Closed geodesic Created 2026-10-07 Updated 2026-10-14
A nonconstant geodesic whose position and tangent velocity repeat after a positive time. On a flat torus it is a straight line modulo the lattice, closing when its displacement is a Euclidean lattice vector. A primitive closed geodesic traverses its image without being an iterate of a shorter periodic geodesic.
Half-turn flat three-manifold 2026-10-14
This closed flat Riemannian manifold is the mapping torus of on a two-dimensional flat torus. Equivalently, take planar lattice translations and the screw in Euclidean three-space. Its square is a unit axial translation, its holonomy is , and its first Betti number is one.
Integrals of an irrational two-frequency oscillator Created 2026-10-07 Updated 2026-10-14
On the regular set of two uncoupled oscillators with irrational frequency ratio, every continuous globally defined integral of motion is constant on each fixed-action flat torus, because every orbit is dense there. A differentiable integral is consequently a function of the two actions, so its differential of a smooth map lies in their span. Local angle combinations can still give extra local integrals; they fail to be single-valued on the torus.
Two isospectral manifolds remain isospectral after taking a Riemannian product with the same closed factor: eigenvalues add, with multiplicities given by the product Laplacian eigenspace decomposition. For nonisometric closed connected factors , choose . Every closed product geodesic of length less than has constant projection to , since a nonconstant closed geodesic shorter than the injectivity radius is impossible. The short coordinate geodesics of the small flat torus span exactly its tangent factor at every point. Any product Riemannian isometry must preserve that intrinsically defined smooth distribution and its orthogonal complement, hence restrict to an isometry of the original factors. Thus these products are still nonisometric. Taking to be four-dimensional tori proves the higher-dimensional stabilization statement without assuming a cancellation theorem for arbitrary products.
Lagrangian torus Created 2026-10-07 Updated 2026-10-14
A Lagrangian submanifold diffeomorphic to an -dimensional flat torus in a -dimensional symplectic manifold. Compact connected regular levels of a completely integrable Hamiltonian system are examples, by the Arnold-Liouville theorem.
Past exam of the mathematics course of the University of Cambridge 2003 iii Paper 28 1 Solution Created 2026-10-03 Updated 2026-10-14
The possible ranks and the corresponding homeomorphism types areThere is no example for or . The uniqueness here is topological, not uniqueness of a Riemannian metric.
To see the restrictions, use the prime decomposition of a closed orientable three-manifold. Its fundamental group is the free product of the prime three-manifold groups. A free product of two nontrivial groups is nonabelian, so at most one summand can have nontrivial fundamental group. The remaining simply connected summands are three-spheres by the Poincaré conjecture. If the nontrivial summand is not irreducible, it is , giving rank one. If it is irreducible and has infinite fundamental group, the sphere theorem for three-manifolds makes it aspherical. Its fundamental group is then a dimension-three Poincare duality group. But is a duality group of dimension , as seen from the torus classifying space or its Koszul resolution, so .
For rank three, the flat case of the geometrization theorem says that a closed irreducible orientable three-manifold with fundamental group admits a flat metric. Write it as . By the Bieberbach theorem, contains a full-rank translation lattice . If commutes with translation by each , then for every lattice vector, forcing . Since itself is abelian, it consists entirely of translations. Its quotient is therefore the three-torus. Rank zero follows directly from the Poincaré conjecture, and rank one was already determined by the prime decomposition of a closed orientable three-manifold. Historically, before the Poincare theorem, the uniqueness conclusions were phrased allowing connected sums with homotopy three-spheres; the modern theorem removes that ambiguity.
The six flat types below mean the six closed orientable flat three-manifold types. There are infinitely many individual flat metrics and there are additional noncompact flat Riemannian manifolds. In each closed case the universal cover is Euclidean space and the deck transformation group is a torsion-free crystallographic group. The holonomy groups of closed orientable flat three-manifolds are .
The first is a flat torus , with trivial holonomy. For each cyclic case choose a planar lattice admitting the relevant rotation , and generate the deck transformation group by translations in and the screwIts th power is unit translation along the axis. A nontrivial holonomy element has nonzero axial displacement modulo integers, so has no fixed point. All transformations preserve orientation. A compact prism fundamental domain gives compactness. Equivalently, the manifold is a torus mapping torus with finite-order monodromy. In planar lattice bases one may takeThe half-turn uses any planar lattice, the quarter-turn a square lattice, and the third- and sixth-turns a hexagonal lattice. These give respectively the half-turn flat three-manifold, third-turn flat three-manifold, quarter-turn flat three-manifold, and sixth-turn flat three-manifold. Their holonomies have orders , and their first Betti numbers are all one, since the only fixed direction of holonomy is the screw axis.
The sixth type is the Hantzsche-Wendt manifold. A concrete group is generated byTheir linear parts are commuting half-turns about perpendicular axes. Moreover , and are translations by , and . Their translation subgroup is the unit cubic lattice and their linear quotient is . In each nontranslation coset, the displacement along the fixed axis is a half-integer, so the action is free. The quotient is compact and orientable. No vector is fixed by both linear parts, so its first Betti number is zero. Bieberbach classification gives precisely these six affine types; the explicit screw descriptions distinguish them without identifying arbitrary finite quotients with holonomy.
Past exam of the mathematics course of the University of Cambridge 2003 iii Paper 28 5 Solution Created 2026-10-03 Updated 2026-10-14
For a connected smooth Riemannian manifold without boundary, let be the group of all global diffeomorphisms with , endowed with its compact-open topology. Equivalently these maps preserve the intrinsic distance; the Myers-Steenrod theorem supplies smoothness of distance-preserving bijections and makes a finite-dimensional Lie group acting smoothly. The group is metric-dependent: a topological manifold can carry a very symmetric metric or one with no continuous symmetries. Connectedness matters here. With infinitely many identical disconnected components, permutations of those components can produce a group which is not a finite-dimensional Lie group.
An isometry is determined by its value and derivative at one point. To prove this, suppose and . Preservation of the Levi-Civita connection giveswhere the exponential map is defined locally. Thus is the identity in a normal neighborhood of . The set where its value and derivative agree with the identity is both closed and, by the same argument, open. Connectedness makes it all of . No completeness is needed for this determination argument.
The stabilizer consequently acts faithfully and orthogonally on , so embeds in . More generally an isometry takes a chosen orthonormal frame at to one at its image point, providing an injective map into the orthonormal frame bundle. It is a smooth embedding, and the full isometry action is proper. The orbit-stabilizer dimension formula therefore givesIsotropy is compact. If is compact then itself is compact, by the same frame description or Arzela-Ascoli applied to isometries and their inverses. Properness also means that the quotient by isometries has well-controlled local orbit structure, even though orbit dimensions may vary.
The infinitesimal equation is the Killing equation. Differentiating at zero givesConversely a vector field satisfying this equation has local flows preserving the metric. A global one-parameter subgroup requires the field to be complete. On a geodesically complete manifold every Killing field is complete: its length is constant along each of its own trajectories, so a finite-time trajectory has bounded speed and stays in a closed bounded ball. The Hopf-Rinow theorem makes that ball compact, and the ordinary differential equation can then be continued. The caveat is real: on the Euclidean interval , the field is Killing but its translations are not globally defined for all time. This illustrates Killing fields on incomplete manifolds need not generate global isometries.
The Killing transport identities determine a Killing field from and the skew-adjoint map . Along a geodesic the field is a Jacobi field, so uniqueness for its second-order differential equation propagates these initial data. They have components, reproducing the dimension bound infinitesimally. The important distinction on an incomplete manifold is between all local Killing solutions and the Lie algebra of complete global isometry flows.
In dimension three the bound is six, and rotational isotropy gives stronger information. The Lie subalgebras of have dimensions zero, one or three: identifying the bracket with the cross product, a two-plane is not closed under brackets. Suppose . Since an orbit has dimension at most three, its stabilizer would have dimension at least two, and hence exactly three. Its connected isotropy acts as all of on the tangent space. But the orbit tangent space is invariant under that action and would have dimension , impossible because the standard rotational representation is irreducible. Therefore dimension five cannot occur.
If the dimension is six, each orbit has dimension three and each stabilizer dimension three. All orbits are open, so connectedness makes the action transitive. At each point the isotropy is transitive on tangent two-planes, forcing sectional curvature to be independent of the plane; transitivity makes its value independent of the point. Thus a maximally symmetric connected three-manifold has constant sectional curvature. Homogeneous Riemannian manifolds are complete, so its simply connected cover is the round sphere, Euclidean space or hyperbolic space after normalization. Conversely these three simply connected metrics attain dimension six:Here preserves the chosen sheet of the hyperboloid; it includes orientation-reversing hyperbolic isometries. Constant curvature alone does not imply that a quotient retains six-dimensional global symmetry.
Dimension four likewise forces transitivity: stabilizer dimension three would give a one-dimensional invariant orbit tangent space, again impossible for full rotational isotropy. The stabilizer thus has dimension one and the orbits are three-dimensional. Typical four-dimensional isometry groups occur for and , where the factor groups have dimensions three and one. The usual Nil metric has three translation symmetries and one rotational isotropy symmetry. In centred Heisenberg coordinates it iswhich exhibits the horizontal rotations. Its distinct vertical and horizontal Ricci eigenvalues bound continuous isotropy by that rotation group, so its full isometry dimension is four. Standard metrics on and nonround Berger spheres also have dimension four. The standard Sol geometry has a three-dimensional translation group and only finite isotropy, giving dimension three. These examples show how a preferred vertical direction reduces rotational symmetry.
The isometry dimensions in dimension three are precisely , all realizable. A compact hyperbolic three-manifold gives zero; the product of a circle and a closed hyperbolic surface gives one. For dimension two take withIts scalar curvature is the constant , but its Ricci eigenvalues are the distinct constants in the two horizontal directions and the vertical direction. Hence connected isotropy is trivial and possible isometries preserve these line distributions. A local isometry preserving them has and would require constant dilations and of the two circle coordinates. A global circle diffeomorphism with such a constant dilation has degree , forcing . The identity component is therefore exactly the two circle translations. A flat three-torus gives dimension three; gives four; a simply connected constant-curvature model gives six. This construction distinguishes local homogeneity from the smaller global isometry group.
For quotients, let a complete simply connected Riemannian manifold cover . Every isometry of lifts to an isometry of which normalizes the deck transformation group. Conversely, every normalizing isometry descends. Two lifts differ by a deck transformation, giving the isometry group of a Riemannian quotient formulaThus one should not replace the isometry group of a quotient by that of its model geometry. For a flat torus, for example,The second factor is finite, whereas translations give the three-dimensional identity component.
There is a useful curvature test for compact manifolds. Contracting and differentiating the Killing equation gives , with the nonnegative rough-Laplacian convention. Integration yields the integrated Bochner identity for Killing fieldsIf Ricci curvature is negative definite, no nonzero Killing field exists. The compact isometry group then has dimension zero and is finite. This explains why closed hyperbolic manifolds have finite global symmetry, despite the six-dimensional isometry group of their universal cover. Compactness is essential to this conclusion.
For a compact flat Riemannian manifold, the same identity makes every Killing field parallel. Parallel fields are exactly the vectors fixed by linear holonomy. ThereforeApplied to the six flat types of Question 1, this gives dimensions three for the torus, one for each of the four nontrivial cyclic types, and zero for the Hantzsche-Wendt manifold. These are identity-component dimensions; disconnected finite symmetries can still be present. Local curvature symmetry, global deck-group constraints, and completeness thus play different roles in determining the isometry group.
Past exam of the mathematics course of the University of Cambridge 2004 iii Paper 18 1 Solution Created 2026-10-03 Updated 2026-10-14
Let be a flat torus, where is a full-rank Euclidean lattice, and use the nonnegative Laplace-Beltrami operator . The dual lattice isFor , the function is periodic under , so it defines a function on , and direct differentiation gives . Integrating characters on a fundamental parallelepiped proves that these functions are orthonormal. After choosing a lattice basis, the ordinary Fourier series theorem on proves completeness, so there are no additional eigenvalues. Thus the spectrum of a flat torus iswith multiplicity equal to the number of dual vectors of the given length; the zero eigenvalue occurs once.
It remains to show that a two-dimensional Euclidean lattice is determined up to an orthogonal transformation by its vector-length multiset. This is the two-dimensional lattice reconstruction from vector lengths. Let be a shortest nonzero vector, of length . It is primitive: with integer would give a shorter nonzero vector. The lattice vectors on its line are therefore precisely the integer multiples of . Subtract two occurrences of each length , , from the length multiset. This removes exactly one primitive lattice line, even when other vectors happen to have the same lengths. The smallest remaining length is the length of a shortest vector independent of .
These two vectors form a basis of the lattice. Otherwise a nonzero coset of has a lattice representative with . It cannot have , since primitivity of would then force . It is therefore independent of , whileThe inequality is strict: if both coefficients are nonzero, the vectors are not parallel and the triangle inequality is strict; if , then . This contradicts the choice of .
The covolume is also determined by the length multiset. If counts lattice vectors in a radius- disk, thenTo justify the estimate, tile the plane by a bounded fundamental parallelogram. Only cells meeting a fixed-width neighbourhood of the disk boundary contribute an error; their total number is . The lattice basis has Gram matrixThus is determined. Replacing by makes , so the Gram matrix is determined entirely. Equal Gram matrices give an orthogonal map between the lattice bases and hence between the lattices.
Two isospectral two-dimensional flat tori consequently have orthogonally equivalent dual lattices. Taking duals shows that their original lattices are orthogonally equivalent too, and the orthogonal map descends to an isometry of the quotient tori. Two isospectral flat tori of dimension two are isometric.
Past exam of the mathematics course of the University of Cambridge 2005 iii Paper 20 1 Solution Created 2026-10-03 Updated 2026-10-14
The central problem of spectral geometry is how much of a Riemannian metric is determined by the eigenvalues, with multiplicities, of its Laplace-Beltrami operator. For a membrane with fixed boundary, the squared vibration frequencies are proportional to the Dirichlet eigenvalues. Kac's 1966 paper brought this inverse problem into focus: the full sequence of frequencies might encode a domain's shape, but recovering a few geometric quantities is much weaker than recovering an isometry class.
The first major line of development extracts geometric information from spectral sums. On a smooth closed -dimensional manifold, the heat trace has the short-time expansionwhere are the nonnegative eigenvalues of and is scalar curvature. The coefficients are heat invariants, formed by integrating local curvature expressions. They recover dimension, Riemannian volume and total scalar curvature. For smooth planar domains with the Dirichlet boundary condition,Thus area and boundary length are audible. The constant term also contains topological information through Gauss-Bonnet theorem, with corner corrections when the boundary is polygonal. The Weyl law is the corresponding leading eigenvalue-counting result. Neither the leading term nor the whole list of local heat coefficients automatically reconstructs global geometry.
In the 1970s, the wave trace introduced a complementary link with global closed geodesics. The distributionis spectral, while its singularities are governed by periodic geodesic motion. The 1975 Duistermaat-Guillemin work made this relationship precise. Its trace formula places singularities at geodesic lengths and computes their coefficients under suitable clean or nondegenerate hypotheses. When a coefficient is nonzero, the corresponding length is audible. Possible cancellation must be addressed before identifying the entire length spectrum with the singular support. This approach led to spectral rigidity questions, rather than just the recovery of integral curvature quantities.
The second main line constructs counterexamples to metric determination. A precedent was the 1964 flat-torus example: different sixteen-dimensional Euclidean lattices could give the same Laplacian eigenvalues. Subsequent lattice constructions lowered the dimension; Conway and Sloane's 1992 work exhibited four-dimensional Euclidean lattices with matching lattice theta series. For a flat torus , the spectrum is the multiset for , so equality of lattice theta series of the dual lattices gives isospectrality even when the Euclidean lattices are not related by an orthogonal transformation.
A systematic geometric construction arrived with the Sunada theorem in 1985. For a common Riemannian cover with finite isometry group , almost conjugate subgroups have the same number of invariant vectors in each Laplacian eigenspace. Their freely acting quotients are therefore isospectral manifolds. This reduces an analytic equality to a finite-group representation identity. It produces many examples on surfaces and in higher dimensions, but nonconjugacy of the subgroups must be supplemented by an argument excluding extra quotient isometries. Transplantation theorem provides an explicit version: piecewise eigenfunctions are recombined by a fixed invertible linear transformation that respects the gluing and boundary conditions.
The planar problem itself was answered negatively by Gordon, Webb and Wolpert in 1992: noncongruent simply connected polygonal plane domains have equal Dirichlet spectra. This is stronger than counterexamples among abstract manifolds or flat tori, since these are genuine two-dimensional planar membranes. It does not settle every restricted version, such as asking for two domains within a specified smooth or convex class.
By the 1990s, isospectrality was also known to coexist with continuous metric variation and changes in local geometry. For example, a 1997 construction gives continuous isospectral families on products of spheres and tori whose members need not be locally isometric; in some examples the maximum scalar curvature changes. Conversely, rigidity and compactness results show that an isospectral class is not unrestricted. The 1988 surface compactness theorem makes fixed-spectrum sets of closed orientable surface metrics compact modulo diffeomorphisms. Compactness does not mean there is only one metric, or even finitely many metrics. By 2005, the subject therefore combined local heat invariants, global geodesic information, group-theoretic constructions, explicit transplantation and geometric rigidity. The spectrum determines substantial geometry, while generally failing to determine the full metric.
Past exam of the mathematics course of the University of Cambridge 2005 iii Paper 20 2 Solution Created 2026-10-03 Updated 2026-10-14
For a smooth function and the Levi-Civita connection , the Riemannian Hessian is the covariant two-tensorThe connection rules show that this is linear over smooth functions in both vector-field arguments, so it is a tensor. Torsion-freeness gives its symmetry:Its metric trace defines the paper's Laplace-Beltrami operator, ; its eigenvalues on a closed manifold are nonpositive. We use when listing nonnegative eigenvalues or writing .
At , choose an orthonormal basis of the tangent space and unit-speed geodesics . Differentiating twice along a curve givesThe covariant acceleration vanishes on a geodesic. Taking the trace at proves the geodesic trace formula for the Laplace-Beltrami operator:The last equality is Taylor's formula along the geodesics, and expresses the Laplacian entirely through nearby function values.
For stabilization, let be the assumed isospectral nonisometric four-dimensional tori, and let . Take the same small flat torus as an extra factor. On a Riemannian product,If and are eigenfunctions with eigenvalues and , then has eigenvalue . Products of orthonormal eigenbases are complete in the product space, so this lists all eigenvalues with multiplicities, including coincidences between different sums. Consequently the two products are isospectral.
One must also prove that multiplying by a common factor has not accidentally made the two manifolds isometric. ChooseThe injectivity radii are positive because the factors are compact. A closed product geodesic of length less than projects to a closed geodesic of length less than in . Such a projected geodesic must be constant: otherwise its initial vector over one period would be a nonzero vector of norm below the injectivity radius mapping back to its starting point under the exponential map. Thus all sufficiently short closed geodesics lie in the directions. Conversely the coordinate circles, of length , pass through every point and span those directions.
It follows that the smooth distribution tangent to the small flat-torus factor is characterized intrinsically by the span of tangent vectors to closed geodesics of length below . A hypothetical product isometry would preserve this smooth distribution and its orthogonal complement. The latter has leaves ; the isometry and its inverse take whole leaves to whole leaves, thereby giving an isometry , a contradiction. This proves isospectral stabilization by a small flat torus:This argument works even without assuming that the original four-dimensional torus metrics are flat.
Past exam of the mathematics course of the University of Cambridge 2006 iii Paper 23 2 Solution Created 2026-10-03 Updated 2026-10-14
The given two-dimensional ternary subspace is the tetracode. Its nonzero words form four one-dimensional subspaces, represented byEvery nonzero word has exactly three nonzero coordinates; in particular every codeword has a zero coordinate. We use this zero coordinate to construct a length-preserving reflection, but then choose unequal coordinate weights to rule out a global isometry.
The parity congruence description of tetracode lattices is especially convenient. Define integer Euclidean latticesThese congruences define additive subgroups of containing , so they are discrete and have full rank. For all-even vectors the last conditions coincide. For all-odd vectors they require respectively coordinate sum zero or two modulo four.
For an explicit basis, setTheir columns reduce to modulo three and satisfy the stated parity congruences, so . Since , . Conversely, the modulo-three condition has index nine, while each allowed set of residues modulo four has sixteen elements out of : eight all-even and eight all-odd residues. The Chinese remainder theorem therefore gives . The inclusions have equal index and are equalities.
For any , choose the first coordinate for which , and reflect that coordinate:Such a coordinate always exists by the codeword description. Its reflection preserves and the common coordinate parity. If is odd, its sum changes by two modulo four, taking the plus condition to the minus condition. If is even, the sum changes by zero modulo four, and the two conditions already coincide. The set of coordinates divisible by three is unchanged, so applying the same rule again recovers . This proves a bijection preserving each absolute coordinate, the coordinate reflection bijection for tetracode lattices.
Now choose positive weights linearly independent over , for instanceThe Euclidean lattices and have the same multiset of vector lengths, because is preserved by . A closed geodesic of a flat torus is the projection of a straight line with displacement a lattice vector. Thus the two tori have identical length spectra, including multiplicities of lattice displacements. Primitive lengths can also be recovered by subtracting proper iterates successively, so they agree as well.
Suppose a torus isometry existed. Its lift to is affine with orthogonal linear part taking onto . Put , so . Since , its image is an integer vector satisfyingRational independence of the weights forces . Hence for every : the rationally independent axis weights obstruct a lattice isometry by forcing it to be a diagonal sign change.
Such a sign change would preserve modulo three. The code has a unique nonzero projective word supported on each triple of coordinates, so preserving each of its four support sets forces the three signs on each triple to agree. All four signs therefore agree, leaving only or . Both preserve rather than exchanging it with . These lattices are distinct: is odd and has sum zero, so is not in . This contradiction provesThe choice of weights is essential; setting all four equal would make the displayed bases orthogonal with equal lengths and would not prove nonisometry.
Past exam of the mathematics course of the University of Cambridge 2009 iii Paper 22 1 Solution Created 2026-10-03 Updated 2026-10-14
The musical isomorphisms are the pointwise maps induced by the Riemannian metric:In local coordinates they are and . The inverse metric makes these inverse bundle maps, hence inverse maps between smooth vector fields and smooth differential one-forms.
For a covariant two-tensor , its metric trace is in a Riemannian orthonormal frame. The contraction is independent of the chosen frame. There is a terminology distinction here: if is literally an alternating differential two-form, this trace is zero. The object needed below is the covariant two-tensor , generally not alternating. Interpreting that tensor as an alternating form would make the stated divergence construction impossible.
Let be the Levi-Civita connection. The Riemannian gradient, divergence and positive Laplace-Beltrami operator areThus , and in coordinatesThis choice makes the Laplacian eigenvalues nonnegative on a closed Riemannian manifold. The alternative convention reverses all eigenvalue signs, but none of the comparisons below.
For the Riemannian submersion, let and choose orthonormal horizontal lifts of a base frame and an orthonormal vertical frame . The horizontal part of projects to the base Levi-Civita connection; this follows directly from its Koszul formula applied to basic horizontal fields, because their inner products and projected brackets are pulled back from the base. Therefore the horizontal trace of the Riemannian Hessian of is the pulled-back base Hessian trace. Since , its vertical terms areWriting for the unnormalised fibre mean curvature vector givesThe totally geodesic submanifold condition on the fibres makes . Consequently eigenfunction pullback by a Riemannian submersion givesHere the Riemannian submersion is onto, so the map is injective. Its image is exactly the members of the upstairs eigenspace constant on each whole fibre: such a function descends smoothly by local sections of the submersion, and the displayed identity forces the descended function to have the same eigenvalue. In particular, whenever these eigenspaces are finite dimensional, the downstairs multiplicity is no larger than the upstairs multiplicity. The upstairs eigenspace may also contain functions varying along fibres.
A flat torus is with the metric induced by the Euclidean metric, where is a full-rank Euclidean lattice. Its dual lattice is . The characters descend to the quotient and form an orthogonal Fourier basis of on it. Direct differentiation gives the spectrum of a flat torus:with multiplicity equal to the number of dual lattice vectors of each length. In particular zero has multiplicity one.
Let the given four-dimensional isospectral manifolds be and . For , take their products with the same circles of length , equivalently the lattices . The dual lattices are , so their Laplacian eigenvalues areThe given equality of base spectra with multiplicities makes these product spectra identical.
Choose smaller than the shortest nonzero vector length in either original Euclidean lattice. By short-vector cancellation for flat tori, the shortest nonzero vectors in the product lattice are exactly in its new factor. To spell out the nonisometry argument, any isometry of the product flat tori lifts to an isometry of their universal Euclidean covers. This is affine, and its orthogonal linear part maps one product lattice onto the other. It must map the span of their shortest vectors, namely the new -dimensional factor, onto itself; orthogonality then preserves the original four-dimensional factor and maps onto . That would give an isometry of the original flat tori, contrary to the assumption. Isospectral nonisometric flat tori therefore exist in every dimension .
Past exam of the mathematics course of the University of Cambridge 2012 iii Paper 12 1 Solution Created 2026-10-03 Updated 2026-10-07
Use the positive Laplace-Beltrami operator, . A flat torus is the quotient , where is a full-rank Euclidean lattice and the metric descends from the Euclidean metric. Its dual lattice isIf is the covolume of , the functions descend to the torus. Direct differentiation gives the spectrum of a flat torus:The multiset includes one entry for each dual vector. In particular zero has multiplicity one. Writing identifies the dual vectors with , . Standard Fourier series on the unit cube then prove that these functions are an orthonormal basis of on the torus. If , its Fourier coefficients satisfy . Thus every eigenfunction is a linear combination of the indicated shell of frequencies, with no additional eigenvalues. Compactness gives the discrete self-adjoint realization, whose eigenfunctions are smooth. For real functions the two characters at become cosine and sine, giving the same real multiplicities. The opposite Laplacian sign would negate all displayed eigenvalues without changing the rigidity conclusion.
To prove two-dimensional spectral rigidity, let and recover its Gram matrix from its vector-length multiset. Let be the shortest nonzero length and choose with . This vector is primitive: a proper integer multiple would have a shorter lattice vector. Hence . From the spectral multiset subtract exactly two occurrences of each length , . The remaining multiset consists precisely of the vectors not on this line, including correct multiplicities at coincident lengths. Its least length is the length of a shortest vector independent of .
The pair is a basis of . Otherwise take a lattice point in a nonzero coset of and reduce its two coefficients into . The resulting nonzero lattice vector is not on , by primitivity of . Butwith strict inequality: when both coefficients are nonzero the triangle inequality is strict for independent vectors, and when one vanishes its length is at most . This contradicts the definition of . Subtract an integer multiple of from and, if necessary, change its sign to arrange ; minimality of makes this reduction possible without decreasing its length.
The covolume of is also spectral. The count of lattice vectors with length at most obeysFor example, translate a bounded fundamental parallelogram of diameter bound : the union of tiles with centres in contains and lies inside . Comparing their areas proves the leading coefficient. The spectrum supplies this count. Since is a lattice basis, its parallelogram area satisfies . ThereforeAll three quantities are determined by the spectrum. Equal Gram matrices give an orthogonal map between the dual lattices; taking duals gives an orthogonal map between the original lattices, descending to a Riemannian isometry of the tori. Isospectral flat two-tori are isometric. This is the two-dimensional lattice reconstruction from vector lengths; it uses lengths with multiplicities, not merely the set of distinct lengths.
Past exam of the mathematics course of the University of Cambridge 2012 iii Paper 20 2 ii Solution Created 2026-10-03 Updated 2026-10-07
Take the separate oscillator energiesThey depend on disjoint canonical coordinate pairs, soThus they are first integrals in involution. Their differentialsare nonzero precisely when and have disjoint supports. Hence they are functionally independent on the specified open set.
To understand all further global integrals of motion, use the actionsDirect differentiation gives , so these are action-angle variables withWhen is irrational, every orbit is dense in its fixed-action two-dimensional flat torus. One elementary proof samples the orbit at times : the first angle returns, and the second undergoes an irrational rotation. These samples are dense in the second circle. Allowing a fixed additional time supplies any desired first angle, proving density in the whole flat torus.
A continuous, globally defined integral of motion must have the same value on an orbit and its closure. Therefore it is constant on every fixed-action flat torus. A such integral has the form , with , by evaluating at a fixed choice of angles. HenceIt cannot be independent of . This proves there are at most two globally independent differentiable integrals. If a third global integral were independent at a point on an excluded coordinate plane, independence would persist in a neighbourhood and hence at nearby points of the regular open set, again a contradiction.
The word global matters: on an angular chart, is a local integral of motion independent of the actions. It is not single-valued on the flat torus when the frequency ratio is irrational. Thus the assertion would be false for arbitrary local, rather than global, integrals.
Past exam of the mathematics course of the University of Cambridge 2012 iii Paper 20 4 ii Solution Created 2026-10-03 Updated 2026-10-07
Fix a smaller angular strip and a momentum ball ; the canonical transformation will be defined there and map into the original domain for small . This domain restriction is natural for a near-identity change of coordinates with a nonzero momentum shift.
First solve the scalar cohomological equation on a Diophantine torusThe preceding result gives a real-analytic periodic on every smaller strip. We need a near-identity angular diffeomorphism and a constant vector such thatHere is a complete analytic construction of this translated conjugacy of a Diophantine vector field.
Write and, for a current approximation, putAssume is close to the identity. For a correction , the linearized defect isThe identity follows by differentiating the definition of , so the term already contains the current error. ChooseThe matrix average is invertible near the identity, and the right-hand side has zero mean. The preceding torus small-divisor estimate therefore solves this vector equation componentwise. Update , . The exact new error isIt is quadratic in the current defect.
For convergence, reserve a fixed outer strip on which is analytic and work on shrinking inner strips whose total loss is less than half the starting width. Let and choose loss parameters with small enough that the fixed multiples of these losses used at each stage fit in the reserved total width. The torus small-divisor estimate and Cauchy estimates give, with constants uniform while stays close to the identity and remains in the reserved strip,andIndeed, the first remainder is bounded using , and the second using the uniformly bounded second derivatives of on the reserved strip. Taking a slightly larger exponent covers all these losses. Start with and , so . If a small constant satisfies and , induction gives . The estimates on and are summable. Choosing still smaller keeps the derivatives close to the identity and the images inside the reserved strip, closing the induction. The real periodic analytic limits solve the conjugacy equation. Since , the very first constant correction is zero; the first new defect is , and the remaining summed constant corrections are . Also the summed change of is . This proves the claimed estimates as well as existence, without leaving an unsolved linear remainder.
Now set and define the symplectic cotangent lift with a closed momentum shiftThe cotangent lift of a diffeomorphism is symplectic. Translation by is also symplectic because the one-form is closed. The constant term need not be exact on the flat torus; closedness is sufficient. Thus their composition is the required near-identity canonical transformation.
Expanding the quadratic kinetic term gives a transformed linear coefficient , which is exactly by the conjugacy equation. The constant-in- part issince . Consequently defineand, for ,In the exact identities, squared Euclidean norms mean the real polynomial , continued bilinearly on complex strips, without complex conjugation. The latter is uniformly bounded and analytic on the retained strip because . At use the identity map, , and . We have the exact requested normal formHere , is genuinely homogeneous quadratic in , and its coefficient matrix differs from the identity by . Thus . The remainder depends only on the angular coordinate. If the printed final uses the old angle , the same term is evaluated at ; this is just the corresponding coordinate convention.
Past exam of the mathematics course of the University of Cambridge 2012 iii Paper 20 4 i Solution Created 2026-10-03 Updated 2026-10-07
Take and . A Diophantine frequency vector satisfiesChanging the norm only changes the admissible . The bound excludes resonances and quantifies the small divisors in the inverse of .
A periodic solution necessarily has zero-mean right-hand side, because the integral of each angular derivative vanishes. Conversely, if , expand its Fourier series:The Diophantine condition makes every denominator nonzero. These are all differentiable periodic solutions: any difference satisfies and has no nonzero Fourier coefficient. Thus a solution exists exactly for zero-mean , and is unique up to a constant; setting fixes it. The analytic conclusion is on every strictly smaller complex strip, not necessarily the original boundary strip.
Shifting each contour in the Fourier coefficient integral toward the appropriate edge of the complex strip givesHence for ,This converges normally, including differentiated series on any still smaller strip, and proves both holomorphic extension and . Real-valuedness follows from and the corresponding identity for .
An explicit bound is obtained by counting lattice shells. The number of with is at mostPut and . For , ; for use the bound . Since , this givesUse in this constant when . The estimate is valid for every positive ; optimality of the exponent is not needed. For a unit-period flat torus, insert the factors in both the Fourier exponent and denominator and adjust the constant.
Past exam of the mathematics course of the University of Cambridge 2013 iii Paper 16 2 Solution Created 2026-10-03 Updated 2026-10-07
Construct the Euclidean lattices in an orthogonal coordinate system whose squared axis lengths areThus an integer coordinate vector has squared norm . These four numbers are linearly independent over : the four sign changes of in their field extension isolate the four coefficients of a rational relation. The independence will distinguish the two flat tori; equality of their length spectra will hold for every positive choice of axis weights.
LetHere and below these integer coordinates are interpreted in the weighted orthogonal axes. Direct multiplication gives , , andIn particular both are full-rank Euclidean lattices. Denote their four generator columns by . Modulo the two matrices have the same columns, and the nonzero words of the given ternary tetracode are exactlyThese are the reductions of , respectively. The first and fourth displayed words are the specified generators, and the middle two arise from their sum and difference, up to sign. Thus reduction modulo maps either Euclidean lattice onto precisely .
To construct a length-preserving correspondence, find the kernels of this reduction. The inverse matrix is . Hence exactly when . These congruences are equivalent toFor clarity, the first congruence is ; subtracting the others gives evenness of each required pair sum. For the matrix the first congruence is ; the remaining congruences again give equality of the four parities. Substitution proves the converse in each case.
Let be the orthogonal reflection changing only coordinate . If the coordinates of are even, this changes their sum by a multiple of ; if all are odd, it changes the sum by modulo . ThereforeEach Euclidean lattice is the disjoint union of its kernel and the eight cosets , since has nine words. Define a map by using on and on each of and . Each piece maps bijectively onto the corresponding coset. Thus is a bijection satisfyingIt also commutes with . The closed geodesics of a flat torus have lengths for nonzero translation vectors in its Euclidean lattice. Consequently the two length spectra agree, including the multiplicities of translation classes and with either consistent orientation convention. If only primitive closed geodesics are counted, the same conclusion follows by removing iterates in increasing order of length: the total multiplicity at is the sum of primitive multiplicities at , and there are only finitely many such contributions below any fixed length. The correspondence need not itself preserve primitiveness to give this conclusion.
It remains to prove the two flat tori are not isometric. An isometry of flat tori lifts to an affine Euclidean isometry; its linear part of an affine map must carry onto . In integer coordinates write this part as . For every , both and have integer coordinates and have equal squared norms. Rational independence of the impliesApply this to for all . Two squared linear forms agreeing on all integer points agree as polynomials; their difference factors, so one form is the other or its negative. Since is invertible, is consequently a diagonal sign change .
Modulo , such a must preserve . Each nonzero word of has exactly three nonzero coordinates, and for each choice of zero coordinate there are just two such words, differing by an overall sign. Preserving these words forces the three signs on each support to agree. The four overlapping supports then force . Thus only and are possible. Neither changes .
Finally : it has the same ternary word as , but their difference is , which fails the congruence defining . Hence and no such linear isometry exists. This completes the tetracode length-isospectral lattice construction.
Past exam of the mathematics course of the University of Cambridge 2016 iii Paper 117 2 Solution Created 2026-10-03 Updated 2026-10-06
A flat torus is the quotient , where is a full-rank Euclidean lattice, with the Riemannian metric induced from the Euclidean metric. Translation by a lattice vector is a Riemannian isometry, so this metric is well defined and has zero curvature. Its volume is the covolume of .
The dual lattice is . For , the function descends to the flat torus, and direct differentiation givesConversely, write and . Ordinary Fourier series on have frequencies , corresponding to . After normalization they form a complete orthonormal basis in , and the Fourier coefficients of a smooth function decay rapidly. Applying the Laplace-Beltrami operator term by term shows that a smooth eigenfunction with eigenvalue has nonzero coefficients only where . Thus the spectrum of a flat torus isThe formula counts multiplicity over either the real or complex numbers: the two frequencies give the real sine and cosine functions.
For the rigidity assertion in dimension two, it suffices to reconstruct a rank-two Euclidean lattice from its vector-length multiset. The multiset first determines its covolume . Indeed, a bounded fundamental parallelogram, and comparison of the cells meeting a disk with slightly larger and smaller disks, give the lattice-point asymptotics by fundamental cellsThe spectrum determines , including multiplicities, so it determines .
Let be the smallest nonzero vector length, and choose with . This vector is primitive: with would contradict minimality. From the length multiset subtract the two vectors at every positive length , for . The shortest remaining length is exactly the length of a shortest vector . This subtraction is legitimate even if several directions have length : it subtracts just two copies, and then . It also proves that is independent of our choice of shortest direction. Replace by to arrangeThe replacement stays outside and cannot be shorter than , so a reduced choice still has length .
The shortest-independent-vector basis lemma says that are a basis of . To prove it, choose coordinates with . Because is primitive, ; let be the smallest positive vertical coordinate in . Write , with after changing its sign. If , a vector of vertical coordinate , reduced horizontally modulo , hasThis contradicts the definition of , since . Hence and generate .
The Gram matrix of this basis is , with , and its determinant is . ThereforeChanging the sign of one basis vector allows . The numbers , all audible from the spectrum, consequently determine the Gram matrix, and hence , up to an orthogonal transformation. Taking dual lattices gives the same conclusion for . Isospectral flat two-dimensional tori are isometric. The argument uses the special basis property of rank-two Euclidean lattices, so it does not claim higher-dimensional rigidity.
Periodic domain Created 2026-09-24 Updated 2026-10-14
A periodic domain identifies opposite faces of a fundamental cell, producing a flat torus. On an interval , smooth periodic boundary conditions require and matching spatial derivatives. These endpoints are one location, rather than two independently controlled walls. Periodic integration by parts has no boundary term, and zero-mean functions have no zero Fourier mode.
Sixth-turn flat three-manifold 2026-10-14
The mapping torus of an order-six rotation of a hexagonal flat torus is a closed orientable flat Riemannian manifold with holonomy . The screw rotation has angle and axial displacement . Its first Betti number is one.