Find structure of Lie Algebra from Lie Group My question is how one, in a general sense, calculates the Lie algebra for a given Lie group. Note, I have an engineering and not mathematics background, as such I am mostly interested in groups like $\mathrm{SO}(n)$ and $\mathrm{SE}(n)$.
Taking for example the group of unit-length complex numbers $z$ with multiplication as the operation, intuition says that the tangent space at identity $1+i0$ is $i \mathcal{R}$, but how does one find this rigorously? As intuition is likely difficult for more complicated spaces.
One very well known source from my domain, https://arxiv.org/abs/1812.01537, proposes that the structure of the Lie algebra can be found by differentiating the group constraint, in this case that would be
\begin{align*}
z^* z = 1 \,, \\
\frac{d}{dt} \Rightarrow z^*\dot{z} + \dot{z^*}z = 0 \,.
\end{align*}
He then goes on to say that an element of the Lie algebra $v^{\wedge}$ (where $v$ is the vector space velocity) is given by
$$
v^{\wedge} = z^*\dot{z} = -\dot{z^*}z
$$
and this equality is what I fail to understand, how the “velocity” is equal to $z^*\dot{z}$.
Perhaps someone can help me with my misunderstanding of this approach, and in my understanding of a general purpose process to follow for finding Lie algebra structure, e.g. for $\mathrm{SO}(3)$ or $\mathrm{SE}(3)$.
 A: I can try to help you with this misunderstanding:

He then goes on to say that an element of the Lie algebra $v^{\wedge}$ (where $v$ is the vector space velocity) is given by
$$
v^{\wedge} = z^*\dot{z} = -\dot{z^*}z
$$
and this equality is what I fail to understand, how the "velocity" is equal to $z^*\dot{z}$.

I think this misunderstanding may come from a confusion on the "reference frame" each variable is referred to.
In general, for an element $\mathcal{X}$ of the manifold $\mathcal{M}$, the elements of the lie algebra are of the form $\mathbf{v}^{\wedge} = \mathcal{X}^{-1} \dot{\mathcal{X}}$ (eq. (9) in Solà's paper).
We can rewrite equation (12) of that paper to explicitly state the reference frame each variable is referred to:
$^\mathcal{E}\dot{\mathcal{X}} =\ ^\mathcal{E}\mathcal{X}\ ^\mathcal{X}\mathbf{v}^{\wedge}$
So  they both  refer to the same velocity but in different reference frames:

*

*$\dot{\mathcal{X}}$ is the time derivative of $\mathcal{X}$ in the global reference frame of the manifold $\mathcal{M}$, whereas

*$\mathbf{v}^{\wedge}$, as you very well said, is defined at the local tangent space at $\mathcal{X}$: $T_{\mathcal{X}}\mathcal{M}$.

Cheers.
A: The group $SO(3)$ sits in the space of $3\times 3$ matrices, i.e. $\mathbb R^9$. A curve in $\mathbb R^9$ is a matrix
$$
\gamma(t)
=
\begin{bmatrix}
a_1(t) & b_1(t) & c_1(t) \\
a_2(t) & b_2(t) & c_2(t) \\
a_3(t) & b_3(t) & c_3(t)
\end{bmatrix}
$$
The curve is in $SO(3)$ if it satisfies the constraint $\gamma(t)\gamma(t)^\intercal = \gamma(t)^\intercal\gamma(t) = I$.
The elements of $\mathfrak{so}(3)$ are vectors tangent to the identity matrix when $t=0$, i.e. $\gamma'(t)$ for some $\gamma$ in $SO(3)$.
So such a vector is a matrix
$$
\gamma'(0)
=
\begin{bmatrix}
a'_1(0) & b'_1(0) & c'_1(0) \\
a'_2(0) & b'_2(0) & c'_2(0) \\
a'_3(0) & b'_3(0) & c'_3(0)
\end{bmatrix}.
$$
Now differentiating the constraint of $SO(3)$ at $t=0$ we find out that an element of $\mathfrak{so}(3)$ satisfies
$$
\gamma'(0)\gamma(0) + \gamma(0)\gamma'(0) = 0
$$
But $\gamma(0) = I$ so this is just $\gamma'(0) = -\gamma^{\intercal}{}'(0)$.
Forget about $\gamma'(0)$: your Lie algebra consists of $3\times 3$ matrices satisfying $A^\intercal = -A$.
For the quotient of the Lie algebras, this is just the Lie algebra of the quotient Lie groups (theorem).
