$\newcommand{\skew}{\operatorname{skew}}$
$\newcommand{\sym}{\operatorname{sym}}$
$\newcommand{\SO}{\operatorname{SO}_n}$
I am interested to know which real matrices $A \in M_n$ can be realized as second derivatives of paths in $\text{SO}_n$ starting at the identity. That is, for which matrices $A$, there exist a smooth path $\alpha:(-\epsilon,\epsilon) \to \text{SO}_n$, such that $\alpha(0)=Id$ and $\ddot \alpha(0)=A$. We denote the space of realizable matrices by $D$.
Question: I prove below that $ (\skew)^2 \subseteq D \subseteq (\skew)^2+\skew $. Does $D=(\skew)^2+\skew$ always hold?
Comment: Note that $(\skew)^2+\skew \subsetneq M_n$, at least for odd $n$: In that case every skew-symmetric matrix is singular, so $(\skew)^2 \subseteq \sym $ consists only of singular matrices, hence does not contain all symmetric matrices.
Edit: I proved below that equality holds in dimension $n=2$.
Proof of $ (\skew)^2 \subseteq D \subseteq (\skew)^2+\skew $:
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Every square of skew-symmetric matrix can be realized: For skew $B$, take $\alpha(t)=e^{tB}$. Then, $\dot \alpha(t)=Be^{tB}$, $\ddot \alpha(t)=B^2e^{tB}$.
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The space of realizable matrices is contained in $(\skew)^2+\skew$: Indeed, since $\dot \alpha(t) \in T_{\alpha(t)}\SO=\alpha(t)\skew$, we have $\dot\alpha(t)=\alpha(t)B(t)$ for some $B(t) \in \skew$, so
$$\ddot \alpha(t)=\dot \alpha(t) B(t)+\alpha(t) \dot B(t)$$ hence $\ddot \alpha(0)=\dot \alpha(0) B(0)+ \dot B(0)= B(0)^2+\dot B(0) \in (\skew)^2 +\skew,$ where the last equality followed from $\dot \alpha(0)=B(0)$ (put $t=0$ in $\dot\alpha(t)=\alpha(t)B(t)$).
Edit 2: When trying to show the converse direction, I hit a wall: we need to show that there exist solutions $\dot\alpha(t)=\alpha(t)B(t)$, where $\alpha(t) \in \SO,B(t) \in \skew$, with arbitrary $B(0),\dot B(0) \in \skew$. A naive attempt would be to define $\alpha(t)=e^{\int_0^t B(s)ds}$ for $B(s)=B(0)+s\dot B(0)$. However, it is not true in general that $\alpha'(t)=\alpha(t)B(t)$; this happens if $B(t)$, $\int_0^t B(s)ds$ commute, which happens if and only if $B(0),\dot B(0)$ commute.
Proof $D = (\skew)^2+\skew$ for $n=2$:
$\alpha(t)$ can always be written as $\alpha(t)=\begin{pmatrix} c(\phi(t)) & s(\phi(t)) \\\ -s(\phi(t)) & c(\phi(t)) \end{pmatrix}$, where $c(x)=\cos x,s(x)=\sin x$, and $\phi(t)$ is some parametrization satisfying $\phi(0)=0$.
Differentiating $\alpha(t)$ twice, we get
$$ \ddot \alpha(t)=-(\phi'(t))^2\alpha(t)+\phi''(t)\begin{pmatrix} -s(\phi(t)) & c(\phi(t)) \\\ -c(\phi(t)) & -s(\phi(t)) \end{pmatrix},$$
so
$$ \ddot \alpha(0)=-(\phi'(0))^2Id+\phi''(0)\begin{pmatrix} 0 & 1 \\\ -1 & 0 \end{pmatrix}.$$
Since we can choose $\phi'(0),\phi''(0)$ as we wish, we conclude that $$ D=\mathbb{R}_{\le 0}Id+\mathbb{R}\begin{pmatrix} 0 & 1 \\\ -1 & 0 \end{pmatrix}=\mathbb{R}_{\le 0}Id+\skew.$$ Since $\skew=\text{span} \{ \begin{pmatrix} 0 & 1 \\\ -1 & 0 \end{pmatrix}\}$, and $\begin{pmatrix} 0 & 1 \\\ -1 & 0 \end{pmatrix}^2=-Id$, we have $\skew^2=\mathbb{R}_{\le 0}Id$, so indeed $D=(\skew)^2+\skew$.
Best Answer
Yes. Given skew-symmetric matrices $B$ and $C,$ define $\alpha(t)=\exp(Bt+\tfrac12 Ct^2).$ Then \begin{align} \alpha(t) &=I+(Bt+\tfrac12 Ct^2)+\tfrac12 (Bt+\tfrac12 Ct^2)^2+O(t^3)\\ &=I+Bt+\tfrac12 (B^2+C)t^2+O(t^3) \end{align} as $t\to 0.$ This shows that $\ddot \alpha(0)=B^2+C.$