differential-topologygeometric-topologygroup-theorylie-groupsprojective-space
Which of the projective spaces
$$\Bbb R\Bbb P^n, \quad \Bbb C\Bbb P^n,\quad\Bbb H\Bbb P^n$$
admits the structure of a topological group/Lie group (compatible with its usual topology)?
Trivially, $\Bbb R\Bbb P^2\cong\Bbb S^1$ does, as it can be interpreted as the unit complex numbers.
According to this answer, the $\Bbb C\Bbb P^n$ do not admit a Lie group structure for any $n\ge 2$. What about a topological group structure?
What about the others?
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).
The question is actually quite hard. Below is a partial answer.
First of all, I am using the "textbook" definition of a manifold (including Hausdorff and 2-nd countable). Thus, for instance, every countable group has a Lie group structure (when equipped with the discrete topology). In view of this, it makes sense to restrict to connected Lie groups: Once you gave an algebraic description of connected Lie groups, a general group $G$ is isomorphic to a Lie group if and only if $G$ contains a normal subgroup $H< G$ of (at most) countable index such that $H$ is isomorphic to a connected Lie groups.
I will also work with real Lie groups.
With this in mind, I will start with few examples.
Lemma 1. ${\mathbb Q}$ (the additive group of rational numbers) can be abstractly defined as a nontrivial torsion-free divisible rank 1 abelian group.
(Recall that an abelian group $A$ is said to be divisible if for every natural number $n$ and every $a\in A$, there exists $b\in A$ such that $nb=a$ (where $nb$ means $b+b+...+b$, sum is taken $n$ times). An abelian group $A$ is said to have rank one if for any two non-neutral elements $a, b\in A$ there exist natural numbers $m, n$ such that $ma=nb$ or $ma=-nb$.)
Lemma 2. ${\mathbb R}$ (the additive group of real numbers) can be abstractly defined as the direct sum of $c$ copies of ${\mathbb Q}$ where $c$ has cardinality of continuum.
Thus, we have a definition of the additive group ${\mathbb R}^n$ as the $n$-fold direct sum of ${\mathbb R}$, but as an abstract group, it is isomorphic to ${\mathbb R}$.
Corollary. 1. The group ${\mathbb S}^1$ (the multiplicative group of unit complex numbers) can be abstractly defined as the as the quotient of ${\mathbb R}$ by an infinite cyclic subgroup. Equivalently, it is isomorphic to ${\mathbb R}\times ({\mathbb Q}/{\mathbb Z})$.
For instance, ${\mathbb C}^\times$, the multiplicative group of complex numbers, can be abstractly defined as the direct product ${\mathbb S}^1\times {\mathbb R}$.
I will also need a description of ${\mathbb R}$ as a field. Let $F$ be a field. Define the relation $\le$ on $F$ by: $x\le y \iff \exists z\in F, (y-x)=z^2$, i.e. $y-x$ is the square of an element of $F$. Then $F$ is isomorphic to ${\mathbb R}$ if and only if the following hold:
a. $\le$ is a total order on $F$.
b. $(F, \le)$ is an ordered field.
c. $(F, \le)$ satisfies the completeness axiom: For every nonempty subset $E\subset F$ which is bounded above, there exists the least upper bound.
Now, the natural thing to do would be to proceed to nilpotent and then solvable Lie groups since these can be described as iterated central extensions and semidirect products (starting with abelian Lie groups). However, I do not know how to do this and I suspect that this is already quite hard. The nilpotent case hinges on the following:
Let $A$ be an abelian topological group and $G$ be a topological group. Central extensions (in the category of topological groups) of the form $$ 1\to A\to \tilde{G}\to G\to 1 $$ (where $A$ embeds in $G$ as a central subgroup) are classified by continuous cohomology classes, elements of $H_{c}^2(G; A)$. Abstract central extensions are described as elements of $H^2(G; A)$. There is a natural homomorphism $$ H_{c}^2(G; A)\to H^2(G; A) $$ and the question is to understand its image (as topology of $G$ and $A$ varies!) in the case when $G$ is a nilpotent Lie group and $A$ is an abelian Lie group. It is conceivable that (as long as we do not insist on having connected Lie groups) that every such extension is realizable.
The problem becomes even harder for solvable group, so I will consider the complementary case case of semisimple Lie groups: Every simply connected Lie group is isomorphic to a semidirect product of a solvable Lie group and a semisimple Lie group.
Up to taking finite central extensions, every connected semisimple Lie group is a finite direct product of connected simple Lie groups. The latter are defined by the property that their Lie algebras are simple. This is "almost the same" as the simplicity condition as abstract groups (the difference comes again from central extensions by finitely generated abelian groups: mostly finite or infinite cyclic). Let me ignore the distinction in order to simplify the matter.
Simple Lie groups are completely classified (see e.g. this wikipedia article) into several infinite families and finitely many exceptional groups. An easy way out is to say that a simple Lie group is one from a list. This is, of course, completely unsatisfactory, so I will give an answer (of sorts) below.
First of all, every simple Lie group has an important numerical invariant, called rank (see for instance this discussion). For instance, $SL_n({\mathbb R})$ has rank $n-1$.
The theory then splits in three very different cases:
$rank(G)=0$, i.e. $G$ is compact.
$rank(G)=1$; for instance, $O(n,1)$ and $U(n,1)$.
Higher rank: $rank(G)\ge 2$.
I have to say that I do not have anything interesting to say about compact groups, so below I will discuss an algebraic description of groups of higher rank. This feels like the "generic case" and it is hard enough.
The key thing to know is that each semisimple Lie group $G$ has a BN-pair structure:
This structure is given by a pair of subgroups $B< G, N<G$ and their intersection $H=B\cap N$. This data should satisfy the 5 axioms listed in the linked Wikipedia page. These axioms involve no topology and are purely group-theoretic.
Given this, one defines the Weyl group $W=N/H$ and its distinguished generating set $S$ ("simple generators"). The cardinality $n$ of $S$ is going to be the rank of $G$.
Many groups which are far from Lie groups will have such a BN-pair structure, we need extra axioms:
Axiom 6. The group $W$ is a finite Coxeter group; in particular, the cardinality $n$ of $S$ is finite.
(This axiom rules out Kac-Moody groups among other things.)
The Coxeter group $W$ has Coxeter-Dynkin diagram $D(W,S)$, which is a decorated graph encoding the relators of this finite group; the vertices of this graph are the simple generators $s\in S$. Let $n_1,...,n_k$ denote numbers of vertices of these connected components; thus, $n=n_1+...+n_k$. This number is the rank of $G$.
The case of simple Lie groups corresponds to $k=1$.
Theorem. (J.Tits) Suppose that each number $n_i$ is $\ge 3$. Then the group $G$ satisfying Axioms 1-6 is of "algebraic origin". Namely, there exists a division ring $F$ and a semisimple algebraic group ${\mathbb G}$ such that $G$ is isomorphic to the group of $F$-points of ${\mathbb G}$, ${\mathbb G}(F)$. (I am lying here a little bit but this little lie will be irrelevant.)
For instance, maybe ${\mathbb G}= SL_{n+1}$, then ${\mathbb G}(F)= SL_{n+1}(F)$: The group of matrices with entries in the field $F$ and unit determinant. As an example, one can take $F$ equal to the division ring of quaternions ${\mathbb H}$.
With a bit more work (one needs an extra "Moufang axiom" which I am not going to define here) this theorem also works for groups of rank 2. However, it utterly fails for groups of rank 1.
Furthermore, one can work out the ring structure of $F$ from the group-theoretic structure of $G$ (I am not going to do this either). The field requirement for $F$ amounts to solvability assumption on $B$. Now, we can use the characterization of ${\mathbb R}$ given earlier. Lastly, the groups ${\mathbb G}({\mathbb R})$ are exactly the connected semisimple real Lie groups (as mentioned above, up to taking some finite index subgroups and central extensions with finitely generated abelian kernels).
Like it or not, this is a characterization that you asked for, except for the fact that it leaves out the case of groups of rank $\le 1$ and restricts to semisimple Lie groups.
See this mathoverflow discussion for more references.
Update: As it turns out, the extension problem is not as bad as I thought. In Theorem 3 in
David Wigner, Cohomology of Topological Groups, Transactions of the American Mathematical Society, Vol. 178 (1973), pp. 83-93
it is proven that the natural map $H^*_c(G,A)\to H^*(G,A)$ is an isomorphism, where $G$ is a connected Lie group, $A$ is a simply connected abelian Lie group and the action of $G$ on $A$ is continuous. Hence, every extension of $G$ by $A$ (as an abstract group) is equivalent to an extension as a Lie group. Thus, indeed, the overall problem of characterizing Lie groups as abstract groups reduces to:
(a) The characterization problem for semisimple Lie groups which is mostly solved by the answer given above.
(b) Characterizing which abstract homomorphisms $\phi: G\to Aut(A)$ from connected Lie groups to abstract automorphism groups of abelian Lie groups (for simplicity, ${\mathbb R}^n$) are equivalent to continuous representations to the group of continuous automorphisms, i.e. continuous representatioins $$ G\to GL(n, {\mathbb R}) $$ This problem might be hard but at least it reduces to a problem of the finite-dimensional representation theory.
Best Answer
A topological group structure lifts to a covering space. Therefore, other than $\mathbb RP^1$ the only real projective space with a group structure is $\mathbb RP^3\simeq SO(3)$.