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In the definition of Lie Group, we require that $$(x,y)\rightarrow x*y \text{ and } x\rightarrow x^{-1}$$ both be smooth. Are there any examples of groups that satisfy only one of these and not the other and are hence are not smooth manifolds? The definition I am using of smooth manifold is the same as we use for topological spaces, i.e. if it is locally diffeomorphism to $\mathbb{R}^n$

The reason I am asking this is because, I am wondering if the latter requirement follows from the former.


Are there any examples of groups that satisfy only one of these and not the other and are hence are not smooth manifolds?

should be:

Are there any examples of groups that satisfy only one of these and not the other and are hence are not Lie groups?

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Do you mean "and are not Lie groups" instead of "and are not smooth manifolds"? – Arctic Char Apr 6 '14 at 10:12
@John, I think it's ok. If they are groups and smooth manifolds, they are Lie groups too. So, for a group not to be Lie Group, it should not be smooth manifold. – user140802 Apr 6 '14 at 11:32
Your comment's not right-as my answer indicates, there are bajillions of group structures on an uncountable set, in particular on any smooth manifold-too many for them all to be smooth. Besides, it's meaningless to ask whether a group that's not a smooth manifold could have smooth multiplication or inversion, since smooth maps are only defined on smooth manifolds. – Kevin Carlson Apr 6 '14 at 11:42
up vote 9 down vote accepted

Here's a set-theoretic family of examples of smooth manifolds that are groups with smooth inversion but not multiplication. Let $(G,e,*)$ be a group of exponent $2$, i.e. $g *g=e$ for every $g\in G$ with the cardinality $\mathfrak{c}$ of $\mathbb{R}$. For instance, $G$ could be a direct product of $\mathfrak{c}$ $\mathbb{Z}_2$s. Let $\phi:\mathbb{R}\to G$ be any bijection and define a group structure on $\mathbb{R}$ by $x\star y=\phi^{-1}(\phi(x)*\phi(y))$. This kind of construction always yields a group. Now since we picked our $G$ to have an inversion map preserved under bijection, inversion is guaranteed to be smooth: for $x\in\mathbb{R}, x^{-1}_\star=\phi^{-1}(\phi(x)^{-1}_*)=\phi^{-1}\phi(x)=x$, i.e. the inversion map $\mathbb{R}\to\mathbb{R}$ is just the identity.

But for the vast majority of choices of $\phi$ the multiplication will not be smooth. For since $\mathbb{R}$ has $\mathfrak{c}^\mathfrak{c}$ self-bijections, we've exhibited $\mathfrak{c}^\mathfrak{c}$ distinct group structures on $\mathbb{R}$ all with smooth inversion. On the other hand, a continuous-in particular smooth-group structure on $\mathbb{R}$ is specified by the maps $x,y\mapsto x\star y$ for $x,y\in \mathbb{Q}$, i.e. by an element of $\mathbb{R}^{\mathbb{Q}\times\mathbb{Q}},$ which has cardinality only $\mathfrak{c}$!

Edit Incidentally, smoothness of multiplication actually implies smoothness of inversion if inversion is continuous. I don't know if there are groups with smooth multiplication and discontinuous inversion-my guess is there are. Re-edit But as Jack Lee's comment below shows, in fact smooth multiplication does imply smooth inversion.

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thanks. I corrected the question as per comment. – user140802 Apr 6 '14 at 11:51
Nice answer!. The proof I know of "smooth multiplication implies smooth inversion" first proves it near $e$ using the inverse function theorem applied as a function $G\times G\rightarrow G$. Then one proves inversion is smooth everywhere by writing inversion at any point as a composition of inversion at $e$ with left and right multiplications by particular elements. Does this argument use the fact that $i$ is continuous? (Edit: Here's a sketch I found in a similar vein – Jason DeVito Apr 6 '14 at 12:36
It seems to me that it does: for instance in Litt's proof to show the map $(g,g^{-1})\mapsto^\pi g$ is a homeomorphism one should know that $N^{-1}$ is open for $N$ an open neighborhood of $g$ to make $\pi^{-1}N$ open as the intersection of $\{(g,g^{-1})\}$ with $N\times N^{-1}$ in $G\times G$. On the other hand maybe you can get this from the lemma that the antidiagonal is a submanifold. – Kevin Carlson Apr 6 '14 at 13:39
If a group $G$ is a smooth manifold with smooth multiplication, then it's a Lie group, without any assumption that inversion is even continuous. This can be proved by considering the map $F\colon G\times G\to G\times G$ defined by $F(g,h) = (g,gh)$. You can show that $F$ is bijective and $dF$ is invertible everywhere, so $F$ is a bijective local diffeomorphism and hence a diffeomorphism. Then the inversion map is easily constructed from $F^{-1}$. – Jack Lee Apr 6 '14 at 14:57
Ah, great, thanks! – Kevin Carlson Apr 6 '14 at 19:34

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