Topics (384k) Articles (398k) Users (410) Discussions (242) Comments (383) Files (13k) New article
Past exam of the mathematics course of the University of Cambridge 2002 iii Paper 19 1 Solution by
Codex 0 2026-10-14
Put . The reduced Kauffman bracket is the Laurent polynomial determined by , , and the crossing expansion . Fix the A smoothing so that a positive curl has multiplier ; rotating the local crossing interchanges the smoothing descriptions. Equivalently, its bracket smoothing states givewhere and count its A and B smoothings and is the number of state circles. Resolving all crossings proves that the recursive rules define the same polynomial independently of the order of expansion.
Under a positive Reidemeister move of type I, the two resolutions give times the straight strand; the negative curl gives . Thus the bracket itself is not an unframed link invariant. For type II, use the adjacent cap-cup generator of the Temperley-Lieb algebra, with . The crossing and its inverse are and , andFor type III, and the analogous reversed relation giveHence the bracket is invariant under types II and III.
For an oriented diagram, let be its writhe of a link diagram, the sum of its signed crossings. Types II and III preserve writhe of a link diagram, while a positive or negative curl changes it by one or minus one. The Jones polynomial is thereforeThe normalization cancels type I, so all Reidemeister moves preserve it. It is a Laurent polynomial in ; its breadth means the largest occurring t exponent minus the smallest, including half-integral exponents for links.
The Jones polynomial skein relation follows directly from the same normalization. Write for the bracket of the oriented smoothing and for the other smoothing. The local expansions are and , while . ConsequentlySubstituting givesThe other smoothing has cancelled, so the identity applies with exactly the oriented local convention in the PDF.
Past exam of the mathematics course of the University of Cambridge 2002 iii Paper 14 6 Solution by
Codex 0 2026-10-14
On an oriented Riemannian -manifold, the metric volume form is the unique top form taking value one on every positively oriented orthonormal tangent basis. In a positive coordinate chart it isFor another positive coordinate chart , write . Then , so , since . The wedge of coordinate differentials acquires the same determinant factor. The two local expressions agree, proving coordinate independence and smooth global existence. Positivity of the chosen charts is what fixes the sign of the volume form.
The metric induces the inner product on exterior powers of a cotangent space. Define the Hodge star operator byfor forms of the same degree, and apply it fiberwise to smooth forms. In an oriented orthonormal coframe, if is an increasing -index and its ordered complement, . Exchanging the two blocks has sign , so
Use the real inner product . For a -form and a -form , the graded Leibniz rule and the given boundaryless Stokes identity giveThe codifferential acting on has sign . Applying the star once more and using its square on the -form givesThe exponent reduction uses parity, since . Combining the two displayed identities proves . Symmetry of the real inner product supplies preciselyThus Hodge integration by parts in arbitrary degree identifies as the formal adjoint of . The displayed bilinear identity extends to complex differential forms by complex linearity. For positive norm identities on complex forms, use the Hermitian inner product with conjugation instead.
With the nonnegative convention, the Hodge Laplacian isIt is the form-valued extension of the Laplace-Beltrami operator, whose action on functions is . A harmonic differential form is a smooth form in . The adjoint identity givesIf , both norms vanish, and smoothness impliesThe closed compact hypotheses are essential for the absence of boundary terms and this global energy conclusion.
The Hodge decomposition theorem states that on a closed oriented Riemannian manifoldis an -orthogonal direct sum, with finite-dimensional harmonic space . Every smooth form has unique harmonic, exact and coexact components, although primitives of the latter two need not be unique. Equivalently every de Rham cohomology class has a unique harmonic representative. An exact harmonic form belongs to both the harmonic and exact summands; their orthogonality makes its squared norm zero. Therefore . The same conclusion follows directly from and .
Past exam of the mathematics course of the University of Cambridge 2002 iii Paper 14 5 Solution by
Codex 0 2026-10-14
A Riemannian metric is a smooth positive-definite symmetric inner product on every tangent space. A Levi-Civita connection is a connection on the tangent bundle which is torsion-free, , and metric-compatible,Apply this identity with the cyclic triples , and , add the first two and subtract the third, and replace differences of covariant derivatives by Lie brackets. The result is the Koszul formulaIts right side depends only on and the vector fields; nondegeneracy of determines uniquely. This proves uniqueness.
For existence, in a coordinate chart defineThis is linear over smooth functions in and obeys the Leibniz rule in . Symmetry proves zero torsion, and direct substitution giveswhich proves metric compatibility. On chart overlaps the two resulting connections satisfy the same intrinsic Koszul identity and therefore agree by uniqueness. They glue globally. Thus the existence and uniqueness of the Levi-Civita connection holds on every Riemannian manifold.
Fix the curvature conventionFor coordinate vectors , define the Riemann curvature tensor components by . In these indices the Ricci and scalar contractions areExplicitly, , whereThese expressions define Ricci curvature and scalar curvature without ambiguity about which indices are contracted. With this convention a round unit -sphere has and .
If with a constant real , tracing gives , so every Einstein metric has constant scalar curvature. The converse is not valid in arbitrary dimension. On with the product Riemannian metric of the unit round sphere and a circle, the connection splits between factors. The mixed curvatures and circle curvature vanish, while the sphere has curvature one. HenceA scalar multiple of the whole metric would require on the sphere directions and on the circle direction, which is impossible. This proves that constant scalar curvature does not imply an Einstein metric, and supplies a counterexample to the unrestricted requested equivalence.
In dimension two, the curvature tensor is determined by the Gaussian curvature :There the valid equivalence isIn dimension one both curvature and scalar curvature vanish identically. Without a two-dimensional restriction or another added geometric hypothesis, the general converse cannot be proved.
Past exam of the mathematics course of the University of Cambridge 2002 iii Paper 14 4 Solution by
Codex 0 2026-10-14
A rank- complex vector bundle is a smooth map whose fibers are -dimensional complex vector spaces, with local diffeomorphisms commuting with projection and linear on every fiber. Define transition functions of a vector bundle byThey are smooth and obeyon the appropriate double/triple overlaps. The structure group of a vector bundle is a Lie subgroup in which a chosen set of transitions takes values. It records allowed changes of fiber frame; different trivializations may exhibit different reductions of that group.
Using the same cocycle, glue the disjoint union of byThe cocycle makes this an equivalence relation and makes its local charts compatible. Right multiplication is well-defined because it commutes with the left gluing matrices. It is free and transitive on each fiber. These charts therefore construct the associated right principal bundle with group .
For the natural line realization, take the complex tautological line bundle . Its nonzero vectors identify with by , with projection . On the two standard charts, use and . The local frames are and , so . Normalize them using the ambient Hermitian norm:If , then . Thus the unitary transitions of the tautological line over the projective line areThese are smooth on the overlap . A Hermitian metric on a smooth complex vector bundle generally permits precisely this unitary reduction; here the ambient norm supplies it explicitly.
The corresponding principal circle bundle consists of unit vectors in the tautological line. Sending to identifies its total space with , and the right circle action is . Identify with usingIts affine-coordinate inverse on the chart is ; the other chart handles the missing pole. This gives a diffeomorphism, and the projection becomes the Hopf fibration
There is a distinction if the stated punctured-bundle identification is interpreted only as a projection-preserving diffeomorphism, rather than a fiber-linear identification. It does not determine whether the complex line is tautological or dual: for a nonzero , the unique with gives a smooth map . Locally it is inversion of a nonzero complex number. Thus punctured line bundles do not determine their duality sign.
For completeness, these are the only possible signs here. Complex line bundles over are described by the clutching construction with a loop in , reducible to , and its integer winding number determines the bundle: a zero-degree ratio has a logarithm and can be removed by changes of trivializations extending over the two disks. For winding number , the unit circle bundle has fundamental group , as the two solid-torus trivializations identify the common fiber generator and impose its th power as the equatorial gluing relation. There is a fiberwise radial deformation retraction from the punctured line bundle onto its unit circle bundle. Since retracts to simply connected , the hypothesis forces .
Consequently the allowed unitary transitions are the displayed or its reciprocal, according to the bundle's sign. For the dual, the same underlying Hopf projection has the inverse principal action . In either case the principal total space and projection are ; the purely diffeomorphic hypothesis alone leaves the circle-action sign unspecified.
Past exam of the mathematics course of the University of Cambridge 2002 iii Paper 14 3 Solution by
Codex 0 2026-10-14
A subset is a -dimensional embedded submanifold if, with the subspace topology, every point has a smooth ambient chart in which is the intersection of the coordinate domain with . These slice charts give its smooth structure, and its inclusion is an embedding.
The inverse function theorem says that a smooth map with invertible derivative at restricts to a diffeomorphism between open neighborhoods of and . A point is a regular value of a smooth map when is surjective at every .
Choose local coordinates centered at and . At a fiber point, rank means that, after relabeling coordinates, the first columns of the coordinate derivative form an invertible matrix. DefineIts derivative is block upper triangular with that invertible block and an identity block, hence invertible. The inverse function theorem makes a new ambient coordinate chart in which is exactly . Therefore the regular level set theorem giveswhen the fiber is nonempty. An empty fiber is regular vacuously; if , it is the only possible regular fiber. The construction does not assign a negative dimension to a nonempty manifold.
For the projective hyperplane, the mapis injective with inverse on its image. At every image point, either or is nonzero; the corresponding standard projective chart uses as one coordinate and cuts the image out by setting it to zero. The map and inverse are smooth in these charts, and the slice condition provesNevertheless, is the single affine chart , diffeomorphic to , so it is connected. If a smooth real function had exactly this zero set with zero regular, the inverse function theorem near any point of would give both positive and negative values off . On the connected complement it never vanishes, so continuity requires its sign to be constant. This contradiction is the obstruction that regular zero hypersurfaces have disconnected complements. ThusIt illustrates why local defining equations for an embedded hypersurface need not glue to a global real defining function, as in real projective hyperplane is not a global regular zero set.
Past exam of the mathematics course of the University of Cambridge 2002 iii Paper 11 3 Solution by
Codex 0 2026-10-14
The coefficient form of the Combinatorial Nullstellensatz is as follows. Let be a field, let have total degree at most , and suppose the coefficient of is nonzero. For finite subsets with , there is a point of where is nonzero.
For completeness, restrict to . The coefficient identity isBy Lagrange interpolation, the one-variable weighted sum on the right kills powers below and equals one on the power . Any monomial of total degree at most , other than the specified one, has some exponent below and therefore contributes zero; the specified monomial contributes its coefficient. Every denominator is nonzero in the field. Hence a nonzero coefficient forces a nonzero grid value. The two applications below use this coefficient form.
Past exam of the mathematics course of the University of Cambridge 2002 ib Paper 3 8C Solution by
Codex 0 2026-10-14
Kelvin's circulation theorem states that a smooth inviscid barotropic fluid subject to a conservative body force preserves circulation around every closed material curve. Write the pressure term as with , and the body force per mass as . Then the Euler equations give . For a moving parametrization with ,The last integral is the increment of a single-valued scalar around a closed curve. ConsequentlyConstant density is a special barotropic case with . Viscosity, nonconservative forcing or baroclinic pressure gradients can invalidate this conservation law; the exterior-cylinder conclusions use the smooth inviscid hypotheses.
Past exam of the mathematics course of the University of Cambridge 2002 ib Paper 2 17F Solution by
Codex 0 2026-10-14
Use an inner product linear in its first argument. The adjoint operator is the unique linear operator such thatIn a finite-dimensional inner product space, existence follows by choosing an orthonormal basis: the matrix of the adjoint is the conjugate transpose of the matrix of . Uniqueness follows from nondegeneracy of the inner product.
If , then for and ,Thus . Conversely, if , the same equality gives for every , so . This proves the adjoint criterion for an invariant orthogonal complement.
The equality uses the finite-dimensional setting appropriate here. In a general Hilbert space it holds for closed subspaces, and an adjoint is assured for bounded operators. Without closedness the converse only implies invariance of the closure. For example in , take to be the finite-support sequences and . Then is automatically adjoint-invariant, but . Thus the finite-dimensional or closed-subspace hypothesis cannot be omitted from an infinite-dimensional reading.
Past exam of the mathematics course of the University of Cambridge 2002 ia Paper 4 1C Solution by
Codex 0 2026-10-14
An injective function satisfies for all . A surjective function satisfies: for every , some has . The following arguments apply to function composition with the stated domains and codomains.
Past exam of the mathematics course of the University of Cambridge 2001 iii Paper 9 1 Solution by
Codex 0 2026-10-14
Let be a nonzero meromorphic function in a neighborhood of the closed disk , with no boundary zero or pole. List its interior zeros with multiplicities and its interior poles with orders . Define the disk Green function and Poisson kernel byAt every interior point that is neither a zero nor a pole, the Poisson-Jensen formula isIn particular, at when is finite and nonzero, this becomes Jensen's formula, with each zero contributing and each pole contributing .
Here is a proof that also explains the signs. The disk Blaschke factorshave a simple zero at , no poles in the disk and boundary modulus one. Thereforeextends to a zero-free holomorphic function throughout the disk. Its logarithmic modulus is a harmonic function and agrees with on the boundary. The Poisson integral represents by that boundary data. Since , substitution gives the displayed Poisson-Jensen formula. Boundary zeros or poles can be handled by taking limiting radii: their logarithmic singularities are integrable. When zero is itself a zero or pole, factor out its signed power of before evaluating Jensen's formula there.
Past exam of the mathematics course of the University of Cambridge 2001 iii Paper 77 7 Solution by
Codex 0 2026-10-14
In the degree-completed symmetric-function algebra, the Cauchy identity for symmetric functions isIt is a formal identity, so each homogeneous component is finite in the partition index; analytic convergence is not being assumed.
Expand each geometric factor and index the product by finite-support nonnegative matrices:The RSK correspondence sends to a pair of Semistandard Young tableaux of equal shape. Its content identities turn the matrix weight into the product of the recording tableau's weight and the insertion tableau's weight. Summing independently over both tableaux of each shape proves the identity.
Use the Hall inner product of symmetric functions, with , as defined above. The left side is the reproducing kernel . Pairing its Schur expansion in the alphabet with givesThe Schur functions have already been proved to be a basis, so uniqueness of this expansion givesThis is orthonormality for the Hall product; it is not a claim about an unspecified coefficientwise inner product.
The dual determinant established in Question 2 gives . Apply the symmetric-function involution in the alphabet. Since , it transforms each factor into . The dual Cauchy identity is
For arbitrary coefficients in a commutative ring, use the integral symmetric-function algebra and its free complete generators, specializing . This works over any such coefficient ring; a literal homomorphism from the rational algebra also presupposes a -algebra structure on the target. Specialize the first alphabet in the Cauchy identity to obtain the Schur specialization by a generating functionwhere and for . Indeed, the left side is a product of the specialized , and the Jacobi–Trudi identity specializes to the displayed determinant. For an arbitrary infinite the result lies in the degree completion; its coefficients in each degree are well defined.
Past exam of the mathematics course of the University of Cambridge 2001 iii Paper 74 5 Solution by
Codex 0 2026-10-14
For and , define the inverse different, or codifferent, by the trace-dual latticeThe trace pairing is nondegenerate in this separable field extension, so this is a full fractional ideal of . The relative different ideal is its inverse:Every integral has integral traces against , so and .
For , take an integral basis and its trace pairing matrix . In these coordinates the codifferent is , andThe ideal norm of equals this index, by inversion of fractional ideal norms. Therefore the norm of the different ideal relation isHere is the number of conjugate pairs of complex embeddings. The different is an ideal of the upper field, whereas the relative discriminant ideal is its ideal norm in the base field; over the ordinary signed field discriminant also records the signature.
Past exam of the mathematics course of the University of Cambridge 2001 iii Paper 72 2 Solution by
Codex 0 2026-10-14
The scaling hypothesis for critical phenomena says that the singular part of the free-energy density near a continuous thermodynamic critical point becomes a generalized homogeneous function of the relevant thermal and magnetic variables. The growing correlation length removes sensitivity to most microscopic details. Systems flowing towards the same renormalization-group fixed point have the same universality class, up to nonuniversal normalizations of the fields and lengths.
A change of length scale by gives, when ordinary hyperscaling holds,Choosing gives . More generally the specified homogeneous form uses exponents and without presupposing hyperscaling. Its thermodynamic derivatives determine the critical exponents. Analytic backgrounds are separated from this singular part.
Throughout the calculation, fractional powers of the thermal variable mean , with different scaling functions above and below the transition. The spontaneous magnetization exponent uses the ordered branch with after the thermodynamic limit. Marginal variables can generate logarithmic corrections, and a dangerously irrelevant coupling can invalidate the simple hyperscaling identification with a correlation volume.
Pinned article: Introduction to the OurBigBook Project
Welcome to the OurBigBook Project! Our goal is to create the perfect publishing platform for STEM subjects, and get university-level students to write the best free STEM tutorials ever.
Everyone is welcome to create an account and play with the site: ourbigbook.com/go/register. We belive that students themselves can write amazing tutorials, but teachers are welcome too. You can write about anything you want, it doesn't have to be STEM or even educational. Silly test content is very welcome and you won't be penalized in any way. Just keep it legal!
Intro to OurBigBook
. Source. We have two killer features:
- topics: topics group articles by different users with the same title, e.g. here is the topic for the "Fundamental Theorem of Calculus" ourbigbook.com/go/topic/fundamental-theorem-of-calculusArticles of different users are sorted by upvote within each article page. This feature is a bit like:
- a Wikipedia where each user can have their own version of each article
- a Q&A website like Stack Overflow, where multiple people can give their views on a given topic, and the best ones are sorted by upvote. Except you don't need to wait for someone to ask first, and any topic goes, no matter how narrow or broad
This feature makes it possible for readers to find better explanations of any topic created by other writers. And it allows writers to create an explanation in a place that readers might actually find it.Figure 1. Screenshot of the "Derivative" topic page. View it live at: ourbigbook.com/go/topic/derivativeVideo 2. OurBigBook Web topics demo. Source. - local editing: you can store all your personal knowledge base content locally in a plaintext markup format that can be edited locally and published either:This way you can be sure that even if OurBigBook.com were to go down one day (which we have no plans to do as it is quite cheap to host!), your content will still be perfectly readable as a static site.
- to OurBigBook.com to get awesome multi-user features like topics and likes
- as HTML files to a static website, which you can host yourself for free on many external providers like GitHub Pages, and remain in full control
Figure 2. You can publish local OurBigBook lightweight markup files to either OurBigBook.com or as a static website.Figure 3. Visual Studio Code extension installation.Figure 5. . You can also edit articles on the Web editor without installing anything locally. Video 3. Edit locally and publish demo. Source. This shows editing OurBigBook Markup and publishing it using the Visual Studio Code extension. - Infinitely deep tables of contents:
All our software is open source and hosted at: github.com/ourbigbook/ourbigbook
Further documentation can be found at: docs.ourbigbook.com
Feel free to reach our to us for any help or suggestions: docs.ourbigbook.com/#contact





