ag.algebraic geometry - parameterizing polynomial loops in $mathbb{C}^times$

This is mostly a series of comments, but guided by the questions you asked.

First of all, I will only talk about $X_n$, interpreting it as the space of non-zero complex polynomials $p$ of degree at most $n$ such that no root of $p$ lies on the unit circle, taken up to non-zero scaling. We may as well think of the polynomials as homogeneous of degree exactly $n$ in two variables, so that each point of $X_n$ defines a subset of $n$ points of the complex projective line $mathbb{CP}^1$ (the set of roots of the polynomial) that is disjoint from the unit circle. Therefore, $X_n$ is certainly a non-empty open subset in the standard topology of the projective space of homogeneous polynomials of degree $n$, and therefore $X_n$ has (real) dimension $2n$. Observe that the unit circle disconnects $mathbb{CP}^1$, and that the points of $X_n$ are likewise distributed into (at least) $n+1$ connected components, corresponding to how the $n$ points in $mathbb{CP}^1$ are distributed with respect to the two halves obtained by removing the unit circle (recall that $mathbb{CP}^1$ is topologically a two-dimensional sphere and that the unit circle can be identified with the equator of the sphere, so that, among the $n$ points we are talking about, there are some that are in one hemisphere and some that are in the other). On the other hand, a non-empty Zariski open subset of an irreducible algebraic variety is connected. Thus, as a subset of the space of homogenous polynomials of degree $n$, the space $X_n$ is certainly not a complex algebraic subvariety.

But, we may decide to analyze further the space $X_n$, zooming in on the locus $X_n^k$ where, for a fixed integer $k$, there are $k$ points on the half containing the origin and $n-k$ points on other half. Clearly, within each hemisphere, the points are free to roam around! Thus $X_n^k$ is homeomorphic (and in fact diffeomorphic with the natural choice of differentiable structure) to the space of ordered pairs $(p_1,p_2)$ where $p_1$ is a monic polynomial of degree $k$ and $p_2$ is a monic polynomial of degree $n-k$: expand each hemisphere to a whole complex plane and "code" the $k$ points on one half by the unique monic polynomial having them as a root (and do the same to the other half). Thus each space $X_n^k$ is connected and in fact diffeomorphic to $mathbb{C}^k times mathbb{C}^{n-k}$. As a complex manifold you can also say that $X_n^k$ is the product of the symmetric product of $k$ copies of the unit disk with the symmetric product of $n-k$ copies of the unit disk. Thus, again using a structure induced from the ambient space of homogeneous polynomials, any complex algebraic subvariety of $X_n$ would be a complex algebraic subvariety of a symmetric product of unit disks: I think that this means that the only complex algebraic subvarieties of $X_n$ are the points.

Finally, let me make a small stab at getting your hand on $X$, by mentioning one description of the "glueing" of $X_n$ inside $X_{n+1}$. From the point of view of non-homogeneous polynomials, this corresponds to simply realizing that a polynomial of degree at most $n$ is also a polynomial of degree at most $n+1$. From the point of view of their homogenizations, the inclusion corresponds to replacing $z^i$ by $x^iy^{n+1-i}$ instead of $x^iy^{n-i}$. Effectively, we are adding the point at infinity as one of our roots (namely the extra root $y=0$). Thus in terms of the description above, we observe that the two hemispheres are not "identical": one of them has a point that is special, namely the point at infinity. The points of $X_{n+1}$ that come from points of $X_n$ are the points that correspond to $(n+1)$-tuples one of whose elements is the point at infinity.