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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.

Edit:

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"? –  John Apr 6 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 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 at 11:42

1 Answer 1

up vote 8 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 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 math.stanford.edu/~dlitt/briefnotes/notes/inversion.pdf) –  Jason DeVito Apr 6 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 at 13:39
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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 at 14:57
    
Ah, great, thanks! –  Kevin Carlson Apr 6 at 19:34

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