Suppose that $\gamma:[0,1]\to \operatorname{SO}(3)$ is a path in the space of orientation preserving rotations of $\mathbb R^3$. It is classical that we can find a corresponding $\omega:[0,1]\to \mathbb R^3$, the angular velocity, such that if $r(t)=\gamma(t)r(0)$, then $r'(t)=\omega(t)\times r(t)$ (where $\times$ is the cross product). Phrased differently, $\frac{d}{dt}\left(\gamma(t)\right)\gamma^{-1}(t)$ is skew-symmetric with respect to an orthonormal basis. The correspondence between these two statements comes from the formula

$$\omega\times v =\pmatrix{0 & -\omega_3 & \omega_2 \\ \omega_3 & 0 & -\omega_1 \\ -\omega_2 & \omega_1 & 0 }v.$$

It is worth pointing out that this formula underlies an isomorphism of Lie algebras $(\mathbb R^3,\times)\cong \mathfrak{so}_3$, given by $x\mapsto ad_x$

Assuming whatever niceness we need (likely just continuity), given $\omega(t)$, there is a unique solution to the differential equation $$\gamma'(t)=\omega(t)\gamma(t);\quad \gamma(0)=I.$$

My question: Is there any nice expression for $\gamma$ in terms of $\omega$, the way there is when solving first order linear differential equations?

What I have tried so far:

This is a first order linear system of differential equations, and so there should be some well established theory regarding the solution of such systems. However, I have only seen the constant coefficient case before, and my initial attempts to generalize the solution have met with some resistance. In particular, the formula $\frac{d}{dt}e^{A(t)}=A'(t)e^{A(t)}$ doesn't hold for matrices unless $[A(t),A'(t)]=0$, being replaced by the more complicated formula $\frac{d}{dt}e^{A(t)}=e^{\operatorname{Ad}_{A(t)}}(A'(t))e^{A(t)}$. As such, a solution of the form $e^{A(t)}$ would have to satisfy


The problem can be simplified somewhat since $A(t)$ must be skew-symmetric. By using the realization of the bracket as coming from the cross product and combining that with properties of the cross product, $e^{\operatorname{Ad}_X}Y$ can be rewritten in more explicit, if not simpler, terms. However, it does not seem to make the problem any more solvable.


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