# Semi group of polynomials which all roots lie on the unit circle

Let $X=\{f\in \mathbb{C}[z]\mid |z| \neq 1 \implies f(z) \neq 0\}$.

The motivation for consideration of such an $X$ is the the concept of Lee-Yang polynomials.

With the standard multiplication, $X$ is an Abelian semigroup with cancellation property.

Let $G$ be the Grothendieck group associated with $X$.

Is there a well known group which is isomorphic to $G$? In other words, is there an alternative formulation of $G$ in terms of some well known group? Is there a natural topology on $G$ which makes it a locally compact topological group?

• Don't know if you'd call it well-known, but an obvious realization of the group of fractions (or "Grothendieck group") is the group of rational fractions with no zero or poles outside the unit circle.
– YCor
Nov 29 '17 at 10:33

A complex polynomial is uniquely determined by its set of roots together with multiplicities. This means that the semigroup of your polynomials is freely generated by the set of point on the unit circle, aka by $\{z-k | k:\mathbb C, |k|=1\}$. Its Grothendieck group is the free abelian group on the continuum of generators. It can be given a nicer structure if you consider the natural topology on the set of polynomials.

In this case it will be the free topological abelian group generated by the unit circle $S^1$. This space is also not particularly nice, but up to homotopy equivalence it is $S^1 \times \mathbb Z$. This follows from Dold--Thom theorem which states that the free topological abelian group of a CW-complex $X$ is homotopy equivalent to the space that represents reduced singular homology $\tilde H_*(X)$.

The reduced homology of $S^1$ is $0, \mathbb Z, 0, \dots$, which corresponds to the homotopy groups of Eilenberg--Maclane space $S^1$. An extra factor of $\mathbb Z$ comes from connected components. Geometrically it means that any set of points on $S^1$ with multiplicities is equivalent to $n\cdot 1$, where $1: S^1$ is the basepoint.

• Since the question asks about all polynomials, not just the monic ones, it seems an extra factor of $\mathbb{C}^\times$ is needed. Nov 29 '17 at 5:06
• This is really nice! I wonder if this Grothendieck group can be in turn naturally identified with the group of rational functions with zeroes and poles confined to the unit circle Nov 29 '17 at 7:10
• @მამუკაჯიბლაძე it's trivially the case, isn't it?
– YCor
Nov 29 '17 at 10:35
• @YCor You are right. At least, a bijection is indeed trivial. Whether topologies match I don't see so quickly. Nov 29 '17 at 10:45
• @მამუკაჯიბლაძე A group isomorphism is trivial. I was not claiming more since I'm not aware of any canonical topology, so saying that "topologies match" does not make any sense to me. No topology was even specified on the semigroup!
– YCor
Nov 29 '17 at 11:21

Every locally compact group topology on $G$ makes the constants $\mathbf{C}^*$ an open subgroup.

Indeed, as observed in Fetisov's answer, $G$ is a direct product $\mathbf{C}^*\times A$ with $A$ free abelian. So $\mathbf{C}^*$ is the intersection of all kernels of homomorphisms $G\to\mathbf{Z}$.

R. Alperin (1980) proved that every homomorphism from a locally compact group into $\mathbf{Z}$ is continuous. It follows that $\mathbf{C}^*$ is closed in $G$.

So, working in the quotient, it is enough to prove: the only locally compact group topology $T$ on a free abelian group $A$ is the discrete one. Indeed, every subgroup of $A$ is free abelian. So $(A,T)$ cannot have a non-trivial compact subgroup (e.g., using again Alperin's result). It follows (by Hilbert's fifth problem, but which was previously known, probably due to Pontryagin) that the zero component $(A,T)^\circ$ is isomorphic to $\mathbf{R}^k$, and again this forces $k=0$, so $A$ is discrete.

Now $\mathbf{C}^*$ admits plenty of exotic locally compact group topologies, but any reasonable assumption (e.g., $\sigma$-compact + evaluation at some point is continuous) will force the topology to be the canonical one.