matrix is positive semidefinite: for all $m$, $x_1,\dots ,x_m\in \mathcal{X}$ and $a_1,\dots ,a_m\in \mathbb{R}$,This is the usual positive-semidefinite kernel convention; strict positivity for distinct points is not required. In particular $k(x,x)\ge 0$. Write $a=k(x,x)$, $b=k(x^{\prime },x^{\prime })$ and $r=k(x,x^{\prime })$. Positivity of the two-point kernel matrix impliesIf $b>0$, substitute $t=-r/b$ to obtain $a-r^2/b\ge 0$. If $b=0$, the affine expression $a+2tr$ can be nonnegative for every $t$ only if $r=0$. Both cases give the Cauchy-Schwarz inequality for positive-semidefinite kernels:The zero-diagonal case is essential when the positive-definite kernel is not strictly positive.