Why do a set of continuous transformations form a manifold?

Question

I am reading Sean Caroll's book on GR, and he defines manifolds to be "a space that may be curved and have a complicated topology, but in local regions looks just like R$^n$. Here by "looks like" we do not mean that the metric is the same, but only that more primitive notions like functions and coordinates work in a similar way" (Carrol). Hence, manifolds are a type of space with certain properties. However, later on he states that "a set of continuous transformations such as rotations in R$^n$ form a manifold." Why are the set of continuous transformations a manifold? In other words, how do we prove that the set of continuous transformations a manifold?

Answer 1

Not all sets of continuous transformations can be given natural manifold structures, but the set $O(n)$ of orthogonal transformations of $\mathbb{R}^n$ can, as can many other familiar groups. A typical proof of this type of result exploits the Rank Theorem from multivariable calculus. (The Wiki link doesn't contain much information, for more I recommend the very good exposition of the topic in Lee's Introduction to Smooth Manifolds.)

In any orthogonal basis, an $n \times n$ matrix $A$ is orthogonal iff $$A^T A = I,$$ which we can interpret as a system of $\frac{1}{2} (n + 1) n$ quadratic equations in the entries $a_{ij}$ of $A$.

Now, for any matrix $A$, $(A^T A)^T = A^T (A^T)^T = A^T A$, so $A^T A$ is symmetric, and we can regard the map $$f: A \mapsto A^T A$$ as a map $$M(n, \mathbb{R}) \to S,$$ where $M(n, \mathbb{R})$ is the space of $n \times n$ orthogonal matrices and $S$ is the space of $n \times n$ symmetric matrices, in which case $O(n)$ is precisely the level set $f^{-1} (I)$.

Now, it's just some easy algebra to show that the tangent map $T_g f$ has full rank at every point $g \in O(n) = f^{-1}(I)$. By the Rank Theorem, this guarantees that there is a local homeomorphism between a neighborhood of any $g \in O(n)$ and some open set in $\mathbb{R}^N$, which is essentially all that a manifold is. (Here, $N$ is the dimension of $O(n)$ as a manifold, which the Rank Theorem gives is $$N = \dim M(n, \mathbb{R}) - \dim S = n^2 - \frac{1}{2}(n + 1) n = \frac{1}{2}n (n - 1).)$$

One can mimic this proof for many other familiar infinite groups, including familiar examples like real and complex special linear groups, unitary groups, real and complex symplectic groups, and so on.

Answer 2

For a more fuzzy account, using Wikipedia's layout,http://en.wikipedia.org/wiki/Rotation_group_SO(3) you can see the space of all rotations in 3-D as a manifold , specifically as $\mathbb RP^3$. The way of making an equivalence between the transformations and the space is this: every rotation is uniquely defined by an axis and an angle of rotation. We start with a 3-ball $B^3$ of radius $\pi$, centered at the origin. Then every axis of rotation is given by a line segment from the origin to a point in the ball; the angle of rotation is given by the length of the segment (notice that the segment has length $\pi$ ). But we consider rotation by an angle $\pi$ the same as rotation by an angle $ -\pi$ , so we identify , for a fixed line segment associated to any rotation by $\pi$ with rotation by $-\pi$(meaning we identify antipodes in the whole of $ \partial B^3 =S^2)$ to a point, and this is $\mathbb RP^3 \simeq B^3/\mathbb Z_2 $ . So we have:

Line segment from $0$ to p $\equiv$ rotation with axis the line segment $0p$ and rotation angle given by $|Op|$ , the length of the line segment from the origin to the point p.

This describes the set of all possible rotations ,considering a rotation by $\pi$ to be equal to a rotation by $-\pi$, and, by the correspondence just described, this gives us the manifold $\mathbb RP^3$. This is an example of how one may turn a collection of maps into a manifold.

In maybe a more trivial way, consider the dual space $V^{*}$ of a finite-dimensional vector space $V$ (In the infinite-dimensional case, one may talk about Banach manifolds, a different issue) over, e.g., $\mathbb R$ or $\mathbb C$. $V^{*}$ is a vector space under $ (f+g)(x)=f(x)+g(x) ; (cf)(x)=cf(x) $ . Then, as a finite-dimensional vector space it is isomorphic to $\mathbb K^n$, e.g., for $\mathbb K:= \mathbb R$, your manifold is equivalent to the manifold $\mathbb R^n$.

Of course, to make all this precise, you could write actual homeomorphisms between the given respective pairs of spaces