9
$\begingroup$

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.

$\endgroup$
  • 2
    $\begingroup$ Your last argument is simpler and correct. (I'm not sure about the first one...my eyes kinda glazed over...) $\endgroup$ – John Hughes Sep 2 '15 at 1:59
  • $\begingroup$ It is actually the same statement: the two-sphere doesn't have a Lie algebra to be associated to it. In the first argument I went over the two possibilities for a two-dimensional Lie algebra. Thank you anyway :-) $\endgroup$ – Giorgio Comitini Sep 2 '15 at 2:00
  • $\begingroup$ I think your first argument is very nice, actually. (But I don't think $\mathfrak{g}$ is semisimple, and I didn't check all your deails of the computation of $R(G)$) $\endgroup$ – Jason DeVito Sep 2 '15 at 2:18
  • $\begingroup$ Thank you Jason. You're totally right about semisimplicity, I'll correct the mistake as soon as possible. Should you check the details, let me know wheter you find errors. I liked the argument too, anyway. It makes concrete use of many elementary results in Lie Theory. It was a good exercise, after all :-) $\endgroup$ – Giorgio Comitini Sep 2 '15 at 3:12
  • 2
    $\begingroup$ The proof is the Morse index theorem: the sum of the indices of the zeroes of a generic VF on a compact mfld is equal to the Euler characteristic. For even spheres, a cellular-decomposition into one 0-cell and one n-cell shows that this Euler characteristic is even, so every vector field has at least one zero of index 2 or two zeroes of index 1. Thus: no left-invariant vector fields, no Lie group structure. $\endgroup$ – John Hughes Sep 2 '15 at 10:56
5
$\begingroup$

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 complex 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.
$\endgroup$
4
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Have you got any reference for the proof for the higher dimensional cases? $\endgroup$ – Giorgio Comitini Sep 2 '15 at 10:33
  • $\begingroup$ Anyway, are you sure that $\mathfrak{g}$ is nilpotent? For sure it is solvable, but what about $\mathfrak{g}_{j}=\text{span}Y$ for any lower central series ideal with $j\neq0$? $\endgroup$ – Giorgio Comitini Sep 2 '15 at 11:01
  • $\begingroup$ @Giorgio: ah, you're right, my apologies. It's only solvable. Here is a proof for spheres: math.stackexchange.com/questions/12453/… $\endgroup$ – Qiaochu Yuan Sep 2 '15 at 16:12
  • $\begingroup$ Nice, and the same person that wrote the answer commented above! Thank you! $\endgroup$ – Giorgio Comitini Sep 2 '15 at 16:19
-5
$\begingroup$

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

$\endgroup$
  • 2
    $\begingroup$ The punctured $2$-sphere is, of course, diffeomorphic to $\mathbb{R}^2$. $\endgroup$ – Qiaochu Yuan Aug 7 '16 at 3:33
  • $\begingroup$ It is R2, so your answer is uninteresting. $\endgroup$ – Giorgio Comitini Aug 7 '16 at 8:41
  • $\begingroup$ In case you didn't know, any two diffeomorphic simply connected abelian Lie groups are isomorphic as Lie groups. This means that, if the group law you wrote down really is abelian (I didn't check), then you have just described the additive Lie group structure on $R^{2}$, in different coordinates. $\endgroup$ – Giorgio Comitini Aug 7 '16 at 9:08

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.