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I am trying to visualize the $SU(3)$ group used in quantum field theory. I have a (reasonably) good understanding of $SU(2)$ as the double cover of $SO(3)$ and also that this is homeomorphic to $S^3$. I have also read from other questions here that $SU(3)$ is "something like" $S^5\times SU(2)$ (would that make it "something like" $S^5\times S^3$?) but I am not sure if the "something like" entails homeomorphism or if it is some more complicated bundle.

Is there a way to visualize $SU(3)$ as a real manifold (ignoring its group structure if necessary to focus on its properties as a manifold)?

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The special unitary groups fit into a sequence of fiber bundles

$$SU(n-1) \to SU(n) \to S^{2n-1}$$

coming from their actions on the unit spheres of $\mathbb{C}^n$ equipped with the standard inner product. For $n = 2$ we get a fiber bundle

$$SU(2) \to SU(3) \to S^5$$

exhibiting $SU(3)$ as a (nontrivial) $S^3$ bundle over $S^5$. This bundle splits rationally in the sense that $SU(3)$ is rationally homotopy equivalent to $S^3 \times S^5$; in particular it has the same rational homology, cohomology, and rational homotopy groups as $S^3 \times S^5$, and so in particular its Betti polynomial is

$$\sum b_i(SU(3)) t^i = (1 + t^3)(1 + t^5).$$

See also this MO question.

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  • $\begingroup$ Thanks very much. I can see that I have a ways to go in order to grasp the non-triviality of this $S^3$ bundle over $S^5$ (at least it appears that there is only one such non-trivial bundle, according to the response on the MO question) but this gives me somewhere to start! $\endgroup$ Nov 12, 2015 at 2:55
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    $\begingroup$ @MattDickau: If you'd like to visualize nontrivial sphere bundles, the Klein bottle $S^1 \widetilde \times S^1$ is a good start. There's also a nontrivial $S^2$-bundle over $S^1$ and a nontrivial $S^2$-bundle over $S^2$. $\endgroup$
    – user98602
    Nov 12, 2015 at 3:38
  • $\begingroup$ The Klein bottle I am familiar with - it is like a torus but the longitudinal circles "invert" when you traverse around the axis. From that I can imagine a non-trivial $S^2$ over $S^1$ - it would be similar but with the longitudinal circles replaced by 2-spheres.Traversing around the axis would return you to the antipode of where you started. Having trouble when the base space is not $S^1$ though. $\endgroup$ Nov 12, 2015 at 6:27

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