# Homeomorphism between base of conifolds and spheres

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?!

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.
• 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! – Alireza May 6 '13 at 17:34