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Let $G$ be a compact connected simple Lie group. It is known that its third homotopy group $\pi_3(G)$ is isomorphic to $\mathbb{Z}$. More precisely, there is a Lie group homomorphism $$\rho:SU(2)\longrightarrow G$$ which induces an isomorphism $$\rho_*:\pi_3(SU(2))\overset{\cong}{\longrightarrow}\pi_3(G).$$ (Recall that $SU(2)\cong S^3$ so $\pi_3(SU(2))=\Bbb Z$.)

Clearly, all conjugates of $\rho$ have the same property. Is there a Lie group homomorphism $\varphi:SU(2)\to G$ such that $\varphi_*=\rho_*$ but $\varphi$ is not conjugate to $\rho$?

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  • $\begingroup$ E.g. $G$ may have outer automorphisms. But you probably would like to exclude these too? $\endgroup$
    – მამუკა ჯიბლაძე
    Commented Feb 13, 2017 at 18:58
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    $\begingroup$ I believe that Jacobson-Morozov applies to compact groups, i.e. maps $\phi$ up to conjugacy are indexed by conjugacy classes of nilpotents in the complex group. And then I think that the $\rho$s that you're interested in are the ones corresponding to short simple roots, so in particular, conjugate. $\endgroup$
    – Allen Knutson
    Commented Feb 13, 2017 at 20:52
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    $\begingroup$ @user104853 Well, any automorphism of $G$ acts by post-composition on $\pi_3(G) = \mathbb{Z}$, and therefore must take $\rho$ to either itself or $-\rho$ in $\pi_3$. $\endgroup$
    – Kevin Casto
    Commented Feb 13, 2017 at 20:56
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    $\begingroup$ If I'm correct, the outer automorphism of $\mathfrak{su}_n$ acts as the identity on $\pi_3$ (I use identification with $H_3$ and then identification of the latter to the (predual of) 1-dimensional space of invariant quadratic forms, working now in $\mathfrak{sl}_n$ as we can complexify). It is true that automorphisms of compact simple Lie groups always act as the identity on $\pi_3$? $\endgroup$
    – YCor
    Commented Feb 14, 2017 at 0:57
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    $\begingroup$ @YCor: yes, if $G$ is a simple simply connected Lie group, then $\pi_3(G)=H_3(G,\mathbb Z)$ is canonically isomorphic to $\mathbb Z$. Similarly, $H^3(G,\mathbb Z)$ is canonically isomorphic to $\mathbb Z$. For Lie algebra cohomology, $H^3(\mathfrak g,\mathbb R)=H^3(G,\mathbb R)$ is canonically $\mathbb R$. At the level of Lie algebra cohomology, one can write down a formula that makes it obvious that the isomorphism is canonical: the canonical generator of $H^3(\mathfrak g,\mathbb R)$ sends $X,Y,Z$ to $\langle [X,Y],Z\rangle$ where $\langle\,,\,\rangle$ is the basic inner product. $\endgroup$
    – André Henriques
    Commented Feb 16, 2017 at 23:38

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No, you cannot have $\varphi_*=\rho_*$ unless $\varphi$ is conjugate to $\rho$.

In fact, as observed by Atiyah-Hitchin-Singer (1978, p. 455), $\varphi_*$ regarded as a map $\mathbf Z\to\mathbf Z$ is just multiplication by the so-called Dynkin index of $\varphi_{\mathbf C}:\text{SL}(2,\mathbf C)\to G_{\mathbf C}$, i.e. "the ratio of the invariant inner products on both Lie algebras, each normalized to make the length of the highest root 2".

And as observed by Dynkin himself (1952, Thm 2.4), all $\varphi$ of index 1 are conjugate (corresponding, as Allen Knutson says, to the minimal nilpotent orbit in $\mathfrak g_\mathbf{C}$): see the translation in (2000, p. 197), or for another proof, Distler-Garibaldi (2010, Lemma 4.5).

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  • $\begingroup$ Conjugate means conjugate by an element of $\mathrm{Aut}(G)$? or of $G$ (inner automorphism)? $\endgroup$
    – YCor
    Commented Feb 14, 2017 at 4:33
  • $\begingroup$ @YCor Of $G$. (Note OP asks for $\varphi_*=\rho_*$, not $\pm\rho_*$.) $\endgroup$
    – Francois Ziegler
    Commented Feb 14, 2017 at 4:40
  • $\begingroup$ What happens then to an outer automorphism of order 3 on $\operatorname{SO}(8)$? $\endgroup$
    – მამუკა ჯიბლაძე
    Commented Feb 14, 2017 at 7:57
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    $\begingroup$ It's reminiscent of the concept of "fusion" of conjugacy classes, where two elements of $H<G$ may be $G$-conjugate without being $H$- or even $N(H)$-conjugate. $\endgroup$
    – Allen Knutson
    Commented Feb 14, 2017 at 14:47
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    $\begingroup$ Actually I've computed for $D_{2n+1}$: outer automorphisms induce $+1$ on $H_3$. use the block matrix $J=((0,I),(I,0))$ as quadratic form. Then $J$ itself has determinant $-1$ and is in $O(J)$, so induces the outer automorphism. But a maximal torus in the Lie algebra is given by diagonal matrices $((D,0),(0,-D))$ and $J$ acts as minus identity on it. So it acts as identity on its second symmetric power. Since the Killing form is nonzero on the maximal torus, it means the Killing form is preserved. $\endgroup$
    – YCor
    Commented Feb 14, 2017 at 18:53

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