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Let $G_1, G_2$ be simple compact Lie groups. Then $h_i := H_3(G_i, \mathbb Z)/ \mathrm{torsion} \cong \mathbb Z$. If $\phi : G_1 \to G_2$ is a homomorphism, then the pushforward $\phi_* : h_1 \to h_2$ carries a generator of $h_1$ to a generator of $h_2$ multiplied by some natural number $\mathrm{ind}(\phi)$, called the Dynkin index of $\phi$. I would like to ask if the value of the Dynkin index is known in the case that $\phi : G_1 \to G_2$ is a universal cover of $G_2$.

It is easy to see that for $G_2 = \mathrm{SO}(3)$ we have $\mathrm{ind}(\phi)=2$. I would like to know what happens for other groups.

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    $\begingroup$ I had not seen this definition of Dynkin index before: the one I am used to is that for simple compact Lie groups, $\pi_3(G_i)\cong \mathbb{Z}$ and the Dynkin index is the map $\mathbb{Z}\rightarrow \mathbb{Z}$ given by multiplication by $\operatorname{ind}(\phi)$. For this definition (which does not give the same answer as your definition), the answer for coverings is always $1$. $\endgroup$
    – Jason DeVito - on hiatus
    Apr 13, 2020 at 13:53
  • $\begingroup$ Dear @JasonDeVito I am not sure to what extent this is a standard definition. I certainly saw Dynkin index defined in terms of homology groups somewhere, but it might be that those authors considered simply-connected Lie groups only, in which case the two definitions do agree. In any case, the question I'm asking now has come up quite naturally in my attempts to understand certain aspects of the theory of characteristic classes. $\endgroup$
    – Blazej
    Apr 13, 2020 at 18:39
  • $\begingroup$ I haven't worked out the details, but here's a thought: If $B$ is the negative of the Killing form on $G$, then $B$ gives a bi-invariant metric on $G$. Then one can form the $3$-form $\omega(X,Y,Z) = B(X,[Y,Z])$. It turns out that $\omega$ generates $H^3(G;\mathbb{R})$. It may be natural enough that the element it represents in $H^3(G;\mathbb{R})$ can be identified as the image of $1\in H^3(G;\mathbb{Z})$ in under the map $H^3(G;\mathbb{Z})\otimes\mathbb{R}\cong H^3(G;\mathbb{R})$. I'm being little sloppy here because $H^3(G;\mathbb{Z}) \cong \mathbb{Z}\oplus T$ for some .... $\endgroup$
    – Jason DeVito - on hiatus
    Apr 13, 2020 at 22:52
  • $\begingroup$ torsion group $T$, and while $T$ is uniquely defined as a subgroup of $H^3(G;\mathbb{Z})$, the $\mathbb{Z}$ isn't. Nonetheless, the image of $(1, t)$ in $H^3(G;\mathbb{R})$ is the same regardless of the value of $t\in T$, so I think it's ok. Anyway, if this form works like that, then you can compute the map $H^3(G_2,\mathbb{Z})\rightarrow H^3(G_1,\mathbb{Z})$ by pulling back the form $\omega$ and integrating. Since $G_1$ and $G_2$ have the same Lie algebra, I think it follows that the Dynkin index is simply the degree of the covering. But admittedly, this is a lot of guess work. $\endgroup$
    – Jason DeVito - on hiatus
    Apr 13, 2020 at 22:54
  • $\begingroup$ Ha, some googling brought up your previous question about properly normalizing this three form. So it seems like a trick like this should work, once someone figures out the normalization. $\endgroup$
    – Jason DeVito - on hiatus
    Apr 13, 2020 at 23:16

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