What does "dR" mean when R is a matrix?

Question

I was trying to derive the Lie algebra $so(3)$ from the corresponding Lie group $SO(3)$. But an intermediate step stopped me. Belows are my derivation process:

1. Specify the identity of SO(3): identity matrix $I$.
2. Specify the inversion of SO(3): $$ R^T \circ R = R \circ R^T = I \Rightarrow R^{-1} = R^T$$
3. Time-differentiate the inversion: $$\dot{R}R^T =-(\dot{R}R^T)^T \Rightarrow \dot{R}R^T = [\omega]_x$$ for some skew-symmetric matrix.
4. Add the time dimension: $$\dot{R}(t)R^T(t) = [\omega]_x(t)$$ By definition, the exponential map $exp$ is just a kind of projection, hence $[\omega]_x(t)$ is constant with regards to $t$. Therefore, we get: \begin{equation} \label{eq:1} \frac{dR}{dt} R^{-1} = [\omega]_x \Rightarrow dR \ R^{-1} = [\omega]_x \ dt \end{equation}
5. Solve the ordinary differential equation(ODE) using Separation of Variables, which is just doing integration on both sides.

My question is:

1. What does it mean $dR$ when $R$ is a matrix?
2. How to solve the equation specified in 4. when the matrix multiplication is not commutative? I.e. Simply put integration sign on both sides: $$\int dR \ R^{-1} = \int [\omega]_x \ dt$$ However, as far as I'm concerned, this form does not make any sense provided the matrix multiplication is not commutative.

Answer 1

$\newcommand{\R}{\mathbb{R}}$Let me extend my comment a bit further.

The Lie algebra of a Lie group $G$ can be identified with the tangent space $T_eG$ at the identity $e\in G$. The Lie group $SO(n)$ is the set of orthogonal $(n\times n)$-matrices of positive determinant and it is an open subgroup of the Lie group $O(n)$. Therefore, the tangent space at the identity of $SO(n)$ is equal to the tangent space of $O(n)$ which we shall compute.

Consider the smooth map $\Phi\colon GL(n) \to \R^{n\times n}, R\mapsto R^TR$. Then $O(n)$ is exactly the level set $\Phi^{-1}(I_n)$ and the tangent space $\mathfrak{o}(n)=T_{I_n}O(n)$ is the kernel of the differential $d\Phi_{I_n}\colon T_{I_n}GL(n) \to T_{I_n}\R^{n\times n}$, i.e. you have to solve the linear equation $d\Phi_{I_n}(X)=0$ for $X\in T_{I_n}GL(n)=\R^{n\times n}$.

One way to calculate the differential $d\Phi_{I_n}$ is as follows. Consider a curve $\gamma\colon (-\epsilon,\epsilon)\to GL(n)$ on the manifold $GL(n)$ which starts at the identity $\gamma(0)=I_n\in O(n)$ with velocity vector $\gamma^\prime(0)=X\in\R^{n\times n}$. In general you would let $\gamma(t)=\exp^G(tX)$ but here you can just take $\gamma(t)=I_n+tX$ (at least for $\epsilon$ small enough). Now you can consider the composition $\Phi\circ \gamma\colon \R\to\R^{n\times n}$ - so that you can use your knowledge from calculus. Then $$d\Phi_{I_n}(X) =\frac{d}{dt}\Big|_{t=0} \Phi\circ \gamma(t) =\frac{d}{dt}\Big|_{t=0} \big(I_n+tX\big)^T\big(I_n+tX\big)=X^T+X.$$ You immediately see that the kernel (and hence $\mathfrak{o}(n)$) is given by the skew-symmetric matrices.

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In regards to your second question about how to calculate the exponential map: The exponential map $\exp^G\colon \mathfrak{g}\to G$ provides a way to go back from Lie algebra onto the Lie group. One way to define it is by considering the ODE $$\gamma^\prime(t)=X_{\gamma(t)}, \gamma(0)=e$$ for an element $X\in \mathfrak{g}$ ( considered now as a left-invariant vector field on $G$). This ODE has a unique solution because the right-hand side $X$ is a smooth vector field. Now define $\exp^G\colon \mathfrak{g}\to G$ by $$\exp^G(X)=\gamma(1)$$ where $\gamma$ is the solution to the above ODE.

For matrix groups $G$ (as in your case) the exponential map is just the regular exponential map of matrices $X\in \mathfrak{g}$: $$\exp(tX)=\sum_{k=0}^\infty \frac{1}{k!}(tX)^k = I_n + tX+ \frac12 t^2X^2+\ldots$$ Note that the exponential map does (in general) not exhibit all the properties of the exponential map for real numbers, see e.g. Baker-Campbell-Hausdorff formula