Let $R:I→SO(3)$, smooth. We know that, for any value of $t∈I$, $R(t)R(t)^T=\mathbf1$, where $\mathbf1$ is the identity matrix. Then, differentiating both sides one finds that $\dot{R}(t)R(t)^T+(\dot{R}(t)R(t)^T)^T=\mathbf0$, meaning that $Z(t)=\dot{R}(t)R(t)^T$ is skew symmetric.
Well, the condition $R(t)R(t)^T=\mathbf1$ may also be written as $R(t)^TR(t)=\mathbf1$, so if instead I differentiate this I'll get that $Y(t)=R(t)^T\dot{R}(t)$ is skew symmetric too.
My question is, what's the relation between the two matrices $Z(t)$ and $Y(t)$? Thanks.
Since $Y(t) = R(t)^T Z(t) R(t)$, it is the result of a unitary change of basis, which preserves skew (and non-skew) symmetry.
Whether this is helpful for understanding $R(t)$ is another matter.