Question: for which \(n\) is \(S^n\) a topological group?
While hanging out with a few friends, we came up with a question that should have a clear direct answer: can the 2-sphere \(S^2\) be given the structure of a Lie group? If not, what is the obstruction?
My initial intuition here was something along these lines: if \(S^2\) were a topological group, we could consider its Lie algebra by taking the tangent space \(T_e\) at the identity \(e\) (where without loss of generality, we can rotate \(S^2\) to identify \(e\) with the North pole.) A standard argument shows that the translation maps \(\tau_g:x\mapsto gx\) are diffeomorphisms, and are “transitive” in the sense that any point \(p_1\) can be carried to \(p_2\) by the translation \(\tau_{p_2 p_1^{-1}}\).
Using this, the heuristic is that we expect the tangent space to “look the same” after performing a series of translations that ultimately maps \(e\) to itself, for example \begin{align*} \tau: S^2 \to S^2 \\ \tau \coloneqq\tau_{p_2^{-1}} \tau_{p_2p_1^{-1}} \tau_{p_1}. \end{align*} Specifically, my hope was that the induced map \(\tau^*: T_e \to T_e\) would not just be an isomorphism, but in fact the identity map. This can be phrased in terms of the parallel transport of a tangent vector \(\mathbf{v} \in T_e\) along geodesics connecting \(e\) to \(p_1\), \(p_1\) to \(p_2\), and finally \(p_1\) to \(e\). However, because the standard metric on \(S^2\) induces nonzero curvature, there is nontrivial holonomy and the vector returns with some rotation. Thus this composition isn’t the identity, a contradiction.
It’s not so clear to me what part doesn’t go through:
- Does nonzero curvature or holonomy on a manifold obstruct it from having a Lie group structure?
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Do all Lie groups have zero curvature and trivial holonomy groups?
- Note that \(\mathrm{Holo}(S^2) = SO(2, {\mathbf{R}})\).
- Should any composition of translations fixing \(e\) induce the identity on \(T_e\), or just a vector space automorphism?
It turns out that the following is true:
Theorem If \(S^n\) is a group, then \(n\) must be odd.
Proof Suppose otherwise and let \(m: (S^n)^2 \to S^n\) be the multiplication. Then \(m_g\) defined by \(m_g(x) = m(g, x)\) has no fixed points. By the Lefschetz fixed point theorem, \(\Lambda_{m_g} = 0\), so use the fact that \(\pi_1S^2 = 0\) to homotope \(m_g \sim \operatorname{id}_{S^2}\), which preserves \(\Lambda_{m_g}\). But \(\Lambda_{\operatorname{id}_X} \chi(X)\) and \(\chi(S^2) = 1 + (-1)^n\), and for this to be odd, \(n\) must be odd.
In fact, more is true:
Theorem \(S^n\) is a topological group \(\iff n=0,1,3\).
My impression was that this should be related to the fact that the only normed division algebras are \({\mathbf{R}}, {\mathbf{C}}, {\mathbb{H}}\), and \({\mathcal{O}}\), but couldn’t find an obvious proof along those lines.
#todo: find a proof