Why $S^1\times S^{2m-1}$ carries a complex structure. Let $S^n$ denotes $n$-sphere, then why $S^1\times S^{2m-1}$ carries a complex structure.
 A: $\newcommand{\C}{\mathbf{C}}$Scalar multiplication by a complex number $\lambda$ of modulus strictly larger than $1$ defines a biholomorphism from $\C^{m}\setminus\{0\}$ to itself. The automorphism group generated by this mapping acts properly discontinuously, so the quotient admits the structure of a holomorphic manifold.
To identify the quotient space, note that
$$
\{z \in \C^{m} : 1 \leq \|z\| \leq |\lambda|\} \simeq S^{2m-1} \times \bigl[1, |\lambda|\bigr]
$$
is a fundamental domain, and the unit sphere (the "inner boundary") is glued to the sphere of radius $|\lambda|$ (the "outer boundary") by the action.
For simplicity, you may as well choose $\lambda$ to be real if you're mainly interested in seeing intuitively why the quotient is a product of spheres. You may also be interested in reading about Calabi-Eckmann manifolds.
(Note that, aside from elliptic curves arising when $m = 1$, none of these manifolds admits a Kähler metric, since $H^{2} = 0$.)
A: Here is an answer I thought of last night which will hopefully be of benefit to someone down the line, even though this question has long since been answered. This construction avoids the use of quotients by a group action and differential topology.
What we will do is emulate the stereographic projection argument we know proves that the sphere is a Riemann surface. So we will have two charts as follows. Regard the $\mathbb{S}^1$ factor as having a north and south pole, $N$ and $S$. Then $\mathbb{S}^3 \times \big( \mathbb{S}^1 - N \big)$ is missing an $\mathbb{S}^3$ fiber. Since the punctured circle is just the line, think of this family of spheres as being spherical shells filling out $\mathbb{C}^2$. We never fill out the origin because we want these to be charts in at atlas. (So really we need the fact that $(0,\infty) \simeq \mathbb{R}$ topologically.) But anyway, this gives a chart on the space which hits everything except one $\mathbb{S}^3$ fiber and has image as the punctured complex plane, which is clearly an open subset of $\mathbb{C}^2$. Do the same thing for the other chart, removing $S$. This gives at atlas with precisely two charts.
In this atlas, the single transition map is inversion in $\mathbb{S}^3$, just like on the Riemann sphere! Inversion in a sphere is (oriented) angle preserving, and so the usual proof that this is holomorphic goes through.
It is clear that this generalizes to arbitrary spheres of odd dimension. I'd like to say a few words about why this is morally correct.
The thing that makes stereographic projection really work is that the 2-sphere can be identified with $\mathbb{CP}^1$. This is how the two pieces glue together to give the sphere - the fact that $\mathbb{CP}^1$ has two natural charts and each can readily be identified with a copy of the complex line. Let's examine what happens if we try to do the same game with $\mathbb{CP}^2$. We naively expect to get $\mathbb{S}^4$ as a result. However, $\mathbb{CP}^2$ does not have quite the same structure. In fact it can be realized as a quotient of the $4-$ball, where we glue points on the boundary $\mathbb{S}^3$ together along the Hopf fibration. Furthermore, $\mathbb{CP}^2$ does not have two natural charts on it, but three. So it is clear that this object is very different. And in fact you can see that they are quite different purely on the level of topology - the computation of De Rham cohomology for $H^2(\mathbb{CP}^2) \neq 0$.
This means that the construction I've presented here is morally the correct analogue of stereographic projection for complex manifolds rather than the thing we would do for ordinary real manifolds.
