Non isomorphic principal $G$-bundles Remark: throughout this post we work in the smooth category, so that all manifolds, bundles, maps etc are assumed to be smooth.
An exercise asks me to show that there exists no principal $S^1$-bundle isomorphism between the Hopf bundle $S^1 \to S^3 \xrightarrow\pi S^2$ (where $S^1$ acts on $S^3$ by multiplication as usual), and the bundle with the same fiber structure but equipped with the "inverted" action $(w_1,w_2) \cdot z = (w_1z^{-1}, w_2z^{-1})$.
Aiming to reach a contradiction I suppose an $S^1$-bundle isomorphism between these two bundles exists. This is a diffeomorphism $\Phi: S^3 \to S^3$ satisfying:
\begin{array}{rc}
\pi \circ \Phi = \pi, & (*)\\[0.2cm]
\Phi(w_1 z,w_2 z) = \Phi(w_1,w_2)z^{-1}. & (**)
\end{array}
As far as I can see, the conditions $(*)$ and $(**)$ alone do not violate each other, since $\pi$ is invariant under multiplication by elements of $S^1$ (and hence their inverses) anyway. So I guess there should be issues with the diffeomorphy of $\Phi$. More particularly, I would guess smoothness/continuity of $\Phi$ (or its inverse) is an issue, since again I don't see why the equivariance condition $(**)$ would contradict the bijectivity of $\Phi$.
But I'm failing to see how to derive a contradiction here. Am I missing something simpler or am I on the right track?
EDIT:  I just realized that the book has solutions to the exercises in the appendix (I don't remember seeing this in an advanced math book so I didn't even notice they were there). I will try to add an outline of the solution to this post once I figure the details out.
EDIT 2: As promised, here is an outline of a solution. Firstly, since $\Phi$ respects the fibers by $(*)$, it must be of the form
$$ \Phi(w) = w \, \phi(w) $$
for some smooth $\phi:S^3 \to S^1$. By $(**)$ we then see that
$$ w\,\phi(wz)z = \Phi(wz) = \Phi(w)z^{-1} = w\,\phi(w)z^{-1}, $$
or in other words,
$$ \phi(w) = \phi(wz)z^2 $$
since the action is free. This is the condition that we are going to use to derive a contradiction. Since $S^3$ is simply connected, we may write
$$ \phi = e^{i\psi} $$
for some smooth $\psi : S^3 \to \mathbb R$. Thus, taking $z = z(\alpha) = e^{i\alpha}$ with $\alpha \in \mathbb R$, we get
$$ e^{i\psi(w)} = \phi(w) = \phi(wz)z^2 = e^{i\psi(we^{i\alpha})+2i\alpha}. $$
This implies that
$$ \psi(we^{i\alpha}) - \psi(w) + 2\alpha = 2\pi n(\alpha) $$
for some integer $n = n(\alpha)$. But the left hand side depends smoothly on $\alpha$, so the right hand side must as well, i.e. $n(\alpha)$ must be constant (since it is integer-valued). In fact, $n(\alpha) = n(0) = 0$. But then
$$ \psi(we^{i\alpha}) - \psi(w) = -2\alpha, $$
which is absurd, since the left hand side is bounded (since $\psi(S^3)$ is compact), while the right hand side is unbounded.
 A: I think the "standard" argument would be that the Hopf fibration is classified by a nontrivial element of $H^1(S^2, S^1) \cong H^2(S^2, \mathbb{Z}) \cong \mathbb{Z}$ (I think more specifically by twice a generator but don't quote me on that) so the inverted bundle is presumably classified by the negative of this element, and since $\mathbb{Z}$ has no $2$-torsion no such bundle is equivalent to its inverse. Probably there's something to learn from trying to show that the diffeomorphism $\Phi$ doesn't exist more concretely though.
A: I apologize in advance for the long post, but, as Qiaochu mentioned, I wanted to learn something from the bare hands approach.
To set notation, I'm going to view $S^3\subseteq \mathbb{C}^2$ as the pairs $(z_1,z_2)$ of unit length vectors, so $|z_1|^2 + |z_2|^2 = 1$.  I'm going to view $S^2\subseteq  \mathbb{C}\oplus \mathbb{R}$ consisting of unit length vectors.
Then the Hopf map $\pi:S^3\rightarrow S^2$ is (according to Wikipedia), given by $\pi(z_1,z_2) = (2z_1\overline{z}_2, |z_1|^2 - |z_2|^2)$.
Let $U = \{(w,t)\in S^2: t > -\epsilon\}$ and $V = \{(w,t)\in S^2: t < \epsilon\}$ where I'm thinking of $\epsilon$ as some fixed very small positive number.  Intuitively, $U$ is the northern hemisphere of $S^2$, except extended slightly below the equation, and $V$ is likewise essentially the southern hemisphere.
Proposition 1:  The open sets $U$ and $V$ form a trivializing cover for the Hopf bundle.
Proof: Let's start with $\pi^{-1}(U)$.  We'll start by finding a section $s_U:U\rightarrow \pi^{-1}(U)$.  So, given $(w,t)\in U$, we want to associate to it $(z_1,z_2)\in \pi^{-1}(U)$.  Let's try making the simplifying assumption that $z_1$ is real and positive.
So, we are solving $(2z_1 \overline{z}_2, z_1^2 - |z_2|^2) = (w, t)$ for $(z_1,z_2)$ under the assumption that $z_1$ is real.  The first equation $2z_1 \overline{z}_2 = w$ can be solved for $z_2$, getting $z_2 = \frac{\overline{w}}{2z_1}$.  Substituting this into the equation $z_1^2 - |z_2|^2 = t$, clearing denominators, we get a quadratic in $z_1^2$.  Using the quadratic formula, together with the fact that $z_1 > 0$, we find $$z_1 = \sqrt{\frac{t+1}{2}}, \text{ and } z_2 =\frac{\overline{w}}{\sqrt{2t + 2}}.$$  Thus, our section $s_U$ is given by $s_U(w,t) = (z_1,z_2)$, with formulas for $z_1$, $z_2$ defined above. I will leave it to you to verify it's a section.
Armed with this section $s_U$, we define $f_U: U\times S^1\rightarrow \pi^{-1}(U)$ by $f_U(w,t,z) = s_U(w,t)z$.  I'll leave it to you to verify that $f_U$ is an $S^1$-equivariant diffeomorphism with inverse $f^{-1}(z_1,z_2) = \left(\pi(z_1, z_2), \frac{z_1}{|z_1|}\right).$
In a similar fashion, we have a section $s_V:V\rightarrow \pi^{-1}(V)$ given by $s_V(w,t) = (z_1,z_2)$ with $$z_1 = \frac{w}{\sqrt{2-2t}} \text{ and } z_2 = \sqrt{\frac{1-t}{2}}.$$  This gives a trivialization $f_V:V\times S^1\rightarrow \pi^{-1}(V)$ given by $f_V(w,t,z) = s_V(w,t)z$ with inverse $f_V^{-1}(z_1,z_2) = \left(\pi(z_1,z_2), \frac{z_2}{|z_2|}\right).$ $\square$
Proposition 2:  The composition $f_V^{-1}\circ f_U$ maps $(w,t,z)$ to $(w,t, \frac{\overline{w}}{|w|} z$.
Proof:  We compute.  \begin{align*} f_V^{-1}(f_U(w,t,z)) &= f_V^{-1}( s_U(w,t)z) \\ &= f_V^{-1}\left(\sqrt{\frac{t+1}{2}}z, \frac{\overline{w}}{\sqrt{2t+2}}z\right) \\ &= \left(\pi\left(\sqrt{\frac{t+1}{2}}z,\frac{\overline{w}}{\sqrt{2t+2}}z\right) , \frac{\frac{\overline{w}}{\sqrt{2t+2}}z}{\left|\frac{\overline{w}}{\sqrt{2t+2}}z\right|}\right) \\&= \left(w,t, \frac{\overline{w}}{|w|} z\right).\end{align*} $\square$
Using this, we can view $S^3$ as $(U\times S^1) \coprod (V\times S^1)/\sim$ where $(w,t,z)\in U\times S^1$ is identified with $(w,t, \frac{\overline{w}}{|w|} z)\in V\times S^1$ for any $t\in(-\epsilon,\epsilon)$.   Using this description, the projection map $\pi$ is simply projection onto the $w$ and $t$ coordinates.
Now, let's show $\Phi$ cannot exist.  To that end, let's assume $\Phi$ does exist.  Note $(\ast)$ implies that $\Phi$ maps fibers to fibers.  In particular, in our above description, $\Phi$ is given by a pair of maps $\Phi_U:U\times S^1\rightarrow U\times S^1$, $\Phi_V:V\times S^1\rightarrow V\times S^1$ which respect $\sim$.
Since $\Phi$ preserves each fiber, $\Phi_U(w,t,z) = (w,t,\phi_U(w,t,z))$ for some function $\phi_U$.
Proposition 3:  The function $\phi_U$ has the property that $\phi_U(w,t,z) = \phi_U(w,t,1)z^{-1}$.
Proof:  Using $(\ast\ast)$, we know that $$\Phi_U(w,t,z) = \Phi_U((w,t,1)z) = \Phi_U(w,t,1)z^{-1} = (w,t,\phi_U(w,t,1))z^{-1} = (w,t,\phi_U(w,t,1)z^{-1}).$$  On the other hand, $\Phi_U(w,t,z) = (w,t,\phi_U(w,t,z))$.  Thus $\phi_U(w,t,1)z^{-1} = \phi_U(w,t,z)$ as claimed.  $\square$
Of course, the above discussion applies equally well to  $\Phi_V$.  In particular, Proposition 3 is also true for $\phi_V$.
We'll now use the fact that $\Phi$ is well defined to find a relationship between $\phi_U$ and $\phi_V$.
Proposition 4:  We have $\phi_U(w,0,1)\overline{w} = \phi_V(w,0,1)w$.
Proof:  Since $(w,t,z)\in U\times S^1$ is identified with $(w,t,\frac{\overline{w}}{|w|}z)$ in $V\times S^1$, we must have $[\Phi_U(w,t,z)] = [\Phi_V(w,t,\frac{\overline{w}}{|w|} z]$ for all $(w,t,z)$ with $t\in (-\epsilon,\epsilon).$  Set $t=0$ (so $|w| = 1$) and set $z=1$.
Now, $\Phi_U(w,0,1) = (w,0,\phi_U(w,0,1))\in U\times S^1$, and so $(w,0,\phi_U(w,0,1))\sim (w,0, \overline{w} \phi_U(w,0,1))\in V\times S^1$.  Since $\Phi_V(w,0, \overline{w} ) = (w,0, \phi_V(w,0, \overline{w}))$, the condition that $\Phi$ respect $\sim$ implies that $$(w,0,\overline{w}\phi_U(w,0,1)) = (w,0,\phi_V\left(w,0,\overline{w})\right).$$
Using Proposition 3 on the last coordinate, we conclude $\phi_U(w,0,1)\overline{w} = \phi_V (w,0,1)w$ as claimed.  $\square$
We are now ready to reach a contradiction.  Namely, we claim that $\phi_U(w,0,1)\overline{w} =\phi_V(w,0,1)w$ is contradictory.  Viewing $\phi_U(\cdot,0,1):S^1\rightarrow S^1$, the degree of this map must be $0$ because $\phi_U$ extends to the disk $U$.  Likewise, the degree of $\phi_V(\cdot,0,1)$ is $0$.  Thus, the maps $\phi_U(\cdot, 0,1)$ and $\phi_V(\cdot,0,1)$ are homotopy to constants.  It now follows from the equation $\phi_U(w,0,1)\overline{w} = \phi_V(w,0,1)w$ that the maps $w\mapsto \overline{w}$ and $w\mapsto w$ are homotopic.   This is absurd since one has degree $1$ while the other has degree $-1$.  This contradiction establishes that $\Phi$ cannot exist.
