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Given any abstract group $ G $, how much is known about which types of topological/Lie group structures it might have?

Any abstract group $ G $ will have the structure of a discrete topological group (since generally, any set can be given the discrete topology), but there are groups that have no smooth structure. An example of this from Wikipedia is the group $ \mathbb{Q} $ with the subspace topology inherited from $ \mathbb{R} $. Which groups can occur as Lie groups? Are there specific families of groups that are known to have no smooth structure?

Similarly, how much can we know about the possible topologies on an abstract group $ G $? For example, which types of abstract groups admit a nontrivial (i.e., not the usual compact case) group structure?

In particular, I am curious about the extent to which properties of the abstract group determine properties of any associated topology/smooth structure.

Does anyone have any good references or a succinct answer for this question?

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I don't know of a good reference, but one example to keep in mind is that $\mathbb{R}^n$ and $\mathbb{R}$ are isomorphic as groups (at least, if you believe in the axiom of choice). This gives countable many topologies/Lie group structures for $\mathbb{R}$. –  Jason DeVito Dec 4 '12 at 1:52
    
Maybe what I should have asked is what can we say about the possible topological/Lie group structures? In the example, $\mathbb{R}$ has lots of structures, but in some sense they are all pretty tame; $\mathbb{R}^n$ is a fairly boring Lie group. Must every Lie group structure on $\mathbb{R}$ be like this? –  rondo9 Dec 4 '12 at 22:19
    
4  
@rondo9, abelian Lie groups are direct products of $\mathbb R^n$ and tori, and tori have torsion, so every Lie group structure on $\mathbb R$ must be one of the $\mathbb R^n$. –  Mariano Suárez-Alvarez Jan 5 '13 at 3:10
    
@Mariano: I believe that this result applies only to connected abelian Lie groups. For example, $ \mathbb{R}^{\times} $ is a $ 1 $-dimensional abelian Lie group that is disconnected, so it cannot be a direct product of $ \mathbb{R}^{n} $’s and tori. :) –  Haskell Curry Jan 8 '13 at 6:39

1 Answer 1

up vote 24 down vote accepted
+50

Section 1

Let us begin with the following theorem.

Theorem 1 Let $ G $ be a topological group. If $ G $ admits a Lie group structure, then this structure is unique up to diffeomorphism.

Proof: Suppose that $ \mathcal{A}_{1} $ and $ \mathcal{A}_{2} $ are smooth structures (maximal smooth atlases) on $ G $ that make it a Lie group. Observe that the identity map $ \text{id}_{G}: G \to G $ defines a continuous homomorphism of Lie groups from $ (G,\mathcal{A}_{1}) $ to $ (G,\mathcal{A}_{2}) $. A basic fact in the theory of Lie groups is that a continuous homomorphism between Lie groups is actually smooth. Hence, $ \text{id}_{G}: (G,\mathcal{A}_{1}) \to (G,\mathcal{A}_{2}) $ is a smooth mapping between smooth manifolds. As $ \text{id}_{G} $ is its own inverse, $ \text{id}_{G}: (G,\mathcal{A}_{2}) \to (G,\mathcal{A}_{1}) $ is also a smooth mapping between smooth manifolds. Therefore, $ (G,\mathcal{A}_{1}) $ is diffeomorphic to $ (G,\mathcal{A}_{2}) $. $ \spadesuit $

Theorem 1 says that if we fix a topology on a group $ G $, then there is at most one Lie group structure on $ G $. What happens when we vary the topology on $ G $ will be investigated in Section 3.


Section 2

In this section, we address the issue of which topological groups admit or do not admit Lie group structures. This issue is very much related to Hilbert’s Fifth Problem, which, in its original formulation, asks for the minimal hypotheses that one needs to put on a topological group so that it admits a Lie group structure. Once again, if a Lie group structure exists, then its uniqueness is guaranteed according to Section 1.

In their attempt to resolve Hilbert’s Fifth Problem, Andrew Gleason, Deane Montgomery and Leo Zippin proved the following deep theorem in the 1950’s.

Theorem 2 (G-M-Z) Let $ G $ be a topological group. If $ G $ is locally Euclidean, then $ G $ admits a Lie group structure.

To appreciate the power of this theorem, realize that it says, “A topological group that is merely a topological manifold is, in fact, a Lie group!” A detailed and insightful proof may be found in this set of notes posted by Professor Terence Tao on his research blog. Upon reading his notes, one will notice that a key concept used in the proof is that of a group having no small subgroups.

Definition A topological group $ G $ is said to have the No-Small-Subgroup (NSS) Property iff there exists a neighborhood $ U $ of the identity element that does not contain any non-trivial subgroup of $ G $.

One can consult this other set of notes by Professor Tao in order to understand the importance of the NSS Property. The Japanese mathematicians Morikuni Gotô and Hidehiko Yamabe were the first ones to formulate this property (in a 1951 paper), which Yamabe then used to recast the G-M-Z solution of Hilbert’s Fifth Problem in a manner that reflects the algebro-topological structure of Lie groups more directly.

Theorem 3 (Yamabe) Any connected and locally compact topological group $ G $ is the projective limit of a sequence of Lie groups. If $ G $ is locally compact and has the NSS Property, then $ G $ admits a Lie group structure.

Theorem 3 implies Theorem 2 because a locally Euclidean group is locally compact (an obvious assertion) and has the NSS Property (a non-trivial assertion). It is far from obvious why both local compactness and the NSS Property should imply locally Euclidean back; indeed, this is the main content of Yamabe’s deep theorem.

Now, a well-known example of a topological group that does not admit a Lie group structure is the group $ \mathbb{Z}_{p} $ of the $ p $-adic integers with the $ p $-adic topology. It is a completely metrizable and compact topological group, but as it is a totally disconnected space, it cannot admit a Lie group structure for obvious topological reasons.

In general, profinite groups (these are topological groups that are compact, Hausdorff and totally disconnected) do not admit Lie group structures. Examples of profinite groups are the discrete finite groups (these are $ 0 $-dimensional manifolds, but we can ignore them as topologically uninteresting), étale fundamental groups of connected affine schemes and Galois groups equipped with the Krull topology (this means that I am referring also to groups that correspond to infinite Galois extensions, not just the finite ones). In fact, every profinite group is an étale fundamental group in disguise, in the sense that every profinite group is topologically isomorphic to the étale fundamental group of some connected affine scheme.

Although the group $ \mathbb{Q}_{p} $ of $ p $-adic numbers is not profinite (it is locally compact, not compact), it is also a totally disconnected space, so it does not admit a Lie group structure.

Until now, we have stayed within the realm of locally compact topological groups. The OP has asked about the non-locally compact case, so here is an attempt at a response.

The Swedish functional-analyst Per Enflo did his Ph.D thesis on Hilbert’s Fifth Problem by investigating to what extent the results of Montgomery and Zippin, formulated only in the finite-dimensional setting, could be carried over to the infinite-dimensional setting. He performed his investigation on topological groups that are modeled on (locally homeomorphic to) infinite-dimensional Banach spaces. The main reason for using infinite-dimensional Banach spaces is due to the following basic theorem from functional analysis.

Theorem 4 A Banach space is locally compact iff it is finite-dimensional.

Citing unfamiliarity with Enflo’s work, we kindly request the reader to consult the references that are provided below.

Note: The OP did say that giving references only was okay! :)


Section 3

In this final section, we shall see how to put different topological structures on an abstract group. Toward this end, let us state the following theorem.

Theorem 5 Let $ G $ be an abstract group and $ H $ a topological group. For any group homomorphism $ \phi: G \to H $, the pre-image topology on $ G $ induced by $ \phi $ makes $ G $ a topological group. If $ \phi $ is further an isomorphism, then $ G $ with the pre-image topology is topologically isomorphic to $ H $.

Proof: The pre-image topology on $ G $ induced by $ \phi $ is defined as the following collection of subsets of $ G $: $$ \{ {\phi^{\leftarrow}}[U] \in \mathcal{P}(G) ~|~ \text{$ U $ is an open subset of $ H $} \}. $$ Pick an open subset $ U $ of $ H $. Then \begin{align} \{ (g_{1},g_{2}) \in G \times G ~|~ g_{1} g_{2} \in {\phi^{\leftarrow}}[U] \} &= \{ (g_{1},g_{2}) \in G \times G ~|~ \phi(g_{1} g_{2}) \in U \} \\ &= \{ (g_{1},g_{2}) \in G \times G ~|~ \phi(g_{1}) \phi(g_{2}) \in U \} \\ &= {(\phi \times \phi)^{\leftarrow}}[\{ (h_{1},h_{2}) \in H \times H ~|~ h_{1} h_{2} \in U \}] \\ &=: {(\phi \times \phi)^{\leftarrow}}[V]. \end{align} Multiplication is continuous in $ H $, so $ V $ is an open subset of $ H \times H $. Therefore, $ {(\phi \times \phi)^{\leftarrow}}[V] $ is an open subset of $ G \times G $ w.r.t. the product pre-image topology. As $ U $ is arbitrary, this implies that group multiplication in $ G $ is indeed continuous w.r.t. the pre-image topology.

Next, we have \begin{align} \{ g \in G ~|~ g^{-1} \in {\phi^{\leftarrow}}[U] \} &= \{ g \in G ~|~ \phi(g^{-1}) \in U \} \\ &= \{ g \in G ~|~ [\phi(g)]^{-1} \in U \} \\ &= {\phi^{\leftarrow}}[\{ h \in H ~|~ h^{-1} \in U \}] \\ &=: {\phi^{\leftarrow}}[W]. \end{align} Inversion is continuous in $ H $, so $ W $ is an open subset of $ H $. Therefore, $ {\phi^{\leftarrow}}[W] $ is an open subset of $ G $ w.r.t. the pre-image topology. As $ U $ is arbitrary, this implies that inversion in $ G $ is indeed continuous w.r.t. the pre-image topology.

The proof of the final statement is easy enough to be left to the reader. $ \quad \spadesuit $

For distinct positive integers $ m $ and $ n $, the groups $ \mathbb{R}^{m} $ and $ \mathbb{R}^{n} $ are isomorphic because they are isomorphic as $ \mathbb{Q} $-vector spaces. To prove the second assertion, first use the Axiom of Choice to deduce the existence of Hamel $ \mathbb{Q} $-bases for $ \mathbb{R}^{m} $ and $ \mathbb{R}^{n} $. Then show that a Hamel $ \mathbb{Q} $-basis $ \beta_{m} $ for $ \mathbb{R}^{m} $ and a Hamel $ \mathbb{Q} $-basis $ \beta_{n} $ for $ \mathbb{R}^{n} $ have the same cardinality, namely $ 2^{\aleph_{0}} $. Any bijection (there are uncountably many) from $ \beta_{m} $ to $ \beta_{n} $ now defines a unique vector-space isomorphism from $ \mathbb{R}^{m} $ to $ \mathbb{R}^{n} $.

Given this vector-space isomorphism, transfer the standard topology on $ \mathbb{R}^{n} $ to $ \mathbb{R}^{m} $. With the pre-image topology, $ \mathbb{R}^{m} $ is a topological group that is topologically isomorphic to $ \mathbb{R}^{n} $ with the standard topology. It follows from Invariance of Domain (a result in algebraic topology) that the new $ \mathbb{R}^{m} $ is not topologically isomorphic to $ \mathbb{R}^{m} $ with the standard topology.

The OP has asked if there is a non-trivial example involving non-Euclidean spaces. Off-hand, I do not have one in mind, but one can carry out the following procedure, which is in the same spirit as the previous example.

(1) Take two known topological groups, $ G $ and $ H $, with different topological properties.

(2) If one can find a discontinuous group isomorphism $ \phi: G \to H $, use $ \phi $ to transfer the topology on $ H $ to $ G $.

(3) Then $ G $ with the pre-image topology is not topologically isomorphic to $ G $ with the original topology.

(4) If $ H $ further admits a Lie group structure, then this Lie group structure can be transferred to $ G $, where there might have been none before.

References

  • Montgomery, D; Zippin, L. Topological Transformation Groups, New York, Interscience Publishers, Inc. (1955).

  • Gotô, M. Hidehiko Yamabe (1923 - 1960), Osaka Math. J., Vol. 13, 1 (1961), i-ii.

  • Yamabe, H. On the Conjecture of Iwasawa and Gleason, Annals of Math., 58 (1953), pp. 48-54.

  • Yamabe, H. A Generalization of a Theorem of Gleason, Annals of Math., 58 (1953), pp. 351 - 365.

  • Enflo, P. Topological Groups in which Multiplication on One Side Is Differentiable or Linear, Math. Scand., 24 (1969), pp. 195–197.

  • Enflo, P. On the Nonexistence of Uniform Homeomorphisms Between $ L^{p} $ Spaces, Ark. Math., 8 (1969), pp. 103–105.

  • Enflo, P. On a Problem of Smirnov, Ark. Math., 8 (1969), pp. 107–109.

  • Enflo, P. Uniform Structures and Square Roots in Topological Groups, I, Israel J. Math., 8 (1970), pp. 230–252.

  • Enflo, P. Uniform Structures and Square Roots in Topological Groups, II, Israel J. Math., 8 (1970), pp. 253–272.

  • Magyar, Z. Continuous Linear Representations, Elsevier, 168 (1992), pp. 273-274.

  • Benyamini, Y; Lindenstrauss, J. Geometric Nonlinear Functional Analysis, Volume 1, AMS Publ. (1999).

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3  
Pretty impressive :-) –  Mariano Suárez-Alvarez Jan 5 '13 at 3:06
    
Maybe this is cheating a bit: All infinite-dimensional separable Banach spaces are isomorphic (as groups) to the vector space of dimension continuum. They all are homeomorphic to $\mathbb{R}^\mathbb{N}$ by a theorem of Anderson. They are isomorphic as topological groups if and only if they are isomorphic Banach spaces. Considering the spaces $\ell^p$ with $1 \leq p\lt \infty$ (for which it is easy to show they are homeomorphic) you obtain a continuum of non-isomorphic topological group structures on the same abstract group. –  Martin Jan 5 '13 at 5:16
    
@Martin: That is very true. I also know of a theorem by David Henderson that classifies separable metric manifolds that are modeled on a separable infinite-dimensional Hilbert space, i.e., $ {\ell^{2}}(\mathbb{N}) $. It states that every such manifold must be homeomorphic to an open subset of $ {\ell^{2}}(\mathbb{N}) $. Banach spaces are much wilder, so one is definitely hard-pressed to find results this nice pertaining to Banach manifolds. I am not sure if Enflo ever dealt with manifolds that are modeled on non-separable Banach spaces. –  Haskell Curry Jan 5 '13 at 5:45
    
this is a really nice answer. thanks a bunch –  rondo9 Jan 5 '13 at 21:37
    
@rondo9: You’re welcome! It was fun undertaking the research behind this post. –  Haskell Curry Jan 5 '13 at 22:27

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