# Topology on the spectrum of the algebra of entire symmetric functions of bounded type on the complex $L_\infty$

### Abstract

It is known that the so-called elementary symmetric polynomials $R_n(x) = \int_{[0,1]}(x(t))^n\,dt$ form an algebraic basis in the algebra of all symmetric continuous polynomials on the complex Banach space $L_\infty,$ which is dense in the Fr\'{e}chet algebra $H_{bs}(L_\infty)$ of all entire symmetric functions of bounded type on $L_\infty.$ Consequently, every continuous homomorphism $\varphi: H_{bs}(L_\infty) \to \mathbb{C}$ is uniquely determined by the sequence $\{\varphi(R_n)\}_{n=1}^\infty.$ By the continuity of the homomorphism $\varphi,$ the sequence $\{\sqrt[n]{|\varphi(R_n)|}\}_{n=1}^\infty$ is bounded. On the other hand, for every sequence $\{\xi_n\}_{n=1}^\infty \subset \mathbb{C},$ such that the sequence $\{\sqrt[n]{|\xi_n|}\}_{n=1}^\infty$ is bounded, there exists $x_\xi \in L_\infty$ such that $R_n(x_\xi) = \xi_n$ for every $n \in \mathbb{N}.$ Therefore, for the point-evaluation functional $\delta_{x_\xi}$ we have $\delta_{x_\xi}(R_n) = \xi_n$ for every $n \in \mathbb{N}.$ Thus, every continuous complex-valued homomorphism of $H_{bs}(L_\infty)$ is a point-evaluation functional at some point of $L_\infty.$ Note that such a point is not unique. We can consider an equivalence relation on $L_\infty,$ defined by $x\sim y \Leftrightarrow \delta_x = \delta_y.$ The spectrum (the set of all continuous complex-valued homomorphisms) $M_{bs}$ of the algebra $H_{bs}(L_\infty)$ is one-to-one with the quotient set $L_\infty/_\sim.$ Consequently, $M_{bs}$ can be endowed with the quotient topology. On the other hand, it is naturally to identify $M_{bs}$ with the set of all sequences $\{\xi_n\}_{n=1}^\infty \subset \mathbb{C}$ such that the sequence $\{\sqrt[n]{|\xi_n|}\}_{n=1}^\infty$ is bounded.

We show that the quotient topology is Hausdorffand that $M_{bs}$ with the operation of coordinate-wise addition of sequences forms an abelian topological group.