Is there a group $G$ with the property that $G$ is a smooth manifold, the multiplication map of $G$ is smooth, but the inversion map of $G$ is not smooth?

1$\begingroup$ I believe I've heard there are infinitedimensional groups for which each left multiplication $g\cdot$ is smooth, but the right multiplications $\cdot g$ aren't all smooth. $\endgroup$ – Allen Knutson Mar 16 '13 at 17:40

7$\begingroup$ For a compact manifold $M$, the group $Diff_{C^k}(M)$ (for $k\ge1$) of all $C^k$diffeomorphisms of $M$ is a smooth manifold. Right translations are smooth, but left translations are only continuous. Inversion is only continuous. The same holds for the group $Diff_{H^s}(M)$ (for $s\ge \frac{\dim(M)}2 +1$), the Sobolev completion of order $s$ of the group of all diffeomorphisms. $\endgroup$ – Peter Michor Mar 17 '13 at 7:05
Robert L. Bryant "An Introduction to Lie Groups and Symplectic Geometry" requires in the definition of a Lie group only that the multiplication map be smooth, and then proves that the inversion map must be smooth also. (Proposition 1, page 14.)

$\begingroup$ Thanks Steve,most of books are gabby,it's sufficient to show that the map G to G*G s,t a to (a,2a) is smooth. $\endgroup$ – R Salimi Mar 16 '13 at 18:38

$\begingroup$ But surely (using your additive notation) proving $a\mapsto 2a$ smooth is just as hard as proving $a\mapsto a$ smooth. $\endgroup$ – Steven Landsburg Mar 16 '13 at 19:52

5$\begingroup$ One could add that the main ingredient in the proof is the Implicit Function Theorem. It's kind of frustrating that this property of Lie groups motivates some people to give the wrong definition of a Lie group. Of course in the end it's all the same, but the concept is wrong (and breaks down for group objects in other categories, for example $\mathsf{Top}$). $\endgroup$ – Martin Brandenburg Mar 16 '13 at 20:27

$\begingroup$ Kind of obvious and pedantic, but it's possibly worth noting that Bryant's proof only works in finite dimensional Lie groups (to be in keeping with Allen Knutson's comment under the question) $\endgroup$ – WetSavannaAnimal Apr 6 '15 at 6:01
Every Čechcomplete paratopological group is a topological group. That means that for Čechcomplete groups you do not have to require the continuity of the inverse, continuity of multiplication suffices. Every manifold is Čechcomplete. Using the affirmative answer to Hilbert’s fifth problem we get that every paratopological group on a manifold is actually a Lie group uniquely determined by the topological group structure.
In the spirit of Martin let me give a correct (I hope I have not forgotten anything) definition which is even wronger than the definition without the inverse:
A topological space $G$ with a function $\cdot\colon G^2\to G$ is called an $n$dimensional Lie group if and only if
$G$ is secondcountable
There exists an injective, open continuous map $\iota\colon \mathbb{R}^n\to G$
For every $g\in G$ the map $x\mapsto g\cdot x$ is continuous and surjective
There exists $e\in G$ such that $x\mapsto x\cdot e$ is the identity
For every $g\in G$ the map $x\mapsto x\cdot g$ is continuous
$\cdot$ is associative