# Can $S^2$ be turned into a topological group?

I know that $S^1$ and $S^3$ can be turned into topological groups by considering complex multiplication and quaternion multiplication respectively, but I don't know how to prove or disprove that $S^2$ can.

This is just a recreational problem for me. Here's the work I have done:

We know that $S^2$ is homogenous and doesn't have the fixed-point property, so those don't help. I know that for any subgroup $H$ of $S^2$ that $S^2/H$ (the coset space with quotient topology) is homogenous. So maybe I can find a subgroup that would have to create a non-homogenous space. The only subgroups that I know would have to exist are cyclic subgroups. But I have no clue what $S^2/\langle x\rangle$ would even look like for any $x\in S^2$.

Any help is appreciated.

EDIT

Mike Miller's answer below certainly makes short work of my problem. But if someone comes up with a more elementary approach, I'd be appreciative.

• what about $SO(3)$ ? Jun 11, 2014 at 1:30
• $SO(3)$ is a Lie group. $S^2$ cannot be made into a Lie group else that would imply a non-vanishing vector field on $S^2$ (contradicting the Hairy ball theorem).
– user123641
Jun 11, 2014 at 1:31
• It would be parallelizable. Jun 11, 2014 at 3:20

We'll show that if $S^n$ is a topological group then $n$ must be odd.

Argument 1: Suppose $m : S^n \times S^n \to S^n$ is a topological group structure on $S^n$. For $g \in S^n$ not equal to the identity, $m(g, -) : S^n \to S^n$ has no fixed points, and so by the Lefschetz fixed point theorem, its Lefschetz trace must be $0$. On the other hand, $S^n$ is path-connected, so $m(g, -)$ is homotopic to $m(e, -) = \text{id}_{S^n}$, and in particular the two have the same action on homology and hence the same Lefschetz trace. But the Lefschetz trace of the identity is the Euler characteristic, which is $1 + (-1)^n$; hence $n$ must be odd. $\Box$

More generally, the same argument shows that if a compact path-connected triangulable space has a topological group structure then its Euler characteristic must be $0$.

Argument 2: I think this is a condensed form of Vladimir Sotirov's argument. Any $H$-space structure on $S^n$ induces a Hopf algebra structure on the cohomology $H^{\bullet}(S^n)$, and in particular a coproduct $\Delta$. If $x$ denotes a generator of $H^n(S^n)$, then for degree reasons and because of the existence of a counit, the coproduct must take the form

$$\Delta(x) = 1 \otimes x + x \otimes 1 \in H^{\bullet}(S^n) \otimes H^{\bullet}(S^n)$$

and hence

$$\Delta(x^2) = (1 \otimes x + x \otimes 1)^2 = 1 \otimes x^2 + (-1)^{n^2} x \otimes x + x \otimes x + x^2 \otimes 1$$

where here we recall that on a tensor product of cohomology rings the cup product takes the form

$$(a \otimes b) \cup (c \otimes d) = (-1)^{\deg(b) \deg(c)} (a \cup c) \otimes (b \cup d)$$

for homogeneous $a, b, c, d$. Since $x^2 = 0$ and $x \otimes x$ is not torsion, the above relation can only hold if $(-1)^{n^2} + 1 = 0$; hence $n$ must be odd. $\Box$

More generally, Hopf proved that over a field, the cohomology of a connected $H$-space with finitely generated cohomology is an exterior algebra on odd generators.

• I think this is about as clean as it gets.
– user98602
Jun 11, 2014 at 3:21
• @Qiaochu Yuan: I don't understand your first argument. The antipodal map acts fixed point freely on $S^n$, so from your argument I can deduce that is homotopic to the identity map on $S^n$. Aug 29, 2014 at 20:07
• @Yassine: I don't see the relevance of my argument to your (false) claim. The antipodal map acts by $(-1)^{n+1}$ on top homology, so it is homotopic to the identity iff $n$ is odd, and for all $n$ it has Lefschetz trace $1 + (-1)^n (-1)^{n+1} = 0$. Aug 30, 2014 at 0:14
• (The argument is that if you have a map $S^n \to S^n$ which both acts fixed point freely and is homotopic to the identity on $S^n$, then $n$ is odd.) Aug 30, 2014 at 1:18
• @MSIS: yes and yes, and the same for smooth structures, not just topologies. Sep 4, 2020 at 1:02

No, it's not possible.

Theorem (von Neumann): A compact locally Euclidean group is a Lie group. Since $S^2$ is compact, if it had a topological group structure, this group structure would provide a Lie group structure; as you mention in the comments, this is impossible by the hairy ball theorem (because Lie groups are parallelizable).

In fact, a more general fact is true: any locally Euclidean group is necessarily a Lie group. This is a theorem of Gleason, Montgomery, and Zippin, as mentioned in the linked reference; Montgomery summarizes their result in the note "Topological Transformation Groups", which also references their original papers.

Using algebraic topology, one can in a relatively straightforward matter the following:

Proposition For even $n>0$, there does not exist a map $\mu\colon S^n\times S^n\to S^n$ such that $$\mu\circ i_1=\mathrm{id}_{S^n}=\mu\circ i_2,$$ where $i_1,i_2\colon S^n\to S^n\times S^n$ are (continuous) inclusions of the form $S^n\to S^n\times\{x_1\}$ and $S^n\to\{x_2\}\times S^n$.

Whether algebraic topology is less elementary than using von Neumann's theorem to pass to Lie groups, giving a parallel vector field, and then going through the hairy ball theorem (or its higher dimension analogs) is... debatable. Nevertheless, the algebraic topology result is more general since it shows the impossibility not only of topological group structure (by taking $x_1=x_2=e$), but of the structure of a topological magma (i.e. an H-space).

Proof. The Künneth formula tells us that the (unital) cohomology algebra $H^*(S^n\times S^n)$ is isomorphic to the tensor product $H^*(S^n)\otimes H^*(S^n)$, with the isomorphism given by $$H^*(S^n)\otimes H^*(S^n)\ni a\otimes b\mapsto \pi_1^*(a)\cup\pi_2^*(b)\in H^*(S^n)$$ where $\pi_1,\pi_2\colon S^n\times S^n\to S^n$ are the canonical projection maps. In particular, if $a$ is a generator for $H^*(S^n)$ (as a unital algebra), then $\pi_1^*(a)$,$\pi_2^*(a)$ are (independent) generators of $H^*(S^n\times S^n)$ as a unital algebra, and of $H^n(S^n\times S^n)$ as a $2$-dimensional vector space.

Now, the continuous inclusions $i_1,i_2\colon S^n\to S^n\times S^n$ are maps such that $$\pi_i\circ i_j=\begin{cases}\mathrm{id}_{S^n}&i=j\\\text{constant}&i\neq j\end{cases}.$$ Therefore the induced homomorphisms $i_1^*,i_2^*\colon H^*(S^n\times S^n)\to H^*(S^n)$ and $\pi_1^*,\pi_2^*\colon H^n(S^n)\to H^*(S^n\times S^n)$ of cohomology algebras are such that $$i_j^*\circ\pi_i^*=\begin{cases}\mathrm{id}_{H^*(S^n)}&i=j\\0&i\neq j\end{cases}.$$ In particular, if $$\sigma=x\pi_1^*(a)+y\pi_2^*(a)\in H^n(S^n\times S^n),$$ then $i_1^*(\sigma)=xa$ and $i_2^*(\sigma)=ya$.

Consequently, if $\mu\colon S^n\times S^n\to S^n$ is such that $\mu\circ i_j=\mathrm{id}_{S^n},$ then the induced map $\mu^*\colon H^*(S^n)\to H^*(S^n\times S^n)$ on cohomology algebras must satisfy $$i_j^*\circ\mu^*=\mathrm{id}_{H^*(S^n)}\colon H^*(S^n)\to H^*(S^n).$$ We have that $i_j^*(\mu^*(a))=a$ for $\mu^*(a)\in H^n(S^n)$, so from the preceding observation $\mu^*(a)=\pi_1^*(a)+\pi_2^*(a)$.

On the other hand, $a\cup a\in H^{2n}(S^n)=0$, so we have \begin{align*}0=\mu^*(0) &=\mu^*(a\cup a) \\ &=\mu^*(a)\cup\mu^*(a) \\ &=(\pi_1^*(a)+\pi_2^*(a))^2 \\ &=\pi_1^*(a\cup a)+\pi_2^*(a\cup a)+\pi_1^*(a)\cup\pi_2^*(a)+\pi_2^*(a)\cup\pi_1^*(a) \\ &=\pi_1^*(a)\cup\pi_2^*(a)+(-1)^n\pi_1^*(a)\cup\pi_2^*(a). \end{align*} Since $$\pi_1^*(a)\cup \pi_2^*(a)\cong a\otimes a\neq0\in H^*(S^n\times S^n)\cong H^*(S^n)\otimes H^*(S^n),$$ this equality holds if and only if $n$ is odd.

Remark. I think (but I have not tried this) that if we assume further that $\mu\colon S^n\times S^n\to S^n$ is an associative multiplication, then applying an analogous argument to $$\mu^{(k)}\colon S^n\times S^n\times\dots\times S^n\to S^n$$ might give extra conditions on the dimension. One could maybe show in this way that $H^*(S^n)$ can have an induced structure as a Hopf algebra (i.e. associative multiplication, and an inversion map) only if the dimension is $n=2^m-1$. Or maybe not, I don't know.

• Even if it's not more elementary, it entirely avoids the smooth category, so I like this. +1.
– user98602
Jun 11, 2014 at 3:03
• This will also require some reading on my part, but I feel like it's more accessible to me. Thanks
– user123641
Jun 11, 2014 at 3:11

An $H$-space is an even more general structure than a topological group. It must have multiplication and inverse map that behave like a group up to homotopy. In fact, the only spheres which are $H$-spaces are $S^0,S^1,S^3$ and $S^7$. According to Wikipedia, this is Adams' Hopf Invariant One Theorem. It's actually not so hard to prove, using the fact that the cohomology ring of an $H$-space is a Hopf algebra, that only dimensions $2^n-1$ work. The hard part is showing that only these first examples are actually realizable. Note that $S^3$ is the unit quaternions and $S^7$ is the unit octonions.

• Can you suggest a proof sketch or a reference that $H^\ast(S^n)$ can only have a Hopf algebra structure if $n = 2^k-1$?
– user98602
Jun 11, 2014 at 2:32
• @MikeMiller: I think this chapter by Jim Lin has a good discussion, although it's behind a paywall. Lin, James P. H-spaces with finiteness conditions. Handbook of algebraic topology, 1095--1141, North-Holland, Amsterdam, 1995. Jun 11, 2014 at 3:33

The only non-trivial group which can acts freely on $S^n$, $n$ even, is the cyclic group of order $2$.

Indeed, if $f$ is a homeomorphism from $S^n$ to $S^n$ without a fixed point, then $f$ and the antipodal map are homotopic via $H(t,x)=f(t,x)/|f(t,x)|$, where $f(t,x)=tf(x)-(1-t)x$. Thus $f$ and the antipodal map induces the same automorphism on the group $H_n(S^n) \cong \mathbb{Z}$, but the group automorphism induced by the antipodal map is $m \mapsto (-1)^{n+1}m=-m$. Now if $G$ is a group which acts freely on $S^n$, we can define a homomorphism $\phi$ from $G$ to $\operatorname{Aut}(\mathbb{Z}) \cong \mathbb{Z}_2$, which map each $g \in G$ to the automorphism induced by $g$ on $H_n(S^n)$. We have seen that if $g$ is not trivial then $\phi(g)$ is not trivial. This shows that $\phi$ is injective and we are done.

Now if $S^n$, $n$ even, has the structure of a topological group, then it acts on itself freely by left translation (as noticed by Qiaochu Yuan), which implies that $S^n$ and $\mathbb{Z}_2$ are isomorphic.

No, every compact Lie group has zero Euler characteristic since it is parallelizable.

Edit: one can avoid existence of a Lie group structure as follows: if $G$ is a connected topological group, so is $H=G\times G$. Now look at any nontrivial closet of the diagonal in $H$. This coset is disjoint and homotopic to the diagonal. Therefore, $G$ has zero Euler characteristic since the latter is the self intersection of the diagonal. (This interpretation of the Euler characteristic is a general fact about closed oriented topological manifolds, although the proof is easier in the smooth case.)

• I'm not looking to turn it into a Lie group--just a topological group.
– user123641
Jun 11, 2014 at 1:33
• Perhaps one could argue that the inversion map of a topological group reverses orientation?
– Neal
Jun 11, 2014 at 1:57
• @Bryan: it is equivalent. If a topological group has a manifold topology then it is a Lie group. This follows from Montgomery and Zippin's work. Jun 11, 2014 at 2:15
• @studiosus Yes, thank you. I wasn't aware of these results previously. I have some reading to do.
– user123641
Jun 11, 2014 at 2:18
• @Neal: inversion reverses orientation iff Lie group has odd dimension. Jun 11, 2014 at 2:21