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$\mathbb S^2$ is the set consists of the points at distance 1 from the origin in $\mathbb R^3$.

Is it possible to define $\mathbb S^2$ as a group? Or, can it be proved that $\mathbb S^2$ can't be a group?

athos
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    The meaning of "defining" $\Bbb S^2$ as a group could use some clarification. Be as it may, the situation is probably simpler than that: there are group structures on $\Bbb S^2$, because you can transfer the structure of any group of cardinality $\beth_1$ on it (say, $(\Bbb R,+)$). Topological group structures, there aren't: https://math.stackexchange.com/questions/829928/can-s2-be-turned-into-a-topological-group – Sassatelli Giulio Oct 09 '22 at 06:48
  • Does this answer your question? Can $S^2$ be turned into a topological group? – Sassatelli Giulio Oct 09 '22 at 06:52
  • @SassatelliGiulio I don't know, I have no knowledge of topology.. or put it this way, if it's confirmed that a set $G$ can't be turned into a topological group, can it still be defined as a group? – athos Oct 09 '22 at 07:01
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    As I said, you have to clarify what you mean by "defining $\Bbb S^2$ as a group". $\Bbb S^2$ is a symbol that usually indicates, depending on context, a topological space that is homeomorphic to the set you've said in your question or a manifold diffeomorphic to it. The set ${x\in\Bbb R^3,:, x^\top x=1}$ itelf is not a subgroup of $(\Bbb R^3,+)$. The topological space $\Bbb S^2$ does not admit continuous group operations $\cdot:\Bbb S^2\times\Bbb S^2\to \Bbb S^2$ and $(\bullet)^{-1}:\Bbb S^2\to\Bbb S^2$ at all. – Sassatelli Giulio Oct 09 '22 at 07:50

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Well, the fact that $\mathbb{S}^2$ has the same cardinality as $\mathbb{R}$, it follows that you can put a (silly) group structure on $\mathbb{S}^2$ (by marking a structure which is formally isomorphic to $\mathbb{R}$ with a bijection $\mathbb{S}^2 \to \mathbb{R}$).

But if you want your group to be smooth (in other words a Lie group), this is impossible. This is because all Lie groups are parallelisable. Which in particular means there exists a nowhere zero vector field on the underlying manifold. This is not true on $\mathbb{S}^2$ because of the Hairy Ball Theorem.

Edit: as the comments suggested, it is also not possible to put (even a) topological group structure on $\mathbb{S}^2$. This discussion of smoothness and continuity is with respect to the subspace topology on $\mathbb{S}^2$. The (silly) group structure mentioned above is also a (silly) Lie group structure (on a silly manifold).

David Perrella
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  • could you pls give an example of a "silly group structure" on $\mathbb R^2$? i don't quite get the meaning of "silly" here... – athos Oct 09 '22 at 07:05
  • @athos $\mathbb R$ is itself a (smooth) real manifold, so transport the manifold structure to $S^2$ using the "silly" bijection that preserves cardinality. Incidentally, this gives $S^2$ (as a set) the structure of $1$-dimensional smooth manifold, diffeomorfic to the real line. –  Oct 09 '22 at 08:17
  • @StinkingBishop sorry for my ignorance -- how would you transport $\mathbb R$ to $\mathbb S^2$? could you pls define the "silly" bijection? I can't figure that out. – athos Oct 09 '22 at 08:19
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    @athos Read the answer above again. It can be proven (I am not going into details) that $S^2$ and $\mathbb R$ have the same cardinality, i.e. the same number of elements, i.e. there is a bijection $f:\mathbb R\to S^2$. Once you have the bijection, you can transport anything on $\mathbb R$ to $S^2$. Say, additive group structure: if $z=x+y$ in reals, define $f(z):=f(x)+f(y)$ in $S^2$. This is unambiguous as $f$ is a bijection. In that group, $f(0)$ is the identity and $f(-x)$ is the inverse of $f(x)$. And so on. –  Oct 09 '22 at 08:25
  • @athos NB There are many ways to construct such a map $f$ too, but you need a few tools from set theory to do that, and the construction will be fairly long and not very pretty. However, for your question it only matters that such a map exists. –  Oct 09 '22 at 08:28
  • @StinkingBishop thanks for the explanation. seems i have to take it that such a mapping exists then, being lack of some set theory knowledge. i'm still not clear though -- i'm sure there's a bijection $f:\mathbb R\rightarrow \mathbb S^2$, but not sure how to make one that is well defined, for example, i can't define $f(1) = e^{i\pi/2}, f(2) = e^{i3\pi/2}$. But i get your point, that there's a long proof in some set theory book. thanks. – athos Oct 09 '22 at 09:20
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    @athos, that explicit result may not appear in set theory books but you can take the follows steps:
    1. Since $\mathbb{S}^2 \subset \mathbb{R}^3$, the cardinality of $\mathbb{S}^2$ is at most that of $\mathbb{R}^3$.
    2. We can easily find injections $\mathbb{R} \to \mathbb{S}^2$, so the cardinality is at least that of $\mathbb{R}^3$.
    3. The cardinality of $\mathbb{R}^3$ and $\mathbb{R}$ is the same.
     – David Perrella Oct 09 '22 at 10:15
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    @athos You need at least the following from the set theory: (a) A finite set added or subtracted doesn't change the size of the infinite set. Cantor-Schröder-Bernstein theorem (c) There is a bijection of $\mathbb R$ to $(0,1)$ (say, $f(x)=(\arctan x+\pi/2)/\pi$); (d) There is a bijection of $[0,1)$ with the set of all "binary" sequences (infinite sequences of $0$'s and $1$'s: use binary and ternary expansion of numbers in $[0,1)$ and (b) above); ... –  Oct 09 '22 at 17:06
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    ... (e) There is a bijection of the set of triplets of binary sequences with just the set of binary sequences (map $((a_1, a_2, a_3,\ldots), (b_1, b_2, b_3,\ldots), (c_1, c_2, c_3, \ldots))$ to $(a_1, b_1, c_1, a_2, b_2, c_2, a_3, b_3, c_3, \ldots)$. So if (as David Parella has said) you can make injections $\mathbb R\to S^2\to \mathbb R^3$ and you can construct a bijection of $\mathbb R\to\mathbb R^3$ using (a)-(e) above,, then you can construct a bijection $\mathbb R\to S^2$ using (b) above. That is about all, but as I said it is not very simple. –  Oct 09 '22 at 17:09