# [Differential Geometry] Matrix of Differential Equations in SO(3)

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1. Oct 9, 2014

### mef51

1. The problem statement, all variables and given/known data
Suppose that $s \to A(s) \subset \mathbb{M}_{33}(\mathbb{R})$ is smooth and that $A(s)$ is antisymmetric for all $s$. If $Q_0 \in SO(3)$, show that the unique solution (which you may assume exists) to
$$\dot{Q}(s) = A(s)Q(s), \quad Q(0) = Q_0$$
satisfies $Q(s) \in SO(3)$ for all $s$.

2. Relevant equations
For $Q \in \mathbb{M}_{33}(\mathbb{R})$, if:
• $Q^TQ = I, \quad Q \in O(3)$
• additionally, if $det(Q) = 1, \quad Q \in SO(3)$
• Note: $Q^TQ = QQ^T$ for $Q \in SO(3)$
For $A \in \mathbb{M}_{33}(\mathbb{R})$,
• $A$ is antisymmetric if $A^T = -A$

3. The attempt at a solution
Since I only know that $Q_0$ is in $SO(3)$, I suspect that $\dot{Q}(s)$ must be 0. In that case, since the derivative is zero, $Q(s)$ must be constant and $Q(s)=Q_0$ for all $s$. So I try to prove that $A(s)Q(s)$ is zero.

Based on a hint someone gave me I'll start by taking the derivative of ${Q_0}^TQ_0 = I$:
$$({Q_0}^TQ_0)' = I' \\ \dot{Q_0}^TQ_0 + {Q_0}^T\dot{Q_0} = 0_{33} \\ {Q_0}^T\dot{Q_0} = - \dot{Q_0}^TQ_0 \\ Q_0{Q_0}^T\dot{Q_0} = - Q_0\dot{Q_0}^TQ_0 \\ Q_0{Q_0}^T\dot{Q_0} = - Q_0\dot{Q_0}^TQ_0 \\ \dot{Q_0} = - Q_0\dot{Q_0}^TQ_0 \\$$
Now I plug in that expression for $\dot{Q_0}$ in the differential equation..
$$- Q_0\dot{Q_0}^T Q_0 = A_0Q_0 = -{A_0}^T Q_0$$
I'm not really sure what this gives me though. I'm also not confident that you can even use the product rule like that. It certainly isn't true for matrices in general.
Any help or pokes in the right direction are appreciated!

mef

2. Oct 10, 2014

### mef51

I made a mistake in my attempt:
Since $Q_0$ is constant, taking the derivative of $Q_0$ is just 0. Should perhaps take the derivative of $A(s)Q(s)$ instead

3. Oct 11, 2014

### mef51

Alright let's take the derivative of $Q^TQ$. If this ends up being zero, then $Q^TQ$ is constant and could be equal to the identity $I_{33}$. This would prove one of the properties we want $Q$ to have to be in $SO(3)$. We need to use $\dot{Q} = AQ$ and $A^T=-A$

$$(Q^TQ)' = \dot{Q^T}Q + Q^T\dot{Q} \\ \qquad = \dot{Q^T}Q + Q^TAQ \\ \qquad = (AQ)^TQ + Q^TAQ \\ \qquad = Q^TA^TQ + Q^TAQ \\ \qquad = -Q^TAQ + Q^TAQ \\ \qquad = 0\\$$

So $Q^TQ$ is constant. Can we just say $Q^TQ = I_{33}$ which is constant?

4. Oct 11, 2014

### mef51

Since $Q^TQ$ is constant $\forall s$,
$Q^TQ = {Q_0}^T Q_0 = I_{33}$ since $Q_0$ is in $SO(3)$. Ok.

I now need to show that $det(Q) = 1 \quad \forall s$, knowing that $det(Q_0) = 1$
I know that $det(Q^TQ) = det(I_{33}) = 1$

So using properties of determinants
$1 = det(Q^TQ) = det(Q^T)det(Q) = (det(Q)) ^2$

This gives me $det(Q) = \pm 1$ but how can I discard the -1 solution??

5. Oct 12, 2014

### mef51

Okokok.

Since $Q(s)$ is differentiable, it must be continuous. $det(Q)$ must also be continuous.

Since $det(Q_0) = 1$, if $det(Q)$ were ever -1, then by the Intermediate Value Theorem there should be an $s$ such that $det(Q(s)) = 0$ (or any other value between 1 and -1).

But since $det(Q)$ can only be either 1 or -1, $det(Q)$ can never be -1 without violating the IVT. So we must have that $det(Q) = 1$ for all $s$.

That in combination with the other posts should prove that $Q \in SO(3) \quad \forall s$.
$\square$

6. Oct 12, 2014

### Dick

Yes, I think you pretty much have it. Good job!