Lie Group Structure on the 2-Sphere: does the following argument hold?

Question

Being inspired by the existence of a Lie group structure on the circle $\Bbb{S}^{1}$, I was looking for a group law that would make the two-sphere $\Bbb{S}^{2}$ into a Lie group. I found out that no such group law exists, and I would like to know whether my argument is a valid one. Here is what I thought.

We know that two simply-connected Lie groups are isomorphic if and only if they have isomorphic Lie algebras. Moreover, we know that the plane $\Bbb{R}^{2}$ is a simply connected abelian, two-dimensional Lie group under the sum between vectors, and that the two-sphere is a simply connected compact manifold, non isomorphic to the plane as a topological space, and thus as a manifold. Finally, we know that there exist only two non-isomorphic two-dimensional Lie algebras: an abelian L.a., and a non-abelian L.a.. This last one is the linear span of two vectors $X$ and $Y$, with Lie product defined as $[X,Y]=Y$.

Let's call this Lie algebra $\mathfrak{g}$. As a consequence of the existence of two and only two non-isomorphic two-dimensional Lie algebras, there can only exist two simply-connected two-dimensional Lie groups up to isomorphisms, one of which, as said, is the abelian $(\Bbb{R}^{2},+)$ with abelian Lie algebra. The two-sphere is not diffeomorphic to the plane, and is itself simply connected. So, if the two-sphere was to admit a Lie group structure, it would necessarily need to have a Lie algebra different from that of $\Bbb{R}^{2}$; namely $\mathfrak{g}$. Moreover, as the two-sphere is connected and compact, the exponential map on $\mathfrak{g}$ should give the whole $\Bbb{S}^{2}$, and thus the entire group structure of $\Bbb{S}^{2}$.

I then tried to look for some features of the simply connected Lie group associated to $\mathfrak{g}$ (I have never read anything about it). So I wrote down the fundamental (real) representation of $\mathfrak{g}$ (on $\Bbb{R}^{2}$), choosing the matrices

$$X=\frac{1}{2}\begin{pmatrix}1&1\\1&1\end{pmatrix}\qquad\quad Y=\begin{pmatrix}-1&1\\-1&1\end{pmatrix}$$ which as you can verify satisfy the commutation relation $[X,Y]=Y$, and moreover have the nice properties

$$X^{k}=X\quad\forall\ k>0\qquad Y^{k}=0\quad\forall\ k\geq 2\qquad X^{k}Y^{n}=Y^{n}\quad\forall\ n>0\\Y^{n}X^{k}=0\quad\forall\ n,k>0$$ with $k,n$ integers. I computed the representation of the Lie group $G$ associated to the chosen representation of $\mathfrak{g}$ on $\Bbb{R}^{2}$ via exponentiation. Here is my result:

$$R(G)=\{g(t,s)\in M(2,\Bbb{R}):\quad(t,s)\in\Bbb{R}^{2}\}$$

where

$$g(t,s)=\exp(tX+sY)=I+\frac{(e^{t}-1)}{2t}\begin{pmatrix}t-2s&t+2s\\t-2s&t+2s\end{pmatrix}$$

Now

$$\lim_{t\to-\infty}g(t,s)=\frac{1}{2}\begin{pmatrix}1&-1\\-1&1\end{pmatrix}\qquad\lim_{t\to+\infty}g(t,s)=+\infty\\\lim_{s\to\infty}g(t,s)=\infty$$

so $R(G)=\exp(\mathfrak{r}(\mathfrak{g}))$, where $\mathfrak{r}$ is the representation mapping of $\mathfrak{g}$ on $\text{End}(\Bbb{R}^{2})$, is not compact. If $R$ were the representation of the compact, simply connected Lie group $G\equiv\Bbb{S}^{2}$ on $\Bbb{R}^{2}$, $R(G)$ would have turned out to be a compact subspace of $M(2,\Bbb{R})$. Since it fails to be so, we conclude that $\mathfrak{g}$ is not the Lie algebra of $\Bbb{S}^{2}$. Since there is no other possibility for $\Bbb{S}^{2}$ other than to have either $\mathfrak{g}$ or the abelian two-dimensional Lie algebra as its Lie algebra, and these are excluded by principle, we can conclude that there exists no Lie algebra associated to a would-be group structure on $\Bbb{S}^{2}$. Thus, as $\Bbb{S}^{2}$ already is a smooth manifold, there must exist no group law making the two-sphere into a Lie group.

I hope that the calculations are alright. I'm still a beginner in the subject, and there could be fallacies in the reasoning that I'm not aware of. Let me know what you think of the argument.

P.S.: Assuming that there really doesn't exist a Lie group structure on $\Bbb{S}^{2}$, wouldn't this have something to do with the Hairy Ball Theorem? Since the dimension of the two-sphere is even, there can't exist a nowhere vanishing vector field on $\Bbb{S}^2$. Thus there cannot exist left-invariant vector fields on $\Bbb{S}^2$ and, again, the two-sphere cannot have a Lie algebra.

Answer 1

In principle, your approach is correct, but it leads to needless complex computations. I would like to suggest three other ways of proving that no Lie structure can be defined on $S^2$.

1. This is actually your own suggestion. Since on any $n$-dimensional Lie $G$ group one can define $n$ vector fields $X_1,X_2,\ldots,X_n$ which are linearly independent everywhere (that is, for each $g\in G$, the vectors $X_1(g),X_2(g),\ldots,X_n(g)$ are linearly independent), the hairy ball theorem allows us to deduce that no Lie group structure can be defined on $S^2$.
2. The space $S^2$ is compact. If it was a Lie group, its Lie algebra $\mathfrak g$ would then be reductive (this is a standard fact within Lie group theory and not hard to prove). But there are only two $2$-dimensional Lie algebras (up to isomorphism). The one spanned (as a vector space) by two elements $X$ and $Y$ such that $[X,Y]=Y$ is not reductive. The other one is the abelian $2$-dimensional Lie algebra, which is the Lie algebra of one and (up to isomorphism) only one simply-connected Lie group, which is $(\mathbb{R}^2,+)$. And $\mathbb{R}^2$ and $S^2$ are not homeomorphic.
3. A slightly different approach, compared with the previous one, consists in finding a simply-connected Lie group $G$ whose Lie algebra $\mathfrak g$ is the non-abelian one from the previous point. And that's easy. Take$$G=\left\{\begin{bmatrix}a&b\\0&\frac1a\end{bmatrix}\,\middle|\,a>0\wedge b\in\mathbb{R}\right\}.$$Again, it is not compact, and therefore $G$ and $S^2$ are not homeomorphic.

Answer 2

The nontrivial $2$-dimensional Lie algebra $\mathfrak{g}$ is nilpotent. It's a general feature of nilpotent Lie algebras that the exponential map $\mathfrak{g} \to G$ to the corresponding simply connected Lie group is a diffeomorphism: in particular, $G$ is necessarily noncompact, and hence cannot be diffeomorphic (or in fact even homotopy equivalent) to $S^2$.

This is a good idea for a proof (I don't think I've actually seen it before), but it doesn't generalize well: it's in fact the case that no sphere $S^n, n \ge 2$ admits the structure of a Lie group except $S^3$, but as $n$ increases this gets harder and harder to prove with Lie theory. By contrast, a generalization of the hairy ball theorem immediately proves that the even-dimensional spheres don't admit the structure of a Lie group.

Answer 3

O.k., the 2-sphere has no GLOBAL Lie group and Lie algebra, BUT it is easy to define a Lie group and algebra on the punctured 2-sphere. Let $a = (a_1,a_2,a_3) and b = (b_1,b_2,b_3)$ be two points on the unit sphere. Define the group product to be $<a,b> = \frac{(a_1 b_1 - a_2 b_2, a_2 b_1 + a_1 b_2, a_3+b_3)}{(1 + a_3 b_3)}$. It can be checked that $e = (1,0,0)$ is the group identity, and $inv(a)=(a_1,-a_2,-a_3)$ is the group inverse of the point a. Finally, it can be checked that the product is associative as is necessary for a group. It is also true that the group is Abelian. URL: http://www.garretstar.com