2

  1. The Hopf bundle is defined as the quotient map of $S^3\subset \mathbb{C}^2$ by the action of $U(1)$ by multiplication $(z_1,z_2)\mapsto (e^{i\varphi}z_1,e^{i\varphi}z_2)$. It's known that it is the circle bundle of the complex line bundle $\mathcal{O}(-1)$ defined over $\mathbb{C}P^1$ by the transition function $g_{0\infty}(z)=z^{-1}$. In other words, it has Chern number $-1$.

  2. On the other hand, the tangent bundle $TS^2\to S^2$ is known to be isomorphic to the Stiefel bundle $SO(3)\to S^2=SO(3)/SO(2)$. It is also known that this bundle has Chern number $2$ (since its transition function is $z^2$).

  3. Thirdly, there is the homogeneous bundle $SU(2)\to S^2=SU(2)/U(1)$. It's not clear to me what its Chern number is, but it's definitely $\pm 1$ because those are the only circle bundles over $S^2$ whose total spaces are $S^3$ and not some lens spaces.

Question: if the Hopf bundle coincides with the third bundle, then it would seem that the second bundle should be nothing but the quotient of the first by the action of $\mathbb{Z}_2=\{\pm I\}$ on $SU(2)$, but clearly that would just square the value of the transition function, not invert it. Does this mean that the Hopf bundle is actually the dual of bundle 3? Where does the sign change come from?

Alex Bogatskiy
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  • How precisely is 3. different from 1.? – Ted Shifrin Apr 23 '24 at 02:34
  • @TedShifrin that's basically the question. Why does the Chern number of (2) have the opposite sign of (1) despite one bundle seemingly being the double cover of the other? – Alex Bogatskiy Apr 23 '24 at 03:38
  • I don't know whether this helps, but here is an explicit calculation of the fibers: https://www.physicsforums.com/insights/journey-manifold-su2mathbbc-part/#4-Hopf-Fibration – Marius S.L. Apr 23 '24 at 15:13
  • Your question is misleading. Since $SU(2)\cong S^3$, the first and third descriptions are identical. That answers your question. But then the unit tangent bundle shows up as item 2. You seem to be wanting to see how to relate it to the Hopf bundle. Yes, $SU(2)$ is a (group-theoretic) double cover of $SO(3)$, but how is that supposed to relate the Chern numbers of the corresponding circle bundles? – Ted Shifrin Apr 23 '24 at 19:37
  • @TedShifrin I don't see how it's misleading. The issue is that the sign of the Chern number depends on a choice of orientation, which doesn't exist in the real case. So the way I've answered this question to myself is that there is no natural isomorphism between $S^2$ and $CP^1$, therefore the Chern number of bundle (2) is defined only up to a sign. Hopefully that's correct. – Alex Bogatskiy Apr 24 '24 at 00:46
  • I disagree. The orientation on $S^2$ is not the issue. Tell me how the double cover gives you the 2. – Ted Shifrin Apr 24 '24 at 04:58
  • @TedShifrin the double cover is a double cover of the circle on every fiber, hence it squares the values of the U(1) transition function. This double cover also doesn't switch the orientation on the circle, so the sign of the Chern number shouldn't change. – Alex Bogatskiy Apr 24 '24 at 23:07
  • For starters, how do you know the fibers of the Hopf fibration map to the unit tangent vectors? – Ted Shifrin Apr 24 '24 at 23:34
  • Assuming the question is still in flux, it might help to be explicit about how $U(1)$ acts on $SU(2)$. It can't be by scalar multiplication, since scalar matrices $U(1) \cdot I_{2}$ do not generally have determinant $1$. – Andrew D. Hwang Apr 25 '24 at 13:39
  • @AndrewD.Hwang I have added a link above where fibers are calculated and action on the fibers are explained. For those who do not like the format, the link contains references to the actual sources. – Marius S.L. Apr 25 '24 at 14:39
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    @MariusS.L. yeah I'm aware that the action on $SU(2)$ isn't diagonal if you want to view it as matrix multiplication. The alternative meaning of "diagonal" is that if SU(2) is parametrized as $\begin{pmatrix}a & -\bar b\ b & \bar a \end{pmatrix}$ by a pair $(a,b)\in\mathbb{C}$ with $|a|^2+|b|^2=1$, then the action is $(e^{i\phi}a,e^{i\phi}b)$. – Alex Bogatskiy Apr 25 '24 at 19:18
  • @TedShifrin I checked the explicit formula for the adjoint representation of SU(2) and multiplying the two complex parameters on $SU(2)$ by a phase is equivalent to an action that preserves one row of the resulting $SO(3)$ matrix. Therefore the fiber in $SU(2)$ gets mapped to the fiber in $SO(3)=S(TS^2)$. – Alex Bogatskiy Apr 25 '24 at 19:22

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