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Hello

Call $Y^4$ a conifold which satisfies the following condition:

$\mathfrak{Y}(z):=\sum_{\alpha=1}^{3}(z_{\alpha})^{2}=0,$

where $z_\alpha \in \mathbb{C}$. Now intersect $Y^4$ with $S^5$ to get a compact manifold called "base", $X^3$, which is $3$ dimensional just like $S^3$. We have a Hopf fibration regarding this base as $S^1\rightarrow X^3\rightarrow S^2.$ Since a similar Hopf map exists between $3$- and $2$-sphere, I doubt that $X^3$ is homeomorphic to $S^3$, or maybe it is not the case! Does $X^3$ really have a non-zero genus?!

Thanks in Advance AB

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  • $\begingroup$ Also, I would call X the link of the singularity rather than the base. $\endgroup$
    – Jonny Evans
    Commented May 6, 2013 at 17:03
  • $\begingroup$ Well string theorists call it "base"! $\endgroup$
    – Alireza
    Commented May 6, 2013 at 17:35
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    $\begingroup$ That's handy to know (someone should write a dictionary). Hopefully the term "link of a singularity" will help you if you need to search the mathematical literature for this kind of stuff in future. $\endgroup$
    – Jonny Evans
    Commented May 6, 2013 at 18:20

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Your space X is $\mathbf{RP}^3$. To see this, blow-up the origin. The proper transform of Y is then the total space of the bundle $\mathcal{O}(-2)\to\mathbf{CP}^1$. The unit circle bundle (your space X) is then $\mathbf{RP}^3$. One can see this by taking the fibrewise double cover to get the bundle $\mathcal{O}(-1)$. On the unit circle bundle this just gives the double cover $S^3\to\mathbf{RP}^3$.

Alternatively you could smooth the singularity slightly to get a smooth real quadric. The real part would be a Lagrangian sphere, and the quadric is diffeomorphic to its cotangent bundle. The unit cotangent bundle of $S^2$ is $\mathbf{RP}^3$, which is again your space X.

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  • $\begingroup$ Oh very nice! I see! Since $\mathbb{RP}^3 \cong SO(3)$ and $SO(3)$ is not simply connected, the genus would be non-zero! Thanks! $\endgroup$
    – Alireza
    Commented May 6, 2013 at 17:34
  • $\begingroup$ You're welcome. Indeed, the fact that SO(3) acts freely on the unit cotangent bundle is another way to see that the unit cotangent bundle is diffeomorphic to RP^3. $\endgroup$
    – Jonny Evans
    Commented May 6, 2013 at 18:18

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