Let's consider a principle $U(1)$-bundle over $S^2$ with the transition function $g_{\infty 0} = z/|z|$ (it is known as the Hopf fibration). There is a simple topological argument showing that this bundle is not trivial by the comparison of fundamental groups. However, it should be clear directly from definitions.

So I would like to prove that this bundle has no global sections (which is equivalent to non-triviality). It means that there is no continuous function $f: \mathbb{C} \to U(1)$ such that there exists $$ \lim_{z \to \infty} \frac{z}{|z|} f(z) $$ How would one prove this?

  • $\begingroup$ Can you link to the topological argument that you refer to? $\endgroup$ – user99914 Aug 18 '18 at 12:23
  • $\begingroup$ The bundle space is just $S^3$ so $\pi_1(S_3) = 0$. If it were a direct product then the fundamental group would be $\pi_1(S_2 \times S_1) \neq 0$. $\endgroup$ – Artem Aug 18 '18 at 12:28
  • 1
    $\begingroup$ I think you could use higher homotopy groups too, although that's probably not the argument you want. I think the point for me at least is that topological arguments such as the one you presented make this problem much much easier. I'm not sure if youd even want to prove it a different way. $\endgroup$ – Sheel Stueber Aug 18 '18 at 15:47
  • $\begingroup$ The point is that these simple and powerful topological techniques definitely allow to solve the problem but at the same time they hide what happens at the level of definitions. Having the purpose of understanding something about fiber bundles I would like to see the fact directly. Maybe the "topological" argument translates to the language I would like to use but I don't see how. $\endgroup$ – Artem Aug 18 '18 at 22:16

Consider the function $g(z) = \frac{z}{|z|} f(z)$ on $\Bbb C \setminus \{0\}$. Your assumption is that this extends to a map on $\Bbb{CP}^1 \setminus \{0\}$. This means that $g$ has degree zero as a map to $U(1)$ (it induces multiplication-by-zero on $H_1$), as it factors through a space with $H_1 = 0$.

For $z \in S^1$, $g(z) = z \cdot f(z)$. Degree is additive under multiplication of functions to $U(1)$, and we see that $\text{deg}(f) = -1$. However, by assumption, $f$ extends across the unit disc, and so should have degree zero. This is a contradiction.

  • $\begingroup$ Very nice ! ${{}}$ $\endgroup$ – Nicolas Hemelsoet Aug 19 '18 at 16:35

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.