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I'm trying to understand the topology of the product of two three dimensional spheres $\mathbb{S}^3\times \mathbb{S}^3$ quotiented by the action of $\pm 1$ sending a pair of points $(x,y)$ to the corresponding pair of antipodal points $(-x,-y)$. My hunch is that this may be better understood by identifying $\mathbb{S}^3$ with the special unitary group $SU(2)$ via the well known homeomorphism and understanding the quotient of the topological group $SU(2)\times SU(2)$ by the group $H=\{I_2\times I_2,-I_2\times -I_2\}$ as a topological space but I haven't yet been successful with that either. I'd like to know if this approach is worthwhile or if I should be looking in another direction to recognize the space $\mathbb{S}^3\times \mathbb{S}^3/\pm1$.

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It's $SO(4)$ by an extension of the argument for $SU(2)/\{\pm 1\} = SO(3)$: en.wikipedia.org/wiki/… –  commenter Oct 17 '12 at 16:37
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Not an answer, but synonyms: $\mathbb{S}^3\times \mathbb{S}^3/\mathbb{Z}_2$ is the double cover, Spin$(4)$, of SO(4). –  c.p. Oct 17 '12 at 17:54
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@Jorge: the double cover of $\text{SO}(4)$ is simply connected. This is just $\text{SO}(4)$ on the nose. –  Qiaochu Yuan Oct 17 '12 at 17:59

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up vote 4 down vote accepted

The quotient $S^3\times S^3/\{\pm (1,1)\}$ is diffeomorphic (but not Lie isomorphic) to $S^3\times \mathbb{R}P^3$ (Here, $\mathbb{R}P^3$ is naturally diffeomorphic to $SO(3)$, so we give it the Lie group structure $SO(3)$ has).

An explicit map is given as follows. Identifying $S^3$ as the unit quaternions, define $\tilde{f}:S^3\times S^3\rightarrow S^3\times \mathbb{R}P^3$ by $\tilde{f}(p,q) = (pq, [q])$ where $[q]$ denotes the class of $q$ in $\mathbb{R}P^3$.

Notice that $$\tilde{f}(-p,-q) = \left((-p)(-q), [-q]\right) = (pq,[q]) = \tilde{f}(p,q)$$ so $\tilde{f}$ descends to a map on $S^3\times S^3/\pm(1,1)$, which I'l call $f$.

The map $\tilde{f}$ is clearly smooth, so $f$ is as well. Further, $f$ has an inverse given by $f^{-1}(p,[q]) = [pq^{-1},q]$. This is also clearly smooth, so we have that $f$ is a diffeomorphism.

Why aren't they isomorphic as Lie groups? I claim that $S^3\times \mathbb{R}P^3$ has a normal subgroup isomorphic to $SO(3)$ while $S^3\times S^3/\pm(1,1)$ doesn't.

First, since $\mathbb{R}P^3$ is isomorphic to $SO(3)$, the subgroup $\{e\}\times \mathbb{R}P^3$ is isomorphic to $SO(3)$ and is normal in $S^3\times\mathbb{R}P^3$.

Second, to see there are no normal subgroups isomorphic to $SO(3)$ in $S^3\times S^3/\pm(1,1)$ note that the Lie algebra of $S^3\times S^3/\pm(1,1)$ is $\mathfrak{so}(3)\oplus\mathfrak{so}(3)$ and $\mathfrak{so}(3)$ is simple, so the only nontrivial ideals are $\mathfrak{so}(3)\oplus 0$ and $0\oplus \mathfrak{so}(3)$. In $S^3\times S^3$, these exponentiate to the two different $S^3$ factors and only can easily check that under the projection $\pi:S^3\times S^3\rightarrow S^3\times S^3/\pm(1,1)$, $\pi$ is injective on each factor.

It follows that the two ideals in $\mathfrak{so}(3)\oplus\mathfrak{so}(3)$ exponentitate to $S^3$s in $S^3\times S^3/\pm(1,1)$, so, in particular, they are not isomorphic to $SO(3)$.

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Since you said you cared about it "as a topological space", I figured "topology like $S^3\times\mathbb{R}P^3$" makes more sense than "topology of $SO(4)$", even if they really are identical, hence the direction of my post –  Jason DeVito Oct 17 '12 at 17:52
    
The answer is actually much nicer than I expected. Thanks! –  Victor Oct 18 '12 at 7:06

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