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I am reading up on Heegaard splittings and need some conceptual help. $S^3$ may be decomposed into two 3-balls with the boundaries of the 3-balls identified. $S^3$ may be also decomposed into two solid 2-tori with boundary points identified.

$\mathbb{R}P^3$ has half as many points as $S^3$. This leads me to suspect that $\mathbb{R}P^3$ may be considered as a solid 2-torus with antipodal points on its boundary identified. Is this interpretation correct?

Thanks.

Evan
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    All positive dimensional manifolds have the same number of points, so absolutely nothing can be deduced from the fact that two manifolds happen to have the same number of points (and certainly not from one having "half" as many as any other...) – Mariano Suárez-Álvarez Jul 04 '17 at 04:07
  • What exactly do you mean by antipodal points? If you're considering $T^2$ embedded in $\mathbb{R}^3$ then you actually get a Klein bottle as shown here https://math.stackexchange.com/questions/2095329/what-is-the-surface-by-identifying-antipodal-points-of-a-2-torus-embedded-in-m Alternatively you can think of $T^2 = S^1 \times S^1$ so that the antipodal map is $f(\theta,\phi) = (-\theta,-\phi)$. – Osama Ghani Jul 04 '17 at 04:13
  • @MarianoSuárez-Álvarez I think that when OP says $\mathbb{R}P^3$ has half as many points as $S^3$, he means that $S^3$ is a double cover of $\mathbb{R}P^3$ – Osama Ghani Jul 04 '17 at 04:15
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    If you think that $RP^3$ is the lie group $SO(3)$, it is easy to describe it as the union of two tori. Note that $SO(3)$ acts on the two sphere, with isotropy group $SO(2)=S^1$, and $RP^3$ is therefor a circle bundle over the two sphere. Writting this 2-sphere as a union of two discs glued along there boundary, we get a decomposition of $RP^3$ as a union of two solid tori (a circle bundle over a disc is trivial). – Thomas Jul 04 '17 at 08:43
  • @Thomas I find your comment enlightening and definitely interesting enough to be written as an answer. This will increase the number of people reading it and make this post more useful. – Amitai Yuval Jul 05 '17 at 01:01

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$\newcommand{\Cpx}{\mathbf{C}}$Think of the unit $3$-sphere sitting inside $\Cpx^{2}$ as the set $$ S^{3} = \{(z_{1}, z_{2}) : |z_{1}|^{2} + |z_{2}|^{2} = 1\}. $$ The standard decomposition into two tori is via the Clifford torus $C$ defined by $|z_{1}|^{2} = |z_{2}|^{2} = 1/2$. The resulting handlebodies are $T_{1} = \{(z_{1}, z_{2}) : |z_{1}| \leq 1/2\}$ and $T_{2} = \{(z_{1}, z_{2}) : |z_{2}| \leq 1/2\}$.

The antipodal map of $S^{3}$ does send $C$ to itself, but the handlebodies are as far as possible from being fundamental domains of this action: Each handlebody is preserved by the antipodal map, and hence double-covers its own image in the quotient space. Particularly, the quotient space is not obtained from one handlebody by boundary identification.

Andrew D. Hwang
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If you think that $RP^3$ is the lie group $SO(3)$, it is easy to describe it as the union of two tori. Note that $SO(3)$ acts on the two sphere, with isotropy group $SO(2)=S^1$, and $RP^3$ is therefor a circle bundle over the two sphere. Writting this 2-sphere as a union of two discs glued along there boundary, we get a decomposition of $RP^3$ as a union of two solid tori (a circle bundle over a disc is trivial). The same argument proves that every circle bundle over the sphere is made by gluing two solid tori along there boundaries.

Thomas
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  • Hi Thomas,

    I thought that $S^3$ could be regarded as the union of two solid tori? How do we get from the two solid tori representation of $S^3$ to the two solid tori representation of $\mathbb{R}P^3$?

     – Evan Jul 06 '17 at 02:18
  • A double cover of a solid torus is a solid torus. – Thomas Jul 06 '17 at 05:05