What does it mean for the group of rotation matrices to have a “manifold structure”

I have just been exposed to rotation matrices, and it is showing extremely strong connection with group theory and differential geometry which I am both totally inept at.

Can someone in simple term explain what it means for a group to have a manifold structure and how this affects the mathematics of rotation group?

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This is the beginning of deep, rich, and powerful subject called Lie theory, and so any answer one could give here would be badly incomplete, but I’ll (1) briefly (and very roughly) describe what a manifold is, and (2) describe one important consequence of having a manifold structure for group theory and rotations in particular.

(1) A manifold is, roughly speaking, a space that around any point looks like a patch of Euclidean space: For example, the usual sphere $\mathbb{S}^2 \subset \mathbb{R}^3$ locally looks like $\mathbb{R}^2$—this is precisely the fact that we exploit when we make flat maps on paper that show sections of the surface of the earth.

So, the statement that the space of rotation matrices (say, in dimension $2$ or $3$) has a manifold structure means that it locally looks $\mathbb{R}^n$ for some $n$.

The space $SO(2)$ of $2 \times 2$ rotation matrices are exactly those of the form
$$\begin{pmatrix} \cos \theta & -\sin \theta\\ \sin \theta & \cos \theta\end{pmatrix}.$$ We can identify these matrices respectively with the points $(\cos \theta, \sin \theta)$, that is, exactly the points on the unit circle, $\mathbb{S}^1$. In other words, we can think of the space of rotation matrices as $\mathbb{S}^1$ itself. The group structure, that is, just concatenation of rotations, is easy to describe here: Any rotation is a rotation by some $\theta$, and if we take rotations by $\theta$ and $\phi$, multiplying them together gives the rotation by $\theta + \phi$. Writing this out in terms of the rotation matrices gives
$$\begin{pmatrix} \cos \phi & -\sin \phi\\ \sin \phi & \cos \phi\end{pmatrix}\begin{pmatrix} \cos \theta & -\sin \theta\\ \sin \theta & \cos \theta\end{pmatrix} = \begin{pmatrix} \cos (\phi + \theta) & -\sin (\phi + \theta)\\ \sin (\phi + \theta) & \cos (\phi + \theta)\end{pmatrix}.$$
Expanding this expression gives a proof of the usual angle sum identities for $\sin$ and $\cos$.

The space $SO(3)$ of $3 \times 3$ rotation matrices is more complicated. But we can describe it in a concrete way: Consider the $3$-sphere $\mathbb{S}^3 \subset \mathbb{R}^4$. We can identify antipodal points on the sphere to produce real projective space $\mathbb{RP}^3$, and this space is homeomorphic to $SO(3)$. There’s a way to describe this group structure in terms of $\mathbb{S}^3$, regarded as the set of unit quaternions, but I won’t say more about that unless someone asks. (Incidentally, we can also regard $\mathbb{S}^3$ as a group in its own right, in which setting we call it $Spin(3)$.)

(2) We can think of the space $M(3, \mathbb{R})$ of $3 \times 3$ matrices as $\mathbb{R}^9$ in the obvious way, and $SO(3)$ is some nicely behaved subset of that space. In particular, we can ask for any point $g$ on that space, say the identity matrix $\mathbb{I}$, about the $3$-plane tangent to $SO(3)$ at that point, and in particular ask for a matrix description of that plane:

A $3 \times 3$ rotation matrix is one that satisfies $A^T A = \mathbb{I}$ and $\det A > 0$. Now, for any vector $X$ tangent to $SO(3)$ at $\mathbb{I}$, there is a curve $\gamma: J \to SO(3)$ such that $\gamma'(0) = X$ (and so $\gamma(0) = \mathbb{I})$.For all time $t$, we have by the above characterization that
$\gamma(t)^T \gamma(t) = \mathbb{I}$. Differentiating at $t = 0$ gives
$$X^T + X = 0;$$
in other words, we can think of the tangent space to $SO(3)$ at the identity $\mathbb{I}$ as exactly the set of $3 \times 3$ antisymmetric matrices; we can think of the elements of this space as “infinitesimal rotations”, and we denote this space by $\mathfrak{so}(3)$. Now, the group multiplication determines a special product on $\mathfrak{so}(3)$ called a Lie bracket, which turns out just to be the matrix commutator:
$$[X, Y] = XY – YX,$$
and which makes $\mathfrak{so}(3)$ a Lie algebra.

So, given the group $SO(3)$, we’ve now produced a vector space $\mathfrak{so}(3)$—which geometrically is much simpler than $SO(3)$ itself, and a product on that vector space which partially encodes the group structure on $SO(3)$. This lets us translate many problems about rotation matrices into questions about linear algebra which a priori may be easier than the original questions about the group itself. Roughly speaking, we can think of the relationship between a Lie group and a Lie algebra as analogous to, but much stronger than, the relationship between a function and its tangent line approximation at a point.

All of this, by the way, can be extended to most familiar matrix groups.

Remark The vector space of antisymmetric $3 \times 3$ matrices is $3$-dimensional (for any such matrix, the diagonal entries are zero, and the entries below the diagonal are determined by those above, which may be prescribed freely), and so we can think of it as $\mathbb{R}^3$. It turns out that in this picture, the Lie bracket is nothing more than the usual cross product from vector calculus.