# Lie Group Isomorphisms

I'm not sure of the difficulty of the question I'm about to ask. If it does not fit the criteria for this site then I apologize in advance, I'm rather new here.

So here it goes: Let $G$ be a Lie group and $Mod(G)$ be the mapping class group of the underlying manifold (by mapping class group I mean homeomorphisms from $G$ to itself up to isotopy). Denote the group of Lie-group isomorphisms as $Aut(G)$.

In general, what is the relationship between $Mod(G)$ and $Aut(G)$?

It is clear that $Aut(G)$ injects the into $Homeo(G)$ (the set of all homeomorphisms of $G$ to itself) and we know $Mod(G)\cong Homeo(G)/homotopy$. However, there are certainly homeomorphisms in $Homeo(G)$ that do not preserve the group structure and moreover, an element of $Aut(G)$ must fix the identity. I guess a specific question of importance would be:

When are two elements of $Aut(G)$ isotopic as homeomorphisms?

Also, would the case change if we were to work over a different category for $Mod(G)$ (i.e. PL, Diffeo, ect.) I hope this all makes sense and meets the standards of the board.

• Well for starters the automorphisms which are conjugation with $g\in G^0$ are isotopic to the identity, so there is a big chunk of $Aut(G)$ that vanishes in $Mod(G)$ (which I consider very bad notation btw) – Johannes Hahn Mar 9 '14 at 17:50
• So should we be looking at $Out(G)$ right off the bat? Also, I'm quite open to suggestions for notation, I'm just used to Farb's usage of $Mod(G)$. Is $\pi_0Homeo(G)$ better? – Joseph Zambrano Mar 9 '14 at 17:54
• The truth might lie somewhere between Aut(G)/Inn^0(G) and Out(G). Consider the following: To each diffeomorphism $\alpha:G\to G$ we can associate the map $x\mapsto\det(D_x\alpha)$. An differentiable isotopy between $\alpha_1\to\alpha_2$ induces a homotopy between these maps. Since $\mathbb{R}\setminus\{0\} \simeq \{\pm 1\}$ we get a sign map $\pi_0(Diffeo(G)) \to Map(G\to\{\pm 1\})$. Are there conjugation automorphisms that have a nontrivial sign? I'm thinking of something like conjugation with a reflection on $O(n)$. – Johannes Hahn Mar 9 '14 at 18:12
• I think, you are confusing homotopy and isotopy, they are not the same, even for the 2-dimensional annulus! Next, restrict to connected groups, otherwise consider the case of finite groups. If you use isotopy, then, I think, for nilpotent connected Lie groups you get surjectivity of the map from out to mod, due to Farrell and Jones. Something similar might be true for compact simply connected groups. – Misha Mar 9 '14 at 18:17
• I searched for Farrell/Jones and came across a conjecture. I am unfamiliar with the literature, I suppose you are referring to a particular solved case? And Johannes Hahn, after considering your question, it does not seem that conjugation automorphisms will have non-trivial sign but I'm still looking at examples. – Joseph Zambrano Mar 9 '14 at 18:52