$\newcommand{\skew}{\operatorname{skew}}$ $\newcommand{\sym}{\operatorname{sym}}$ $\newcommand{\SO}{\operatorname{SO}_n}$

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 $:

  1. Every square of skew-symmetric matrix can be realized: For skew $B$, take $\alpha(t)=e^{tB}$. Then, $\dot \alpha(t)=Be^{tB}$, $\ddot \alpha(t)=B^2e^{tB}$.

  2. 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

$$ \ddot \alpha(t)=-(\phi'(t))^2\alpha(t)+\phi''(t)\begin{pmatrix} -s(\phi(t)) & c(\phi(t)) \\\ -c(\phi(t)) & -s(\phi(t)) \end{pmatrix},$$


$$ \ddot \alpha(0)=-(\phi'(0))^2Id+\phi''(0)\begin{pmatrix} 0 & 1 \\\ -1 & 0 \end{pmatrix}.$$

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 $D=(\skew)^2+\skew$.


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 $\ddot \alpha(0)=B^2+C.$


I believe that the construction you're looking for is called a Dyson Series (Wikipedia). In detail, suppose we are given $B,C \in \mathrm{Skew}(n)$ and we want to construct a $\gamma: (-\varepsilon,\varepsilon) \to \mathrm{SO}(n)$ such that $\gamma(0)=1$ and $\ddot{\gamma}(0)=B^2+C$. I claim that \begin{align*} \gamma(t) & :=\sum_{n=0}^{\infty} \left[\int_0^t \int_0^{t_0} \cdots \int_0^{t_{n-1}} \left(\prod_{k=0}^n (B+t_{n-k} C) \right) \mathrm{d} t_n \cdots \mathrm{d} t_0\right] \\ & = 1 + \int_0^t (B+t_0C) \mathrm{d}t_0 + \int_0^t \int_0^{t_0} (B+t_1C)(B+t_0C) \mathrm{d} t_1 \mathrm{d}t_0 + \\ & \hspace{1cm}\int_0^t \int_0^{t_0} \int_0^{t_1} (B+t_2C)(B+t_1C)(B+t_0C) \mathrm{d}t_2 \mathrm{d}t_1 \mathrm{d}t_0 + \cdots \end{align*} is a well-defined solution to the problem. Indeed, if we let $$m:=\max_{s \in [0,t]} \lVert B+sC \rVert_{L^2},$$ then we have that $$\lVert \gamma(t) \rVert_{L^2} \leq e^m,$$ so that $\gamma$ is defined by a convergent sequence. Moreover, we can compute that $\dot{\gamma}(t)=\gamma(t)(B+tC)$, evincing both that the image of $\gamma$ (which a priori lies in the space of $n \times n$ matrices) in fact lies in $\mathrm{SO}(n)$ and also that $\ddot{\gamma}(0)=B^2+C$.

  • $\begingroup$ Thanks, this is very interesting. However, I am not sure how do you deduce that $\gamma(t) \in SO$, even for sufficiently small $t$; I tried showing that the derivative of $\gamma(t)^T\gamma(t)$ is zero, but I got stuck: Since $\dot \gamma(t)=\gamma(t)D(t)$. where $D(t)$ is skew-symmetric, we have $\frac{d}{dt}(\gamma(t)^T\gamma(t))=(\dot \gamma(t))^T\gamma(t)+\gamma(t)^T\dot \gamma(t)=(\gamma(t)D(t))^T\gamma(t)+\gamma(t)^T\gamma(t)D(t)=-D(t)\gamma(t)^T\gamma(t)+\gamma(t)^T\gamma(t)D(t)$ which is zero if and only if $\gamma(t)^T\gamma(t)D(t)=D(t)\gamma(t)^T\gamma(t)$... $\endgroup$ – Asaf Shachar Dec 26 '18 at 12:47
  • $\begingroup$ , i.e. $\gamma(t)^T\gamma(t)$ and $D(t)$ commute. In our case $D(t)=B+tC$, and I don't see an immediate reason why this commutativity should hold. $\endgroup$ – Asaf Shachar Dec 26 '18 at 12:47
  • $\begingroup$ You're right, of course -- it would seem I miscalculated as to how easily it would follow that $\mathrm{im}(\gamma(t)) \subseteq \mathrm{SO}(n)$. I think some argument like this might work (though i haven't thought out the details). Notice that $\gamma(t)^T \gamma(t)$ is analytic and that its derivatives vanish to every order at the origin. It follows that $\gamma(t)^T\gamma(t)$ is constant, as needed. Do you think that might work? I'll hopefully have more time to think about this later, in any event. $\endgroup$ – Or Eisenberg Dec 26 '18 at 21:47
  • $\begingroup$ Thanks. Your idea sounds correct. It also turns out that this is a general fact; see here:math.stackexchange.com/questions/1576871/…. By the way, do you have any explanation for how you thought of using this Dyson Series here? Thanks. $\endgroup$ – Asaf Shachar Dec 27 '18 at 15:40
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    $\begingroup$ Asaf: Thanks for the reference about foliations -- this was my original intuition for why $\mathrm{im}(\gamma) \subset \mathrm{SO}(n)$, but I certainly didn't have enough intuition to explain it as well as Mike Miller did! Regarding your question of how I thought of the Dyson series, this question is verging on philosophical, I think. Indeed, where do any math ideas come form? =). I was just trying to solve the equation $\dot{\gamma}=\gamma D$ and remembered vaguely that the physicists had figured out how to do this, so I started googling. Thanks for the interesting problem! $\endgroup$ – Or Eisenberg Dec 27 '18 at 17:20

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