Let $\mathbb C^n = \mathbb C \times \mathbb C \times \dots \times \mathbb C$ be the complex space. We define the quotient space $\mathbb C^n_{sym}$ by $\mathbb C^n / \sim$ where $x \sim y$ if there exists some $\sigma \in \mathbb S_n$ such that $x_{\sigma} = (x_{\sigma(1)}, \dots, x_{\sigma(n)}) = (y_1, \dots, y_n)$. I am wondering whether the set $\mathcal E = \{x \in \mathbb C^n_{sym}: x \text{ has distinct components}\}$ is connected.
In $\mathbb R^2$, the set would be $\mathbb R^2$ without the line $x=y$ and we identify the two half spaces divided by this line. It seems to me this set is connected. I am not sure how to prove for $\mathbb C^n_{sym}$.
Let $L = \langle (1, \dots, 1) \rangle$ and $x,y \in \mathbb{C}^n \setminus L$. Now, the (complex) line $L_{xy}$ joining $x$ and $y$ is not $L$, because for example $x \in L_{xy} \setminus L$. Since $L_{xy}$ is a $1$-dimensional affine subspace, $L_{xy} \cap L$ will be finite. Since $L_{xy} \simeq \mathbb{C}$ and $\mathbb{C}$ without a finite amount of points is path connected, there is a path $\gamma$ from $x$ to $y$ (which is contained in $L_{xy}$) so that $\gamma \cap L = \emptyset$. Now, if $q$ is the quotient map to $\mathbb{C}^n_{sym}$, the function $q\gamma$ is an arc from $[x]$ to $[y]$ which does not intersect $\mathcal{E}^c$. Thus, $\mathcal{E}$ is actually path connected.