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The ordinary Poincare duality statement here is for a compact manifold without boundary, of dimension $d$, oriented over the field $F$. Its fundamental class $[M{]}_F$ induces isomorphismsfor every $q$. For a manifold with boundary the appropriate statement is Poincare-Lefschetz duality with relative groups; the ordinary pairing need not be nondegenerate. Over a field, the universal coefficient theorem for cohomology identifies $H^{d-q}(M;F)$ with the full dual of $H_{d-q}(M;F)$. The cap-cup evaluation identity and duality therefore makea perfect pairing. Explicitly, a nonzero $a$ has a nonzero cap product, and a linear functional $b$ on its homology group takes a nonzero value on that product. The same argument in the other variable proves nonsingularity. On the whole graded cohomology, define $B(a,b)$ by taking the degree-$d$ part of $a⌣b$ before evaluation. For a nonzero component $a_q$ choose $b$ homogeneous of degree $d-q$ to pair nontrivially with it. All other components contribute zero in degree $d$. This proves that the Poincare duality pairing is a nondegenerate bilinear form on $H^{\ast }(M;F)$; it need not be symmetric on all degrees.

Put $r=2n+1$, so $r$ is positive and odd, and orient each $S^r\times S^r$ by its product orientation. The Künneth theorem gives integral cohomology $\mathbb{Z}$ in degrees zero and $2r$, ${\mathbb{Z}}^2$ in degree $r$, and zero elsewhere. Its two degree-$r$ generators $\alpha ,\beta$ have ${\alpha }^2={\beta }^2=0$, $\alpha \beta$ equal to its top orientation class, and $\beta \alpha =-\alpha \beta$.

For a connected sum of oriented manifolds, excision and the long exact sequence for deleting a ball show that deleting a ball removes the top homology class and leaves all lower positive homology groups unchanged. The boundary sphere represents zero in the punctured manifold: it is the boundary of its relative fundamental chain. In the Mayer–Vietoris sequence for the two punctured pieces joined along the separating sphere, the orientation class of the connected sum maps onto that sphere's class. The intermediate positive groups are consequently the direct sums of the groups of the two original manifolds. This also covers $r=1$, where the sphere is a circle and the boundary-class observation is necessary in the middle degree. Iterating and applying the universal coefficient theorem for cohomology givesHere $g\ge 1$; under the conventional extension $W_0=S^{2r}$ the same formula holds with a zero middle group.

Let $\omega$ be the top cohomological orientation class. The connected-sum pinch map to the wedge of the $g$ sphere products gives degree-one projections to each summand. Pull back the two factor classes from summand $i$ to obtain ${\alpha }_i,{\beta }_i$. Their product is $\omega$, since the projection has degree one. Classes from different wedge summands have zero positive-degree cup products, and graded commutativity of the cup product supplies the reversed sign. Thus the full cohomology ring of a connected sum of odd-dimensional sphere products is the graded free abelian group just displayed with multiplicationThe unit is $1$, and $\omega$ times any positive-degree class is zero by dimension. This describes all products, including ${\omega }^2=0$.

The smooth involution is a diffeomorphism, since it is its own inverse. Its fixed set is closed; discreteness and compactness therefore make it finite. The supplied positivity of $\det (I-D{f}_x)$ makes every fixed point nondegenerate with local index $+1$. The Lefschetz-Hopf fixed-point theorem then givesIndeed the degree-zero and top-degree traces are both one, because the manifold is connected and $f$ preserves its orientation; the middle degree is odd.

On $V=H^r(W_g;\mathbb{R})$, the form $\Omega (a,b)=⟨a⌣b,[W_g]⟩$ is a nondegenerate alternating bilinear form by Poincare duality and the oddness of $r$. Thus it is a symplectic vector space of dimension $2g$. Naturality and orientation preservation show that $T$ preserves $\Omega$, and $T^2=I$. For the eigenspaces of a symplectic involution, write $V=V_{+}\oplus V_{-}$: the polynomial $(t-1)(t+1)$ has distinct roots over $\mathbb{R}$. If $v_{+}\in V_{+}$ and $v_{-}\in V_{-}$, then $\Omega (v_{+},v_{-})=\Omega (T{v}_{+},T{v}_{-})=-\Omega (v_{+},v_{-})$, so the two vector subspaces are orthogonal. Each restricted form is nondegenerate, since a vector annihilating its own vector subspace also annihilates the other and hence all of $V$. Their dimensions are therefore even, say $\dim V_{+}=2a$, $\dim V_{-}=2b$, with $a+b=g$. ConsequentlyThis proves the fixed-point congruence for an involution on an odd-sphere connected sum:

An orientation of a smooth $n$-dimensional manifold can be specified by a nowhere-vanishing smooth $n$-form $\omega$: an ordered tangent basis is positive precisely when $\omega$ evaluates positively on it. Multiplication by a positive smooth function preserves the orientation. This is the differential-form formulation of the equivalent formulations of orientability of a smooth manifold.

Use the standard convention that the compact connected manifold is nonempty and has no boundary. Choose a positive volume form $\omega$ for its orientation. It is a closed differential form, because there are no nonzero $(n+1)$-forms. For $n>0$, compactness makes its integral finite, and positivity on a coordinate neighborhood together with nonnegativity elsewhere givesIf $\omega =d\eta$, the Generalized Stokes theorem would give ${\int }_M\omega ={\int }_{\partial M}\eta =0$, a contradiction. ThereforeFor $n=0$, a nonempty connected manifold is a point, and $H_{\mathrm{dR}}^0(M)=\mathbb{R}$ directly. The boundary hypothesis matters: a closed interval is an oriented compact connected manifold with boundary with $H_{\mathrm{dR}}^1=0$. Thus the result is about ordinary boundaryless manifolds, not an unrestricted version including boundary.

First work at the level of differential forms. The fiber description implies $\pi \circ f=\pi$, so every pullback ${\pi }^{\ast }\beta$ is $f$-invariant. Since $\pi$ is a surjective local diffeomorphism, its pullback on forms is injective: at any $y\in N$, choose a lift $x$ and use the inverse of $d{\pi }_x$ to evaluate the form on arbitrary tangent vectors at $y$.

Conversely let $\alpha$ be an $f$-invariant form on $M$. For a local inverse $s:V\to M$ of $\pi$, define $\beta {|}_V=s^{\ast }\alpha$. If two inverses select the same lift they agree locally; if they select different lifts, the fiber condition identifies the second with $f$ composed with the first. Locally these alternatives are stable, using disjoint inverse neighborhoods at distinct lifts. The equality $f^{\ast }\alpha =\alpha$ therefore makes the local definitions agree. They glue to a unique smooth form $\beta$ on $N$ with ${\pi }^{\ast }\beta =\alpha$. Thusis an isomorphism of complexes, with the exterior derivative on both sides.

It remains to compare the cohomology of the invariant complex with invariant de Rham cohomology classes. Put $P=(I+f^{\ast })/2$ on forms. If $f^{\ast }[\alpha ]=[\alpha ]$ and $d\alpha =0$, then $P\alpha$ is invariant, closed and represents the same class, because its class is the average of two equal classes. If an invariant form is exact, $\alpha =d\eta$, then $\alpha =d(P\eta )$ with an invariant primitive. This proves both surjectivity and injectivity on cohomology. For $p=0$ there are no negative-degree primitives, and the same conclusion follows directly from closed functions.

Combining the two steps givesThis is the de Rham cohomology of a finite quotient argument specialized to an involution. It does not require orientation. The hypotheses do not explicitly exclude $f=\mathrm{id}$; then all fibers are singletons, $\pi$ is a bijective local diffeomorphism and the conclusion is the ordinary diffeomorphism invariance. If $f$ has one fixed point, local injectivity of $\pi$ makes its fixed set open as well as closed; connectedness of $M$ forces this identity case. Otherwise the fibers have two points and $\pi$ is the usual double covering.

Use the double covering map $\pi :S^n\to {\mathbb{RP}}^n$ and its deck involution, the antipodal map $a(x)=-x$. The given sphere cohomology vanishes outside degrees $0,n$, so only those degrees can survive as invariant classes. The degree-zero action is the identity because the sphere is connected for $n\ge 1$.

For the top degree, orient $S^n$ as the boundary of the unit ball. A positive tangent basis $(v_1,\dots ,v_n)$ at $x$ means that $(x,v_1,\dots ,v_n)$ has positive determinant in ${\mathbb{R}}^{n+1}$. Under the antipodal map every one of these $n+1$ vectors changes sign, so the orientation sign is $(-1{)}^{n+1}$. Equivalently, the sphere volume form isIts cohomology class is nonzero by the integral argument and therefore generates the one-dimensional top de Rham cohomology. Thus the top class is invariant precisely when $n$ is an odd integer. Applying the quotient result yieldsIn particular ${\mathbb{RP}}^1$ has one copy of $\mathbb{R}$ in each of degrees zero and one. These are real cohomology groups; integral torsion is not detected by de Rham cohomology.

Write $T=\dot{\gamma }$ and $D_t={\nabla }_T$. A Jacobi field is a vector field along the geodesic satisfyingThe curvature convention is that used in Question 1. A geodesic variation is a smooth map $F:(-ϵ,ϵ)\times [0,1]\to M$ with $F(0,t)=\gamma (t)$ whose $t$-curves are affinely parametrized geodesics. Torsion-freeness gives $D_s{\partial }_{t}F=D_t{\partial }_{s}F$. Differentiating $D_t{\partial }_{t}F=0$ and commuting covariant derivatives gives the Jacobi equation for ${\partial }_{s}F{|}_{s=0}$.

For the converse, let $a=J(0)$ and $b=D_{t}J(0)$. Choose a curve $p(s)$ with $p(0)=\gamma (0)$ and $p^{\prime }(0)=a$. parallel transport along it identifies its tangent spaces with $T_{\gamma (0)}M$. Set $v(s)=P_s(T(0)+sb)$, so $v(0)=T(0)$ and $D_{s}v(0)=b$. The geodesics with initial data $(p(s),v(s))$ give a variation $F(s,t)$. Smooth dependence on initial conditions and compactness of the original time interval ensure that this family is defined for all $t\in [0,1]$ after shrinking $ϵ$, even if $M$ is incomplete. Its variation field has the initial values $a,b$, so uniqueness for the linear Jacobi equation identifies it with $J$. This proves the realization of Jacobi fields by geodesic variations.

The endpoint-vanishing pointwise normal fields form a vector space, since their conditions and equation are linear. If $\gamma$ is nonconstant, the map $J\mapsto D_{t}J(0)$ is injective because $J(0)=0$ and zero initial derivative force the zero solution. Differentiating $⟨J,T⟩=0$ gives $D_{t}J(0)\perp T(0)$. Hence the endpoint-vanishing normal Jacobi fields have dimension at most n-1 bound isFor a constant geodesic, $D_t^{2}J=0$, and both zero endpoint values force $J=0$, so the bound still holds.

On the unit round $S^n$, let $\gamma (t)=\cos (\pi t)e_0+\sin (\pi t)e_1$. Its speed is $\pi$. The constant ambient vectors orthogonal to $e_0,e_1$ give $n-1$ independent parallel normal fields $E_j(t)$. Since $R(J,T)T={\pi }^{2}J$ on normal fields, the fields $J_j(t)=\sin (\pi t)E_j(t)$ satisfy the Jacobi equation and vanish at both antipodal endpoints. They attain dimension $n-1$.

For the last clause, interpret a closed geodesic as a nonconstant smoothly periodic geodesic. A constant loop has length zero and cannot be shortened. Parametrize ${\gamma }_0$ on $[0,1]$ at constant nonzero speed. parallel transport once around it fixes $T(0)$. Because the manifold is orientable, this transport preserves orientation; on the normal space of dimension $n-1$, which is odd, it lies in $SO(n-1)$. An odd-dimensional special orthogonal transformation has a fixed vector: nonreal eigenvalues pair with their conjugates, real eigenvalues are $\pm 1$, and determinant one in odd dimension forces an eigenvalue $+1$.

Transport such a nonzero normal fixed vector around the loop. It gives a periodic parallel normal field $V$, with $D_{t}V=0$ and $V(1)=V(0)$. Its jets also agree at the seam, so it is smooth as a field on the parametrizing circle. For small $s$, $F(s,t)={\exp }_{{\gamma }_0(t)}(sV(t))$ is a smooth variation through closed curves, providing their homotopy to ${\gamma }_0$.

For energy $E(s)=\frac{1}{2}{\int }_0^1|{\partial }_{t}F{|}^{2}dt$, the first variation vanishes at the closed geodesic. The permitted second variation of geodesic energy has no endpoint term for this periodic variation, soThus $E(s)<E(0)$ for small nonzero $s$. The Cauchy-Schwarz inequality gives $L(F_s{)}^2\le 2E(s)$, while constant speed gives $L({\gamma }_0{)}^2=2E(0)$. Consequently the instability of a closed geodesic in positive even-dimensional curvature yieldsThe deformation need not remain a geodesic or an embedded curve.

At each point, use the inverse metric on the cotangent space. It induces an inner product on $p$-forms byEquivalently, wedges of distinct members of an orthonormal coframe form an orthonormal basis. Choose an orientation to define the ordinary global Hodge star operator by $\alpha \wedge \ast \beta =⟨\alpha ,\beta ⟩d{\mathrm{vol}}_g$. It maps $p$-forms to $(n-p)$-forms and obeys ${\ast }^2=(-1{)}^{p(n-p)}$. We use real forms; complex forms are obtained by complex-linear extension. Without an orientation an ordinary global star is not available without twisting the target, so orientation is implicit in this first part.

With the codifferential ${\delta }_p=(-1{)}^{n(p+1)+1}\ast d\ast$ for $p>0$ and ${\delta }_0=0$, define the Hodge Laplacian $\Delta =d\delta +\delta d$. On functions it is the nonnegative-sign Laplace-Beltrami operator, $\Delta f=-\mathrm{div}\mathrm{grad}f$. A harmonic differential form satisfies $\Delta \alpha =0$.

For $\alpha$ of degree $p$, the displayed sign formula and star square giveApplying the identities again, at degrees $p-1$ and $p+1$, proves the Hodge star commutes with the Hodge Laplacian relationSince star is invertible, $\alpha$ is harmonic if and only if $\ast \alpha$ is harmonic. This part does not need compactness; zero-degree and top-degree terms are interpreted as zero when their degrees are outside the range.

For the remaining cohomological statements assume a compact oriented manifold without boundary. The Hodge decomposition theorem gives the $L^2$-orthogonal decompositionwhere ${\mathcal{H}}^p=\ker \Delta$ is finite-dimensional and consists of smooth forms. Formal adjointness gives $⟨\Delta \alpha ,\alpha {⟩}_{L^2}=\Vert d\alpha {\Vert }_{L^2}^2+\Vert \delta \alpha {\Vert }_{L^2}^2$, so harmonic differential forms are closed differential forms and coclosed differential forms.

If $\Delta f\ge 0$, formal adjointness and $\Delta 1=0$ give ${\int }_M\Delta f=0$. A continuous nonnegative function with zero integral vanishes, so $\Delta f=0$. Then $\Vert df{\Vert }_{L^2}^2=⟨\Delta f,f⟩=0$. Thus the a function with nonnegative Laplacian on a closed manifold is locally constant assertion isIt is globally constant when $M$ is connected. The PDF does not explicitly include connectedness here; on two disjoint circles one may take different constants, so literal global constancy needs that qualification. Boundaryless is the standard manifold convention in this argument; otherwise boundary conditions are necessary.

For a closed differential form $\alpha$, write its Hodge decomposition as $\alpha =h+d\beta +\delta \gamma$. Since $d\alpha =dh=0$, we have $d\delta \gamma =0$. Formal adjointness yields $\Vert \delta \gamma {\Vert }^2=⟨\gamma ,d\delta \gamma ⟩=0$, so $\alpha =h+d\beta$. Thus $h$ represents its de Rham cohomology class. If two harmonic representatives differ by $d\beta$, their difference $h$ has $\Vert h{\Vert }^2=⟨d\beta ,h⟩=⟨\beta ,\delta h⟩=0$. Therefore

Finally let $h\in G$. A connected Lie group admits a smooth path $h_t$ from the identity to $h$. The maps $R_{h_t}$ give a smooth homotopy between the identity and $R_h$. The homotopy invariance of de Rham cohomology says their pullbacks agree on cohomology. The maps $R_{h_t}$ preserve orientation, since their Jacobian signs cannot change from the identity along the path. As an orientation-preserving isometry, $R_h$ commutes on forms with star, exterior differentiation, codifferentiation and consequently the Hodge Laplacian. Hence $R_h^{\ast }\alpha$ is harmonic and represents the same cohomology class as a harmonic $\alpha$. Uniqueness now proves the harmonic forms are fixed by a connected isometric group action conclusionThe argument also applies componentwise when $M$ is disconnected.

Use the nonnegative Hodge-Laplacian convention from Question 4. The Bochner-Weitzenbock formula for one-forms isHere $\nabla$ is the Levi-Civita connection induced on the cotangent bundle, and the rough Laplacian in a local orthonormal frame isIt is the composition of covariant derivative with its formal adjoint. The Ricci endomorphism is defined by $g({\mathrm{Ric}}^{♯}X,Y)=\mathrm{Ric}(X,Y)$ and acts on one-forms by $(\mathrm{Ric}\cdot \alpha )(X)=\alpha ({\mathrm{Ric}}^{♯}X)$. The musical isomorphism defines ${\alpha }^{♯}$ by $g({\alpha }^{♯},X)=\alpha (X)$. The associated scalar formula isThese signs make the integrated rough-Laplacian term $\Vert \nabla \alpha {\Vert }^2$ on a closed manifold.

For a connected manifold, the full Riemannian holonomy group at $p$ is the subgroup of $O(T_{p}M,g_p)$ consisting of parallel transports around all piecewise smooth loops based at $p$. Its natural action on $T_{p}M$ is the holonomy representation; it induces actions on cotangent spaces and all tensor spaces. The holonomy representation is an irreducible representation when it has no nonzero proper invariant subspace.

The fundamental principle of Riemannian holonomy identifies parallel tensor fields with tensors at $p$ fixed by full holonomy. A parallel field returns to its value under every loop. Conversely, transport a fixed tensor along a path from $p$ to each point. Any two paths differ by a loop, so the result is independent of the path; local smooth parallel transport yields a smooth parallel field. Evaluation and construction are inverse. Full holonomy, not merely the contractible-loop subgroup, is required for this global correspondence.

On a compact manifold without boundary and with nonnegative Ricci curvature, a harmonic one-form satisfies the integrated Bochner identityBoth terms are nonnegative; therefore $\nabla \alpha =0$. This proves that harmonic one-forms are parallel under nonnegative Ricci curvature. Integration uses the Riemannian volume density and does not require an orientation.

If $n=\dim M\ge 2$, a nonzero such form would give a nonzero fixed tangent vector via metric duality and hence a proper invariant line, contradicting irreducibility. This is precisely the irreducible holonomy in dimension at least two has no parallel one-form criterion. Since the question permits harmonic representatives of all classes, it gives $b_1(M)=0$, where $b_1$ is the first Betti number.

The covering $\pi :X\times T^k\to M$ is finite: its fibre over a point is a closed discrete subset of the compact total space and hence finite. Use the explicitly permitted equality of the Betti numbers of $M$ and its finite cover, in particular $b_1(X\times T^k)=b_1(M)=0$. This equality is a permission of this question, not a general theorem about finite covers.

The $k$ angular one-forms on $T^k$, pulled back to $X\times T^k$, represent linearly independent de Rham classes: their periods on the $k$ coordinate circles are the standard basis vectors. exact differential forms have zero periods. Thus $b_1(X\times T^k)\ge k$, forcing $k=0$. Now $\pi :X\to M$ is a finite universal covering map because $X$ is simply connected. The fundamental group acts freely and transitively on a fibre, or equivalently loop-lifting identifies its elements with the finite set of possible endpoints upstairs. Consequently the intended conclusion isThe printed statement needs the dimension qualification under the usual definition of irreducibility. Take the standard circle $M=S^1$, with $X$ a point, $k=1$, and the identity covering $X\times T^1\to S^1$. Its Ricci curvature is zero. parallel transport fixes its global unit tangent, so its holonomy representation is the trivial representation on a one-dimensional real vector space, which is irreducible. The allowed Betti-number equality holds for this identity cover, yetThus there is no proof of the unqualified literal assertion in dimension one. If “irreducible holonomy” is instead intended to exclude the one-dimensional trivial representation, the preceding intended proof applies. A connected zero-dimensional manifold is a point and has trivial fundamental group.

First use the nonzero loci, as required by the standard SL2R action on differentials. A nonzero holomorphic one-form has, away from its zeros, flat coordinateswhose changes of coordinate are translations. A nonzero holomorphic quadratic differential similarly has local flat coordinates $w=\int \sqrt{q}$, with $q=d{w}^2$ and changes of coordinate $w_j=\pm w_i+c_{ij}$. These are respectively translation surfaces and half-translation surfaces.

Identify a flat coordinate with a vector in ${\mathbb{R}}^2$. For $A\in {\mathrm{SL}}_2(\mathbb{R})$, replace every flat coordinate by $W=Aw$. Since $A$ preserves orientation and commutes with multiplication by $-1$, the new changes of coordinate areThey are holomorphic in the new coordinates, and so define a new complex structure. Define ${\omega }_A=dW$ or $q_A=d{W}^2$ in that structure. The forms glue because translations preserve $dW$, and the extra signs preserve $d{W}^2$.

The zeros also extend. A zero of order $m$ of a holomorphic one-form has cone angle $2\pi (m+1)$; a zero of order $m$ of a holomorphic quadratic differential has cone angle $(m+2)\pi$. The real-linear deformation preserves the corresponding winding multiplicity. Filling the cone in a local coordinate $\zeta$ gives $W={\zeta }^{m+1}$ in the first case, or a local branch of $W={\zeta }^{(m+2)/2}$ in the second. Thus the resulting forms are constant multiples of ${\zeta }^{m}d\zeta$ or ${\zeta }^{m}d{\zeta }^2$ and have the same zero orders. This verifies extension across the missing points, rather than merely producing an atlas on the punctured surface.

An isomorphism preserving the original differential identifies its flat coordinates up to the permitted translations or signs; applying $A$ identifies the deformed atlases too. Hence the construction descends to the corresponding moduli spaces. Applying $B$ after $A$ replaces $w$ by $BAw$, soThe area of a quadratic differential, and the analogous area of a holomorphic one-form, are preserved because $\det A=1$.

For $q={\omega }^2$, the flat coordinates obtained from $\omega$ already give the required half-translation surface atlas for $q$. The same replacement $w\mapsto Aw$ therefore constructs both deformations, and

The printed sets include identically zero differentials. They have no flat coordinates, so the customary geometric group action is defined on the nonzero loci. One can obtain a set-theoretic action on the displayed entire sets by declaring $A\cdot (X,0)=(X,0)$; the same equivariance identity then holds at zero. This extension is generally not continuous: as $t\omega \to 0$, the deformed underlying surface is the same $X_A$ for every real $t>0$, and can differ from $X$. Thus a claim about the standard continuous geometric group action requires the nonzero convention.

The analogous flat-coordinate construction does not define a full ${\mathrm{SL}}_2(\mathbb{R})$ action on arbitrary cubic differentials. Away from zeros, a nonzero holomorphic cubic differential has flat coordinatesTheir changes of coordinate are $w_j=\zeta w_i+b_{ij}$ with ${\zeta }^3=1$. A branch change is therefore a rotation through $2\pi /3$, not merely a sign. Applying a real-linear map $A$ changes its linear part to $AR{A}^{-1}$, where $R$ is that rotation. In general this is not a complex-linear map, so the proposed changes of coordinate are not holomorphic and cannot define the required deformed complex structure.

For example, take $A=\mathrm{diag}(2,1/2)$ and $R=\left(\begin{array}{cc}\cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{array}\right)$ with $\theta =2\pi /3$. Thenwhose off-diagonal entries fail the condition for a complex-linear map. This dependence on the choice of cube-root coordinate is the obstruction even when considering descent from the locus $c={\omega }^3$: the three possible roots need not lead to the same deformation of the cubic pair.

The real matrices preserving orientation that normalize the order-three rotations are precisely the matrices of complex-linear maps; intersecting with ${\mathrm{SL}}_2(\mathbb{R})$ leaves $\mathrm{SO}(2)$. Indeed a nonreal rotation determines its complex structure, and conjugation to its inverse would reverse that structure's orientation. Thus there is a natural rotation action,The conclusion concerns the geometric group action analogous to that for translation surfaces and half-translation surfaces; it does not rule out artificial group actions unrelated to these atlases.

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.

In local holomorphic coordinates, write the two positive singular values of $Df$ as $s_1\ge s_2$. Preservation of orientation gives $J_f=s_1{s}_2$, and quasiconformality gives $s_1/s_2\le K$.

For an admissible conformal metric $\rho |dw|$ on $Y$, define a density on $X$ by $\sigma (z)=\rho (f(z))s_1(z)$. Along each path,so $L_{\rho }(f(\Gamma ))\le L_{\sigma }(\Gamma )$. The change of variables formula and $s_1^2\le K{J}_f$ giveAlso $s_1^2\ge J_f$, so $A_{\sigma }\ge A_{\rho }>0$. Thus $\sigma$ has positive finite area and is admissible, andTaking the supremum over $\rho$ provesApplying the same argument to $f^{-1}$, which has the same bound on its maximal dilatation, also gives $K^{-1}\lambda (\Gamma ,X)\le \lambda (f(\Gamma ),Y)$. The inequalities remain valid for extended extremal lengths; no extremizing density need exist.

Use the closed genus $g\ge 2$ version of Teichmüller's uniqueness theorem. Let $f_0:X\to Y$ be a Teichmüller map: for a nonzero holomorphic quadratic differential $q$, normalized by ${\int }_X|q|=1$, and $0<k_0<1$,The value at a zero of $q$ is irrelevant to the Beltrami coefficient, which is defined almost everywhere. For every quasiconformal map $f:X\to Y$ preserving orientation and homotopic to $f_0$, the conclusion isThe normalized $q$ is also uniquely determined when $k_0>0$.

We use the following analytic input, stated with its hypotheses. The Reich–Strebel inequality says that if $f_0$ has the displayed Beltrami coefficient and a competitor $f$ has the same target and homotopy class, then, for $\mu ={\mu }_f$,This is the standard fundamental inequality for an integrable holomorphic quadratic differential; it also holds on a finite-type punctured surface with the homotopy fixing the punctures. Its formulation is given in Gardiner and Hu, §5. We quote this analytic inequality as the lecture result used in the proof.

Put $k=\Vert \mu {\Vert }_{\infty }$. Pointwise, away from the isolated zeros of $q$,Integrating against the probability density function $|q|$ proves $K_0\le K(f)$. If $K(f)=K_0$, all these inequalities are equalities almost everywhere. The strictly increasing last function forces $|\mu |=k_0$ almost everywhere; equality in the triangle inequality then forces $\mu q/|q|=k_0$. ThereforeTwo quasiconformal maps with the same Beltrami coefficient differ by postcomposition with a biholomorphism, by the local chain rule for the Beltrami equation. Consequently $h=f\circ f_0^{-1}$ is a biholomorphism from $Y$ to itself homotopic to the identity.

For completeness, such a biholomorphism is the identity when $g\ge 2$. Apply the uniformization theorem and choose the lift of the homotopy to $\mathbb{H}$ starting at the identity. Its endpoint lift $\tilde{h}$ commutes with every deck transformation. It is a real Möbius transformation. The compact quotient has no parabolic Möbius transformations in its deck group, and freeness excludes elliptic Möbius transformations. Two distinct hyperbolic axes exist: a discrete free group preserving just one axis would be cyclic, contradicting the fundamental group of a closed surface of genus at least two. Commutation with two deck transformations represented by hyperbolic Möbius transformations having distinct axes makes it fix their boundary endpoints individually; there are at least three such endpoints. A Möbius transformation fixing three points is the identity. Hence $h=\mathrm{id}$ and $f=f_0$.

If $q_1$ is another unit-area differential for the same map, then $|q_1|/q_1=|q|/q$, so $q_1/q$ is positive real wherever defined. This meromorphic function is constant by the open mapping theorem, and normalization makes the constant one. Without normalization, positive multiples of $q$ describe the same map. At $k_0=0$, uniqueness of the map still holds in genus at least two, but there is no distinguished $q$. In genus one, translations supply nontrivial biholomorphisms homotopic to the identity, so equality determines the map only up to those biholomorphisms; the genus hypothesis cannot be omitted.