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In this paper of Mimura and Toda, tables are given for low-dimensional homotopy groups of $SU(3)$, $SU(4)$ and $Sp(2)$. As far as I understand it, Theorem 6.1 gives the generator of $\pi_3(SU(4))$ as $i_\ast \varepsilon_3$. Here $i\colon SU(2) \hookrightarrow SU(4)$ is the inclusion such that $SU(4)/SU(2) \simeq S^5\times S^7$. Here $\varepsilon_3$ should be some known element in $\pi_3(SU(2)) = \pi_3(S^3) = \mathbb{Z}$, but I can't figure out from information in the paper what it is (there are other elements labelled $\varepsilon$, but they live elsewhere with no obvious analogues in this dimension). Any ideas?

To contrast, the generator of $\pi_3(SU(3))$ is given $i_\ast\iota_3$, where here $i\colon SU(2) \hookrightarrow SU(3)$ is the inclusion in the top left corner, so that $SU(3)/SU(2) \simeq S^5$, and $\iota_3$ is the isomorphism $S^3\stackrel{\sim}{\to} SU(2)$.

I know that the generator of $\pi_3$ of (suitable) Lie groups is given by an inclusion of $SU(2)$ as a subgroup, constructed involving longest roots in the Lie algebra, how do I reconcile this description with that in the first paragraph?

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    $\begingroup$ The standard inclusions $SU(n)\to SU(n+1)$ induce isomorphisms on $\pi_3$ for $n\geq 3$, as one can see from the long exact sequence of the fibration $SU(n)\to SU(n+1)\to S^{2n+1}$. Thus, if you take either generator of $\pi_3(SU(2))\simeq\mathbb{Z}$ and apply $i_*$ to it, you will get a generator for $\pi_3(SU(n))$. Does this answer your question, or are you concerned about the $\pm$-sign? $\endgroup$
    – Neil Strickland
    Commented May 29, 2014 at 8:05
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    $\begingroup$ @NeilStrickland - ah, that was much easier than the Lie theoretic argument. I'm happy to not think too much about ±1. $\endgroup$
    – David Roberts ♦
    Commented May 29, 2014 at 8:28
  • $\begingroup$ In both your maps $i$ you have $SU(2) \to G \to G/SU(2)$ where the target is $3$-connected. Hence both are giving isomorphisms $\pi_3(SU(2)) \to \pi_3(G)$. This is a slightly more specific version of Neil's answer. $\endgroup$
    – Allen Knutson
    Commented May 29, 2014 at 14:49
  • $\begingroup$ @AllenKnutson yep, I got that. $\endgroup$
    – David Roberts ♦
    Commented May 29, 2014 at 23:00

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Assembling comments of Neil Strickland and Allen Knutson, we have that $\varepsilon_3$ is just the standard generator $\iota_3\colon S^3 \stackrel{\sim}{\to} SU(2)$, since the inclusion $SU(2) \to SU(4)$ induces an isomorphism on $\pi_3$. This is shown using the long exact sequence in homotopy groups, and the fact $S^5\times S^7$ is 3-connected.

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