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I am studying Lie theory and just thought of this random question out of curiosity. Can any manifold be turned into a Lie group?

More precisely, given a manifold $G$, can we always construct (or prove the existence of) some smooth map $m:G\times G\to G$ that makes $G$ into a Lie group? If not, is there an easy counterexample?

I could imagine a construction going something like this: pick an arbitrary point $e\in M$ to be the identity, and define $m(e,g)=m(g,e)=g$ for all $g\in G$. Then we already have the elements of the Lie algebra given as the tangent space at the identity $T_eG$, and maybe we can use these to extend $m$ to all of $G$?

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    $\begingroup$ So far, you've only constructed multiplication by a single element, the identity. So you have a long way to go. :-) But you'll find that a Lie group must be parallelizable, and many manifolds (e.g., $S^2$) are not. $\endgroup$ Dec 18, 2019 at 22:53
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    $\begingroup$ Parallelizability is the major obstacle, but there's a result from Wolf that a compact, simply-connected manifold is parallelizable iff it's the product of a Lie group and copies of $S^7$. $\endgroup$
    – anomaly
    Dec 22, 2019 at 4:27
  • $\begingroup$ Somehow related: mathoverflow.net/questions/5262/lie-groups-and-manifolds $\endgroup$
    – C.F.G
    Mar 1, 2020 at 8:49
  • $\begingroup$ anomaly: Are you missing a hypothesis? There are many more examples of parallelizable manifolds. The simplest is probably $S^2\times S^3$. $\endgroup$ Mar 6, 2020 at 17:09

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There is an easy counterexample: $S^2$ cannot be given a Lie group structure (this is a consequence of the hairy ball theorem). The problem with your construction is that it doesn't offer how to define $m(g,h)$ for any two nonidentity elements $g$ and $h$.

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Lie groups as manifolds, are very special, owing to the group operations. Basically, "what happens at the identity" determines what happens everywhere. And this means that the tangent bundle $T G$ is always trivializable: here is a sketch of the proof, based on what I remember from Lee's book.

Take any basis $\{v_i\}^n_{i=1}$ for $T_eG$. Since left multiplication $L_g:G\to G:h\mapsto gh$ is a diffeomorphism, it induces an isomorphism $dL_g:T_eG\to T_gG.$ Now, define vector fields $\{V_i\}^n_{i=1}$ by $(V_i)_g:=dL_g(v_i)$and show that they are smooth. Then, since $dL_g$ is an isomorphism, $\{dL_g(v_i)\}^n_{i=1}$ is a basis for $T_gG$, so the vector fields $\{V_i\}^n_{i=1}$ are a global frame for $TG$.

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    $\begingroup$ Can you add a final sentence along the lines that this shows that $S^2$ cannot be made into a Lie group because this would imply the existence of a nowhere vanishing vector field which doesn't exist by the hairy ball theorem. $\endgroup$
    – quarague
    Dec 19, 2019 at 7:53
  • $\begingroup$ I'm not sure how this relates to my question, other than being about Lie groups. $\endgroup$
    – WillG
    Dec 19, 2019 at 20:14
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    $\begingroup$ If every manifold could be realized as a Lie group then every manifold would have a trivial tangent bundle, which is not true, as the other answers and comments point out. $\endgroup$ Dec 19, 2019 at 20:19
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    $\begingroup$ @quarague I guess your comment is enough ...+1 $\endgroup$ Dec 19, 2019 at 20:40
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To add to the previous answers, topological groups have abelian fundamental groups.

$G$ is Topological $\implies$ $\pi_1(G,e)$ is Abelian

Orientable surfaces of genus at least two are not parallelizable, but this is another way to show that they can't be Lie (even topological) groups. The Klein bottle is parallelizable (edit: no it's not), but its fundamental group is not abelian, so it can't be a group either.

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    $\begingroup$ The Klein bottle is non-orientable, so the tangent bundle is nontrivial. $\endgroup$ Dec 20, 2019 at 17:39
  • $\begingroup$ @JasonDeVito you're right, oops! $\endgroup$
    – Moisés
    Dec 21, 2019 at 0:04
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The answers so far are great, but I wanted to add some more obstructions. Suppose $M$ is a manifold which can be given the structure of a Lie group. Then $M$ has the following properties...

  1. $\pi_1(M)$ acts trivially on $\pi_n(M)$
  2. Each $\pi_n(M)$ is finitely generated.
  3. $\pi_2(M) = 0$.
  4. $\pi_{2k}(M)$ is a finite abelian group for all $k\geq 1$.
  5. $\pi_3(M)$ contains no torsion.
  6. If $M$ is compact, then at least one of $\pi_1(M)$ and $\pi_3(M)$ contains $\mathbb{Z}$ as a subroup.
  7. If $M$ is non-compact, then $M$ must be diffeomorphic to $\mathbb{R}^k\times N$ for some compact Lie group $N$.
  8. If $M$ is simply connected, then it can only torsion of order $2$, $3$, or $5$ in its cohomology groups.

There are still many manifolds which pass all these obstructions (as well as all the obstructions in the other answers!) - for example, $M = S^3\times S^5$. However, this $M$ isn't a Lie group (though the only way I know to show this is using the classification. It's simply connected and dimension $8$, so the only Lie group $M$ could be diffeomorphic to is $SU(3)$. However, $\pi_4(M) = \mathbb{Z}/2\mathbb{Z}$ while $\pi_4(SU(3)) = 0$.)

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    $\begingroup$ Do you have a reference for where I can read about these properties? I've never heard of most of these but am very interested. $\endgroup$
    – kamills
    Dec 20, 2019 at 18:29
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    $\begingroup$ @Hood: $S^5$ is not parallelizable, but $S^3 \times S^5$ is. More generally, a product of $n\geq 2$ spheres is parallelizable iff at least one of the spheres is odd dimensional. See, e.g. ams.org/journals/proc/1967-018-03/S0002-9939-1967-0219082-6/… $\endgroup$ Dec 21, 2019 at 1:33
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    $\begingroup$ @kamills: For 1 and 4, Algebraic Models in Geometry by Felix, Oprea, and Tanre has the results (but maybe not the proofs). 2 is a general fact of algebraic topology and applies to anything (and more) satisfying 1. The key word is "simple space". 3 and 5 can be found in Milnor's Morse Theory book. 6 works as follows: if $\pi_1(M)$ doesn't contain a $\mathbb{Z}$, then finite generation implies $\pi_1(M)$ is finite, so the universal cover of $M$, $\hat{M}$ is a simply connected compact Lie group. Then $H^3(\hat{M};\mathbb{R})\neq 0$, which can be seen using the bi-invariant form... $\endgroup$ Dec 21, 2019 at 1:40
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    $\begingroup$ $\omega(x,y,z) = \langle [x,y],z\rangle$ where $\langle \cdot, \cdot \rangle$ is a bi-invariant metric on $\hat{M}$ and $x,y,z$ are left invariant vector fields. It then follows that $H^3(\hat{M})$ contains $ \mathbb{Z}$. Now use Hurewicz (noting that $\pi_1(\hat{M}) = \pi_2(\hat{M} = 0$). 7. Is called the Iwasa decomposition. I don't study/use non-compact Lie groups, so I don't know of a reference (also, you may need to assume $M$ is connected for 7 to be true...). Finally, 8 follows from the classification of simply connected Lie groups: $M$ must be isomorphic to a product with ... $\endgroup$ Dec 21, 2019 at 1:42
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    $\begingroup$ factors of the form $SU(n)$, $Spin(n)$, $Sp(n)$ (where $n$ can vary) or factors of the form $G_2, F_4, E_6, E_7, E_8$. But the torsion in each case is $2,3,$ or $5$-torsion. $\endgroup$ Dec 21, 2019 at 1:46
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As many people are giving intersting counterexamples I thought I should also add one . Any surface( compact orientable hausdorff 2 manifold) with non zero Euler characteristics cannot be a Lie group because from standard theorem in differential topology , Euler's characteristic of compact orientable lie group is zero. For instance it's 2 for 2 sphere so it can't be a Lie group.

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If $G$ is a Lie group, then $G$ is homogeneous, meaning that for any $g,h\in G$, there is a diffeomorphism $F:G\to G$ such that $F(g)=h$, namely $F(x)=hg^{-1}x$. This gives an easy way to find manifolds that cannot be Lie groups. For instance, let $G$ be the disjoint union of $\mathbb{R}$ and $S^1$. Then no diffeomorphism of $G$ can map a point of $S^1$ to a point of $\mathbb{R}$ (since a diffeomorphism must preserve the property of being in a compact connected component), so $G$ cannot be a Lie group.

(Note though that actually any connected manifold is homogeneous, whereas most connected manifolds do not admit a Lie group structure. So admitting a Lie group structure is actually much stronger than just being homogeneous.)

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    $\begingroup$ So I guess the question would be more interesting if it were about connected manifolds only. $\endgroup$
    – WillG
    Dec 20, 2019 at 16:10
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    $\begingroup$ Right. But you asked for an easy counterexample, and this is the easiest one I think! $\endgroup$ Dec 20, 2019 at 16:13

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