# Which matrices can be realized as second derivatives of orthogonal paths?

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

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

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

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

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 $$\ddot{\gamma}(0)=B^2+C$$.
• 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 $\gamma(t)^T\gamma(t)D(t)=D(t)\gamma(t)^T\gamma(t)$... – Asaf Shachar Dec 26 '18 at 12:47
• , 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. – Asaf Shachar Dec 26 '18 at 12:47
• 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. – Or Eisenberg Dec 26 '18 at 21:47
• 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! – Or Eisenberg Dec 27 '18 at 17:20