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To show that a certain set is a manifold.

caffeinemachine

Well-known member
MHB Math Scholar
Mar 10, 2012
834
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)
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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.

Can anybody see how to do that?