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The Seifert-van Kampen theorem states that if a space is covered by path-connected open sets $U,V$ with path-connected intersection containing a basepoint, then its fundamental group is the free product ${\pi }_1(U)\ast {\pi }_1(V)$ modulo the normal subgroup generated by identifying the two images of each loop in $U\cap V$. More generally it is the corresponding pushout of groups.

A wedge of two circles has fundamental group $F(a,b)$, the free group on their two loops: apply the theorem to thickened circle neighborhoods whose intersection is contractible and use ${\pi }_1(S^1)=\mathbb{Z}$. Represent the torus by a square with opposite edges identified. Remove an interior disk; the remainder deformation retracts onto this wedge. Replacing the disk attaches a two-cell whose boundary follows $ab{a}^{-1}{b}^{-1}$. A disk neighborhood and the punctured torus have annular intersection, and van Kampen kills precisely that boundary loop. HenceA torus with a point removed retracts to a torus with a small open disk removed, and then to the same wedge. ThusFor the connected sum, remove a disk from each torus and glue their boundary circles with reversed orientations. The two pieces have free groups $F(a,b)$ and $F(c,d)$ and their overlap is an annulus. Its boundary generators correspond to $[a,b]$ and $[c,d{]}^{-1}$, where $[a,b]=ab{a}^{-1}{b}^{-1}$. ThereforeThe boundary relation is essential: this group is neither the free group on four generators nor the abelian group ${\mathbb{Z}}^4$.

For $G=U(n)$ take the diagonal maximal torusThe spectral theorem for unitary matrices says that every element is conjugate to an element of $T$. A diagonal element with pairwise distinct eigenvalues has centralizer exactly $T$. Its conjugates therefore determine its eigenvalues up to permutation, and the normalizer quotient is $W=N_G(T)/T\cong S_n$. The character lattice of a torus $T$ is ${\mathbb{Z}}^n$, with $e^{\lambda }(z)=z_1^{{\lambda }_1}\cdots z_n^{{\lambda }_n}$. Conjugation on the matrix unit $E_{ij}$ has weight $e_i-e_j$, so these are the roots. Choose $e_i-e_j$ positive for $i<j$; thenDominant integral weights are precisely the integer tuples ${\lambda }_1\ge \cdots \ge {\lambda }_n$. Negative entries are allowed: they correspond to determinant twists, so polynomial representations alone do not exhaust the irreducible representations of $U(n)$.

Normalize Haar measure on $G$ and $T$ to have mass one, and putThe Weyl integration formula for U(n) for a continuous class function isFor a general continuous function, replace $f(t)$ in the torus integral by ${\int }_{G/T}f(gt{g}^{-1})d(gT)$, using the normalized invariant quotient measure.

To prove the formula, consider the conjugation map $q:G/T\times T\to G$, $q(gT,t)=gt{g}^{-1}$. On regular torus elements its fibres have exactly $n!$ points: the possible orderings of the distinct eigenvalues. Equip the group with an invariant inner product on its Lie algebra and split $\mathfrak{u}(n)=\mathfrak{t}\oplus {\mathfrak{t}}^{\perp }$. After translating the derivative back to the identity, its torus directions contribute the identity, while its transverse directions contribute ${\mathrm{Ad}}_{t^{-1}}-I$. Each pair $i<j$ supplies a real two-dimensional off-diagonal plane. On that plane ${\mathrm{Ad}}_t$ is multiplication by $z_i/z_j$, regarded as a plane rotation, so the absolute real determinant of the transverse derivative isThe change-of-variables theorem, divided by the covering multiplicity, now gives the desired weighted integral up to the constant relating the normalized invariant measures. The nonregular locus has measure zero: repeated eigenvalues are the zero-set of the nonzero discriminant, and the torus coincidence sets likewise have measure zero. Thus its omission causes no change to the integral. To determine the normalization, expand $\Delta$ as a sum of $n!$ distinct torus characters. Fourier orthogonality gives ${\int }_T|\Delta {|}^{2}dt=n!$. Applying the formula to $f=1$ fixes the coefficient to $1/n!$. This proves both the class-function and conjugacy-averaged forms.

We can now derive the character formula from integration, rather than merely quote it. Use the basic highest-weight theorem: a finite-dimensional irreducible representation has a unique highest weight $\lambda$ with a one-dimensional highest-weight space; all its other weights are $\lambda$ minus a nonnegative sum of positive roots. Its character is Weyl invariant. Also, character orthogonality for compact groups gives ${\int }_G|{\chi }_{\lambda }{|}^2=1$. This orthogonality follows by averaging the representation on $\mathrm{End}(V_{\lambda })$: the integral projects onto intertwining operators, whose dimension is one by the Schur lemma.

Let $\delta =(n-1,n-2,\dots ,0)$ and, for a strictly decreasing integer tuple $\eta$, put $A_{\eta }(z)=\det (z_i^{{\eta }_j})$. The finite Laurent polynomial $\Delta {\chi }_{\lambda }$ is alternating. Every alternating Laurent polynomial is a linear combination of the $A_{\eta }$: collect its monomials by permutation orbits, and observe that an orbit with repeated exponents has zero coefficient by antisymmetry. Different $A_{\eta }$ are orthogonal on $T$ and each has squared norm $n!$.

The coefficient of $z^{\lambda +\delta }$ in $\Delta {\chi }_{\lambda }$ is exactly one. Indeed, a contribution from the term $z^{w\delta }$ would require a weight $\lambda +\delta -w\delta$ in $V_{\lambda }$. For $w\ne 1$, the vector $\delta -w\delta$ is a nonzero nonnegative combination of the positive simple roots, so that weight lies above $\lambda$ and is impossible. The identity permutation supplies the highest-weight coefficient one. ConsequentlyApply the integration formula to $|{\chi }_{\lambda }{|}^2$. Torus orthogonality givesAll the other coefficients vanish, provingThe quotient extends across repeated eigenvalues because its product with the denominator is the already defined continuous character. Equivalently, alternant divisibility makes it a Laurent polynomial invariant under permutations; multiply by a determinant power if negative exponents must first be removed.

Finally, $\delta -\rho =\frac{n-1}{2}(1,\dots ,1)$ is fixed by the Weyl group, so its common factor cancels in the quotient. Since $A_{\rho }=e^{\rho }{\prod }_{\alpha >0}(1-e^{-\alpha })$, the determinant expression is exactlyUsing $\delta$ in the derivation avoids treating a possibly half-integral $\rho$ as an actual character of the torus. At the identity, taking the leading Vandermonde term of the two determinants gives the accompanying dimension formula

Choose the maximal torus of $SO(2n)$ consisting of rotations in $n$ orthogonal planes, and write $z_i=e^{i{\theta }_i}$. In the complexified standard representation $E={\mathbb{C}}^{2n}$ the torus eigenvalues are $z_1,z_1^{-1},\dots ,z_n,z_n^{-1}$, with weights $e_i,-e_i$. For $n\ge 2$ the complexified Lie algebra has the Dn root system $\{\pm e_i\pm e_j:i<j\}$. The root-space decomposition theorem splits this Lie algebra into its Cartan subalgebra and one-dimensional root spaces. The Cartan subalgebra contributes the zero weight with multiplicity $n$. Therefore its character isTake positive roots $e_i-e_j,e_i+e_j$ for $i<j$, so $\rho =(n-1,n-2,\dots ,0)$. For $n>2$ the root $\theta =e_1+e_2$ is a highest weight: its root vector is killed by every positive root vector, since $\theta +\alpha$ is never a root for positive $\alpha$. To verify irreducibility without relying on an unstated simplicity theorem, apply complete reducibility and compare dimensions. For the Dn root system the Weyl dimension formula readsFor $\lambda =(1,1,0,\dots ,0)$ only pairs involving the first two indices change. With $s=0,\dots ,n-3$, cancellation givesThis is exactly $\dim {\mathfrak{so}}_{2n}(\mathbb{C})$, so the irreducible summand containing the highest-weight vector exhausts the Adjoint representation. Thus for $n>2$ the complexified Adjoint representation is irreducible, with highest weight $e_1+e_2$.

For $n=2$, the roots split into the two independent systems $\{\pm (e_1+e_2)\}$ and $\{\pm (e_1-e_2)\}$. The Adjoint representation splits into two irreducibles of dimension three, with respective highest weights $(1,1)$ and $(1,-1)$. Equivalently these are the self-dual and anti-self-dual exterior two-forms in four dimensions. Their characters areThey add to the displayed adjoint character. This agrees with the Chiral decomposition of the complexified so4 Lie algebra.

For a direct character comparison, put $s(z)={\sum }_i(z_i+z_i^{-1})$. The eigenvalue formula for an exterior square givesThis is precisely ${\chi }_{\mathrm{ad}}$. More intrinsically, let $B$ be the nondegenerate symmetric bilinear form preserved by $SO(2n)$ and defineThis endomorphism is skew with respect to $B$, so $\Psi$ maps ${\Lambda }^{2}E$ to $\mathfrak{so}(E,B)$. In an orthonormal basis the bivectors $e_i\wedge e_j$ map to a basis of skew skew-symmetric matrices, proving that it is an isomorphism. Since $B$ is invariant,Thus the exterior square and the complexified Adjoint representation are naturally isomorphic, explaining the character equality without a weight calculation.

The symmetric square is not irreducible. The inverse metric tensor $\Omega$ is a nonzero invariant vector, and contraction with $B$ splits off its trivial line:The contraction sends $\Omega$ to $2n$, so this is indeed a direct sum. A highest-weight vector $f_1\otimes f_1$, where $f_1$ has torus weight $e_1$, lies in the traceless part because $B(f_1,f_1)=0$, and has highest weight $2e_1$. The same dimension formula gives, for $n\ge 2$,This equals $\dim {\mathrm{Sym}}^{2}E-1=n(2n+1)-1$. Complete reducibility therefore proves that the traceless part is irreducible, including the dimension-nine case for $SO(4)$. In particular,If the circle case $n=1$ is included, the adjoint character is $1$ and the symmetric square instead has three one-dimensional weights $2,0,-2$.

The printed identity has a sign error. With the defined Kostant partition function and the exponent $e^{-\nu }$, the cancelling factors must be $1-e^{-\alpha }$. This is already visible for $SU(2)$ with positive root $\alpha$: the formal series ${\sum }_{k\ge 0}{e}^{-k\alpha }$ multiplied by $1-e^{\alpha }$ is $-e^{\alpha }$, not $1$.

For the corrected formula, a partition of $\nu$ is a tuple of nonnegative integers $(k_{\alpha }{)}_{\alpha >0}$ satisfying $\nu ={\sum }_{\alpha >0}{k}_{\alpha }\alpha$. The order of summands is not counted. Multiplying the formal geometric series yieldsThis product is coefficientwise meaningful. Choose a linear functional strictly positive on every positive root. For a fixed $\nu$ it bounds each $k_{\alpha }$, so only finitely many tuples contribute. We are working in the completion supported on the negative positive-root cone, not claiming convergence of an ordinary Fourier series on the torus.

Each geometric series cancels its factor $1-e^{-\alpha }$, provingEquivalently, reversing all the exponential signs gives $({\sum }_{\nu }p(\nu )e^{\nu }){\prod }_{\alpha >0}(1-e^{\alpha })=1$ in the opposite completion. More generally, if $N$ is the number of positive roots, the expression literally printed in the paper equalssince ${\prod }_{\alpha >0}(1-e^{\alpha })=(-1{)}^N{e}^{2\rho }{\prod }_{\alpha >0}(1-e^{-\alpha })$. Thus the correction is substantive, not merely a choice of Fourier convention.

Use the corrected generating series of the Kostant partition function and the permitted Weyl character formula:For a fixed $w$, the coefficient of $e^{\mu }$ arises exactly whenThere are only finitely many Weyl-group terms, and the partition function vanishes outside the cone of nonnegative root combinations. Coefficient extraction is therefore legitimate in the formal completion used in the preceding part. Summing the signed contributions provesThis proves the Kostant multiplicity formula for every compact connected Lie group, not just the unitary groups. In the presence of a central torus, roots have zero central component, so the partition function automatically gives zero for any incompatible central weight. Also $\rho$ need not itself be an integral weight for the given global group: $w\rho -\rho$ always belongs to the root lattice, so every argument in the formula is nevertheless integral whenever $\lambda$ and $\mu$ are integral group weights.

As a normalization check, for $SU(2)$ write a highest weight as the nonnegative integer $\ell$, with positive root $2$ and $\rho =1$. The two Weyl-group terms giveSince $p(a)=1$ precisely when $a$ is even and nonnegative, this is one for $\mu =\ell ,\ell -2,\dots ,-\ell$ and zero otherwise, including the cancellation below the lowest weight. This recovers the full familiar weight string.

A symplectic blowup replaces a small symplectic ball around the point by an exceptional divisor ${\mathbb{CP}}^{n-1}$, for ${\dim }_{\mathbb{R}}X=2n$. One can construct its form by symplectic reduction. In a Darboux chart, consider ${\mathbb{C}}^n\times \mathbb{C}$ with the circle actionOn ${\mu }^{-1}(R^2/2)$ the action is free because $z\ne 0$. Its quotient therefore carries a smooth symplectic form. Where $w\ne 0$, choose the representative with $w$ positive real; the reduced form is the original form on $|z|>R$. Where $w=0$, the quotient is the Hopf fibration quotient of $|z|=R$, namely ${\mathbb{CP}}^{n-1}$. The local quotient is the tautological complex line bundle near its zero section, hence the local blowup. Glue this model to the exterior of the ball to obtain a symplectic form on the blowup.

The area of a projective line in the exceptional divisor is $\epsilon =\pi R^2$. For a closed $X$, changing this size changes the symplectic volume:Consequently the blowup is not determined up to symplectomorphism without a size choice. The underlying smooth blowup is fixed, but the symplectic construction has a parameter, as in symplectic blowup size changes volume.

For the blowdown map $\pi$ and exceptional divisor $E$, the First Chern class isIn real dimension four this becomes ${\pi }^{\ast }{c}_1(TX)-\mathrm{PD}[E]$, the first Chern class formula for a symplectic blowup.

The tangent bundle of the standard four-torus is a trivial complex rank-two bundle, so its symplectic canonical class is zero. For a generic fiber $F$ of a symplectic Lefschetz pencil, the symplectic adjunction formula gives $2g-2=F^2$. Two generic fibers intersect precisely at the base points, each with local intersection number one in the pencil model $[z_1:z_2]$. HenceThis is base-point count for a Lefschetz pencil on a symplectic four-torus.

Work first in the parameter rectangle of the parametric surface. Set $g(u,v)=f(F(u,v))$. With $F_u,F_v$ linearly independent, the chain rule givesThe intersection is transversal precisely when the implicit normal vector $\nabla f(F)$ and the parametric normal vector $F_u\times F_v$ are not parallel. Consequently $(g_u,g_v)\ne (0,0)$, and the implicit function theorem makes $g=0$ a locally regular parameter curve. This is the setting for transversal surface intersection tracing by a predictor-corrector continuation method.

Find one seed on the required component. Subdivide parameter patches, enclosing their images by bounding volumes and testing whether a certified range of $f$ excludes zero. For a positive-weight NURBS, the active control-point convex hull supplies a convenient enclosure; interval arithmetic can tighten the range of the composed function. Search the surviving patches for a bracketed root or a local corrected seed. Merely checking signs at four corners is inadequate: a small closed intersection can lie wholly inside a patch with the same corner signs. An adaptive or certified patch search, or a supplied seed, avoids this failure. For a sphere centered at $C$, use $f(P)=\Vert P-C{\Vert }^2-r^2$ and $\nabla f=2(P-C)$; for a torus, use its implicit polynomial and its differentiated polynomial in the same algorithm.

At a corrected seed ${\eta }_k=(u_k,v_k)$, an unnormalized parameter tangent vector and a unit spatial tangent vector areChoose the sign consistently with the previous tangent vector. Normalize $q=q_0/\Vert DF{q}_0\Vert$ and predict ${\eta }_p={\eta }_k+hq$, so $h$ is approximately a physical arc length step. Correct the predictor by Newton method applied toThe Jacobian has rows $(g_u,g_v)$ and $(T_k\cdot F_u,T_k\cdot F_v)$. At the starting point its first row vanishes along $q$, whereas its second row applied to $q$ is one. The two rows are therefore independent. The transverse-plane condition prevents the corrector from sliding arbitrarily along the intersection. Restrict correction to a neighborhood of the predictor, use damping, and reduce $h$ when correction fails or its displacement becomes excessive; this prevents jumps to another nearby branch.

To obtain a sufficiently dense sequence, impose a physical point-spacing bound, a bound on tangent turning, and a chord-error estimate. For local curvature $\kappa$, the small-step sagitta is approximately $\kappa h^2/8$; estimate curvature from successive tangent vectors and reduce $h$ until this and the desired spacing are acceptable. The correction residual should be small relative to the drawing tolerance. Uniform steps in $u$ or $v$ alone do not ensure spatial density because the surface metric can vary considerably.

For an open component, trace in both tangent directions from the seed. On reaching a parameter boundary, locate the endpoint by solving $g=0$ together with that boundary's parameter equation, rather than accepting an overshoot. For a closed component, stop only after nontrivial travel and return to the initial parameter location with compatible tangent orientation; near-coincidence in three-dimensional space alone can mistake a self-intersection or neighboring branch for closure. Cross periodic seams using the appropriate chart identification. Adaptive physical stepping, local correction, and component-aware termination give the required point sequence. A vanishing parameter gradient signals a tangency or singularity outside the transversal hypothesis and must be reported rather than passed through by this regular algorithm.

Use the usual cell structure on the torus: one vertex, two one-cells $a,b$, and one two-cell attached by the word $ab{a}^{-1}{b}^{-1}$. Take the meridian being collapsed as $a$. The quotient has one vertex, the surviving circle $b$, and a two-cell whose attaching word reduces to $b{b}^{-1}$ and is null-homotopic. Attaching a two-cell to a circle by a null-homotopic loop gives a space homotopy equivalent to $S^1\vee S^2$.

Alternatively its cellular chain groups are $C_2=C_1=C_0=\mathbb{Z}$, and both boundary maps vanish: the two-cell traverses $b$ once in each direction, and the one-cell's endpoints coincide. Thus the integral homology groups areThe pinched torus is connected, so reduced $H_0$ is zero.

Identify the pointed torus with ${\mathbb{R}}^2/{\mathbb{Z}}^2$ with $p$ at the origin. Its fundamental group is ${\mathbb{Z}}^2$, generated by the two coordinate loops. A homeomorphism fixing $p$ induces an automorphism of this group, givingFunctoriality makes it a homomorphism. It is nontrivial, indeed onto: every integer invertible matrix defines a pointed torus homeomorphism by $x\mapsto Ax$ modulo the Euclidean lattice.

If $f_{\ast }=I$, lift $f$ to a continuous $F:{\mathbb{R}}^2\to {\mathbb{R}}^2$ with $F(0)=0$. Its deck-transformation law is $F(x+n)=F(x)+n$ for every $n\in {\mathbb{Z}}^2$, because the induced map on coordinate loops is identity. Thenis well defined: adding $n$ changes the bracketed expression by exactly $n$. It is a continuous pointed homotopy from $f$ to identity. Hence every kernel element is homotopic to the identity, although the intermediate maps need not be homeomorphisms.

For a regular parametrized surface $\sigma (u,v)$, the first fundamental form is the induced metricChoose the unit normal $N=({\sigma }_u\times {\sigma }_v)/|{\sigma }_u\times {\sigma }_v|$. With the convention $I{I}_{ij}=N\cdot {\sigma }_{ij}$, the second fundamental form isThe Gaussian curvature, the product of the principal curvatures, isFor a compact smooth embedded surface without boundary, the Gauss-Bonnet theorem giveswhere $e(S)$ is the Euler characteristic. The no-boundary condition is needed for this form; a smooth embedded surface with boundary has the additional boundary geodesic curvature integral.

For the surface of revolution, useThe parameters are periodic, with local open parameter rectangles providing the coordinate charts. Put $R=r+a\sin u$ and $D=\sqrt{a^2{\cos }^{2}u+b^2{\sin }^{2}u}$. Since $r>a>0$ and $b>0$, both $R$ and $D$ are positive. The tangent vectors areTheir inner products are $E=D^2$, $F=0$, and $G=R^2$. ThusTheir cross product is $R(b\sin u\cos v,b\sin u\sin v,a\cos u)$, so its normalization gives the outward unit normalNext,Taking their inner products with $N$ yields $e=-ab/D$, $f=0$, and $g=-bR\sin u/D$. Hence the fundamental forms of an elliptic ring torus giveChoosing the inward unit normal changes both displayed coefficients of $II$ to their negatives and leaves $I$ and $K$ unchanged. As a consistency check, $K=a{b}^2\sin u/(R{D}^4)$; its integral over the torus is zero, since $KdA=a{b}^2\sin ududv/D^3$ cancels under $u\mapsto u+\pi$. This agrees with $e(S)=0$ in the Gauss-Bonnet theorem.

On the torus ${\mathbb{T}}^n=(\mathbb{R}/2\pi \mathbb{Z}{)}^n$, choose normalized Haar measure, so that $e_k(x)=e^{ik\cdot x}$, $k\in {\mathbb{Z}}^n$, is an orthonormal basis of $L^2$. A periodic Sobolev space is defined, for any real $s$, byFor negative $s$, the coefficients describe distributions rather than necessarily functions. For a nonnegative integer $s$, the weight is comparable to ${\sum }_{|\alpha |\le s}|k^{\alpha }{|}^2$, so this norm is equivalent to the sum of the squared $L^2$ norms of all weak derivatives through order $s$. Fourier truncation proves that smooth trigonometric polynomials are dense. The $L^2$ pairing identifies the continuous dual of $H^s$ with $H^{-s}$, giving Sobolev duality.

Two Fourier estimates explain the usefulness of these spaces. If $s>m+n/2$, the Cauchy-Schwarz inequality givesUniform convergence of the differentiated Fourier series proves the Sobolev embedding $H^s\hookrightarrow C^m$. If $\delta >0$, a bounded subset of $H^{s+\delta }$ has uniformly small high-frequency tails in $H^s$:Finite-dimensional Fourier truncation therefore proves the compact embedding $H^{s+\delta }\hookrightarrow H^s$, the periodic version of Rellich-Kondrachov compactness theorem. In addition, for $s>n/2$ multiplication is continuous $H^s\times H^s\to H^s$, making it a Sobolev algebra. Indeed the weighted coefficient of a product is bounded by the sum of the convolutions of $⟨k{⟩}^s|\hat{u}(k)|$ with $|\hat{v}(k)|$ and of $|\hat{u}(k)|$ with $⟨k{⟩}^s|\hat{v}(k)|$; Cauchy-Schwarz puts unweighted coefficients in ${\ell }^1$, and the ${\ell }^1\ast {\ell }^2\to {\ell }^2$ case of Young's convolution inequality finishes the estimate.

For eigenfunction expansions, distinguish ellipticity from self-adjointness. An arbitrary non-self-adjoint elliptic operator need not have an orthonormal basis of eigenfunctions. A useful definite setting iswith smooth real periodic coefficients, $a_{ij}=a_{ji}$, real $V$, and uniform ellipticityTake the self-adjoint realization in $L^2$ with domain $H^2$. Its construction and spectral properties can be seen from the associated form on $H^1$,For a sufficiently large constant $C$, $b_C=b+C(\cdot ,\cdot )$ is a positive coercive form: $b_C(u,u)\ge c_1\Vert u{\Vert }_{H^1}^2$, and $L+C\ge I$. The Lax-Milgram theorem supplies the unique weak solution of $(L+C)u=f$ for $f\in H^{-1}$; the form is Hermitian, which gives symmetry of its inverse on $L^2$.

The regularity input is the periodic elliptic estimateHere is why an elliptic operator gains two derivatives. With coefficients frozen at a point, its principal symbol $\sum a_{ij}{\xi }_i{\xi }_j$ dominates $c|\xi {|}^2$, and the estimate follows by multiplying Fourier coefficients by that symbol. On a sufficiently small coordinate patch, the difference of the leading coefficients from these constants has arbitrarily small supremum norm. In the integer-order differentiated estimate, this absorbs the highest derivatives into the left side; derivatives falling on coefficients contribute only lower-order terms. A partition of unity reduces the variable-coefficient problem to such patches. Its commutators with $L$ have order one, controlled bywhich follows by splitting low and high Fourier modes. Absorbing these terms gives the displayed estimate. For a weak $H^1$ solution, the same localization with difference quotients first bounds its second weak derivatives for $f\in L^2$; differentiating and repeating then gives $f\in H^s\Rightarrow u\in H^{s+2}$ for nonnegative integer $s$. Interpolation and duality extend the scale to real orders. In particular $(L+C{)}^{-1}:L^2\to H^2$ is bounded.

Let $R=(L+C{)}^{-1}$. Compactness of $H^2\hookrightarrow L^2$ makes $R$ a compact operator. Symmetry of the form makes $R$ self-adjoint, and positivity of $L+C$ makes $R$ positive and injective. Its range is dense because $\ker R^{\ast }=\ker R=0$. The spectral theorem for compact self-adjoint operators consequently gives an orthonormal basis ${\varphi }_j$ withElliptic regularity bootstraps every ${\varphi }_j$ to a smooth function. Each eigenvalue has finite multiplicity. Thus a self-adjoint uniformly elliptic second-order operator on the torus has a complete smooth orthonormal eigenfunction expansion with discrete eigenvalues tending to infinity.

Writing $u_j=(u,{\varphi }_j)$, every $u\in L^2$ has $u={\sum }_j{u}_j{\varphi }_j$ in $L^2$, with $\Vert u{\Vert }_2^2={\sum }_j|u_j{|}^2$. Iterating the elliptic estimate shows that the graph norm of $(L+C{)}^m$ is equivalent to the $H^{2m}$ norm. Interpolation and duality give the more general spectral descriptionThe comparison constants depend on the operator and $s$. In particular smooth functions have coefficients decaying faster than every power of ${\lambda }_j+C$, and their expansions converge in every Sobolev norm and hence in every $C^m$ norm.

This gives a practical way to solve elliptic equations. The equation $Lu=f$ is solvable if and only if $f_j=0$ whenever ${\lambda }_j=0$; then one may take $u_j=f_j/{\lambda }_j$ on the other eigenspaces and add an arbitrary vector in $\ker L$. The Sobolev description proves the two-derivative gain and the Fredholm alternative, rather than leaving them as formal series manipulations. Similarly the solution of the heat equation ${\partial }_{t}u+Lu=0$ isFor every $t>0$ the exponential decay at high eigenvalues places this solution in every Sobolev space, hence it is smooth.

For the Laplacian $-\Delta$ the eigenfunctions are precisely $e^{ik\cdot x}$ with eigenvalues $|k{|}^2$. The number with $|k{|}^2\le R$ is the number of lattice points in a radius-$\sqrt{R}$ ball, asymptotic to its volume by comparison with unit cubes. For the general self-adjoint operator above, bounds on its form above and below by positive multiples of $\Vert \nabla u{\Vert }_2^2$ plus constants, together with the Courant–Fischer min-max principle, imply ${\lambda }_j+C\asymp j^{2/n}$. Thus the growth reflects dimension and the two-derivative order. Compactness of the torus is essential: on ${\mathbb{R}}^n$, translated bumps prevent the compact embedding, and the Laplacian instead has continuous spectrum described by the Fourier transform.

Label four face colors by ${\mathbb{F}}_2^2$. Give each edge the sum of its two adjacent face labels. This is nonzero because adjacent faces differ. At a cubic vertex the three face labels are pairwise distinct, so their three edge sums are the three different nonzero elements. This is a Tait coloring.

Conversely, label the three edge colors by the nonzero elements of ${\mathbb{F}}_2^2$, whose sum is zero. Choose a base face with label zero and label any other face by summing crossed edge labels along a path in the dual graph. The result is independent of the path: a simple closed dual path encloses vertices, and summing the zero incident sums at these vertices leaves exactly its crossed boundary edges, since interior edges occur twice. General closed paths decompose into such cycles. Adjacent labels differ by a nonzero edge label, giving a four-color face coloring. The assumption that every edge borders distinct faces excludes bridges in a graph and ensures this map coloring is the relevant one.

The equivalence fails on a torus. Use the Heawood torus map. On seven vertices indexed modulo seven, take the triangles $A_i=(i,i+1,i+3)$ and $B_i=(i,i+3,i+2)$. Each edge of $K_7$ occurs in exactly two oppositely oriented triangles; each vertex link is a six-cycle, for example $1,3,2,6,4,5,1$ at zero. This is a connected closed orientable surface with Euler characteristic $7-21+14=0$, hence a torus.

Its dual graph is cubic, with $A_i$ adjacent to $B_i,B_{i+1},B_{i-2}$. The three offsets partition its edges into perfect matchings, giving a Tait coloring. Its seven faces are discs and their adjacency graph is $K_7$, so they need seven colors. The obstruction to the planar proof is a possible nonzero edge-label sum around a noncontractible cycle.

Parametrize the torus by $F(u,v)=((2+\cos u)\cos v,(2+\cos u)\sin v,\sin u)$. Its outward Gauss map isThe derivative $N_u$ has unit length, while $N_v$ has length $|\cos u|$ and is perpendicular to $N_u$. Thus rank drops precisely at $u=\pi /2$ or $3\pi /2$. These are the two circles of height $z=1$ and $z=-1$ and radius two, and their images are the north and south poles. Therefore the critical values are $(0,0,\pm 1)$.

For any nonpolar target point, its third coordinate determines two values of $u$, one with positive and one with negative $\cos u$. For each, the longitude fixes a unique $v$ modulo $2\pi$, with a shift of $\pi$ when $\cos u$ changes sign. Thus every regular value has exactly two preimages, one on the outer and one on the inner side of the torus. ConsequentlyThis explicit two-sheet description justifies the count without relying on a sketch.

We consider finite-dimensional continuous complex representations of SU(2). Its maximal torus is $\mathrm{diag}(e^{i\theta },e^{-i\theta })$. The irreducible representations areThey may be realized as homogeneous polynomials of degree $n$ in two variables, with the induced linear substitution action. Their torus weights are $n,n-2,\dots ,-n$, each once. The centre $-I$ acts by $(-1{)}^n$, so precisely the even-$n$ representations descend to SO(3). In physics $n=2j$ labels integral or half-integral spin $j$.

Here is a classification argument. Average any positive definite Hermitian form over the compact group using normalized Haar measure. The averaged form is invariant, so the orthogonal complement of an invariant subspace is invariant. Repeatedly splitting gives complete reducibility. Differentiate an irreducible representation and complexify to the Lie algebra ${\mathfrak{sl}}_2$, with generators $H,E,F$ satisfying $[H,E]=2E$, $[H,F]=-2F$, and $[E,F]=H$. Torus periodicity gives integer $H$-weights. Choose a vector $v$ of maximal weight $n$; then $Ev=0$. Induction on $k$ using the commutators givesThe chain terminates: if $F^{p}v\ne 0$ and $F^{p+1}v=0$, the displayed formula forces $n=p\ge 0$. The vectors $v,Fv,\dots ,F^{n}v$ have distinct weights, span an invariant subrepresentation, and hence span the irreducible representation. Their generator action agrees with ${\mathrm{Sym}}^n{\mathbb{C}}^2$. Invariance under the Lie algebra is equivalent to invariance under the connected group, completing the classification. Conversely, a nonzero invariant subspace of a symmetric power contains a weight vector by projection onto torus weights. Raising it to the highest weight and then lowering generates every monomial, proving that the symmetric power is irreducible.