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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 $G$.

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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
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@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
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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.

See Onishchik, "Topology of transitive transformation groups", §3.10 and §17.2 for details on this.

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