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Consider the action of $S^1$ on the product of 3-spheres $S^3\times S^3$ defined by:

$$e^{it}.(z_1, z_2)=(e^{2it}z_1, e^{3it}z_2)$$

where $z_1, z_2\in S^3$. Here we understant $e^{2it}z_1$ as the multiplication by $e^{2it}$ in each component of $z_1$ when we look at $S^3$ as a subspace of $\mathbb{C}^2$ (more precisely, $S^3=\{(a,b)\in \mathbb{C}^2: |a|^2+|b|^2=1\}$).

So my question is what is (topologically, for instance) the quotient of $S^3\times S^3$ by this action? I'm convinced that this should give $S^3\times S^2$ because this action is in some sense a "twisted" Hopf fibration, but I've not been able to show this (at least in an explicit way).

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    $\begingroup$ Not sure if this is helpful, but there is a surjective map from the quotient to $\mathbb{C}P^1=S^2$ given projection onto the first copy of $S^3$ followed by projectivization, and the fiber over each point is $S^3/\{\pm1\}=\mathbb{R}P^3$. $\endgroup$ – Julian Rosen Jan 2 '17 at 19:32
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    $\begingroup$ @JulianRosen It is a theorem that $\text{Diff}(\Bbb RP^3)$ has fundamental group $\Bbb Z/2$, and $\Bbb{RP}^3$ bundles are classified by $B\text{Diff}(\Bbb{RP}^3)$, so we get bundles over $S^2$ in bijection with $\Bbb Z/2$. On the other hand, I believe the nontrivial one is indeed $S^2 \times S^3$, since one may calculate that it's simply connected (though I don't immediately see how to fit that into the appropriate fibration). $\endgroup$ – user98602 Jan 2 '17 at 20:35
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    $\begingroup$ Modulo details, this space appears to be the total space of the circle bundle over $S^{2} \times S^{2}$ whose first Chern class is $(-2, -3)$. In other words, if $\mathcal{O}(p)$ denotes the holomorphic line bundle of degree $p$ over $\mathbf{CP}^{1} \simeq S^{2}$, this space looks like the unit circle bundle in the holomorphic line bundle $$\pi_{1}^{*}\mathcal{O}(-2) \otimes \pi_{2}^{*}\mathcal{O}(-3) \to \mathbf{CP}^{1} \times \mathbf{CP}^{1}.$$I don't have 1. A proof offhand, or 2. A more explicit description. If anyone can provide either (or a refutation of the claim!), please post. :) $\endgroup$ – Andrew D. Hwang Jan 2 '17 at 23:51
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    $\begingroup$ @AndrewD.Hwang That's correct and the total space of that line bundle is $S^2 \times S^3$. 1) It's simply connected because it's a fiber bundle over $S^2$ with fiber $\Bbb{RP}^3$ or, using the second factor, $L(3,1)$, and therefore has fundamental group a quotient of both $\Bbb Z/2$ and $\Bbb Z/3$. 2) It has $\pi_2 = \Bbb Z$. 3) It's spinnable. (Use the fiber bundle $S^1 \to M \to S^2 \times S^2$, and the decomposition $w(M) = w(T_v M)\pi^*w(S^2 \times S^2)$, and note that the first is a line bundle on an s.c. manifold.) Thus we conclude from the classification. $\endgroup$ – user98602 Jan 3 '17 at 6:39
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    $\begingroup$ Unfortunately, this is extremely inexplicit. Note that there is only one (two, counting orientation) line bundle over $S^2 \times S^3$ with total space $S^3 \times S^3$, so there must be some diffeomorphism of $S^3 \times S^3$ that conjugates this action to the standard action (or itself with the opposite orientation). I have absolutely no idea what it is. $\endgroup$ – user98602 Jan 3 '17 at 6:41
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Ivy's answer is completely correct, but one would have to show that $S^3\times S^3/S^1$ is spin for this particular action of $S^1$ on $S^3\times S^3$. It is, and, in fact, for any homogeneous action of of $S^1$ on $S^3\times S^3$, the quotient is diffeomorphic to $S^2\times S^3$. (Here, homogeneous means the action can be written as $z(p,q) = (z^a p, z^b q)$ for $p,q\in \mathbb{H}$, $a,b\in \mathbb{Z}$, $\operatorname{gcd}(a,b) = 1$).

As Ivy says, this follows from the Barden-Smale classification of simply connected $5$-manifolds. In fact, for most choices of $(a,b)$, no one knows how to write down an explicit diffeomorphism. (The only exceptions I know are $(a,b)\in \{\pm (1,0), \pm (0,1), (\pm 1,\pm 1)\}$). My advisor, Wolfgang Ziller, has offered to pay for dinner at any restaurant in Philadelphia for anyone who figures out such a diffeomorphism for general $(a,b)$.

Such a diffeomorphism must use the fact that the action is homogeneous, because (as Mike mentioned in the comments) there are free linear actions of $S^1$ on $S^3\times S^3$ whose quotients are not diffeomorphic to $S^2\times S^3$. That said, only one other diffeomorphism type can arise: the unique non-trivial linear $S^3$ bundle over $S^2$.

In case you care, here is a summary of the results:

A general linear action of $S^1$ on $S^3\times S^3$ is equivalent to one of the form $z\ast(p,q) = (z^a p z^c, z^b q z^d)$ for integers $(a,b,c,d)\in \mathbb{Z}^4$. If $\operatorname{gcd}(a,b,c,d) \neq 1$, we have an ineffective kernel we can divide out, so we assume $\operatorname{gcd}(a,b,c,d) = 1$. The action is effectively free iff $\operatorname{gcd}(a^2 - c^2, b^2 - d^2) \in \{1,2\}$. If this $\operatorname{gcd}$ is $1$, the quotient is $S^2\times S^3$ and other wise, the quotient is non-trivial $S^3$ bundle over $S^2$.

And I apologize for the self promotion, but all of this was worked on in my thesis - the relevant paper can be found at https://arxiv.org/abs/1304.1770.

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Actually, for a general action $e^{it}.(z_1,z_2)=(e^{ikt}z_1,e^{ilt}z_2)$ where $(k,l)=1$, the quotient is $S^2\times S^3$. Just use Smale's classification results of simply connected spin $5$-manifolds. For more details, see for example

Einstein metrics on principal torus bundles, by M. Wang and W. Ziller.

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What about using quaternions ? Identify $S^3$ with the Lie group quaternions of norm 1, we see that the manifold $X$ is the homogeneous space $T $ \ $ G $ where $T$ is a circle, and $G=S'\times S"$; $S'$ and $S"$ being two samples of $S$. Now the group $S$ freely acts on the right on $G$ by the digaonal action, and this action pass to the quotient on a free right action of $S$, making $X$ a $S$ principal bundle over something $T $ \ ($ S'\times S" $) \ $S$. To identify this quotient with $S^2$ note that this manifold is homogeneous under the right action of the simply connected group $ S' \times S"$ with a connected isotropy group $T\times S$. Now a principal $G$ bundle over $S^2$ is trivial if $G$ is simply connected, therefore the manifold is the trivial $S$ bundle over $S^2$. It seems rather non trivial to get an explicit trivialization as $S\times S$ viewed as a $T\times S$ bundle over $S^2$ is not trivial.

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