# 3rd homotopy group of a compact Simple Lie Group

Suppose $G$ is a compact simple Lie group with Lie algebra $\mathfrak g$. Then we know that $\pi_3(G)=Z$. Now suppose that $H_\alpha$ is a co-root vector in correspondence with a root $\alpha$. So it means that there are $X_\alpha$ and $Y_\alpha$ such that $span${$H_\alpha, X_\alpha, Y_\alpha$} is a sub-Lie algebra of $\mathfrak g$ isomorphic to $\mathfrak{su}(2)$. It induces a map of Lie groups $\phi:SU(2) \to G$. I'm wondering what's the image of this map as an element of $\pi_3(G)$ in terms of $G$.

-
Just out of curiosity, why do we know $\pi_3(G)\cong\mathbb{Z}$? Is there a simple reason, like maybe it has $S^3$ or $S^2$ as universal cover or something? –  William Mar 3 '12 at 17:48
@William: No. But see the answer to mathoverflow.net/questions/8957/homotopy-groups-of-lie-groups –  Dylan Wilson Mar 3 '12 at 20:36
I don't know the answer, but of course it will depend on how you are making the identification $\pi_3(G) \simeq \mathbb{Z}$. If that identification is made in terms of the root system of $\mathfrak{g}$ then perhaps you can do something. But if you don't have an explicit isomorphism then probably it will be difficult to say much. –  MTS Mar 3 '12 at 21:26
This number is called the index of the map $\phi: SU(2)\to G$. It can be defined for any homomorphism $\phi:H\to G$ where $H$ is simple. Algebraically it can be computed as follows. Since $\mathfrak h$ is simple the restriction of the Killing form of $\mathfrak g$ to $\mathfrak h$ is a constant multiple of the Killing form of $\mathfrak h$. That constant is the index of $\phi$. In the specific case you are asking about for a simple root $\alpha$ the index can also be written as $\frac{(\alpha_{max},\alpha_{max})}{(\alpha,\alpha)}$ where $\alpha_{max}$ is the longest simple root of $\mathfrak g$. Note that from the classification of compact simple Lie groups this can only be equal to 1,2 or 3.