I'm working on $X = \mathbb{R}^n$ and I'm considering the set of unordered sequence of points of $X$. Considering $F(p) = \lbrace (x_1, \dotsc, x_p) \in X^p ; i \neq j \implies x_i \neq x_j \rbrace$ the set of ordered sequences on $X$ (which is an open set) and the symmetric group $S_p$, we can show that the quotient $B(p) = F(p) / S_p$ is a manifold.
My question is the following: is it possible to determine the dimension of this manifold? Thanks!
The space $F(p)$ is an open submanifold of $X^p$, so it has the same dimension*, namely $np$. The group $S_p$ acts freely and properly discontinuouly on $F(p)$ and is discrete, thus $B(p) = F(p) / S_p$ has the same dimension as $F(p)$, again $np$.
* If you want to be convinced, each $x \in F(p)$ has an open neighborhood contained entirely in $F(p)$; $X^p$ being a manifold of dimension $np$, it has a neighborhood basis of open subsets homeomorphic to $\mathbb{R}^{np}$. Thus $x$ has an open neighborhood in $F(p)$ homeomorphic to $\mathbb{R}^{np}$.