TITLE: Isomorphism from $\mathbb{Z}$ to third homotopy group of compact simple Lie group
QUESTION [7 upvotes]: 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 $\rho$?

REPLY [11 votes]: 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).