Past exam of the mathematics course of the University of Cambridge 2001 iii Paper 1 4 Solution Created 2026-10-03 Updated 2026-10-14
We prove uniform approximation, not just approximation in . The analytic facts used are: a continuous square-integrable kernel gives a compact integral operator on ; the spectral theorem for compact self-adjoint operators decomposes the closure of its range into its finite-dimensional nonzero eigenspaces; and continuous functions are dense in for normalized Haar measure on a compact Lie group. Normalized Haar measure on a compact group is both left- and right-invariant and is preserved by inversion.
Choose a continuous nonnegative function of integral one, supported in a sufficiently small symmetric neighborhood of the identity, with . Such functions are obtained from a local bump and its inverted bump followed by normalization. Define the convolution operatorIts kernel is continuous on , so is compact. The symmetry of makes it self-adjoint. It commutes with the left translations by a change of variable .
By Cauchy-Schwarz inequality and invariance of Haar measure,The same integral formula and continuity of the kernel show is continuous for every . Moreover, rewriting gives . Uniform continuity on the compact group therefore makes arbitrarily small when the support of is small. Also because and has integral one.
Fix and take so . ThenLet be the orthogonal projection onto increasing finite sums of the nonzero eigenspaces of . Since lies in the closure of the range, the spectral theorem gives in . Applying the smoothing estimate givesThus some satisfies .
Each nonzero eigenspace is finite dimensional, invariant under , and consists of continuous functions: if with , then . The finite sum containing therefore carries the finite-dimensional unitary representation . It is continuous, since translating each of its finitely many continuous basis functions depends continuously on in the uniform and hence the norm.
Let be evaluation at the identity on , and let the dual representation on be . For the linear functional on defined by ,The rank-one endomorphism of satisfies . ThereforeThis is the requested Peter-Weyl theorem, proved via the convolution proof of uniform Peter-Weyl approximation. The extra convolution after the spectral truncation is what upgrades the estimate to a uniform one.
To deduce a faithful finite-dimensional representation, first note that these representations separate points. For , choose a continuous function taking different values at and , and approximate it closely enough by a trace coefficient to retain that difference. The representation appearing in that trace must satisfy .
There is an identity neighborhood containing no nontrivial subgroup. Here is the Lie groups have no small subgroups argument: choose an injective exponential chart on a Lie-algebra ball of radius , and take with . If a subgroup contained a nonidentity in , choose the first integer with . Then , so its power lies outside , a contradiction. For a zero-dimensional group, take .
For every , choose a representation nontrivial at . The sets on which these representations are not the identity are open and cover the compact set . A finite subcover gives finitely many representations, whose direct sum has kernel contained in . A subgroup contained in is trivial, so this direct sum is faithful. Averaging its Hermitian form makes it unitary. Thus the finite faithful representation from point-separating representations yieldsUsing the no-small-subgroups neighborhood is essential; alone need not be compact, so a finite-subcover argument on that punctured set would not be valid.
Past exam of the mathematics course of the University of Cambridge 2004 iii Paper 20 2 i Solution Created 2026-10-03 Updated 2026-10-14
Interpret as a continuous finite-dimensional complex group representation, so its values lie in , and interpret approximation in the uniform norm. Assume the usual Hausdorff convention for the compact topological group. The following argument proves the required form of the Peter-Weyl theorem without first assuming that is a Lie group.
Start with normalized Haar measure. It is both left- and right-invariant on a compact group. Average any Hermitian inner product to make a finite-dimensional representation unitary. The left regular action is already unitary on .
Choose a continuous nonnegative function supported in a small symmetric identity neighbourhood, with and . Such functions are obtained from a nonzero continuous bump near the identity, symmetrizing and normalizing. DefineThe convolution operator commutes with left translations. Its continuous kernel is Hermitian, hence it is self-adjoint. It is compact on : approximate the continuous kernel uniformly by finite sums of separated functions of and , using the Stone-Weierstrass theorem on the compact product. The corresponding operators have finite rank and converge in operator norm. It also maps into and satisfiesby the Cauchy-Schwarz inequality; continuity follows from the uniform continuity of the translated kernel.
The spectral theorem for compact self-adjoint operators decomposes the orthogonal complement of its kernel into finite-dimensional eigenspaces with nonzero eigenvalues. Every such eigenspace is left-translation invariant, and all its elements are continuous, since there. If is one of these invariant spaces, write . Evaluation at the identity givesThus is a matrix coefficient of the dual representation. This identifies each finite spectral sum with finite-dimensional representation data.
To get uniform rather than just approximation, let project onto increasing finite sums of the nonzero eigenspaces. For , is orthogonal to the kernel, so in . Applying the displayed smoothing bound givesMeanwhile, choosing the support of sufficiently small makes uniformly close to , by uniform continuity under small translations. Since , it also makes uniformly close to . Every belongs to a finite sum of invariant eigenspaces, hence is a finite sum of matrix coefficients. This completes the convolution proof of uniform Peter-Weyl approximation.
A coefficient equals for the rank-one endomorphism . Finite sums of coefficients can be combined using a direct-sum representation and a block-diagonal . Therefore the conclusion has exactly the requested form:
For a class function , apply conjugation averaging on a compact group,It fixes and is a contraction in the uniform norm, so averaging an approximation preserves its error bound. A trace coefficient averages to , whereThis endomorphism commutes with the representation. A finite-dimensional unitary representation splits into irreducibles by repeatedly taking invariant orthogonal complements. On each irreducible diagonal block, Schur lemma makes its average a scalar multiple of the identity, namely its trace divided by the block dimension. Off-diagonal blocks do not contribute to the total trace. Thus the averaged coefficient is a finite linear combination of irreducible characters. Consequently irreducible complex characters span a uniformly dense subspace of the continuous class functions.