Let $G$ be a connected Lie Group. We will denote the outer automorphism group of $G$ by $Out(G)$. Consider $\pi_0(Diffeo(G))$, the mapping class group of the underlying manifold of $G$, and the map,

$k:Out(G)\rightarrow \pi_0(Diffeo(G))$

This map is well defined as follows: Given $\phi_1, \phi_2\in Aut(G)$ such that $\phi_1=g\phi_2g^{-1}$, we may construct a path $\alpha: [0,1]\rightarrow G$ such that $\alpha(0)=g$ and $\alpha(1)=e$, the identity in $G$. Then $H(w,t)=\alpha(t)\phi_2(w)\alpha(t)^{-1}$ is an isotopy with $H(w,0)=g\phi_2g^{-1}=\phi_1$ and $H(w,1)=\phi_2$.

Consider the case where $G=T^2$, the $2$-torus. Then $Out(T^2)\cong GL_2(\mathbb{Z})$ and $\pi_0(Diffeo(T^2))\cong Out(\pi_1(T^2))\cong GL_2(\mathbb{Z})$ via the Dehn-Nielsen-Baer Theorem. In this case, $k$ is an isomorphism.

Another case is the $3$-sphere, in which $Out(S^3)\cong 0$ and $\pi_0(Diffeo(S^3))\cong\mathbb{Z}_2$. In this case, $k$ is injective.

Is there a case where $k$ is surjective? If not, what is the obstruction? Moreover, are there higher dimensional examples of the bijective case? It seems that for higher dimensions, the smooth structures may make this problem computationally intractable.

  • $\begingroup$ Why do you think $Out(SU(2))=1$? And what is exactly your question? Do you mean to ask for an example of noninjective but surjective $k$? $\endgroup$ – Misha Sep 6 '16 at 0:29
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    $\begingroup$ For $1$-connected compact $G$, $Out(G)$ is the automorphism group of its Dynkin diagram, which for $SU(2)$ (but not $SU(n>2)$) is indeed $1$. $\endgroup$ – Allen Knutson Sep 6 '16 at 2:10
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    $\begingroup$ @Misha yes, I felt it was clear that by asking for a surjective map I meant a non-bijection. $\endgroup$ – Joseph Zambrano Sep 6 '16 at 11:26

Your question is largely intractable with present technology. Little is known about smooth mapping class groups of high-dimensional manifolds. But in dimensions 3 and below, there is considerable knowledge. . . unfortunately there are not many Lie groups in those dimensions.

The mapping class group of $SO_3$ is trivial. I am not certain of who proved it originally, but the uniqueness of the genus 1 Heegaard splitting would be one way to prove it. This approach would be due to Bonahon.

So for $SO_3$ you have surjectivity.

I believe there are no other Lie Groups where there any kind of upper bound on the elements of $\pi_0 Diff(G)$ known, so to get any results of the kind you are interested in would require some substantial work.


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