Why is $T^*S^3$ a conifold? So, I was reading the famous Gopakumar Vafa paper, and they mention that $T^*S^3$ is a conifold. Why is this the case? I would naively expect $T^*S^3$ to be basically the same everywhere ($S^3$ is a maximally symmetric space, so I kind of expected its cotangent bundle to be nice as well). So where are the conical singularities coming from?
 A: The conifold singularity is the quadric hypersurface singularity given in complex coordinates by
$x_1^2+x_2^2+x_3^2+x_4^2=0$
It is also known as the 3-fold ordinary double point. You can smooth this singularity by perturbing the equation:
$x_1^2+x_2^2+x_3^2+x_4^2=\epsilon$
(for some $\epsilon\neq 0$). This variety is smooth ("looks the same everywhere" in the sense that it's a complex manifold). It is diffeomorphic to the cotangent (or equivalently tangent) bundle of $S^3$: the zero section is just the real locus (assuming $\epsilon$ is real). You can write down a diffeomorphism explicitly, e.g.
https://math.stackexchange.com/q/1784898
So $TS^3$ is a smoothing ("the Milnor fibre") of the conifold singularity.
This singularity also admits a small resolution, which is the complex manifold given by the total space of the bundle $\mathcal O(-1)\oplus \mathcal O(-1)\to \mathbb{P}^1$. The process of degenerating the quadric to become singular and then resolving is often called a conifold transition.
