Past exam of the mathematics course of the University of Cambridge 2018 iii Paper 303 3 c Solution Created 2026-10-03 Updated 2026-10-05
Assume for stability. With equal quadratic coefficients and , the internal symmetry is the full orthogonal group . In two dimensions at positive temperature, with short-range interactions, the Mermin-Wagner theorem forbids spontaneous symmetry breaking of this continuous symmetry. Thus in both phases in the infinite system; a nonzero average order parameter does not distinguish them.
Instead measure the correlation functionThe low-temperature phase has quasi-long-range order and algebraic correlations; the high-temperature phase has a finite correlation length and exponential correlations. More precisely,A finite order parameter amplitude allows . Its long-distance Goldstone-mode effective free energy is . Restoring the Boltzmann constant and using the renormalized phase stiffness, the thermal phase fluctuations giveThe Berezinskii–Kosterlitz–Thouless transition separates vortex-antivortex binding from unbound vortex defects. At the boundary, the universal stiffness jump gives and . The low-temperature phase stiffness is nonzero; it vanishes at long distances in the disordered phase. The exact transition location is fluctuation dependent and cannot be obtained just by setting the bare to zero.
Past exam of the mathematics course of the University of Cambridge 2019 ia Paper 3 2D Solution Created 2026-09-24 Updated 2026-10-03
Every eigenvalue of an orthogonal matrix has modulus one. A real three-by-three matrix has at least one real eigenvalue, and nonreal eigenvalues occur in a complex conjugate pair. If has such a pair , their product is one, so the remaining eigenvalue is . If all three eigenvalues are real, each is and their product is one; an odd number of three signs with product one must include . Hence every element of has an eigenvector of eigenvalue one and represents a rotation in three dimensions about its span.
It is false that every element of is either a rotation or a plane reflection. For example,is an improper orthogonal transformation combining a rotation with a reflection. Its eigenvalues are and , whereas a plane reflection has eigenvalues and a rotation has determinant one.
Past exam of the mathematics course of the University of Cambridge 2019 ii Paper 1 26H b Solution Created 2026-09-24 Updated 2026-09-29
Let be the vector space of real symmetric matrices and considerAt ,This derivative is surjective: for , choosing gives . Thus is a regular value and the orthogonal group is a submanifold of codimensionThe determinant takes only the values on , so the special orthogonal group is an open-and-closed submanifold of . Therefore
The tangent space is the kernel of :where the Special orthogonal Lie algebra is the vector space of skew-symmetric matrices. This proves the special orthogonal group as a submanifold description.
Past exam of the mathematics course of the University of Cambridge 2022 iii Paper 102 2 a Solution 2026-09-28
A finite root system in a real Euclidean vector space is a finite spanning set such that, for every , the root reflectionpreserves , and the Cartan integer is an integer for all . It is reduced when the only scalar multiples of in are and . Its Weyl group is the subgroup of the orthogonal group generated by the reflections . A base of a root system is a basis of such that every root is an integer combination of elements of whose nonzero coefficients all have the same sign.
The coroot of isLet be the Weyl chamber determined by :The roots and are positive scalar multiples, so their reflecting hyperplanes and their positive half-spaces are identical. The same chamber therefore defines positivity in the coroot system . Its walls correspond exactly to the rays for . Hence its simple roots arewhich is therefore a base of .
Past exam of the mathematics course of the University of Cambridge 2022 iii Paper 115 3 c Solution Created 2026-09-28 Updated 2026-10-07
Suppose first that the left-invariant 1-form is closed. The scalar function is constant for every , because both the form and vector field are left-invariant. Cartan's magic formula therefore givesPart b says that is locally constant. Since is connected, it is constant, and its value at the identity is . Hence for every , so is bi-invariant.
Conversely, if is bi-invariant, then all these Lie derivatives vanish. Cartan's formula and the constancy of give . The left-invariant vector fields span every tangent space, so .
Connectedness cannot be omitted. The orthogonal group has an abelian Lie algebra, so every left-invariant 1-form is closed by the Maurer-Cartan equation in a Lie-algebra basis. Conjugation by a reflection acts as on its one-dimensional Lie algebra, so a nonzero left-invariant 1-form is not right-invariant. This is the standard obstruction recorded by the closed left-invariant 1-form criterion.
Past exam of the mathematics course of the University of Cambridge 2023 iii Paper 302 1 a Solution 2026-09-28
The matrices are rotations about the third coordinate axis. Direct multiplication givesand . Thus they form a one-parameter subgroup of the orthogonal group , isomorphic to the circle group.
Past exam of the mathematics course of the University of Cambridge 2024 iii Paper 303 3 a i Solution Created 2026-09-24 Updated 2026-09-25
When all masses are equal, the free energy depends on the real vector only through inner products, so its symmetry is the orthogonal group . The mean-field potential isFor , its unique minimum is , which preserves . For , the minima form the sphereChoosing one minimum leaves the subgroup that fixes its direction, so spontaneous symmetry breaking gives .