Finding the flow of vector field with lie group (orthogonal group) If $X_S (A) = AS$ defines a smooth vector field, where $A \in O(n)$ (matrices with $A^{-1} = A^{T}$) and $S$ belonging to the space of skew matrices (matrices with $S^{T} = -S$). How would I calculate the flow of $X_S$ here? 
I'm using the definition of flow as one whose flow domain is all of $\mathbb{R} \times O(n)$, where the flow is the map $\theta : O(n) \times \mathbb{R} \to O(n)$ such that for all $s,t \in \mathbb{R}$ and $A \in O(n)$ the following holds:
$$\theta(A,0) = A \space\ \space\ \text{and} \space\ \space\ \theta(\theta(A,t),s) = \theta(A, s+t)$$
I'm only familiar with finding flow via solving an ODE (or system of them). Need help understanding how this works with Lie Groups.
 A: The general construction for Lie groups that's typically used is that of a one-parameter subgroup.  Recall that the exponential map $\exp: \mathfrak{g} \to G$ is a local diffeomorphism, and in the case of the orthogonal group we have that $\mathfrak{o}(n)$ is the vector space of skew-symmetric matrices.  A one-parameter subgroup is the image of a Lie group homomorphism $\gamma: \mathbb{R} \to G$, hence the homomorphism property enforces that $\gamma(s+t) = \gamma(s)\gamma(t)$.  In particular, for matrix Lie groups these take the form of
$$
\theta(t) \;\; =\;\; \exp(tS)
$$
where $S\in \mathfrak{g}, \; t \in \mathbb{R}$, and the matrix exponential map is given by
$$
\exp(Q) \;\; =\;\; \sum_{k=0}^\infty \frac{1}{k!}Q^k.
$$
To obtain the flow through $A$ with initial tangent vector $AS$ we would simply need to left translate the above group to the orbit:
$$
\theta_A(t) \;\; =\;\; A\exp(tS).
$$
It should be clear that the image of $\theta_A$ will not in general be a subgroup but will have the desired properties needed to be a flow.  Because $tS$ and $rS$ commute with one another then we know by the Baker-Campbell-Hausdorff formula that $\exp(tS)\exp(rS) = \exp((t+r)S)$.  This relates to ODEs as well since this is the unique solution to the initial value problem $\theta_A(0) = A$ and $\dot{\theta}_A(0) = AS$.  I'd recommend taking a look at chapters 8, 9, and 20 of this book.
