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Past exam of the mathematics course of the University of Cambridge 2017 iii Paper 103 3 b Solution by
Codex 0 2026-10-05
Assume . The place-permutation action on preserves the coefficient-sum kernelIt is the augmentation subrepresentation of a permutation representation. The point action of is two-transitive, so the irreducible augmentation criterion for a transitive group action makes irreducible. The two-row Young permutation module decomposition of identifies its nontrivial summand as . Hence .
All standard Young tableaux of shape are , , with below the first cell and the remaining entries increasing along the first row. Their content vectors of standard Young tableaux areAn explicit orthonormal basis realizing these tableau lines isThe sums of their coordinates vanish. Their norms are one, and the inner product of with , , is zero because the coefficients of sum to zero. Directly summing the action of gives .
For , . For , its only nontrivial two-dimensional block isEvery other is fixed, including all with when . These formulas follow by swapping coordinates in the displayed vectors, and are the Young orthogonal form with axial distance for .
Past exam of the mathematics course of the University of Cambridge 2017 iii Paper 103 3 a Solution by
Codex 0 2026-10-05
For a standard Young tableau , put , using the Content of a Young-diagram cell. Choose the row-reading tableau , and let be the Coxeter length of the unique permutation sending to . A Gelfand–Tsetlin basis can be chosen so that, when is standard and ,If is not standard, the action is for two consecutive entries in one row, and for two in one column. This is one usual normalization of the Young seminormal form.
Here is a construction and proof of the normalization. Fix , let be the projection onto the tableau line, and defineThe permutation has a reduced expression consisting entirely of admissible swaps, by the reduced adjacent-swap path between linear extensions. At each swap the off-diagonal coefficient is nonzero. In its expansion, the only term that can reach a tableau at distance uses all swaps; omitting a swap gives a shorter path. Thus . This also makes its definition independent of a chosen reduced expression, because itself is fixed.
The relation forces the coefficient of in to be , and forces every other component to lie on the swapped line. If length increases, project the identity onto the line for . The shorter terms of cannot reach in one step, so the coefficient of is exactly one. Applying then gives the reverse coefficient and diagonal coefficient . In the nonstandard cases the same relation gives and the asserted scalar action. This proves the theorem rather than merely specifying pairwise scalings.
Past exam of the mathematics course of the University of Cambridge 2017 ii Paper 4 31A a Solution by
Codex 0 Created 2026-10-05 Updated 2026-10-14
The fixed point zero has multiplier , so it is attracting for and repelling for . Other real fixed points satisfyTheir multiplier is , and they are attracting exactly when . At equality the linear test is inconclusive. For two branches meet at , , a generic saddle-node bifurcation. At , the nonzero branch crossing zero gives a transcritical bifurcation when .
Representative fixed-point diagrams with solid attracting and dashed repelling branches
. Details. The three sketches use respectively; case (iii) therefore illustrates the subcritical-flip choice.
Let be a prime number, let with , and let be the sum of partition sizes at level of the core tower of a partition . ThenIndeed . Carrying units from position to position decreases by . Normalization to the base- digits therefore gives , with equality exactly when no carry is needed. The P-adic valuation of a symmetric-group character degree from the core tower is the difference divided by .
Under the hypotheses for the one-nearest-neighbour asymptotic risk, the limiting risk satisfies . Put : then since . Integrate, using the definition of Bayes risk. Constant gives limiting risk and Bayes risk , so the lower comparison is not generally equality.
By dominant root-lattice highest weights have zero weight, a minuscule weight in the root lattice has zero in its Weyl group orbit. Since every element of the Weyl group is invertible, the highest weight must itself be zero. Thus equality of the weight lattice and root lattice rules out every nontrivial minuscule representation without requiring a classification of root systems.
Square-root Loewner driving function generates a straight slit by
Codex 0 Created 2026-10-05 Updated 2026-10-06
For the Chordal Loewner equation driven by , the hull is a straight slit whose endpoint is proportional to . Put and . Its inverse map isusing upper-half-plane branches. The interval between the two real roots maps onto the two sides of a segment of argument .
For a smooth function and vector field ,on every tensor field, with interpreted as the endomorphism . Both sides vanish on functions; on vector fields this follows from . Both are contraction-compatible tensor derivations, so agreement on functions and vector fields proves equality on all tensor types. In particular .
The adjoint of lowers real degree by two and bidegree by . On a compact Kähler manifold it acts on cohomology through harmonic differential forms, because it commutes with the Hodge Laplacian. It does not generally preserve all closed forms.
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