parameterizing polynomial loops in $\mathbb{C}^\times$

Question

Say $L\mathbb{C}^\times$ is the loop group of smooth maps $S^1 \to \mathbb{C}^\times$. There is a submonoid $L_{poly}\mathbb{C}^\times$ of loops that look like $w_0 + w_1z +w_2z^2 + \cdots + w_nz^n$ where $z = e^{i\theta}$ (as Andrew notes below this is not a group because its not closed under taking inverses). Equivalently $L_{poly}\mathbb{C}^\times$ as a set is just polynomials $p(z) \in \mathbb{C}[z]$ such that $p(z) \ne 0$ for $|z| = 1$.

If we mod out by scaling and rotation then the set of polynomial loops describe a subset $X$ of $\mathbb{P}(\oplus_{n \in \mathbb{N}} \mathbb{C})$ (by identifying a loop with its vector of coefficients). I want to look at $X$ from an algebro-geometric point of view, but I have no intuition about how bad or nice $X$ may be; i.e. can it be a variety?

The way I've been thinking about it is that $X = \cup_{n \in \mathbb{N}} X_n$ where $X_n \subset \mathbb{P}^n$ are the loops of degree at most $n$. I think $X_1$ is the image under the projection $\mathbb{C}^2 - 0 \to \mathbb{P}^1$ of the set {$(w_0,w_1):|w_0|\ne |w_1|$}. So it seems describable as the complement of a hypersurface in $\mathbb{R}^4$ but probably its not a complex variety.

But already trying to figure out what $X_2$ is seems difficult. Also I feel I don't have any `sophisticated' way of thinking about this stuff meaning my attempts to describe $X_2$ seems to always degenerate to just fumbling around with planar geometry.

Some specific questions regarding this setup:

0) What is the dimension of $X_n$?

1) Which if any of the $X_n$ or $X$ are a variety over $\mathbb{C}$ or $\mathbb{R}$?

2) If $X$ or $X_n$ are not varieties can you find any positive dimensional varieties contained in them?

3) Can you suggest any tools that might be useful for answering any of the previous questions?

Of course if any of this seems to easy you are welcome to replace $\mathbb{C}^\times$ with $GL(n,\mathbb{C})$, polynomials with rational functions or with power series convergent in an annulus containing $|z| = 1$.

An extra thought: So $L_{poly}\mathbb{C}^\times$ is not a group, but it seems you do get a group if you look at convergent series in non positive powers of $z$; i.e. loops that look like $\sum_{j \in \mathbb{N}} c_j z^{-j}$. I wonder if there's anything interesting you can say about the $c_j$.

Answer 1

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.

Answer 2

(This is perhaps more of a comment than an answer, though I hope it also gives you some ideas where to look further.)

I have difficulty with the fourth word of the second sentence: "subgroup". The space that you describe is not a subgroup of $L \mathbb{C}^\times$. It is a submonoid, but is not closed under taking inverses. Even if (as is usual) you allow Laurent polynomials, it is still not a subgroup. So exactly what answer will satisfy you depends on whether you really meant "subgroup" and so need to change the rest, or you really meant the space you describe and are happy to drop the requirement that it be a subgroup.

If the former (that you really want a subgroup), then I have some pointers for you. If the latter, I don't.

To get a polynomial subgroup of a loop group, you need to be working with a compact Lie group. The simplest example being $U_n$. A polynomial loop in $U_n$ is simply one that looks like a polynomial when viewed as a loop in $M_n(\mathbb{C})$. These do have polynomial inverses since one simply takes the adjoint of the coefficients to produce the inverse and that is still a polynomial. (For a more general compact Lie group, embed it analytically in some $U_n$.) For $\mathbb{C}^\times$, this produces polynomial loops in $S^1$ which is ... not a lot! You get $\{z^k : k \in \mathbb{Z}\}$. Note, however, that this is homotopy equivalent to the much larger group $L\mathbb{C}^\times$!

If this is what you are interested in, then the place to look is Loop Groups by Pressley and Segal. Another reference, also by Segal, is Loop groups and harmonic maps. From memory (it's been a while since I last looked at it), the latter article deals with issues close to what you're asking about.

Answer 3

0) Under most reasonable notions of dimension, the (real) dimension would be $2n$. For a polynomial up to scaling is the same as an element of $\mathbb{C}^n$ (i.e. a collection of $n$ roots). We can mod out by scaling by a positive real factor, giving something of real dimension $2n-1$. When doing this modding out, the fibre of each element of the quotient contains at most $n$ (hence a finite number) of polynomials which are not in $X_n$, implying that the real (topological) dimension should still be $2n$.

As for whether it's a variety, you're looking at the intersection of infinitely many Zariski closed sets defined by $\sum_{i=0}^n w_i z^i$ for each $z \in S^1$, which I would not necessarily expect to be a variety. It is, however, a definable real set (in the sense of model theory), which would imply that it's at least almost a kind of quasiprojective real variety. In other words, the infinitely many inequation relations satisfied by elements of $X_n$ lie on a real variety themselves, meaning they probably have some sort of nice structure. Someone who knows more than I do should try to answer the rest of the question.