Past exam of the mathematics course of the University of Cambridge 2011 ii Paper 1 25I Solution Created 2026-09-24 Updated 2026-09-24
Past exam of the mathematics course of the University of Cambridge 2011 ii Paper 3 30A a Solution Created 2026-09-24 Updated 2026-09-24
Past exam of the mathematics course of the University of Cambridge 2012 iii Paper 19 4 Solution Created 2026-10-03 Updated 2026-10-07
The multiplicative norm first rules out zero divisors: if and , then forces . If the dimension is zero the claimed bound already holds, so suppose it is positive.
It is worth checking that the printed description of a ring with an underlying real vector space really permits differentiation of multiplication. This follows from the norm identity even if compatibility with real scalars is not separately stated. For fixed , the map is additive by distributivity, andIt is therefore continuous. Additivity gives rational linearity. Approximating a real number by rationals and using continuity proves for every real . The same argument holds in the other variable. This is automatic real bilinearity from a multiplicative norm, so multiplication is a bilinear map between finite-dimensional vector spaces, and hence is smooth.
Let be the unit sphere of the given positive-definite bilinear form. It is a copy of . The squaring mapis well defined because , and is smooth. Commutativity gives its derivativeFor and , the norm identity gives , so the derivative is injective, including on the tangent space . It takes tangent vectors into since maps the sphere to itself. Alternatively, polarization of gives , so directly.
The only identifications under squaring are antipodal:The factorization uses commutativity, and the last implication uses the absence of zero divisors. Conversely . It follows that induces an injective smooth mapThe antipodal quotient map is a local diffeomorphism, so the injectivity of implies injectivity of . Since Real projective space is compact, the stated compact injective immersion is an embedding theorem makes a closed embedding.
For its source and target have the same dimension . The inverse function theorem makes its image open as well as closed in the connected sphere . It is nonempty, so the image is the whole sphere; consequently is a diffeomorphism. But for this is impossible:Indeed, the antipodal two-sheeted covering map has simply connected total space in this range, and is its universal cover, with deck group . A sphere of dimension at least two is simply connected. These fundamental groups cannot be those of homeomorphic spaces.
We have obtained the dimension bound for a commutative Euclidean normed algebra:The real numbers and complex numbers with their usual products and Euclidean norms attain dimensions one and two. In dimension two, the induced map is entirely compatible with the argument: the circle squaring map identifies exactly the two antipodal preimages. We neither assume a multiplicative identity nor use a classification of real division algebras; the squaring argument proves the bound from the supplied hypotheses.
Past exam of the mathematics course of the University of Cambridge 2012 iii Paper 20 1 iii Solution Created 2026-10-03 Updated 2026-10-07
Under the given rank hypothesis, the necessary and sufficient local condition isVanishing only at the single point would not suffice: an extension preserves the coordinate Poisson brackets throughout its domain. Necessity follows from the coordinate criterion just proved.
For sufficiency, put and . Because the symplectic form is a nondegenerate bilinear form at every point, the independent differentials give independent fields . They commute, sinceMoreover , so they span the tangent spaces of the -dimensional fibres of the submersion . Choose a local section transverse to those fibres. Its joint local flow defines coordinates byThe inverse function theorem makes a local diffeomorphism, and commuting flows give .
In these coordinates . Each flow preserves the symplectic form, by Cartan's magic formula. Thus its coefficients are independent of , andwhere is a closed differential two-form on the base. On a small ball, the Poincare lemma gives , with . DefineThenConsequently the original coordinates are extended to the required canonical transformation , with the prescribed first half . Its Jacobian matrix is invertible by construction. This is the canonical completion of independent commuting functions. For the printed data, the flows and resulting canonical transformation can be taken , which is sufficient to preserve the symplectic form; smooth data give smooth coordinates.
Past exam of the mathematics course of the University of Cambridge 2012 iii Paper 20 3 ii Solution Created 2026-10-03 Updated 2026-10-07
The Arnold-Liouville theorem concerns a connected compact component of a regular common level of commuting independent smooth functions on a -dimensional symplectic manifold. It states that is a Lagrangian torus, and that a neighbourhood of has action-angle variablesin which every depends only on . If or, more generally, , its Hamiltonian flow satisfiesAngles have period here. The theorem is local near this regular compact component; it does not assert globally defined action-angle variables across singular levels or over an entire base with monodromy.
First, has dimension of a manifold by the submersion theorem. The Hamiltonian vector fields are independent, tangent to , and commute. They span . Moreoverso is a Lagrangian submanifold. Compactness makes these restricted vector fields complete. Their joint flow is an -action on . Its orbits are open because the fields span , so connectedness gives one orbit. The stabilizer of a point is discrete by the inverse function theorem, andCompactness forces to be a full-rank Euclidean lattice: otherwise an unbounded linear coordinate transverse to its span would descend to the quotient. Thus is an -torus.
Next choose a small ball of regular values near and a neighbourhood of this component which is a product family of compact tori . This local trivialization follows directly by choosing transverse vector fields with and lifting short radial paths in the base; compactness of gives a uniform neighbourhood where their flows exist. Hence deformation retracts onto . The commuting joint flows on nearby fibres have smoothly varying full-rank Hamiltonian period lattices. A basis of their periods can be chosen smoothly on this small ball: continue the return equations from a basis at the central fibre, using their nonsingular vertical flow derivatives and the implicit function theorem. No global choice of lattice basis is needed.
Because and retracts onto , its closed symplectic form is exact on . Choose a one-form with . Let be the smoothly continued cycles represented by the period basis and define the action integralsFor a transverse variation of the fibre, differentiating a cycle integral and using Cartan's magic formula eliminates the integral of the exact term. With the cycle parametrized by the joint-flow time , this givesThusThe period matrix is nonsingular, so gives coordinates on the base. The corresponding Hamiltonian vector fields satisfyTheir time- flows return along the basis cycles. They commute because each is a function of the commuting . Consequently they give a free torus action on .
Choose a section over the action-coordinate ball and use this torus action to define angles . Since , the symplectic form takes the formThe closed base two-form is exact on the ball: write , , by the Poincare lemma. Changing the angular origins by removes this term, since . This completes the construction of action-angle variables. The functions are constant on the fibres, so are functions of alone, and Hamilton's equations give the claimed straight-line motion. This proves both the topological and symplectic parts of the Arnold-Liouville theorem.
Past exam of the mathematics course of the University of Cambridge 2012 iii Paper 55 2 Solution Created 2026-10-03 Updated 2026-10-07
The topological degree measures an oriented net number of sheets of a map. Its simplest integer-valued setting is a map between connected oriented closed manifolds of the same positive dimension . The hypotheses matter: an integer sign count needs orientations; compactness prevents preimages from escaping; and the same dimension makes a regular fiber discrete. A fundamental class gives the definition for continuous maps:Here and are generated by their chosen fundamental classes. Reversing either orientation changes the sign of the degree of a map between oriented manifolds. Disconnected sources contribute a sum over components; disconnected targets require a degree for each target component rather than one universal integer.
For a smooth map, choose a regular value . The Sard theorem guarantees plentiful choices. Each point of has an invertible derivative, so the inverse function theorem makes it an isolated point. The fiber is finite by compactness of . An oriented coordinate chart defines , and the degree as a sum of local degrees becomesAn empty fiber contributes zero. The sign is independent of the chosen oriented charts. Thus the mapping degree counts sheets with signs, not simply the number of preimages: orientation-preserving and orientation-reversing sheets cancel.
A differential-form argument both proves that this count is independent of and relates it to the topological definition. Around a regular value, take a small neighborhood whose full preimage is a disjoint union of neighborhoods mapped diffeomorphically onto . Compactness rules out additional preimages approaching from elsewhere. Choose a smooth top-degree form supported in , with . The change of variables formula givesAlthough need not be everywhere positive, it represents the same normalized top cohomology class as any normalized volume form. The top de Rham cohomology of a compact connected oriented manifold says that two top forms with equal integrals differ by . Their pullback integrals agree, because the Generalized Stokes theorem gives . Therefore the degree by integration of a pullback volume form isfor every smooth top form . Pairing de Rham cohomology with fundamental classes identifies this integer with the homological definition. This proof also shows that every regular value gives the same signed sum.
The homotopy invariance of mapping degree makes it stable under continuous deformation. Homologically this follows from homotopic maps inducing the same map on homology. For a smooth homotopy , a direct proof uses and the Generalized Stokes theorem on the cylinder to getwith the cylinder orientation chosen to give the indicated boundary signs. Smooth approximation extends the geometric description to continuous maps, while the fundamental class definition already applies without differentiability. The multiplicativity of mapping degree follows from functoriality and givesA constant map has degree zero when . If the mapping degree is nonzero, the map is onto: a point outside its image is a regular value with empty preimage. An orientation-preserving diffeomorphism has degree , and an orientation-reversing one has degree . An orientation-preserving -sheeted covering map has degree ; the signs of sheets must be included for other orientation conventions.
Examples show both the strength and the limits of the invariant. On the circle, has degree , including negative integers and zero; it is the winding number. On the sphere , an ambient reflection has degree , and the antipodal map has degree , because the determinant of on the ambient is . On the torus, an integer matrix induces a map of degree , obtained by pulling back the constant top form. However, degree does not classify general manifold maps: the identity and an integer shear on the two-torus both have degree one but induce different maps on , so are not homotopic.
The mapping degree gives useful obstruction and existence arguments. There is no retraction of a closed ball onto its boundary sphere: such a retraction would make the degree-one identity of the boundary null-homotopic, since it would extend across a contractible ball, contradicting homotopy invariance and the degree-zero constant map. This is the degree obstruction underlying the Brouwer fixed-point theorem. Also an everywhere nonzero tangent vector field on an even-dimensional sphere would, after normalization, produce the homotopy from the identity to the antipodal map. Orthogonality makes every value a unit vector, but the endpoint degrees are and , impossible. This proves the corresponding Hairy ball theorem obstruction.
There are appropriate extensions rather than an unrestricted integer degree for every pair of manifolds. For a proper map between connected oriented manifolds without boundary of equal dimension, the regular-value count remains finite and is invariant; compactly supported top forms or locally finite homology replace the compact fundamental-class description. Without orientations, degree modulo two counts preimages modulo two and uses mod-two fundamental classes. For manifolds with boundary one uses relative mapping degree for maps of pairs, keeping the boundary away from the chosen target value. Maps of unequal dimensions, or nonproper maps where preimages can escape to infinity, do not in general admit this same signed sheet-count invariant. The central principle is an integer signed count stable under homotopy, with its hypotheses explicitly preserved.
Past exam of the mathematics course of the University of Cambridge 2014 iii Paper 52 1 a Solution Created 2026-10-03 Updated 2026-10-06
Use geometrized units and metric signature , with . In the exterior of Schwarzschild spacetime, put . The Schwarzschild tortoise coordinate satisfiesThe retarded and advanced null coordinates and then give . The logarithmic divergence of suggests exponentiating these null coordinates. In the right exterior define the Kruskal–Szekeres coordinatesTheir product eliminates :Since and , the Schwarzschild metric becomesHere is an implicitly defined function of . The derivative of the right side with respect to is , which is nonzero at . The inverse function theorem therefore makes smooth across that surface, and the coefficient of tends to . Thus the Schwarzschild event horizon is a coordinate singularity of the original chart, while this Lorentzian metric remains regular there.
Extend the Kruskal–Szekeres coordinates to all real with . The signs give two exterior regions, and , a future black hole region , and a past white hole region . The event horizons are or , intersecting at the bifurcation surface. The boundary has and is a genuine Schwarzschild singularity, as the Kretschmann scalar diverges there.
Finally, and give and a radial metric proportional to . Hence radial null geodesics have slopes , the event horizons are , and the singular boundaries are . This constructs the maximal Kruskal extension; a black hole produced by collapse need not contain the second exterior or the white hole of that eternal extension.
Past exam of the mathematics course of the University of Cambridge 2015 iii Paper 15 1 Solution Created 2026-10-03 Updated 2026-10-06
The degree and its local signs. Orient and let generate its top reduced homology . The mapping degree is the integer characterized byFor this is the usual top homology definition using the fundamental class; using reduced homology also covers .
Suppose is smooth and is a regular value. Each has invertible tangent map , so the inverse function theorem makes discrete. It is also closed in the compact sphere, hence finite. Choose disjoint small neighborhoods of these points on which is a local diffeomorphism. The Excision theorem identifies the source local homology with the direct sum of one copy of for each inverse image. The induced local map is multiplication by or according as preserves or reverses orientation. The map from the global fundamental class to these local orientation classes therefore proves the degree as a sum of local degrees formula:Here the determinant is computed in positively oriented tangent bases. Thus the mapping degree counts inverse images with signs, rather than just their cardinality.
The quotient map. Put and write . Give its standard orientation and its boundary the induced orientation. The connecting homomorphismis an isomorphism: the disk has zero positive reduced homology. It sends the relative fundamental class to . If the map on relative homology induced by multiplies this class by , naturality givesThus . Collapsing the boundary gives a sphere , and the quotient map identifies its top reduced homology with the top relative homology of the disk pair. Give this quotient sphere the orientation determined by that identification. The relation now showsThis is the quotient-sphere degree identity.
The graph intersection. Orient by the product orientation, orient by its first factor, and orient the graph of a function by . An intersection is precisely a zero of , and no such zero lies on the boundary because . At an intersection, transverse intersection means that is surjective, hence invertible. The zeros are consequently isolated and finite.
For the smooth intersection number use the ordered tangent spaces first and second. Relative to the product basis their concatenated basis has matrixIts determinant is , so the intersection sign is . By the same local Excision theorem argument, now in relative homology at the interior point , the sum of these signs equals the multiplier of on . Therefore the graph intersection formula for mapping degree isIf the tangent spaces are ordered first and second, every sign changes by ; the order above specifies the appropriate convention.
Three transverse graph intersections with signs plus, minus, plus and total degree one
. The one-dimensional model on illustrates the graph intersection formula for mapping degree: its three zeros have signs , and its endpoint map has mapping degree on reduced homology.
Past exam of the mathematics course of the University of Cambridge 2015 iii Paper 17 4 Solution Created 2026-10-03 Updated 2026-10-06
Write for the Levi-Civita connection and define its Christoffel symbols by . For the Riemannian metric matrix and its inverse ,A curve with affine parameter is a geodesic when , equivalentlyThe connection is metric compatible and torsion free; in particular . The coordinate velocities are , and the displayed coefficients are smooth functions of position.
For whose initial-value geodesic exists through time , with and , define . The domain is open and contains a neighborhood of zero. Indeed the geodesic equation is a smooth first-order ODE in ; the constant solution with initial velocity zero exists through time , and existence on a compact time interval and smooth dependence persist for nearby initial data. Reparameterization and uniqueness give wherever defined. Consequently the differential of the exponential map at zero isThe inverse function theorem gives a neighborhood of zero on which is a diffeomorphism onto a neighborhood of . Shrink to a ball in and choose an orthonormal basis there. Its linear coordinate functions composed with are the geodesic normal coordinates on . This establishes well-defined local coordinates without assuming geodesic completeness.
In these geodesic normal coordinates, every radial geodesic has coordinate expression . Its equation at reads for every . The coefficients are symmetric in by the torsion-free connection property. Evaluating at the coordinate unit vectors and their pairwise sums, or using the polarization identity, therefore givesThis is the fact that Christoffel symbols vanish at the center of normal coordinates. Also , since is the identity in an orthonormal basis.
For sufficiently small , the geodesic sphere is . It is a smooth hypersurface because is a diffeomorphism on the ball. It equals the local distance sphere for sufficiently small radius, as the length comparison below shows; no assertion of global smoothness for large radii is needed.
The full Gauss lemma isfor in a star-shaped domain of and . To prove it, set , and , for near zero. Every -curve is a geodesic and has squared speed . The torsion-free connection property and the commuting parameter fields give ; metric compatibility and then giveAt , , so . Integration from to followed by setting proves the boxed identity, because and . For any in the domain, a small neighborhood of its compact radial segment suffices for the variation. The zero vector is covered directly.
In particular, if , the images of the radial and spherical tangent directions are orthogonal; taking also proves preservation of radial length. Thus in geodesic polar coordinates the Riemannian metric has radial part and no mixed radial-angular term. Every curve in a small normal ball from its center to radial coordinate has length at least , by integrating the absolute radial derivative; the radial geodesic realizes this length. To exclude shortcuts leaving the ball, choose a larger normal ball of radius with compact closure in and restrict to : an escaping curve first reaches radial coordinate and already has length at least . This proves the asserted small-radius distance interpretation and orthogonality of radial geodesics to geodesic spheres.
Radial geodesics and a geodesic sphere on the round unit sphere, with orthogonal radial and angular tangent vectors
. Past exam of the mathematics course of the University of Cambridge 2015 iii Paper 2 6 Solution 2026-10-06
A real Lie group is a finite-dimensional real smooth manifold with a group structure whose multiplication and inversion are smooth. Its tangent space at the identity element is the vector space of velocities of smooth curves through . For the general linear group it identifies with . The matrix exponential and local matrix logarithm areThe exponential is globally defined, and each matrix is invertible with inverse . The logarithm here is a local map near ; it is not a globally defined inverse on every real invertible matrix. The differential of at zero is the identity, and the two power series are inverse locally.
For a general , the assumed Exponential map of a Lie group with has a smooth local inverse by the inverse function theorem, giving a local exponential chart. Define the Lie bracket from local group commutators byFor small the group commutator lies in that chart. More explicitly, the map is smooth and has . Its mixed second differential at is a bilinear map of , proving bilinearity. Swapping the two group elements inverts the commutator, and near . Thus and . In the matrix group, expansion to the mixed term gives the familiar commutator .
For , let and define the Adjoint representation of a Lie group by . It is smooth and satisfies . Its derived representation is . Naturality of the Exponential map of a Lie group under conjugation givesAt , the derivative of the logarithm of the commutator is : the derivative of multiplication at adds tangent vectors, and . Differentiate in to concludeTo prove the Jacobi identity without assuming it in the bracket construction, first establish that the differential of a Lie group homomorphism preserves Lie brackets. For a homomorphism , the allowed exponential identity gives locally. Apply this identity to the group commutator and take the mixed derivative to obtain . In particular givesApplying both sides to yields . Rearranging with antisymmetry provesNeither injectivity of nor the existence of a global logarithm was used.
Finally let be a normal Lie subgroup of , with its corresponding Lie algebra . For each , conjugation restricts to , so its differential preserves : . Differentiate along to get for every and . Thus the Lie algebra of a normal Lie subgroup is an ideal of a Lie algebra. The assertion concerns a subgroup carrying the corresponding Lie-subgroup structure, for example any closed subgroup; no arbitrary abstract subgroup is being assigned a tangent space.
Past exam of the mathematics course of the University of Cambridge 2015 iii Paper 5 1 3 b Solution 2026-10-06
The inverse function theorem applied to makes smooth. Since , . Monotonicity of the characteristic flow map and the vanishing of outside implyThus the solution is a smooth function with compact support at each such time. The fact that some interior characteristic speeds differ from causes no difficulty: they cannot cross the two exterior characteristic curves before .
Past exam of the mathematics course of the University of Cambridge 2015 iii Paper 52 1 a Solution Created 2026-10-03 Updated 2026-10-06
Choose local coordinates on the hypersurface , and let be the local flow of the vector field . The map follows the integral curves of a vector field starting at . At , its differential maps the direction to and the remaining directions to a basis of . Transversality makes these vectors linearly independent, so the inverse function theorem makes a local manifold chart.
Set and transport the unchanged along each integral curve of a vector field. Then the adapted coordinates satisfyThe flow-box theorem gives this construction locally. The parameter is the flow parameter, and need not be arc length; a global manifold chart would require additional hypotheses.
Past exam of the mathematics course of the University of Cambridge 2015 iii Paper 52 4 b iv Solution Created 2026-10-03 Updated 2026-10-06
Choose a basis of and use the affine exponential map to define affine normal coordinates. Its differential at zero is the identity, so the inverse function theorem gives a chart near . By part (iii), radial geodesics have coordinates . Their geodesic equations at imply for every . The polarization identity therefore gives .
The disformation tensor is symmetric in , whereas is antisymmetric. Also . Taking the symmetric part of the corrected affine connection decomposition consequently gives the normal-coordinate identityThus the final identity in the PDF is correct after the half-factor repair in part (i). In general in the paper's convention, rather than zero. Neither nor the first derivatives of the metric tensor need vanish in these affine normal coordinates. With the Levi-Civita connection, both torsion and nonmetricity vanish and the usual vanishing of the Christoffel symbols is recovered.
Past exam of the mathematics course of the University of Cambridge 2016 iii Paper 308 2 Solution 2026-10-06
Orient both spheres in the standard way and normalize their area forms to total area . The degree of a map between oriented manifolds can be obtained in two ways. For a regular value , use the degree as a sum of local degrees:Each inverse image is isolated by the inverse function theorem, and compactness makes the set finite. The determinant is computed in consistently oriented local coordinates. A second method is spherical degree by area pullback:where the last formula represents as a unit vector in . The pullback of a differential form already contains the signed Jacobian determinant; no extra is to be inserted in that last coordinate expression.
To relate the methods, replace by a smooth top-degree differential form with the same total integral supported in a small neighbourhood of a regular value. Two such top-degree forms with equal integral differ by an exact differential form on , by its top-degree de Rham cohomology. Their pullbacks therefore have the same integral by Stokes theorem. Over the chosen neighbourhood, splits into local inverse branches; the change of variables formula makes the contribution of each branch its orientation sign times . Their sum is precisely the first formula. Thus the area integral is an integer and agrees with the signed inverse-image count.
For a nonconstant rational map, first use common-factor reduction of a rational map so and are coprime. Write for these reduced polynomials. A generic finite target value has inverse images at the roots of : avoiding exceptional values makes its degree and its roots simple. The fundamental theorem of algebra supplies roots. A holomorphic map has positive real Jacobian determinant at a regular point, so every local sign is . HenceThe source leaves coprimality implicit. In an unreduced representation the answer is , including degree zero for a constant reduced map. For example extends to and has degree one, although the unreduced maximum degree is two. Exceptional inverse images at infinity or multiple roots do not change the degree of a rational map of the Riemann sphere.
For the rational map approximation for Skyrmions, use stereographic projection and the unit target vectorCombine this rational map with a radial profile to form a special unitary group field:where are the Pauli matrices. The endpoint values make independent of angle and . Appropriate radial behaviour gives an admissible finite-energy field configuration. With , choose the topological baryon number in the Skyrme model conventionSeparating the radial and angular factors givesThus the degree of a rational map of the Riemann sphere supplies the Skyrmion charge.
In conventional dimensionless massless Skyrme model units, its static energy reduces towith the angular Jacobian of a rational mapThe Cauchy-Schwarz inequality gives . These formulas follow from the radial strain and the two equal angular strains : the quadratic energy sums their squares and the quartic Skyrme term sums their pairwise products of squares. Minimize the angular integral in the rational map approximation over degree- maps, then minimize the remaining radial energy with the stated endpoints. This replaces a three-dimensional field minimization by finitely many map coefficients and an ordinary differential equation for .
The method constructs a charge- variational approximation, with topology built in and with rotational symmetry of a rational map translated into combined spatial and isospin rotations. It is efficient for identifying shapes and providing initial data for unrestricted numerical relaxation. Its restrictions are equally concrete: it uses one radial profile and a holomorphic angular map independent of radius, so it cannot represent arbitrary radial-angular correlations, separated clusters, or all deformations. Apart from the degree-one Skyrmion hedgehog ansatz, it generally does not solve the full field equation exactly. Massive-pion terms can be included in the radial functional but do not remove these restrictions, and multi-shell or unrestricted fields may be needed for larger charges. Approximate energy minima and a final collective-coordinate quantization are distinct steps.
Past exam of the mathematics course of the University of Cambridge 2017 iii Paper 114 2 Solution Created 2026-10-03 Updated 2026-10-05
For the sphere homology calculation use singular homology, homotopy invariance of homology, and the reduced Mayer–Vietoris sequence, not cellular homology. For a point, its chain group for singular homology is in every nonnegative degree, and its boundary operator is multiplication by , hence alternately zero and one in positive degrees. Its homology is therefore in degree zero and zero above. Singular chains of a disjoint union split into the summand chain complexes. Start with , the disjoint union of two points: , all positive groups vanish, and . For , remove the north and south poles to obtain an open cover of . Each member is contractible, while has a deformation retraction to , including the disconnected intersection when . The reduced exact sequence therefore suppliesAlso is connected, so its reduced degree-zero group vanishes. Induction givesThe separate calculation avoids conflating its two degree-zero generators.
Choose an orientation of , with , and let be its fundamental class. The mapping degree is the integer determined by . To define the local degree at , set and require to be isolated in . Choose an open neighborhood containing no other point of that fiber. Then defines a map of pairsThe degree- relative homology groups on both sides are , by a manifold chart and excision. The local orientation of a manifold chooses their generators. The induced map multiplies these generators by an integer, denoted . Excision and naturality show independence of the choice of . Differentiability is unnecessary; if is smooth and is invertible, the inverse function theorem gives relative to the orientations. Without isolation in the fiber this pointwise definition need not apply.
For the degree as a sum of local degrees, assume has finite fiber . Disjoint coordinate neighborhoods and excision identifyThe absolute fundamental class maps to the tuple of local orientation generators. Under the relative map induced by , the th generator goes to times the generator at , and the resulting map from the direct sum adds these contributions. On the other hand, naturality says that first applying the absolute map and then passing to the local group at gives times that generator. ThusIf the fiber is empty the sum is zero: the map factors through the contractible punctured sphere and has degree zero. If every point of a fiber is isolated, compactness makes that closed fiber finite, so the theorem applies. In particular, a regular value of a smooth map gives the signed count of its inverse images. For , define a mapping degree on the reduced generator : the identity has mapping degree one, the swap of the two points minus one, and either constant map zero. With local orientation signs positive at and negative at , the same sum formula holds. The connected-sphere convention above is the standard positive-dimensional one.
Past exam of the mathematics course of the University of Cambridge 2017 iii Paper 131 2 Solution Created 2026-10-03 Updated 2026-10-05
For in the domain of the exponential map at , let be the unique geodesic with initial data and define . The domain consists of initial velocities whose geodesics exist on ; it is open, contains zero, and need not be the whole tangent space. The ordinary differential equation theorem used here says that a smooth vector field has unique maximal integral curves, with an open flow domain and smooth dependence on initial point and time. Applied to the geodesic equation on , it gives smoothness of the exponential map. Rescaling the parameter gives and hence . The inverse function theorem makes a diffeomorphism from a neighbourhood of zero to one of . Coordinates in an orthonormal tangent basis transported by this map are geodesic normal coordinates.
The geodesic sphere of radius is . For small , it is , a smooth compact hypersurface; larger distance spheres need not be smooth. The Gauss lemma statesIn particular radial and angular directions are orthogonal and the radial coordinate measures arc length.
For a piecewise smooth path , put . The Riemannian distance is over such paths from to . Connectedness makes it finite. Reversal and concatenation give symmetry and the triangle inequality; positivity and compatibility with the manifold topology follow locally from geodesic normal coordinates and the Gauss lemma.
Choose so that is a diffeomorphism on a neighbourhood of the closed tangent ball of radius . For , radial geodesics realize the distance from on that ball. Indeed, Gauss lemma bounds any path remaining in the normal ball below by its radial change, while a path leaving the radius- ball already has length at least . Thus the radius- sphere is the compact exponential image described above.
Given , take . The continuous function attains its minimum at some . Every path from to crosses that sphere, since the distance from is continuous. Splitting at a crossing gives length at least . Taking the infimum over paths, and using the triangle inequality in the other direction, proves the distance splitting through a small geodesic sphere equalityThis compact local argument does not assume completeness or the existence of a globally minimizing geodesic to .
For the Lie-group clause, a bi-invariant Riemannian metric has adjoint-invariant identity inner product. Differentiating this invariance gives . The Koszul formula for left-invariant vector fields has no metric-derivative terms, and this skew-adjointness reduces it toThis is the Levi-Civita connection of a bi-invariant metric.
Let be the integral curve through the identity of the left-invariant field with identity value . Uniqueness and left translation show wherever the local curves are defined. The field is complete: a fixed local existence interval about the identity translates to an equally long interval at every point of the group. Restarting the curve before any proposed finite endpoint extends it beyond that endpoint. Thus exists on , and uniqueness now gives the group law for every .
Since , this integral curve is a geodesic. Conversely any geodesic starting at the identity has the same initial data as one of these curves and agrees with it by uniqueness. Therefore the geodesics of a bi-invariant metric are one-parameter subgroups conclusion isThe zero velocity gives the constant, trivial subgroup. Here is the Exponential map of a Lie group; for this metric it agrees at the identity with the Riemannian exponential.
Past exam of the mathematics course of the University of Cambridge 2017 ii Paper 4 24I Solution 2026-10-05
For sufficiently small, let be the geodesic with , , and define . The domain consists more generally of initial velocities whose geodesic exists through time one. Uniqueness and affine rescaling give near zero, soThus the differential is the identity under the natural identification of tangent spaces, and the inverse function theorem makes the exponential map a local diffeomorphism near zero.
A punctured plane has geodesics reaching its missing point in finite time, so its exponential map need not be global; adding the point remedies this example. But an extension cannot always remedy the failure. On the punctured circular cone , radial generators are geodesics and reach the missing point in finite intrinsic length. Any globally defined extension would have to include their limiting point . The cone's tangent planes have different limits as varies; no smooth embedded surface through that point can contain this punctured circular cone. Hence no such extension exists for this .
For an oriented surface and a closed topological disc with smooth positively oriented boundary, the Gauss-Bonnet theorem with boundary saysHere is signed geodesic curvature with the disc on the left. A smooth boundary has no corner terms; piecewise smooth boundaries require exterior-angle terms.
For the flat disc in the problem this gives . Conditions (i) and (ii) mean that the proposed replacement disc has the same boundary and agrees with the original smooth surface from the outside. Smoothness therefore gives the same tangent planes, metric derivatives and boundary geodesic curvature on the replacement, with consistent orientation. Applying Gauss-Bonnet to that disc forcesThus these gluing conditions fix the total curvature of any smooth replacement disc, regardless of its detailed interior shape.
The nonnegative continuous Gaussian curvature on the compact replacement disc, together with its zero integral established by Gauss-Bonnet, forces it to vanish at every point of that disc. Outside it the replacement surface agrees with the original flat surface, so its curvature also vanishes there; smoothness covers the common boundary. Consequently no point of the replacement surface can have positive Gaussian curvature. Therefore no surface can satisfy all three conditions. A disc with some positive curvature could only have compensating negative curvature or alter the smooth boundary gluing, both excluded here.
Past exam of the mathematics course of the University of Cambridge 2018 ib Paper 1 11F a Solution Created 2026-09-24 Updated 2026-10-03
The map is differentiable at when there is a linear map such thatIt is when it is differentiable everywhere and is continuous. The definition immediately gives , so differentiability implies continuity.
The inverse function theorem says that if is near and is invertible, then restricts to a diffeomorphism between neighbourhoods of and .

