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This may be a fairly simple question. Suppose G is a (T0) topological group. Assume that G is path-connected, locally path-connected, and semilocally simply connected, so that covering space theory applies.

Question: Is it true that for any element of $\pi_1(G,e)$ (where e is the identity element of G), there exists a [ADDED: continuous] homomorphism from $S^1$ to $G$ having that element of $\pi_1(G,e)$ as its homotopy class?

Another way of formulating this is that there is a set map:

$$\operatorname{Hom}_{cts}(S^1,G) \to \pi_1(G,e)$$

The subscript cts is to indicate continuous.

(when G is abelian, the left side has a group structure too [ADDED: under pointwise multiplication], and the Eckmann-Hilton principle tells us that we get a group homomorphism).

  1. Is the set map surjective in all cases (regardless of whether G is abelian)?
  2. Does the image of $\operatorname{Hom}(S^1,G)$ generate $\pi_1(G,e)$ as a group (this is equivalent to surjectivity when $G$ is abelian)?
  3. Does surjectivity work for Lie groups? Compact Lie groups?
  4. Does the weaker formulation (2) work for Lie groups?

I have a sketch of an argument/proof that may show (4) (basically, using properties of one-parameter subgroups), but I'm hoping somebody will have a clean proof that works in general for topological groups.

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Do you mean for the homomorphisms to be continuous (stupid question, but necessary). What group structure do you want to Hom(S^1,G)? Don't you get another monoid structure from convolution or similar? (S^1 is compact after all) This would then interact with the monoid structure given by pointwise multiplication, and I'm pretty sure they share the same identity. If interchange holds, then we know that Hom(S^1,G) is an abelian monoid. Whether your map is a homomorphism is an interesting question (to me at least). And as a warning, the proof for (4) will probably fail for Frechet Lie groups ... –  David Roberts Jan 14 '11 at 22:49
    
...as the exponential map g -> G is not surjective on any neighbourhood of the identity. But maybe (2) will spring to your aid in that case. –  David Roberts Jan 14 '11 at 22:50
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David, thanks for noting the requirement of continuity. I've edited the question accordingly. Indeed, if we don't specify that the homomorphisms are to be continuous, the map to pi_1(G) is not well-defined, so I've added that stipulation. In the case of Lie groups, of course, continuous is equivalent to being smooth and in fact being a one-parameter subgroup that loops back. –  Vipul Naik Jan 14 '11 at 22:56
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It is probably far from what you're looking for, but you can find counter examples in symplectic geometry. Let $(M,\omega)$ be a symplectic manifold such that $M$ doesn't admit any circle action then there are no homomorphism from $S^1$ to $Ham(M,\omega)$ (the group of hamiltonian diffeomorphisms). However you can find plenty of $4$-dimensionnal example where $\pi_1(Ham(M,\omega)$ is non-trivial (blow ups of $K3$ surfaces for instance). –  Noz Jan 14 '11 at 23:11
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2 Answers

up vote 22 down vote accepted

No. A continuous homomorphism $S^1\to G$ yields a map $BS^1\to BG$. The space $BS^1$ is homotopy equivalent to $\mathbb CP^\infty$. There is a topological group $G$ such that $BG$ is homotopy equivalent to the sphere $S^2$. A map corresponding to a generator of $\pi_1G=\pi_2BG=H_2S^2$ would give an isomorphism $H^2BG\to H^2BS^1$, but this is incompatible with the cup product.

EDIT: This example is universal in the following sense: A standard way of making a Kan loop group for the suspension of a based simplicial set $K$ is to apply (levelwise) the free group functor from based sets to groups. The realization of this is then the universal example of a topological group $G$ equipped with a continuous map $|K|\to G$. Apply this with $K=S^1$.

EDIT: Yes in the Lie group case. It suffices to consider compact $G$ since a maximal compact subgroup is a deformation retract. Now put a Riemannian structure on $G$ that is left and right invariant, and use that the geodesics are the cosets of the $1$-parameter subgroups, and that in a compact Riemannian manifold every loop is freely homotopic to a closed geodesic.

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Is it obvious that "there is a topological group $G$ such that $BGA$ is homotopy equivalent to $S^2?$ –  Igor Rivin Jan 15 '11 at 2:01
    
Not obvious, but it more or less follows from standard things. $G$ can be the realization of a simplicial group: a "Kan loop group" for a simplicial model for $S^2$. I confess that I am ignoring the detail that the realization of a simplicial group is not quite a topological group, because realization is not product-preserving. (Not unless you take products in the compactly generated sense.) But I trust that someone out there can close this gap somehow. –  Tom Goodwillie Jan 15 '11 at 2:25
    
Tom: Thanks for your solution sketch. If you (or somebody else) can fill in the missing details into your main answer, I'll check mark and accept it. –  Vipul Naik Jan 15 '11 at 17:50
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I find that there is no gap. The realization of a countable simplicial group is in fact a topological group. (They say that for a product of two countable CW complexes the product topology equals the CW topology.) –  Tom Goodwillie Jan 16 '11 at 3:34
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An alternative argument for the counterexample: Milnor showed that loopspaces are homotopy equivalent to topological groups, so consider $\Omega S^2$. The Pontryagin ring $H_*(\Omega S^2)$ is a polynomial ring on a generator in degree 1, so there can be no H-space homomorphism $S^1 \to \Omega S^2$ inducing an isomorphism on $H_1$. –  Allen Hatcher Jan 16 '11 at 22:44
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Here's an abelian counterexample: Let $G$ be the following topological group homotopy equivalent (as a space) to $\mathbb RP^\infty$. It is the realization of the simplicial group that corresponds by Dold-Kan to the chain complex in which the only nontrivial chain group, the first, has order $2$. To put it another way, if a group $A$ is abelian then the usual simplicial model for $BA$ is itself a group (in fact, abelian); let $G$ be $BA$ where $A$ has order $2$.

In this topological group, every nontrivial element has order $2$. Therefore there is no nontrivial homomorphism (continuous or not) from $S^1$.

Note that the reasoning here is very different from what it was in my other example. It is not that there is no homotopically nontrivial map $BS^1\to BG$, and it is not (as in Allen Hatcher's comment) that there is no nontrivial $H$-space map (homomorphism up to homotopy) $S^1\to G$.

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Very nice. . –  David Roberts Jan 17 '11 at 22:57
    
Ah, of course! This exponent-2 group was also considered here: mathoverflow.net/questions/43002/… –  Todd Trimble Jan 18 '11 at 2:16
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