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I am aware that principal $SU(N)$ bundles over 3-manifolds are trivial.

Consider, however, a principal $PSU(N)$ bundle over $\mathbb R^3$. Is the topology of this bundle classified by $$\pi_2(SU(N)/U(1)^{N-1})=\pi_1(U(1)^{N-1})=\mathbb Z^{N-1}$$? (The motivation for this hypothesis comes from the physics literature.)

dennis
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Every fiber bundle over a contractible space is trivial. This is a consequence of the following result:

Theorem: If $p : E \to B$ is a fiber bundle with fiber $F$ and $f_0 : X \to B$ and $f_1 : X \to B$ are homotopic, then $f_0^*E \cong f_1^*E$.

See Theorem $2.1$ of these notes for a proof.

If $B$ is contractible, then the identity map $\operatorname{id}_B: B \to B$ is homotopic to a constant map $c : B \to B$. So for any fiber bundle $p: E \to B$, we have $E \cong \operatorname{id}_B^*E \cong c^*E$ which is trivial.

To see that $c^*E$ is trivial, note that

$$c^*E = \{(b, e) \in B\times E \mid c(b) = \pi(e)\} = \{(b, e) \in B\times E \mid \pi(e) = b_0\} = B\times E_{b_0}$$

where $b_0$ denotes the constant value of the map $c$ and $E_{b_0} = p^{-1}(b_0)$. The projection map $B\times E_{b_0} \to B$ is just projection onto the first factor, so after choosing an isomorphism between $E_{b_0}$ and $F$, we obtain an isomorphism between $c^*E$ and the trivial bundle.

Michael Albanese
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  • Thanks Michael. Do you know anything about the topology of PSU(N) bundles over $S^3$? (or should I ask a separate question?) – dennis Jan 15 '23 at 12:02
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    They're all trivial, see here. – Michael Albanese Jan 15 '23 at 12:40
  • Thanks. I think the way to get the physics result is to remove the origin from $\mathbb R^3$. In that case PSU(N) bundles over $\mathbb R^3/{0}$ are non trivial and characterized by $\pi_1(PSU(N))=\mathbb Z_N$. – dennis Jan 15 '23 at 19:44