I am in the process of learning about Lie Algebras and Lie groups, specifically for $SO(3)$ and $SE(3)$. I've been reading a tutorial here: http://www.ethaneade.org/lie.pdf, but I'm getting stuck at the derivation of the Adjoint for $SO(3)$ (equations 21 - 25 from the link). There's a derivation that looks like so: enter image description here

After going through the adjoint derivation, they use it to derive the Jacobian of a product of rotations (equations 33-37).

I'd appreciate any help in getting my head around these derivations. Thanks!!!


2 Answers 2


You didn't specify exactly what you don't understand, but I'll try to provide some statements that may help.

Loosely speaking, a Lie algebra is just a vector space equipped with a Lie bracket operation. One common example is the set of all $n \times 1$ matrices ("column vectors") equipped with the cross-product, but another example is the set of all skew-symmetric matrices equipped with the matrix commutator. They are equally valid representations of $\text{so3}$, and Ethan somewhat confusingly switches between them at his convenience, calling $\omega \in \text{so3}$ for the "column with cross product" representation and $\omega_{\times} \in \text{so3}$ for the skew-symmetric matrix representation. I needed to emphasize that because of this simple relation that is otherwise confusing, $$\log(e^{\omega}) = \omega_{\times}$$

It should be clear to you that the appearance/disappearance of the $\cdot_{\times}$ subscript is a change in representation preference (perhaps for computational purposes) and not a change in mathematical identity. I would prefer if he did away with $\omega$ and only used $\omega_{\times}$ because it makes the exponential $e^{\omega_{\times}}$ more obviously a matrix exponential (from a computational perspective) instead of just the abstract notion of the exponential mapping, which is "needed" when we write $e^{\omega}$. Basically, apply the $\cdot_{\times}$ subscript to $\text{so3}$ members at your leisure.

Now then, for simplicity lets view eq.20 as the definition of the $\text{Adj}$ operator. What does that equation say about it? Recall that the Lie group $\text{SO3}$ is not commutative. The left side of eq.20 is the composition of two members of $\text{SO3}$, one of which happens to be represented as the exponential of some member of $\text{so3}$. To be clear, we could say $R_1 := e^{\omega}$ and $R_2 := R$ so that the left side of eq.20 is $R_2 R_1$. We know that, $$R_2 R_1 \neq R_1 R_2\ \ \implies\ \ R e^{\omega} \neq e^{\omega} R$$ but what eq.20 tells us is that the $\text{Adj}_R$ operator is what relates $R e^{\omega}$ to its commutation. Also, it should be noted that it is a mapping from the Lie algebra to the Lie algebra, i.e. $\text{Adj}_R:\text{so3}\to\text{so3}$, or expressed another way, $\text{Adj}_R \omega \in \text{so3}$.

The next steps are to "solve for" the $\text{Adj}_R$ operator, or really, express its action in terms of other operators that have already been defined (like $R$ or $\omega_{\times}$). Eq.21 follows directly from eq.20: simply right-multiply both sides by $R^{-1}$. He then takes the logarithm of both sides, where from the series definition of the matrix logarithm it is easy to show that, $$ \log(VAV^{-1}) = V\log(A)V^{-1}$$

There was really no reason to bring in the mention of generators, since it is obvious that $\log(e^{\omega}) = \omega_{\times}$. I feel like he should have written eq.22 as, $$(\text{Adj}_R\omega)_{\times} = R \omega_{\times} R^{-1}$$

Note that on the left I emphasize that $\text{Adj}_R \omega \in \text{so3}$ is represented as a skew-symmetric matrix. Going from eq.23 to eq.24 is just matrix identities (remember that for $R \in \text{SO3}$, $R^{-1}=R^T$). Finally, the last step makes a lot more sense when you consider that the left side of eq.24 is $(\text{Adj}_R\omega)_{\times}$.

Hope that helps!

  • 1
    $\begingroup$ wonderful - thank you for this detailed walk-through @jnez71! $\endgroup$
    – trdavidson
    Sep 10, 2018 at 17:11

To expand on the existing answer: for me the tricky part was actually going from eq. 23 to eq. 24. There's probably an easier way to see it, but here's how I did it.

Recall that for a vector $\mathbf u$, the matrix $\mathbf u_{\times}$ is defined as the linear transformation

$$ \mathbf u_\times :: \mathbf v \mapsto \mathbf u \times \mathbf v. $$

So the identity $\mathbf R \cdot \boldsymbol\omega_\times \cdot \mathbf R^{-1} = (\mathbf R \boldsymbol\omega)_\times$ can be intuitively interpreted as "[unrotating, then crossing with $\boldsymbol\omega$, then rotating] is the same as [crossing with a rotated $\boldsymbol \omega$]". We can prove this identity by letting the RHS of the identity act upon an arbitrary vector $\mathbf v$:

$$ (\mathbf R \boldsymbol \omega)_\times \mathbf v = (\mathbf R \boldsymbol \omega) \times \mathbf v = (\mathbf R \boldsymbol \omega) \times (\mathbf R \mathbf R^{-1}\mathbf v) = \mathbf R \left[\boldsymbol \omega \times (\mathbf R^{-1} \mathbf v)\right] = \mathbf R \boldsymbol \omega_\times \mathbf R^{-1} \mathbf v, $$

where we have used the fact that for any rotation matrix $U$ and vectors $\mathbf a, \mathbf b$ we have

$$ (U \mathbf a) \times (U \mathbf b) = U (\mathbf a \times \mathbf b). $$


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