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Lie's second theorem says that if $G$ is a simply connected Lie group, then every isomorphism $\Phi$ of its Lie algebra $\mathfrak{g}$ lifts to an isomorphism $\phi$ of $G$, i.e. such that $d\phi_e = \Phi$ where we identify $\mathfrak{g}$ with $T_e G$, the tangent space to $G$ at the identity.

Now consider $G = \mathrm{SO}(3)$, which is not simply connected. Is there an explicit counterexample to Lie's second theorem in this case? That is, can we write down a Lie algebra isomorphism $\Phi$ of $\mathfrak{so}(3)$ which is not the differential of any isomorphism $\phi$ of $\mathrm{SO}(3)$?

I feel like, if $(\eta_1, \eta_2, \eta_3)$ is the usual basis of $\mathfrak{so}(3)$, where $[\eta_i, \eta_j] = \eta_k$ cyclically, then a map like $\Phi(\eta_1)=\eta_2$, $\Phi(\eta_2)=\eta_1$, $\Phi(\eta_3)=-\eta_3$ ought to work, but I can't figure out how to prove there is no such group isomorphism $\phi$. I've been trying to find $t_1, t_2, t_3 \in \mathbb{R}$ and a relation involving $(e^{t_1 \eta_1}, e^{t_2 \eta_2}, e^{t_3 \eta_3})$ that is not satisfied by $(e^{t_1 \eta_2}, e^{t_2 \eta_1}, e^{-t_3 \eta_3})$ but I can't come up with one.

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  • $\begingroup$ Btw, modulo conventions and if I did not miscalculate, the exponentials of your basis are given by $\pmatrix{1&0&0\\ 0&-\cos(t)&-\sin(t)\\ 0&\sin(t)&-\cos(t)}$, $\pmatrix{\cos(t)&\sin(t)&0\\ -\sin(t)&\cos(t)&0\\0&0&1}$, $\pmatrix{\cos(t)&0&\sin(t)\\0&1&0 \\-\sin(t)&0&\cos(t)}$, and your $\Phi$ lifts to the inner automorphism given by conjugation with $\pmatrix{0&0&-1\\ 0&-1&0\\ -1&0&0}$. $\endgroup$ Commented Feb 9, 2019 at 5:50

3 Answers 3

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There is an isomorphism $r: \text{Aut}(SU(2)) \to \text{Aut}(SO(3))$, because any automorphism of $SU(2)$ necessarily preserves the center and $SO(3) = SU(2)/Z(SU(2))$. To see that $r$ is injective, observe that if $r(\rho) = \text{Id}$, then necessarily $\rho(g) = \pm g$ for each $g$. By continuity, the sign is independent of $g$, and because $\rho$ is a homomorphism, the sign is 1, as desired.

To see that $r$ is surjective, observe that taking the derivative at the identity gives an embedding $\text{Aut}(G) \hookrightarrow \text{Aut}(\mathfrak g)$. By Lie's theorem, the map $\text{Aut}(SU(2)) \to \text{Aut}(\mathfrak{su}(2))$ is an isomorphism, and in particular we have factored this isomorphism as the composite of two injective maps. Both maps must therefore be isomorphisms.

In fact, $$\text{Aut}(SU(2)) = \text{Inn}(SU(2)) \cong \text{Inn}(SO(3)) = SO(3).$$

José's answer gives an example of a map of Lie algebras which may not be lifted to a map of Lie groups, but we see here that if you're looking to find an automorphism of $\mathfrak{so}(3)$ which does not lift to an automorphism of $SO(3)$, you are out of luck. You'll have to try something more complicated.

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    $\begingroup$ 1. Does math.stackexchange.com/q/1020865/96384 imply more generally that the inner automorphisms are the same for every group "in between" the simply connected and the adjoint group, thus can never work as counterexample? 2. Would e.g. something like the triality morphism on $\mathfrak{so}_8$ work, as lifting to $Spin(8)$ but not to $SO(8)$? $\endgroup$ Commented Feb 8, 2019 at 1:06
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    $\begingroup$ @TorstenSchoeneberg Your first sentence seems correct to me. I'm not sure about your second sentence: I think that there is an identification of $\text{Out}(G)$ with automorphisms of the Dynkin diagram in more generality than simply connected Lie groups, but I'm not sure at what generality it is true. I vaguely recall that it is at least true for $SO(n)$. In general, I am not so comfortable with the high-dimensional stuff. I'd love to see an example worked out of a group where an automorphism of $\mathfrak g$ doesn't lift to $G$. $\endgroup$
    – user98602
    Commented Feb 8, 2019 at 1:41
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    $\begingroup$ @Torsten is right: the triality automorphism of $\mathfrak{so}8$ does not lift to an automorphism of $SO(8)$. If I remember, I can write an answer tomorrow morning. $\endgroup$ Commented Feb 8, 2019 at 5:54
  • $\begingroup$ @JasonDeVito: Please do! Trying to look it up, I figured I possibly learned the idea from an old comment of yours here: math.stackexchange.com/a/59919/96384, and I would love to see that fleshed out. Further, if I understand that answer correctly (and extend this answer's argument with the centre to the case of $PU(n)$), it seems to me that that those outer automorphisms of order $3$ and $6$ of $\mathfrak{so}_8$ would be the only counterexamples among the classical simple compact connected Lie groups --? $\endgroup$ Commented Feb 8, 2019 at 6:22
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    $\begingroup$ Proposition 25.15 of "The Book of Involutions" says that for any connected semisimple Lie group G, the group Inn(G) agrees with Inn(G/Z(G)); there is always an injection Out(G) to the automorphisms of the Dynkin diagram, which is an isomorphism of G is either simply connected or centerless. So one needs something in the middle of those two cases, if I haven't messed up. I believe the center of $SO(4k)$ is non-trivial, so this should indeed be a promising case to look at. $\endgroup$
    – user98602
    Commented Feb 8, 2019 at 6:57
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As discussed in the comments to Mike's answer, if $G$ is a classical group other than $SO(8)$, then Lie's second theorem holds. For $G = SU(n)$ or $Sp(n)$, it holds because they are simply connected. For $SO(2n+1)$ there are no outer automorphisms (because the Dynkin diagram has no non-trivial symmetries). For $SO(2n)$ with $n\geq 5$, there is a unique interesting symmetry of the Dynkin diagram, but comes from a map on $SO(2n)$ given by conjugation by $\operatorname{diag}(-1,1,1....,1)$.

The group $SO(8)$ is special because its Dynkin diagram has more symmetry than every other simple Lie group: the Dynkin diagram for $D_4$ has an order $3$ rotational symmetry.

As every automorphism of the Dynkin diagram extends to an automorphism of the Lie algebra, there is a special automorphism $t:\mathfrak{so}(8)\rightarrow \mathfrak{so}(8)$ known as the triality automorphism.

I claim that there is no automorphism $T:SO(8)\rightarrow SO(8)$ which induces $t$, that is, Lie's second theorem fails for this map.

I'll write a roadmap below; proofs can be found in this paper by Varadarajan. The linked .pdf has no "Propositions", so below, I'll write, e.g. Proposition 1 (Theorem 3) to indicate that my Proposition 1 is Theorem 3 in the paper.

Proposition 1 (Theorem 3 and Lemma 4): There are precisely two conjugacy classes $\Sigma_1, \Sigma_2$ of $Spin(7)\subseteq SO(8)$. This is precisely one conjugacy class $\Sigma_0$ of $SO(7)\subseteq SO(8)$. $\square$

Now, let $\widetilde{T}:Spin(8)\rightarrow Spin(8)$ induce $t$. (The map $\widetilde{T}$ exists by Lie's second theorem.) Since $t^3$ is the identity (up to conjugacy), so is $\widetilde{T}^3$.

Proposition 2 (Theorem 5): The lifts of each of the $\Sigma_i$ to $Spin(8)$ form three distinct conjugacy classes $\widetilde{\Sigma_i}$ in $Spin(8)$. Further, $\mathbb{Z}/3\mathbb{Z} = \{Id_{Spin(8)}, \widetilde{T}, \widetilde{T}^2\}$ acts transitively on $\{\widetilde{\Sigma_i}\}$. $\square$

With these in hand, it's easy to prove the claim above, that there is no $T:SO(8)\rightarrow SO(8)$ inducing $t$. Namely, since powers of $\widetilde{T}$ act transitively on $\{\widetilde{\Sigma_i}\}$, powers of $T$ must act transitively on $\{\Sigma_i\}$. However, the elements of $\Sigma_1$ are isomorphic to $Spin(7)$, so are simply connected, while the elements of $\Sigma_0$ are isomorphic to $SO(7)$, so aren't. Thus, there is no diffeomorphism (or even homotopy equivalence!) which can move $\Sigma_0$ to $\Sigma_1$, even if we remove the restriction that maps be homomorphisms.

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    $\begingroup$ In case anyone is wondering why conjugation by $A:=\operatorname{diag}(-1,1....,1)$ is an inner automorphism of $SO(2n+1)$ even though its outer for $SO(2n)$, just note that since $-I$ is in the center, conjugating by $A$ and conjugating by $-IA$ are the same operation. For $SO(2n+1)$, $\det(-I A) = 1$, so $-I A \in SO(2n+1)$. $\endgroup$ Commented Feb 8, 2019 at 19:15
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See $SU(2)$ as the group of quaterninons with norm $1$. Let$$\operatorname{Im}\mathbb{H}=\{ai+bj+ck\in\mathbb{H}\,|\,a,b,c\in\mathbb{R}\}$$and, for each $q\in SU(2)$, let $\varphi(q)\colon\operatorname{Im}\mathbb{H}\longrightarrow\operatorname{Im}\mathbb{H}$ be the map defined by $\varphi(q)(v)=q^{-1}vq$. Then $\operatorname{Im}\mathbb{H}\simeq\mathbb{R}^3$ and $\varphi$ is a surjective group homomorphism from $SU(2)$ onto $SO(3,\mathbb{R})$, whose kernel is $\pm1$. Then $D\varphi_1\colon\mathfrak{su}(2)\longrightarrow\mathfrak{so}(3,\mathbb{R})$ is a Lie algebra isomorphism and its inverse is a representation of $\mathfrak{so}(3,\mathbb{R})$ in $\mathbb{C}^2$. But you cannot lift it to $SO(3,\mathbb{R})$, since $SU(2)$ and $SO(3,\mathbb{R})$ are not isomorphic.

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