Definitions and Notation:
Let us write $\underbrace{\mathbb R^n\times \cdots\times\mathbb R^n}_{m \text{ times}}$ as $(\mathbb R^n)^m$.
A rigid motion in $\mathbb R^n$ is a function $L:\mathbb R^n\to \mathbb R^n$ such that $||L(x)-L(y)||=||x-y||$ for all $x,y\in \mathbb R^n$.
Let $U$ and $V$ be open sets in $\mathbb R^n$. A function $h:U\to V$ is said to be a diffeomorphism if it is differentiable and has a differentiable inverse.
A subset $M$ of $\mathbb R^n$ is said to be a $k$-dimensional manifold in $\mathbb R^n$ if for every point $x\in M$ we have:
There is an open set $U$ containing $x$, an open set $V\subseteq \mathbb R^n$, and a diffeomorphism $h:U\to V$ such that $h(U\cap M)=V\cap(\mathbb R^k\times\{0\})=\{y\in V:y_{k+1}=\cdots=y_n=0\}$.
(The above definition is taken from Spivak's Calculus on Manifolds)
The Problem:
It is known that any rigid motion in $\mathbb R^n$ can be written as a translation composed with an isometry.
Let $p=(p_1,\ldots,p_m)\in(\mathbb R^n)^m$.
A point $q=(q_1,\ldots,q_m)\in(\mathbb R^n)^m$ is said to be congruent to $p$ if there exists a rigid motion $L$ in $\mathbb R^n$ such that $q_i=L(p_i)$ for $1\leq i\leq m$.
Let $M(p)$ be the set of all the point in $(\mathbb R^n)^m$ which are congruent to $p$.
I need to show that $M(p)$ is a smooth manifold.
Right now I am just trying to show that $M(p)$ is a manifold. I think the isometries and translations are themselves manifolds and that may be the key to solve this problem.
I don't know anything about manifolds except for the definition I have given above and I am new to these things.
Can anybody see how to do that?
Best Answer
Let $G$ denote the group of isometries of $ R^n$ and let $H$ denote the stabilizer of your point configuration $p$. Then $M(p)$ is naturally diffeomorphic to $G/H$. Note that both $G$ is a Lie group and $H$ is its closed Lie subgroup. Hence, $G/H$ is a smooth manifold.