Past exam of the mathematics course of the University of Cambridge / 2003 / iii / Paper 11 (Topological Groups)

Assume first that the topological group is Hausdorff. This separation condition is necessary: an indiscrete group with two elements has the one-element neighbourhood basis consisting of the whole group, but its topology cannot come from a metric.

Translations are homeomorphisms, since multiplication by a fixed element is continuous and its inverse is multiplication by the inverse element. Inversion is likewise a homeomorphism. If $W$ is an neighbourhood of the identity element, continuity of the triple product at $(e,e,e)$ supplies neighbourhoods of the identity element $A,B,C$ with $ABC\subset W$. Their intersection with their inverses contains a symmetric open neighbourhood of the identity element $V$, and $V^3\subset W$. This proves the neighbourhood shrinking property needed below rather than assuming it.

Let $(B_n)$ be a countable neighbourhood basis at the identity element, choosing open representatives inside its members. Starting with $U_0=G$, recursively choose symmetric open neighbourhoods of the identity element such thatThey are still a basis. Their intersection is $\{e\}$: for $x\ne e$, the Hausdorff condition gives an neighbourhood of the identity element not containing $x$, and some $U_n$ lies inside it.

Here is the dyadic product estimate for group neighbourhoods. If $g_j\in U_{n_j}$ and ${\sum }_j{2}^{-n_j}\le 2^{-n}$, then the ordered product lies in $U_{n-1}$, for $n\ge 1$. Prove this by induction on the number of factors. A single factor has $n_j\ge n$ and hence belongs to $U_n\subset U_{n-1}$. For a longer word, choose the factor crossing the half-total-cost point. The words preceding and following it each have cost at most half the total, hence at most $2^{-n-1}$. The induction hypothesis puts both in $U_n$, and the middle factor also lies in $U_n$. Their threefold product is in $U_{n-1}$. Empty side words cause no problem because $e\in U_n$.

Give an edge from $x_{j-1}$ to $x_j$ cost $2^{-n_j}$ whenever $x_{j-1}^{-1}{x}_j\in U_{n_j}$, and defineChains exist because $U_0=G$. Reversing a chain proves symmetry, concatenating chains proves the triangle inequality, and translating every vertex on the left leaves all edge differences unchanged. Thus $d$ is left invariant. The product estimate also givesThe first implication follows by choosing a chain with total cost below the bound and applying the estimate with index $n+1$. If $d(x,y)=0$, the first implication puts $x^{-1}y$ in every $U_n$, so $x=y$. Hence $d$ is a genuine left-invariant group metric. These two inclusions show that its balls and the original neighbourhoods of the identity element refine one another; translations then show equality of the two topologies everywhere. More explicitly, every metric ball contains a sufficiently small translated $U_n$, and every open neighbourhood contains a sufficiently small metric ball. This proves the sufficient direction of the Birkhoff-Kakutani theorem.

Conversely, if a compatible metric exists, the balls $B_d(e,1/n)$ form a countable neighbourhood basis at the identity element. Invariance is not even needed for this direction. Thus, with the necessary Hausdorff convention, a compatible left-invariant metric exists exactly when the identity element has a countable neighbourhood basis.

For the requested counterexample, take the orientation-preserving affine group of the real line, represented by $(a,b)$ with $a>0$ and product $(a,b)(a^{\prime },b^{\prime })=(a{a}^{\prime },b+a{b}^{\prime })$. Its topology is metrizable, for instance by the ordinary Euclidean distance in coordinates $(\log a,b)$, and the product and inverse are continuous. Let $D_a=(a,0)$ and $T_b=(1,b)$. Direct multiplication gives $D_a{T}_b{D}_a^{-1}=T_{ab}$. A compatible bi-invariant group metric would make every conjugation an isometry fixing $e$. ThereforeBut $T_{1/n}\to e$ in the group's topology, a contradiction. This is the affine conjugation obstruction to a bi-invariant group metric: the group is metrizable, and even admits a compatible left-invariant metric, but admits no compatible metric invariant on both sides.

There is an essential qualification to the printed hypothesis. If compactly generated group means merely generated algebraically by a compact subset, the claim is false. In the additive group $\mathbb{Q}$ with its usual topology,is compact, being a convergent sequence together with its limit, and generates every rational number since any integer denominator divides some factorial. The group is metrizable. However, an invariant Radon measure has a common mass $c$ on each singleton. If $c>0$, the compact set $K$ has infinite measure, contradicting finiteness on compact sets. If $c=0$, countable additivity makes the entire countable group have measure zero. Thus it has no nonzero Haar measure. This proves that compact generation does not imply local compactness and disproves the literal unrestricted statement.

The intended existence statement is valid when local compactness is included, either explicitly or in the convention that the group has a compact generating neighbourhood. Here is a construction for a locally compact, compactly generated, metrizable Hausdorff group. No invariant measure is assumed during its construction.

Enlarge a compact generating set to a symmetric compact neighbourhood of the identity element $S$. Then $G={\bigcup }_n{S}^n$, and each $S^n$ is compact. Moreover $S^n$ lies in the interior of $S^{n+1}$, since multiplication by the open neighbourhood of the identity element contained in $S$ supplies an open neighbourhood of every point of $S^n$. This gives a compact exhaustion. Every compact metric space has a countable dense subset: take finite $1/k$-nets and unite them. Their union over this exhaustion is dense in $G$, so the metric space is separable and has a countable open basis.

We use nonnegative compactly supported continuous functions and cutoffs equal to one on specified compact sets. Their existence is covered by the source's permission concerning auxiliary functions. Fix a nonzero such function $h$. For nonzero $u\ge 0$ define the weighted covering quantityThe quantities are finite: an open set where $u$ is bounded below has translates covering the compact support of $f$, so finitely many sufficiently large weighted translates dominate it. The denominator is positive, since evaluation at a point where $h>0$ gives $(h:u)\ge h(y)/\Vert u{\Vert }_{\infty }>0$.

Each $I_u$ is nonnegative, homogeneous, monotone, left invariant and subadditive; these follow respectively by rescaling, comparing or combining covers, and translating their centers. Also $I_u(h)=1$. Composing a cover of $f$ by translates of $h$ with a cover of $h$ by translates of $u$ givesThis is a bound independent of $u$. On functions supported in a fixed compact set $C$, choose a cutoff $k\ge 1$ on $C$. Monotonicity and subadditivity implyThus the covering functionals are uniformly Lipschitz on each fixed support.

Choose a nonzero $u_n\ge 0$ with support in an neighbourhood of the identity element shrinking to $e$. The nonnegative functions supported in each compact member of the exhaustion form a separable metric cone in the supremum norm, as a subspace of continuous functions on that compact metric space. Select a countable dense family from each cone and include $h$. The uniform scalar bounds permit a diagonal subsequence along which $I_{u_n}$ converges on their countable union. The displayed Lipschitz estimate extends convergence to every nonnegative member of $C_c(G)$. Denote the limit by $I$. It inherits positivity, homogeneity, monotonicity, subadditivity, invariance and $I(h)=1$.

The crucial remaining step is additivity. Fix $f_1,f_2\ge 0$ with combined support $C$, and fix a compact neighbourhood of the identity element $V_0$. Choose a cutoff $k$ equal to one on $C{V}_0^{-1}{V}_0$. For $\delta >0$, putsetting the ratio to zero off the denominator's nonzero set. These ratios are continuous: $k=1$ on a neighbourhood of the supports, and off those supports the numerator vanishes. Their sum is at most one. Compactness and continuity give, for each $\eta >0$, an neighbourhood of the identity element $V$ small enough that each $r_i$ oscillates by at most $\eta$ on any translate of $V$ meeting $C$. To justify this uniform choice, such translates lie inside the fixed compact set $C{V}_0^{-1}{V}_0$ when $V\subset V_0$; continuity of $(y,v)\mapsto r_i(yv)$ and a finite cover of that compact set give a single small right-increment neighbourhood, then choose $V^{-1}V$ inside it.

If $u$ has support in $V$ and $F\le {\sum }_j{c}_j{T}_{x_j}u$, discard translates not meeting $C$ when covering $f_i$. Let $a_{ij}$ be the supremum of $r_i$ on the remaining translate's support. Then $f_i\le {\sum }_j{a}_{ij}{c}_j{T}_{x_j}u$, and comparing both suprema with the ratios at one common point gives $a_{1j}+a_{2j}\le 1+2\eta$. Taking covering infima yieldsAlong the subsequence this holds eventually, since its supports shrink. Pass to $I$, then let $\eta \downarrow 0$ and $\delta \downarrow 0$; the uniform bound on $I_u(k)$ gives $I(f_1)+I(f_2)\le I(f_1+f_2)$. Subadditivity gives the reverse inequality. Thus $I$ is additive.

Extend $I$ linearly to real functions using positive and negative parts, and then complex-linearly to $C_c(G)$. The compact-support bounds above make it a positive, locally bounded functional. The Riesz representation on compactly supported continuous functions gives a Radon measure $\mu$ with $I(f)=\int fd\mu$. Translation invariance of $I$ and uniqueness in that representation give $\mu (xE)=\mu (E)$ for Borel $E$. It is nonzero because $I(h)=1$, and finite on compact sets because a cutoff dominating their indicators has finite integral. Finally every nonempty open set has positive measure: otherwise its translates cover the compact support of $h$, a finite subcover would give that support measure zero, contradicting $I(h)=1$. Hence $\mu$ is a Haar measure under the stated local-compactness qualification. This is Haar measure from normalized covering functionals, with the limiting additivity proved explicitly.

Let $G$ be locally compact, Abelian and Hausdorff, and fix a Haar measure. Write $T_{x}f(y)=f(x^{-1}y)$ and use convolutionThe L1 group algebra is a commutative Banach algebra. Here a group character means a continuous homomorphism $\chi :G\to \mathbb{T}$, not the trace of an arbitrary group representation. A Gelfand character is a nonzero multiplicative complex-linear functional on the convolution algebra. We use the conventionWith the alternative standard Fourier convention involving $\overline{\chi }$, one simply conjugates every group character in this identification.

For such $\chi$, the functional is bounded by $\Vert f{\Vert }_1$, and Fubini's theorem, the substitution $y=xz$, and the homomorphism law giveIt is nonzero: if $u\ge 0$ is nonzero in $C_c(G)$, then $f=\overline{\chi }u$ has ${\Phi }_{\chi }(f)=\int u>0$.

Conversely let $\Phi$ be an algebra character. It is automatically bounded, with $|\Phi (f)|\le \Vert f{\Vert }_1$. Indeed, extend it to the unitization by $\tilde{\Phi }(f+\lambda 1)=\Phi (f)+\lambda$. If $|\Phi (f)|>\Vert f{\Vert }_1$, the Neumann series makes $f-\Phi (f)1$ invertible, whereas its character value is zero, contradicting multiplicativity of an invertible element and its inverse.

Choose $g$ with $\Phi (g)\ne 0$ and put $\chi (x)=\Phi (T_{x}g)/\Phi (g)$. Commutativity of the group gives $(T_{x}f)\ast g=f\ast (T_{x}g)$; applying $\Phi$ yieldsThus the ratio is independent of the chosen $g$. Applying the relation successively to $T_x{T}_{y}g=T_{xy}g$ gives $\chi (xy)=\chi (x)\chi (y)$, and $\chi (e)=1$. In particular it never vanishes.

The translation map $x\mapsto T_{x}g$ is continuous in $L^1$. For $g\in C_c(G)$, translated supports near a fixed $x$ lie in one compact set; continuity on that set and a finite cover give uniform convergence of the translated values, hence $L^1$ convergence because the compact set has finite Haar measure. For general $g$, approximate by $C_c(G)$ and use $\Vert T_{x}g{\Vert }_1=\Vert g{\Vert }_1$. Therefore $\chi$ is continuous. For all integers $n$,Positive powers exclude modulus greater than one; negative powers exclude modulus less than one. Hence $|\chi (x)|=1$.

Finally, the $L^1$-valued convolution formula $f\ast g=\int f(x)T_{x}gdx$ and boundedness of $\Phi$ giveCancel $\Phi (g)$ to obtain the displayed formula for every $f$. Different characters give different functionals: if their continuous difference is nonzero at a point, integration against a suitably phased compactly supported test function in a small neighbourhood detects that difference. Thus the correspondence is bijective.

It also identifies the character space of an Abelian L1 group algebra topologically with the Pontryagin dual group. For a net converging uniformly on compact subsets in $\hat{G}$, convergence of the functionals follows first for compactly supported $f$, then for all $L^1$ functions using the common norm bound one. Conversely suppose the functionals converge in the Gelfand topology. Choose $g$ whose limiting character value is nonzero; its values are eventually bounded away from zero. For compact $K\subset G$, the set $\{T_{x}g:x\in K\}$ is compact in $L^1$. Convergence of uniformly bounded linear functionals is uniform on this compact set: a finite norm-net reduces it to convergence at finitely many functions, with the common bound controlling the approximation error. The ratio formula then gives uniform convergence of the group characters on $K$. This proves equality with the compact-open topology, without assuming first countability of $G$.

For later use, these transforms form a self-adjoint subalgebra dense in $C_0(\hat{G})$. The set of algebra characters together with the zero functional is weak-star compact by Banach-Alaoglu theorem and the closed multiplicativity equations. Removing zero gives a locally compact Hausdorff character space; each evaluation $\Phi \mapsto \Phi (f)$ vanishes at infinity, since its modulus-at-least-$ϵ$ set is compact and avoids zero. Transforms separate characters, and none vanishes identically at a given character. They are closed under conjugation because $f^{\ast }(x)=\overline{f(x^{-1})}$ satisfies ${\Phi }_{\chi }(f^{\ast })=\overline{{\Phi }_{\chi }(f)}$. The locally compact Stone-Weierstrass theorem therefore gives the claimed density.

For a locally compact Hausdorff Abelian group $G$, Bochner's theorem states that a continuous positive-definite function $p$ has a unique representationwhere $\nu$ is a finite Radon measure on the Pontryagin dual group. Conversely every such measure gives a continuous positive-definite function. Here positive definiteness means ${\sum }_{i,j}{c}_i\overline{c_j}p(x_i{x}_j^{-1})\ge 0$ for all finite choices. The Abelian and continuity hypotheses are part of the theorem; the representation is not claimed for arbitrary non-Abelian groups or discontinuous functions.

First prove existence without presupposing the representation. The elementary consequences of positive definiteness give $p(x^{-1})=\overline{p(x)}$, $p(e)\ge 0$ and $|p(x)|\le p(e)$. If $p(e)=0$, then $p=0$ and take $\nu =0$. Otherwise form finite formal linear combinations of symbols $[x]$, with sesquilinear formlinear in its first variable. Positive definiteness makes this form positive semidefinite. The Cauchy-Schwarz inequality for such a form shows that null vectors are orthogonal to everything, so quotienting by them gives an inner product space. Complete it to a Hilbert space $H$. Since $G$ is Abelian, translation $U_a[x]=[ax]$ preserves the form and extends to a unitary operator. Its inverse is $U_{a^{-1}}$, so $(U_a)$ is a unitary representation. With $v=[e]$,The translated copies of $v$ span a dense subspace. MoreoverFinite sums and density, together with $\Vert U_a\Vert =1$, give strong continuity on all of $H$. This constructs the cyclic unitary representation of a positive-definite function directly.

For $f\in L^1(G)$ define the integrated unitary representation $\pi (f)=\int f(x)U_{x}dx$, with the integral applied to each vector. Its norm is at most $\Vert f{\Vert }_1$. Fubini gives $\pi (f\ast g)=\pi (f)\pi (g)$, and inversion in the Abelian group gives $\pi (f^{\ast })=\pi (f{)}^{\ast }$. Thusis a commutative C-star algebra. It is nondegenerate: choose nonnegative compactly supported functions $u_V$ of Haar integral one with supports shrinking to the identity element. Strong continuity impliesfor every $w\in H$. These form an approximate identity in the representation. Neighbourhoods may form a net, since no metrizability is being assumed here.

The Commutative Gelfand--Naimark theorem identifies $\mathcal{A}$ with $C_0(X)$, where $X$ is its character space. The positive vector functional $\omega (B)=⟨Bv,v⟩$ becomes a bounded positive functional on $C_0(X)$, so the Riesz-Markov-Kakutani representation theorem supplies a finite positive Radon measure $\rho$ withThese general representation theorems are applied to the explicitly constructed algebra; they are not an assumption of Bochner's conclusion.

For each $\theta \in X$, the functional $f\mapsto \theta (\pi (f))$ is a nonzero algebra character of $L^1(G)$, since $\pi (L^1(G))$ is dense in $\mathcal{A}$. By Question 3 it has the form $\int f(x){\chi }_{\theta }(x)dx$. The resulting map $\iota :X\to \hat{G}$ is continuous: pointwise convergence of $\theta$ on $\mathcal{A}$ gives convergence on all $\pi (f)$, and the topological character identification in Question 3 gives compact-open convergence of ${\chi }_{\theta }$. Push $\rho$ forward under $\iota$ to obtain $\nu$. This finite measure is Radon: compact subsets approximating the measure of any inverse-image Borel set have compact images, giving inner regularity; finiteness and complements then give outer regularity.

To recover $p(x)$, note that $U_x\pi (u_V)=\pi (T_x{u}_V)\in \mathcal{A}$. HenceThe left side tends to $p(x)$. On the right, $|{\Phi }_{\chi }(u_V)|\le 1$ and ${\Phi }_{\chi }(u_V)\to 1$ uniformly on compact subsets of $\hat{G}$. For completeness, the evaluation $(\chi ,x)\mapsto \chi (x)$ is jointly continuous: near $x_0$ restrict $x$ to a compact neighbourhood, control the character uniformly there, and use continuity of the fixed character. A finite cover of a compact set of characters then makes those characters equicontinuous at $e$, proving the stated uniform convergence. Continuity of $\iota$ gives uniform convergence on compact subsets of $X$, and the finite Radon measure $\rho$ makes the remaining tails arbitrarily small. Thus passage to the limit is valid even for a net, andTaking $x=e$ also gives the required total mass $p(e)$.

Conversely, a finite positive Radon measure givesIts transform is continuous. Indeed, choose a compact set of characters with arbitrarily small measure outside it; joint continuity of evaluation gives uniform continuity in $x$ near the specified point on that compact set, while the outside contribution is bounded by twice its measure. This proves continuity without using an unjustified dominated-convergence assertion for arbitrary nets.

Finally suppose two finite Radon measures have the same transform. Multiply by $f\in L^1(G)$ and integrate over $G$. Fubini shows that the measures agree on every function $\chi \mapsto {\Phi }_{\chi }(f)$. Question 3 showed that these functions are uniformly dense in $C_0(\hat{G})$. Boundedness of both measure functionals extends equality to all of $C_0(\hat{G})$; uniqueness in the Riesz representation theorem gives equality of the measures. This proves existence, uniqueness and the converse, completing Bochner's theorem. In particular $p(e)=1$ corresponds exactly to a probability measure on the dual group.