A remark on embedding of a cylinder on a real commutative Banach algebra

Abstract: Let $A$ be a real commutative Banach algebra with unity. Let $a_0\in A\setminus{0}$. Let $\mathbb Z a_0:={na_0}_{n\in \mathbb Z}$. Then, $\mathbb Z a_0$ is a discrete subgroup of $A$. For any $n\in \mathbb Z$, the Frechet derivative of the mapping $$x \, \in \, A \ \ \ \mapsto \ \ \ x+na_0 \, \in \, A$$ is the identity map on $A$ and, especially, an $A$-linear transformation on $A$. So, the quotient group $A/(\mathbb Z a_0)$ is a $1$-dimensional $A$-manifold and the covering projection $$x \, \in \, A \ \ \ \mapsto \ \ \ x+\mathbb Z a_0 \, \in \, A/(\mathbb Z a_0)$$ is an $A$-map. We call $A/(\mathbb Z a_0)$ the $1$-dimensional $A$-cylinder by $a_0$. Let $T$ be a compact Hausdorff space. Suppose that there exist $t_1\in T$ and $t_2\in T$ such that $t_1\not=t_2$ holds. Then, the set $C(T;\mathbb R)$ of all real-valued continuous functions on $T$ is a real commutative Banach algebra with unity and $\mathbb R \, \subsetneq \, C(T;\mathbb R)$ holds. In this paper, we show that there exists $a_0 \, \in \, C(T;\mathbb R)\setminus \mathbb R$ such that for any $k\, \in \, \mathbb N$, the $1$-dimensional $C(T;\mathbb R)$-cylinder $(C(T;\mathbb R))/(\mathbb Z a_0)$ by $a_0$ cannot be embedded in the finite direct product space $(C(T;\mathbb R))^k$ as a $C(T;\mathbb R)$-submanifold.