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Using the fact that $\operatorname{Diff}^+(S^2)$ deformation retracts onto $SO(3)$, one can show that over any connected, closed, smooth surface, there are two orientable $S^2$-bundles, the trivial one and a non-trivial one. Moreover, the two bundles can be distinguished by their second Stiefel-Whitney class.

Let $X$ be the total space of the non-trivial orientable $S^2$-bundle over $T^2$. If we pullback $X \to T^2$ by any double covering $T^2 \to T^2$, we obtain the trivial $S^2$-bundle $S^2\times T^2 \to T^2$. It follows that $X$ is double covered by $S^2\times T^2$.

What is the free $\mathbb{Z}_2$ action on $S^2\times T^2$ which has quotient $X$?

It would be enough to realise the total space of an orientable rank two bundle $E \to T^2$ with $w_2(E) \neq 0$ as a $\mathbb{Z}_2$ quotient of $T^2\times\mathbb{R}^2$.

Note that $X$ can be realised as the sphere bundle of $\gamma_1\oplus\gamma_2\oplus(\gamma_1\otimes\gamma_2)$ where $\gamma_i$ is the pullback of the non-trivial line bundle on $S^1$ by projection onto the $i^{\text{th}}$ factor. It is easy to see pulling back by $(z, w) \mapsto (z^2, w)$ (or $(z, w) \to (z, w^2)$), the bundle becomes trivial (as it must). I was hoping these explicit descriptions would help, but so far I have been unsuccessful.

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  • $\begingroup$ It is easier to write down it as a quotient by $Z_2\times Z_2$-action. $\endgroup$ Dec 24 '20 at 14:50
  • $\begingroup$ @MoisheKohan: I'd be interested in seeing the details of this. In particular, it's not clear to me why such a description doesn't allow you to realise $X$ as a $\mathbb{Z}_2$ quotient. $\endgroup$ Dec 26 '20 at 16:11
  • $\begingroup$ Maybe tomorrow. It does, just the description is not as clean. $\endgroup$ Dec 26 '20 at 16:21
  • $\begingroup$ @MoisheKohan: Thanks, whenever you have the time. $\endgroup$ Dec 26 '20 at 16:23
  • $\begingroup$ @MichaelAlbanese: I am not even beginner in theory of bundles but I am curious how one can deduce something about number of orientable $S^2$-bundles by $\operatorname{Diff}^+(S^2)$? $\endgroup$
    – C.F.G
    Aug 21 at 6:13
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For flat $SO(3)$-bundles over a (connected oriented) surface $\Sigma$ the 2nd SW-class is the obstruction to lifting the associated representation $\rho: \pi_1(\Sigma)\to SO(3)$ to a representation $\tilde\rho: \pi_1(\Sigma)\to SU(2)$. Now, if $\Sigma=T^2$ with generators of the fundamental group $a, b$, the standard example of a non-liftable representation $\rho: \pi_1(\Sigma)\to SO(3)$ is the one whose image is $Z_2\times Z_2$ (sending the generators $a, b$ to the generators of the direct factors). If one thinks of $S^2$ as the Riemann sphere and $SO(3)$ as acting by linear-fractional transformations, WLOG, the representation $\rho$ is described as $$ \rho(a): z\mapsto -z, \rho(b): z\mapsto z^{-1}. $$ Consider, therefore, the product $Y=T^2\times S^2$ and $Z_2\times Z_2$ (with generators $\alpha, \beta$) acting on $Y$ via: $$ \alpha\cdot (u, v) \times z= (-u, v) \times (-z), $$ $$ \beta\cdot (u, v) \times z= (u, -v) \times (z^{-1}). $$ (Here $u, v\in S^1$.) The quotient space of this action is $X$, the total space of your oriented nontrivial $S^2$-bundle over $T^2$.

One can (in principle) extract from this description the one for a 2-fold covering space $$ T^2\times S^2\to X. $$ Namely, first divide $Y$ by the action of the first factor of $Z_2\times Z_2$. Then, one has to trivialize the quotient bundle $W=Y/Z_2\to T^2$. This can be done by observing that the map $Y\to S^2$, $(u,v)\times z\mapsto z^2$ is $\alpha$-invariant, hence, descends to a map $W\to S^2$ and, therefore, can be used for a trivialization. Then using the resulting (biholomorphic) diffeomorphism $h: W\to T^2\times S^2$ one obtains the (holomorphic) action of the involution $\beta$ on $T^2\times S^2$ by conjugation via $h$. From this, one concludes for instance that there exists a degree 2 holomorphic covering map $T^2\times S^2\to X$. Writing an explicit holomorphic action of $\beta$ on $T^2\times S^2$ from this is doable but feels unpleasant. If you are sufficiently motivated, I am sure you can write down an explicit formula.

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  • $\begingroup$ This is very nice. I had a conversation with Mike Miller about this problem and he arrived at something similar (I did not make the connection with flat bundles that the two of you did). For my own sake, let me record an explanation for why the representation doesn't lift. The center of $SO(3)$ is $\{I\}$ so the image of $\mathbb{Z}_2\times\mathbb{Z}_2$ contains a non-central element of order two, while the only element of $SU(2)$ which has order two is $-I$, which is central. $\endgroup$ Dec 28 '20 at 3:07
  • $\begingroup$ If I am not mistaken, the natural product metric on $T^2\times S^2$ descends to the quotient. (This was part of the motivation for my question.) $\endgroup$ Dec 28 '20 at 3:10
  • $\begingroup$ @MichaelAlbanese Yes, it does. $\endgroup$ Dec 28 '20 at 3:12
  • $\begingroup$ (For my own future reference) The preimage of $\mathbb{Z}_2\times\mathbb{Z}_2$ under the covering $SU(2) \to SO(3)$ gives rise to a non-trivial central extension of $\mathbb{Z}_2\times\mathbb{Z}_2$ by $\mathbb{Z}_2$. There are five groups of order 8, namely $\mathbb{Z}_8$, $\mathbb{Z}_4\times\mathbb{Z}_2$, $\mathbb{Z}_2^3$, $D_4$, and the quaternion group $Q_8$. It follows from the argument in my first comment that there cannot be more than one element of order two, which leaves $\mathbb{Z}_8$ and $Q_8$. However, the quotient of a cyclic group is cyclic, so the preimage is $Q_8$. $\endgroup$ Dec 28 '20 at 3:41
  • $\begingroup$ I plan to use the construction in your answer in an upcoming paper, and I would like to acknowledge you appropriately. How would you like me to do that? You can send me an email if you'd rather not discuss this publicly. $\endgroup$ Aug 20 at 20:35

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