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A continuous map $F$ of a compact metric space is uniquely ergodic if it has exactly one invariant measure that is a Borel probability measure. The uniqueness requirement ranges over all invariant Borel probability measures.

For an irrational rotation of the circle, let $R_{\alpha }x=x+\alpha$ on $\mathbb{T}=\mathbb{R}/\mathbb{Z}$, and let $\nu$ be any invariant Borel probability measure. For each integer $r$, set ${\chi }_r(x)=e^{2\pi irx}$. Invariance givesFor $r\ne 0$, irrationality forces $e^{2\pi ir\alpha }\ne 1$, so the corresponding Fourier coefficient is zero. For $r=0$ it is one. These are exactly the Fourier coefficients of normalized Lebesgue measure $m$. By the Stone-Weierstrass theorem, trigonometric polynomials are uniformly dense in the continuous functions on the circle group; hence $\nu$ and $m$ integrate every continuous function equally and are the same Borel probability measure. Since $m$ is invariant, the irrational rotation of the circle is uniquely ergodic, with unique measure $m$.

For the irrational skew shift on the two-dimensional torus, writeThe map is invertible, with $T^{-1}(x,y)=(x-\alpha ,y-x+\alpha )$ modulo one, and it preserves $m_2$ as allowed in the question. We first prove the ergodic transformation property by the invariant-function characterization of ergodicity.

Let $F\in L^2(m_2)$ satisfy $F\circ T=F$. Its expansion in the Fourier basis is $F={\sum }_{r,s}{c}_{r,s}{e}_{r,s}$ in $L^2$. Direct calculation of the Koopman operator givesUniqueness of the Fourier coefficients therefore impliesFor $s\ne 0$, the magnitudes of the Fourier coefficients along all distinct indices $(r+js,s)$, $j\in \mathbb{Z}$, are equal. By the Bessel inequality they are square summable, so every such coefficient must be zero. When $s=0$, the relation becomes $c_{r,0}=e^{2\pi ir\alpha }{c}_{r,0}$, which forces $c_{r,0}=0$ for $r\ne 0$. Only $c_{0,0}$ remains. Thus every invariant $L^2$ function is constant, and

To prove unique ergodicity, we will establish uniform averages for every continuous function directly. No theorem on unique ergodicity of skew products is needed. Induction on $n$, using the old first coordinate in the second coordinate of $T$, yieldsIn particular, for a Fourier basis element,If $s=0$ and $r\ne 0$, this is a geometric series in $n$, anduniformly in $(x,y)$.

For $s\ne 0$, we give the finite Van der Corput inequality for finite scalar sequences and its proof. For $|u_n|\le 1$, extend $u_n$ by zero outside $0\le n<N$. Fix an integer $1\le H\le N$ and set $C_h={\sum }_{n=0}^{N-h-1}{u}_{n+h}\overline{u_n}$. Every original summand appears $H$ times in the identityThe Cauchy-Schwarz inequality, followed by expansion of the squared window sums, givesConsequently the Van der Corput inequality for finite scalar sequences isThe finite prefactor is important; the order of limits will be $N\to \infty$ with $H$ fixed, followed by $H\to \infty$.

For the irrational skew shift character sequence above, differencing cancels the quadratic term:For every fixed $h\ge 1$, $s\alpha h$ is irrational, so another geometric series estimate givesuniformly in $(x,y)$. With fixed $H$, the Van der Corput inequality for finite scalar sequences therefore impliesLetting $H\to \infty$ proves that the averages of every nonconstant Fourier basis element converge uniformly to zero. The constant character has average one. This establishes uniform equidistribution of an irrational skew shift on all trigonometric polynomials.

The Stone-Weierstrass theorem makes these trigonometric polynomials uniformly dense in $C({\mathbb{T}}^2)$. If $p$ approximates a continuous $g$ with $\Vert g-p{\Vert }_{\infty }<\epsilon$, thenTaking $N\to \infty$ and then $\epsilon \downarrow 0$ proves uniform convergence to $\int gd{m}_2$ for every continuous $g$.

Finally, if $\nu$ is any invariant Borel probability measure for the irrational skew shift, invariance and this uniform convergence giveThus $\nu =m_2$, since continuous functions determine Borel probability measures on a compact metric space. We concludeThe uniform-average proof also shows that every starting point has the same limiting continuous-function averages, a stronger conclusion than the almost-everywhere assertion provided by the pointwise ergodic theorem.

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

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

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

Label the polygon vertices $v_0,\dots ,v_{2n-1}$ cyclically. Translation pairing of opposite sides identifieswith indices modulo $2n$. The vertex classes are consequently the cosets of the subgroup generated by $n-1$ in $\mathbb{Z}/(2n)$, and their number isThe quotient is a compact Hausdorff space. An interior point has a disk neighbourhood, a paired-side point has two half-disks joined to a disk, and each vertex class has its incident sectors cyclically joined to a cone, topologically a disk. Thus it is a connected closed surface with an orientation. Its Euler characteristic is $V-n+1$, since there are $n$ paired edges and one face. Hence

The polygon interior and paired-side charts are translation surface charts with ${\omega }_n=dz$. Each corner angle is $(n-1)\pi /n$. For even $n$, all $2n$ corners meet, giving cone angle $2\pi (n-1)$; for odd $n$, each of the two classes contains $n$ corners, giving cone angle $(n-1)\pi$. Both are integral multiples of $2\pi$. A cone of angle $2\pi (k+1)$ has the local uniformizing coordinate $\zeta$ with $w={\zeta }^{k+1}$; the holomorphic one-form is $(k+1){\zeta }^{k}d\zeta$. Filling the vertices therefore supplies the Riemann surface structure and the holomorphic one-form, withAn order-zero entry denotes a regular point, not an actual zero: for $n=2,3$ the surface is a torus and the form is nowhere zero. The zero orders otherwise sum to $2g-2$, as a check against the degree of the canonical bundle.

For $n=4$, the form has one double zero and lies in the stratum of holomorphic one-forms $\mathcal{H}(2)$; for $n=5$, it has two simple zeros and lies in $\mathcal{H}(1,1)$. The SL2R action on differentials preserves zero multiplicities, as the local cone argument shows. ThereforeTheir equal genus and area do not distinguish the orbits; their different strata of holomorphic one-forms do.

The given $\gamma$ is a simple closed curve. Assume it is essential; for a contractible class both infimal lengths are already zero. Write ${\ell }_n={\ell }_{X_n}(\gamma )$. The collar lemma supplies an embedded annulus with coordinates $r\in (-w_n,w_n)$ and $t\in \mathbb{R}/\mathbb{Z}$, and metricSet $y(r)={\ell }_n^{-1}{\int }_0^r\mathrm{sech}sds$. This changes the metric to a positive scalar multiple of $d{t}^2+d{y}^2$, so the conformal modulus of an annulus isThe extremal length of its core curves is $1/M_n$. Allowing all curves homotopic to $\gamma$ in $X_n$ can only decrease infimal lengths, while the area of a metric on all of $X_n$ is at least its area on the collar. Thus $\lambda (\gamma ,X_n)\le 1/M_n$. Use the particular conformal metric $|q_n{|}^{1/2}$, whose area of a quadratic differential is one:This proves the required implication uniformly over all the area-one holomorphic quadratic differentials on these surfaces.

The converse is false. Here is an explicit slit connected sum of translation tori. Start with square flat copies of a torus $T_1=\mathbb{C}/(\mathbb{Z}+i\mathbb{Z})$ and $T_{\delta }=\mathbb{C}/(\delta \mathbb{Z}+i\delta \mathbb{Z})$, where $\delta \to 0$. Cut a horizontal slit of physical length ${\delta }^2$ in each, centred in an interior coordinate disk, and cross-glue the banks by translation. The resulting surface has genus two. Its two slit endpoints have cone angle $4\pi$, so the locally defined $dz$ extends to a holomorphic one-form ${\omega }_{\delta }$ with two simple zeros. Its flat area is $1+{\delta }^2$. Setwhich is a holomorphic quadratic differential of area one. Let $\gamma$ be a horizontal generator in the small torus, taken away from the slit and fixed by the marking of this small handle. Then

To verify that its hyperbolic length does not tend to zero, construct a uniform lower bound on extremal length. On the unit square torus choose a disk $D$ about the eventual slit centre and a smooth cutoff function $\chi$ equal to one on a smaller disk and supported in $D$. Let $x$ denote a local real coordinate on $D$; in the first term below, $dx$ is the globally defined torus one-form, while $\chi x$ is extended by zero outside $D$. The real closed differential formis globally defined, vanishes on the smaller disk, and has period one on the horizontal generator. Pull it to $T_{\delta }$ by the rescaling map $z\mapsto z/\delta$, and extend it by zero across the slit and over the other torus. For sufficiently small $\delta$, the slit is inside the region where the form vanishes. Hence this extension ${\alpha }_{\delta }$ is a smooth closed differential form on the connected sum, with ${\int }_{\gamma }{\alpha }_{\delta }=1$.

Define a nonnegative conformal metric density by the pointwise norm of ${\alpha }_{\delta }$ relative to the flat metric. Two-dimensional scale invariance givesindependently of $\delta$. For every representative $\tilde{\gamma }$ homotopic to $\gamma$,because the period of a closed differential form is unchanged by homotopy. This is the extremal length lower bound from a closed one-form; therefore $\lambda (\gamma ,X_{\delta })\ge 1/C>0$. If ${\ell }_{X_{\delta }}(\gamma )$ tended to zero along any subsequence, the collar estimate would force $\lambda (\gamma ,X_{\delta })\to 0$, a contradiction. In fact its hyperbolic lengths are uniformly bounded away from zero. ThusFor every fixed $g>2$, replace $T_1$ by the area-one translation surface $X_{2g-2}$ constructed in the polygon argument; its genus is $g-1$. Cut its slit inside a nonsingular flat coordinate disk. The new connected sum has genus $g$, and the area normalization, small-handle length bound and closed-one-form energy argument are unchanged. Thus the converse fails at every fixed genus $g\ge 2$.

View the torus as the topological group $S^1\times S^1$ under coordinatewise multiplication, and let $x_0$ be the common basepoint. From the given free homotopy constructThis is continuous, has $K(s,1)=x_0$, and at $s=0,1$ agrees with $\alpha ,\beta$ because $H(0,1)=H(1,1)=x_0$. Thus it exhibits the requested based homotopy explicitly.

Algebraically ${\pi }_1(S^1\times S^1)\cong {\mathbb{Z}}^2$ is Abelian, so conjugate elements are equal. Free and based homotopy classes of loops with a specified common basepoint therefore coincide here.

Use singular homology with integer coefficients. The Künneth theorem for triangulable spaces gives a split short exact sequence with tensor terms ${⨁}_{i+j=k}{H}_i(X)\otimes H_j(Y)$ and torsion terms ${⨁}_{i+j=k-1}{\mathrm{Tor}}_1(H_i(X),H_j(Y))$. When one factor's homology groups are free abelian, the torsion terms vanish and the tensor map is an isomorphism.

The circle has $H_0(S^1)=H_1(S^1)=\mathbb{Z}$ and all other groups zero. Inductively, the homology groups of a torus are free, andPascal's identity givesGeometrically the generators correspond to choosing $k$ of the $n$ circle factors and taking their product, consistent with the cellular model having one $k$-cell for each such subset and cancelling opposite boundary faces.

Give $V_n$ the action $(g\cdot P)(x,y)=P((x,y)g)$, where $(x,y)$ is a row vector. Substitution preserves homogeneous degree, and composition agrees with matrix multiplication, so this is a complex representation of SU(2) group. The diagonal torus $\mathrm{diag}(z,z^{-1})$ acts on $x^{n-j}{y}^j$ with weight $z^{n-2j}$; these weights are distinct.

A nonzero invariant subspace is invariant under the torus, so finite-dimensional Fourier projection onto its weight spaces supplies a nonzero monomial. Differentiating the group action and complexifying supplies the operators $E=x{\partial }_y$ and $F=y{\partial }_x$. Their repeated application moves between every successive monomial, with nonzero coefficients until an endpoint is reached:The invariant subspace therefore contains the whole monomial basis. ThusFor $n=0$ this is the one-dimensional trivial representation. Torus weight projections are legitimate because an invariant complex subspace is closed in finite dimension.

The Clebsch-Gordan decomposition for SU2 statesRealize the tensor product as polynomials of bidegree $(m,n)$ in $(x_1,y_1)$ and $(x_2,y_2)$. The determinant $D=x_1{y}_2-y_1{x}_2$ is invariant under the simultaneous SU2 action. For each $r$, the vectorhas torus weight $q=m+n-2r$ and is killed by $E=x_1{\partial }_{y_1}+x_2{\partial }_{y_2}$. Applying $F=y_1{\partial }_{x_1}+y_2{\partial }_{x_2}$ produces $q+1$ nonzero vectors of distinct weights, with $F^{q+1}{v}_r=0$. The relations $[E,F]=H$ and $H{v}_r=q{v}_r$ give $E{F}^j{v}_r=j(q-j+1)F^{j-1}{v}_r$, so their span is an irreducible copy of $V_q$.

These irreducible submodules have distinct highest weights and form a direct sum: a sum of earlier ones cannot contain an irreducible module inequivalent to all its summands. Equivalently use complete reducibility, obtained by averaging an inner product over the compact group. Assuming $m\le n$, their total dimension iswhich equals the dimension of the tensor product. Hence they exhaust it, proving the theorem.

Use the paper's Fubini-Study form normalization ${\tilde{\omega }}_{FS}=\frac{i}{2}\partial \overline{\partial }\log (1+|z{|}^2)$, the Hamiltonian vector field convention ${\iota }_{X_H}\omega =dH$, and the Lie algebra generator whose circle action is $e^{i\theta }$. Write $D=|z_0{|}^2+|z_1{|}^2+|z_2{|}^2$. The normalized moment map for the torus action isThe moment map image is the filled triangleIndeed, the three quantities $|z_j{|}^2/D$ are nonnegative and add to one, and any such triple is realized by choosing their square roots as homogeneous coordinates.

For the diagonal circle action, a point in the fixed-point set satisfies $[z_0,t{z}_1,t{z}_2]=[z_0,z_1,z_2]$ for every $t\in S^1$. If $z_0\ne 0$, the projective scaling must be one, forcing $z_1=z_2=0$. If $z_0=0$, all coordinates are scaled together. ThereforeBy part (b), the diagonal moment map isThe factor $1/2$ follows from the specified Fubini-Study form, not the integral normalization in the linked general article. For example, in one affine complex coordinate $w=r{e}^{i\theta }$ the symplectic form is $r(1+r^2{)}^{-2}dr\wedge d\theta$, whose interior product of a differential form with ${\partial }_{\theta }$ is $-r(1+r^2{)}^{-2}dr=d[-r^2/(2(1+r^2))]$. Choosing the generator $e^{2\pi it}$ rescales all moment maps by $2\pi$; reversing the defining sign reverses their signs. The negative level in parts (d) and (e) uses the convention displayed here.

Use the meridian of a knot $\mu$ and Seifert longitude $\lambda$ of the unknot on its boundary torus. The unknot's knot exterior is a solid torus whose disk-bounding curve is $\lambda$. A genus-one Heegaard diagram is thereforeHere the horizontal coordinate is $\mu$ and the vertical coordinate is $\lambda$. The drawing uses a small translate of $\alpha$ to avoid intersections on the edge of the square; opposite edges are identified. The algebraic intersection number of curves on an oriented surface has absolute value five.

Starting with $\Sigma \times [0,1]$, attach a three-dimensional two-handle along $\alpha \times \{0\}$, using its surface framing, and cap the resulting sphere with a three-handle. This gives the solid torus on that side. Attach a two-handle along $\beta \times \{1\}$, again using the surface framing, and cap its sphere with a three-handle. The second solid torus has its disk-bounding curve identified with $5\mu +3\lambda$, so the resulting oriented lens space is precisely the specified $5/3$ Dehn filling.

Label the four crossings $A,B,C,D$ in top, left-middle, right-middle, bottom order. Label the bounded regions $d$ (left), $e$ (right), $a$ (upper middle), $b$ (central bigon), $c$ (lower middle), and the unbounded region $o$. The drawing preserves the crossings and the orientation of the supplied figure-eight knot diagram. Its overpassing strands are the outer-left/upper-right strand at $A$, the upper-left/lower-central strand at $B$, the upper-central/lower-right strand at $C$, and the lower-left/outer-right strand at $D$.

Here is the Dehn presentation Heegaard diagram specified by this drawing. Replace the underlying four-valent planar graph $G$ by a small three-dimensional regular neighborhood $N(G)$ and put $\Sigma =\partial N(G)$. Since $G$ has four vertices and eight edges, $\Sigma$ has genus $8-4+1=5$. On $\Sigma$, take the five $\alpha$ curves bordering the bounded faces, pushed slightly onto the boundary of $N(G)$. They bound a complete disk system in the complementary handlebody $S^3\setminus \mathrm{int}N(G)$; their labels are ${\alpha }_d,{\alpha }_e,{\alpha }_a,{\alpha }_b,{\alpha }_c$.

The four $\beta$ curves bound the disks dual to the four short crossing tunnels in $N(G)\setminus \mathrm{int}\nu K_8$. Lift the crossing circles shown in red onto the upper and lower sheets of $\Sigma$, routing them along the corresponding face boundaries; the crossing's overpassing strand determines the lift. Orient the $\alpha$ disk normals and the $\beta$ curves so that their signed successive intersections with the $\alpha$ curves areAt $A,D$ the missing fourth corner is the unbounded region $o$, whose generator is omitted. This sheet-lifting prescription, together with the projected graph, specifies the curves on the Heegaard surface; the red circles alone are only their local projections.

To recover the knot exterior from $\Sigma \times I$, attach five two-handles along the $\alpha$ curves on one side and cap the resulting sphere with a three-handle. On the other side attach four two-handles along the $\beta$ curves. Compressing that side leaves the boundary torus of the knot exterior, which is retained. In particular, the counts are five $\alpha$ compressions and four $\beta$ compressions, rather than two complete genus-five disk systems.

Write $V=M_{S^2}(p^{\ast }/q,\ast )$. Filling one end of $S^1$ times an annulus gives a solid torus, so $V$ is a solid torus. The prescribed algebraic intersection number of curves on an oriented surface convention gives ${\tilde{f}}_2\cdot {\tilde{h}}_2=1$. HenceIn $H_1(V;\mathbb{Z})$, the first Dehn filling imposes $p^{\ast }f-q{h}_1=0$, while $h_1+h_2=0$. ThereforeThe class ${\lambda }_2$ is primitive because $\gcd (p^{\ast },q)=1$, a consequence of $p{p}^{\ast }-q{q}^{\ast }=1$. It is therefore the disk-bounding meridian of a solid torus of $V$. Since ${\mu }_2$ meets it once, Dehn filling $V$ along ${\mu }_2$ glues two solid torus pieces with disk boundaries meeting once. This is the standard genus-one Heegaard splitting of $S^3$, soTo identify the regular fiber geometrically, put it on their common boundary torus. Invert the displayed basis change:Here ${\mu }_2$ bounds a disk in the newly attached solid torus and ${\lambda }_2$ bounds a disk in $V$. Thus the curve has winding numbers $p$ and $q$ in the two complementary solid torus pieces. Both coefficients are positive and $\gcd (p,q)=1$, so it is the positive $(p,q)$ torus knot. This identifies the torus knot directly from the genus-one splitting, without invoking a theorem specifically about torus knots.

All classes in the nontrivial homological equalities are taken in $H_1(X_{(p,q)};\mathbb{Z})$, where $X_{(p,q)}=M_{S^2}(p^{\ast }/q,-q^{\ast }/p,\ast )$. The two internal gluing torus components still have inclusion maps into this knot exterior. Interpreting the maps instead as maps into the closed $S^3$ would make every displayed class zero.

Before the two Dehn fillings, $H_1(S^1\times \text{pair of pants};\mathbb{Z})$ has generators $f,h_1,h_2,h_3$ and relation $h_1+h_2+h_3=0$. The Dehn fillings add $p^{\ast }f-q{h}_1=0$ and $-q^{\ast }f-p{h}_2=0$. Put $m=-h_3$. Eliminating the relations using $p{p}^{\ast }-q{q}^{\ast }=1$ givesFor completeness, the three relation rows in generators $(f,h_1,h_2)$ are $(p^{\ast },-q,0)$, $(-q^{\ast },0,-p)$, $(0,1,1)$ when computing the quotient by $m$. Their determinant is $p{p}^{\ast }-q{q}^{\ast }=1$, so $m$ really generates the entire first homology group, and no finite torsion or index is hidden in the elimination.

Orient each exceptional-fiber longitude by ${\mu }_i\cdot {\lambda }_i=1$. In the boundary basis $(m_i,{\tilde{f}}_i)=(-{\tilde{h}}_i,{\tilde{f}}_i)$, write ${\mu }_i={\alpha }_i{m}_i+{\beta }_i{\tilde{f}}_i$ and ${\lambda }_i=u_i{m}_i+v_i{\tilde{f}}_i$, where ${\alpha }_i{v}_i-{\beta }_i{u}_i=1$. ThenThe filled meridian of a solid torus maps to zero, and $({\alpha }_1,{\alpha }_2)=(q,p)$, yieldingAdding multiples of ${\mu }_i$ to ${\lambda }_i$ has no effect. The word “any” therefore means any longitude with this compatible orientation; negating a longitude would negate the corresponding equality.

To compute Turaev torsion, use the general product formula $\tau (S^1\times F)\doteq (1-[f]{)}^{-\chi (F)}$ and multiplicativity under gluing along torus components. For a pair of pants, $\chi (F)=-1$, so the unfilled piece contributes $1-[f]$. The two filling solid torus pieces contribute $(1-[{\lambda }_1]{)}^{-1}$ and $(1-[{\lambda }_2]{)}^{-1}$. With $t=[m]$, the equalities above give $[f]=t^{pq}$, $[{\lambda }_1]=t^p$, and $[{\lambda }_2]=t^q$. ThereforeHere $\doteq$ allows the unit $\pm t^n$: a fully refined Turaev torsion requires an Euler structure and a homology orientation, neither of which the statement specifies. The displayed rational function is the representative whose expansion at $t=0$ starts with $1$, equivalently the standard nonnegative-exponent normalization.

The closed string already contains the particle required for quantum gravity. For bosonic string theory in its critical dimension of string theory, the mass-shell condition and closed-string level matching areAt $N=\tilde{N}=1$, the states ${ϵ}_{mn}{\alpha }_{-1}^m{\tilde{\alpha }}_{-1}^n|0;k⟩$ are massless. After imposing the Virasoro constraints and quotienting null string states, the transverse polarization tensor decomposes into symmetric trace-free, antisymmetric and scalar pieces under $SO(D-2)$. They describe a graviton, the Kalb–Ramond field and the dilaton. In particular, the symmetric trace-free sector is a massless spin-two graviton. Its number of degrees of freedom is $D(D-3)/2$.

The state–operator correspondence associates each physical string state with a string vertex operator. For example, the matter part of a massless closed-string vertex operator isIts conformal weights are $(1,1)$, so its integrated form $\int d^{2}z{V}_{ϵ,k}$ is invariant under changes of string worldsheet coordinates. At fixed insertion positions it is accompanied by $c\bar{c}$ from the worldsheet ghost fields. Physical vertices represent BRST cohomology classes; longitudinal changes of the polarization tensor are BRST-exact operators and decouple from physical scattering amplitudes. For the graviton, this string-state gauge redundancy becomes the linearized target-space diffeomorphism $h_{mn}\mapsto h_{mn}+{\partial }_m{\xi }_n+{\partial }_n{\xi }_m$.

The Polyakov path integral computes a closed-string scattering amplitude by inserting the external string vertex operators, integrating their unfixed positions and the inequivalent worldsheet moduli, and including the worldsheet ghost fields. On a Riemann sphere three positions are fixed by the residual conformal group. For example, contractions of tachyon vertex operators produce the Koba-Nielsen factor, whose position integral gives the Virasoro–Shapiro amplitude.

The string dual resonance property means that one crossing-symmetric scattering amplitude has equivalent expansions in the different channels: its poles exhibit intermediate string states in each channel, rather than separate channel contributions being added again. The scattering-amplitude factorization at a pole identifies the intermediate particle and its couplings. Set $8\pi T=1$, so ${\alpha }^{\prime }=4$ and the closed-string tachyon has $M^2=-1$. For four such external states the Mandelstam variables obey $s+t+u=-4$. The Virasoro–Shapiro amplitude has generic $s$-channel poles at $s=-1,0,1,\dots$, exactly the bosonic string mass spectrum in these units.

The massless pole gives an especially direct check. With overall normalization $C$ and $u=-4-s-t$, the Gamma function recurrence givesThusThe quadratic residue in $t$ contains a spin-two exchange. A scalar exchange alone could not give this angular dependence. Combined with the known massless spectrum and scattering-amplitude factorization, it identifies exchange of the graviton, with possible scalar contributions from the dilaton; the antisymmetric Kalb–Ramond field does not couple to two identical scalar tachyons here. Consequently the graviton participates in interactions, rather than being an isolated free state.

Decoupling longitudinal graviton polarizations forces a universal coupling to the conserved stress-energy tensor. Consistency of this massless spin-two gauge invariance extends the linearized coupling to the nonlinear dynamics of general relativity. At distances large compared with $\sqrt{{\alpha }^{\prime }}$, the gravitational sector of the effective action begins with the Einstein-Hilbert action, alongside dilaton and Kalb–Ramond field terms and higher-derivative string corrections. The infinite tower of string states supplies the short-distance completion of this quantum gravity expansion.

Interactions are organized by string worldsheet topology. A constant dilaton gives the string coupling $g_s=e^{{\Phi }_0}$, and a connected oriented surface of genus $g$ has Euler characteristic $\chi =2-2g$. Its weight is $g_s^{-\chi }$; with $n$ normalized external vertices,The sphere is tree order, the torus is one loop, and each additional handle adds a factor $g_s^2$. Each coefficient is an integral over worldsheet moduli, so this is string perturbation theory in $g_s$, with an independent low-energy expansion in ${\alpha }^{\prime }$.

For the one-loop vacuum contribution, torus modular invariance identifies $\tau$ with $(a\tau +b)/(c\tau +d)$, $ad-bc=1$. Integration is over the standard fundamental domain of the modular group,The potential short-proper-time region ${\tau }_2\to 0$, responsible for a point-particle ultraviolet divergence, is absent. It would count metrics already represented elsewhere in $\mathcal{F}$. Torus modular invariance is therefore the geometric reason for one-loop ultraviolet finiteness. More explicitly, in $D=26$ the vacuum integrand is proportional to $d^2\tau {\tau }_2^{-14}|\eta (\tau ){|}^{-48}$, with $\eta$ the Dedekind eta function; the complete expression is invariant under the modular group.

Ultraviolet finiteness does not make the bosonic one-loop vacuum energy finite. The remaining long-tube region ${\tau }_2\to \infty$ is an infrared divergence from the tachyon: $|\eta (\tau ){|}^{-48}\sim e^{4\pi {\tau }_2}$. It signals instability of the bosonic vacuum. Tachyon-free consistent backgrounds remove this particular obstruction, though other infrared effects must still be treated. The ultraviolet improvement comes from the extended string and the complete spectrum together with the worldsheet gauge identifications; the genus expansion remains a perturbative description of quantum gravity.

A smooth surface in ${\mathbb{R}}^3$ is a subset locally parametrized by a smooth map of two variables whose derivative has rank two and which is a homeomorphism onto its image.

Writing $\rho =\sqrt{x^2+y^2}$, the first surface equation becomesIt is the torus of major radius $R=2\sqrt{2}$ and minor radius $r=1$, with smooth parametrizationSince $R>r$, this is an embedding. The Gaussian curvature of a torus is