Past exam of the mathematics course of the University of Cambridge / 2008 / iii / Paper 9 / 1

A topological group is a group with a topology for which multiplication $G\times G\to G$ and inversion $G\to G$ are continuous. A Hausdorff axiom is not built into this definition. For neighbourhood bases ${\mathcal{B}}_x$ at each point, the precise neighbourhood criterion for a topological group isNecessity follows from continuity: the inverse image of a neighbourhood of $xy$ contains a product neighbourhood of $(x,y)$, and the inverse image of a neighbourhood of $x^{-1}$ contains a neighbourhood of $x$. Conversely, these inclusions give exactly those neighbourhood conditions for continuity of multiplication and inversion, since the rectangles $V\times W$ form a basis for the product topology.

An equivalent identity-level formulation is useful. Bases must be obtainable by translation, ${\mathcal{B}}_x=x{\mathcal{B}}_e$, and for every $U\in {\mathcal{B}}_e$ one must haveThese conditions are necessary because translations are homeomorphisms, multiplication is continuous at $(e,e)$, inversion is continuous at $e$, and each conjugation is a homeomorphism. For sufficiency at $(x,y)$, first choose $D$ with $D^2\subseteq U$, then $V$ with $y^{-1}Vy\subseteq D$ and take $W=D$. This gives $(xV)(yW)\subseteq xyU$. For inversion near $x$, first choose $H$ with $xH{x}^{-1}\subseteq U$, then $V$ with $V^{-1}\subseteq H$; thus $(xV{)}^{-1}=x^{-1}(x{V}^{-1}{x}^{-1})\subseteq x^{-1}U$. The pointwise criterion is therefore satisfied everywhere. If Hausdorffness is also required, the additional identity-neighbourhood condition is ${\bigcap }_{U\in {\mathcal{B}}_e}U=\{e\}$: it gives separation of distinct points by translating sufficiently small neighbourhoods.

False. In the additive topological group $\mathbb{R}$, the subgroup $\mathbb{Z}$ is closed: every noninteger has a sufficiently small interval disjoint from it. It is not open, because no neighbourhood of an integer consists only of integers. Thus a closed subgroup need not be an open subgroup.

True. If $H$ is an open subgroup, every left coset $gH$ is open because left translation is a homeomorphism. The other cosets partition the complement:That complement is open, so $H$ is a closed subgroup.

False as a whole: the identity component is normal, but need not be open. Write $G^{\circ }$ for the identity component. The continuous function $(x,y)\mapsto x{y}^{-1}$ maps the connected space $G^{\circ }\times G^{\circ }$ to a connected set containing $e$, hence into $G^{\circ }$. Thus $G^{\circ }$ is a subgroup. Conjugation by any $g$ is a homeomorphism fixing $e$, so it maps $G^{\circ }$ onto itself; therefore it is a normal subgroup. It is also closed, since the closure of a connected set is connected and a connected component is maximal.

For failure of openness, take the additive group $\mathbb{Q}$ with its usual topology. Any two distinct rational numbers can be separated inside $\mathbb{Q}$ by an irrational cut between them, so every connected subset is a singleton. Its identity component is consequently $\{0\}$, which is not open because every neighbourhood of zero contains nonzero rational numbers.

False. The additive group $\mathbb{Q}$ with its usual metric is a Hausdorff space but not a locally compact space. If a compact set were a neighbourhood of zero, it would contain all rationals in some interval $(-ϵ,ϵ)$. Choose an irrational $a$ in that interval and rational numbers $q_n\to a$ in the real line. All sufficiently late $q_n$ would lie in the compact neighbourhood, so a subsequence would converge in $\mathbb{Q}$. Its real limit would have to be $a$, a contradiction.

The other suggested example is the additive Hilbert space ${\ell }^2$. A compact neighbourhood of zero would contain a closed ball of some positive radius $r$. The vectors $r{e}_n$ in that ball have pairwise distance $r\sqrt{2}$, and thus no convergent subsequence. This again contradicts metric compactness. Both examples are topological groups whose operations are continuous, so Hausdorffness cannot supply local compactness.

True. Choose a compact neighbourhood $K$ of the identity in the locally compact group and put $S=K\cup K^{-1}\cup \{e\}$. Inversion is a homeomorphism, so $S$ is compact; it is symmetric and still a neighbourhood of $e$. DefineThe identity belongs to $S$, so these powers are increasing. Symmetry gives $H^{-1}=H$, and $S^m{S}^n=S^{m+n}$ gives closure under multiplication; hence $H$ is a subgroup. Each $S^n$ is compact, being the continuous image of the finite product of compact spaces. Thus $H$ is a sigma-compact space. Finally $S$ contains some open identity neighbourhood $U$, and $H={\bigcup }_{h\in H}hU$ is open. This constructs the required sigma-compact open subgroup of a locally compact group. The argument also works without Hausdorffness if local compactness means possession of compact neighbourhoods.

False. Give the two-element group $C_2$ the indiscrete topology and take $G=\mathbb{R}\times C_2$ with the product topology. The indiscrete factor is a topological group, since every map into an indiscrete space is continuous, and the product group operations are continuous. Points $(0,e)$ and $(0,a)$, $a\ne e$, have identical neighbourhoods, so $G$ is not a Hausdorff space. However, its continuous projection onto $\mathbb{R}$ is surjective. If $G$ were compact, its image $\mathbb{R}$ would be compact, which it is not.

True. Given a compatible left-invariant group metric $d$, definePullback by the bijection $x\mapsto x^{-1}$ preserves all metric axioms, and because inversion is a homeomorphism, $d_R$ induces the same topology. For right translation by $a$, left invariance givesThus $d_R$ is the required compatible right-invariant group metric. No metrizability theorem is needed.

False. On the additive topological group $\mathbb{R}$, put $d(x,y)=|x-y|$ and $d^{\prime }(x,y)=\sqrt{|x-y|}$. Both are invariant under left and right translations and induce the usual topology. The triangle inequality for $d^{\prime }$ follows from $\sqrt{a+b}\le \sqrt{a}+\sqrt{b}$ for $a,b\ge 0$. Butcan be arbitrarily close to zero and arbitrarily large. Consequently no $K>0$ can satisfy both proposed inequalities. Compatibility of the two invariant metrics is topological equivalence, which need not be bilipschitz equivalence.