# Show that if $\Omega_1,\Omega_2$ are tangent vectors of curves in $SO(3)$ then $[\Omega_1,\Omega_2]\in\mathfrak{so}(3)$

Let $so(3)$ be the Lie Algebra of $SO(3)$ and $R\in SO(3); \Omega_1,\Omega_2 \in so(3)$ and $\Omega_n = \frac{d}{dt}R_n(t)$ at the point $t=0$. So $\Omega_n$ is the tangent vector of the curve $R_n(t)$ at $t=0$.

To show that the Lie bracket $[\Omega_1,\Omega_2]$ is an element of $so(3)$ my professor wrote:

$$\tag{1}[\Omega_1,\Omega_2] = \left.\frac{d}{dt}R_1(t)\Omega_2 R_1(t)^{-1}\right|_{t=0}$$

I am trying to understand that, but I cannot see why this shows that the commutator is an element of $so(3)$.

I understand that for this tangential vector to be in $so(3)$, the product $R_1(t)\Omega_2 R_1(t)^{-1}$ has to be a parametrized curve in $SO(3)$.

But I do not see why it is a curve in $SO(3)$.

• I want to be clear about what you are asking :) Do you want to know why the product $R_1(t)\Omega_2R_1(t)^{-1}$ "has to be a parametrized curve" living completely with $SO(3)$. Or, are you asking if it is a parametrized curve" living completely with $SO(3)$, then you can using it in the above formula to calculate $[\Omega_1, \Omega_2]$. – muaddib Aug 11 '15 at 22:16
• I want to see the proof that the Lie bracket $[\Omega_1,\Omega_2]$ is an element of $so(3)$. In other words, this is the proof that the tangent space of $SO(3)$ at the unity matrix is a Lie algebra, if you use the commutator as the Lie bracket. My professor wrote the above formula $(1)$ as proof, but I don't understand it. – Bass Aug 12 '15 at 8:13
• Edited question for clarification. – Bass Aug 12 '15 at 8:16

We want to show that $$\tag{1}[\Omega_1,\Omega_2] = \left.\frac{d}{dt}R_1(t)\Omega_2 R_1(t)^{-1}\right|_{t=0}$$

We can start doing this by computing: \begin{eqnarray*} \left.\frac{d}{dt}\left(R_1(t)\Omega_2 R_1(t)^{-1}\right)\right|_{t=0} &=& \left.\left[\frac{d}{dt}R_1(t)(\Omega_2 R_1(t)^{-1})\right]\right|_{t=0} + \left.\left[(R_1(t)\Omega_2) \frac{d}{dt}R_1(t)^{-1}\right]\right|_{t=0} \\ &=& \left[\Omega_1\Omega_2 R_1(0)^{-1}\right] + \left.\left[R_1(t)\Omega_2 \frac{d}{dt}R_1(t)^{-1}\right]\right|_{t=0} \\ &=& \left[\Omega_1\Omega_2 R_1(0)^{-1}\right] + \left.\left[R_1(t)\Omega_2 (-R_1(t)^{-1})\left(\frac{d}{dt}R_1(t)\right)R_1(t)^{-1})\right]\right|_{t=0} \\ &=& \left[\Omega_1\Omega_2 R_1(0)^{-1}\right] - \left.\left[R_1(t)\Omega_2 R_1(t)^{-1}\left(\frac{d}{dt}R_1(t)\right)R_1(t)^{-1})\right]\right|_{t=0} \\ &=& \left[\Omega_1\Omega_2 R_1(0)^{-1}\right] - \left[R_1(0)\Omega_2 R_1(0)^{-1}\left.\left(\frac{d}{dt}R_1(t)\right|_{t=0}\right)R_1(0)^{-1})\right] \\ &=& \left[\Omega_1\Omega_2 R_1(0)^{-1}\right] - \left[R_1(0)\Omega_2 R_1(0)^{-1}\Omega_1R_1(0)^{-1})\right] \\ \end{eqnarray*}

See Derivatives of Inverse Matrix for details on the third line's derivation.

There isn't much further we can go without knowing more details about $R_1$. I actually think it might have been defined as $$R_i(t) = e^{\Omega_it}$$ But it's only necessary to require $R_i(0) = I$. Then the final line above gives $\Omega_1\Omega_2 - \Omega_2\Omega_1$.

The curve is in $SO(3)$

This is not always true. A Lie group is a subgroup of $GL$ but a Lie algebra always contains the $0$ matrix. It maps to the identity of the Lie group. So if $\Omega_2 = 0$ in the above, then the curve $C(t) = 0$ does not lie in $SO(3)$.

Show $[\Omega_1, \Omega_2]$ is in $\mathfrak{so}(3)$

This is true independently of the above considerations. $\Omega_1, \Omega_2$ are elements of $\mathfrak{so}(3)$ which is closed under addition and multiplication (matrices of the form $X^T = -X$).

• I'm sorry if I didn't express myself clearly. I see that formula (1) is correct, you showed it using the "product rule of differentiation". However, for $[\Omega_1,\Omega_2]$ to be an element of $so(3)$, this is not sufficient. You also need to show that $C(t) = R_1(t)\Omega_2 R_1(t)^{-1}$ is a curve in $SO(3)$ with $C(0)=\mathbb{1}$. – Bass Aug 12 '15 at 13:28
• @BastianTreichler - Updated – muaddib Aug 12 '15 at 13:43
• Thanks, but your last remark is not true. $\mathfrak{so}(3)$ is not closed under multiplication: $\begin{bmatrix}0 & 0 & 1\\0 & 0 & 0\\-1 & 0 & 0\end{bmatrix}\begin{bmatrix}0 & 1 & 0\\-1 & 0 & 0\\0 & 0 & 0\end{bmatrix} = \begin{bmatrix}0 & 0 & 0\\0 & 0 & 0\\0 & -1 & 0\end{bmatrix}$ The Lie bracket IS the multiplication in $\mathfrak{so}(3)$, and this does not seem trivial. – Bass Aug 12 '15 at 14:11
• @BastianTreichler - Yes, by "multiplication", I meant the one for the lie-algebra, namely the bracket. What I am saying is $[\Omega_1, \Omega_2] = \Omega_1\Omega_2 - \Omega_2\Omega_1$ is part of the definition of the Lie Algebra. – muaddib Aug 12 '15 at 14:13
• @BastianTreichler hmm, well once you obtain the formula from the above that $\Omega_1\Omega_2 - \Omega_2\Omega_1$ is the commutator, you can show it is an element of $\mathfrak{so}(3)$ as follows. Elements of $\mathfrak{so}(3)$ are those matrices that satisfy $X^T = -X$. Then $[X, Y]^T = (XY - YX)^T = Y^TX^T - X^TY^T = -[X^T, Y^T] = -[(-X), (-Y)] = -[X, Y]$. – muaddib Aug 12 '15 at 14:19

Formula (1) is a bit strange indeed, $R_1(t)\Omega_2 R_1(t)^{-1}$ is not necessarily in $SO(3)$ ($\Omega_2$ could have a null determinant for example).

One way to prove the result, is to consider that (for $t \to 0$):

$$R_1(t) = I + t \Omega_1 + t^2 W_1 + o(t^2)$$ $$R_2(t) = I + t \Omega_2 + t^2 W_2 + o(t^2)$$

Then:

$$R_1(t)R_2(t) = I + t (\Omega_1 + \Omega_2) + t^2 (W_1 + W_2 + \Omega_1 \Omega_2) + o(t^2)$$ $$R_2(t)R_1(t) = I + t (\Omega_1 + \Omega_2) + t^2 (W_1 + W_2 + \Omega_2 \Omega_1) + o(t^2)$$

The you can consider $L(t) = R_1(t)^{-1}R_2(t)^{-1}R_1(t)R_2(t)$ which is a curve in $SO(3)$. Since $R_2(t)R_1(t)L(t) = R_1(t)R_2(t)$, you can develop term by term each side and find that:

$$L(t) = I + t^2([\Omega_1, \Omega_2]) + o(t^2)$$

If you re-parametrise L with $s = t^2$ you have $L(s) = I + s([\Omega_1, \Omega_2]) + o(s)$ Thus you have found a curve in $SO(3)$ whose tangent is $[\Omega_1, \Omega_2]$.

• Thanks, but the last term $L(s)=I+s([\Omega_1,\Omega_2])+O(s)$ still contains that $O(s)$ term. What if that term was $sM+O(s^2)$ with some matrix $M$. Then the tangent of $L(s)$ is $M+[\Omega_1,\Omega_2]$, which is different from $[\Omega_1,\Omega_2]$. – Bass Aug 12 '15 at 10:05
• Why is $R_2(t)^{-1}R_1(t)^{-1}L(t) = R_1(t)R_2(t)$ ? I think the inversions are wrong, but you didn't accept my edit. – cheesus Aug 12 '15 at 11:05
• @BastianTreichler: Thanks for the remark. My $O(.)$ are indeed $o(.)$, this was a typo. I edited it. – Tantto Aug 12 '15 at 14:47
• @cheeesus: I think the way wrote is correct. If $L(t) = R_1(t)^{-1}R_2(t)^{-1}R_1(t)R_2(t)$ then you need to multiply from the left by $R_1(t)$ and then by $R_2(t)$ to obtain the second expression. – Tantto Aug 12 '15 at 14:47
• Exactly, but you are left-multiplying by $R_1(t)^{-1}$ and then by $R_2(t)^{-1}$. – cheesus Aug 12 '15 at 16:09

Found a solution, but it has nothing to do with $(1)$:

\begin{align} [\Omega_1,\Omega_2] & = \Omega_1\Omega_2-\Omega_2\Omega_1 = (-\Omega_1^T)(-\Omega_2^T) - (-\Omega_2^T)(-\Omega_1^T) \\ & = \Omega_1^T\Omega_2^T-\Omega_2^T\Omega_1^T=(\Omega_2\Omega_1)^T-(\Omega_1\Omega_2)^T = (\Omega_2\Omega_1-\Omega_1\Omega_2)^T \\ & = -[\Omega_2,\Omega_1] \end{align}

Here is a possibility as to how the professor was arriving at the result that $[\Omega_1, \Omega_2]$ is an element of the Lie algebra. Suppose that we take the above formula as the definition of the Lie Bracket: $$\tag{1}[\Omega_1,\Omega_2] = \left.\frac{d}{dt}R_1(t)\Omega_2 R_1(t)^{-1}\right|_{t=0}$$ There is a proposition that for any Lie group element $A$ and Lie algebra element $X$ (of its associated Lie algebra), that $AXA^{-1}$ is also in the Lie algebra. (Proof from Hall: $e^{t(AXA^{-1})} = Ae^{tX}A^{-1}$ and the latter is in the Lie group).

This means in the above, the curve $$C(t) = R_1(t)\Omega_2 R_1(t)^{-1}$$ lies entirely in the Lie algebra. Hence its derivative at $t = 0$ is also in the Lie algebra (because a Lie algebra is a vector space). So the misconception from the original question was that $C(t)$ lives in $\mathfrak{so}(3)$ instead of $SO(3)$.

• Could you elaborate a bit more about the Proof from Hall part? If you read my question, this was exactly the thing I wanted to see. I can't yet see why $C(t)$ must be a curve in $SO(3)$. – Bass Aug 12 '15 at 15:59
• What misconception do you mean? I never asked why $C(t)$ is in $so(3)$. My original question was: why is $C(t) \in SO(3)$ for an interval around $t=0$. – Bass Aug 12 '15 at 16:01
• It is not a curve in $SO(3)$ it is a curve in $\mathfrak{so}(3)$. Regarding the proof, the RHS is an element of the group because it is the product of three group elements. Then by definition of the Lie Algebra, if $e^{t(AXA^{-1}}$ in the Lie group for all $t$, then $AXA^{-1}$ is in the Lie algebra. – muaddib Aug 12 '15 at 16:03
• Yes, what I am saying is that is where you went wrong. – muaddib Aug 12 '15 at 16:03
• Oh now I see... I always expected $C(t)$ to be in $SO(3)$ so that $(1)$ defines a tangential vector and thus a member of $so(3)$. Okay, so now that this is sorted out, do you have more information about that "Proof of Hall"? When I google it, I only find pages about Hall's theorem, which is something else. – Bass Aug 12 '15 at 16:06