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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 $\mathrm{SE}(3)$.

Morten Nissov
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  • Take $SO(n)$. It sits in the larger space of all matrices, which is just $\mathbb{R}^{n^2}$. So you're looking into curves in Euclidean space with a constraint that they're actually in $SO(n)$. The derivative is just differentiation in $\mathbb R^{n^2}$. Does this help? – Daniel Teixeira Oct 06 '22 at 14:59
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    This technique works for any matrix Lie group, especially if they are defined by a simple equation, but it fumbles for general Lie groups. – Daniel Teixeira Oct 06 '22 at 15:01
  • Can you form it as an example, maybe it's asking too much but I'm really just looking for an explanation of the general methodology and then an example were it's applied to some group, e.g. SO(3). Having a hard time seeing the logic behind what the linked article is doing. – Morten Nissov Oct 06 '22 at 15:02
  • I'm almost entirely interested in matrix Lie groups, is there a specific reason why this approach works only for matrix groups? – Morten Nissov Oct 06 '22 at 15:03
  • because in general the derivative won't be just the derivative in some $R^m$ – Daniel Teixeira Oct 06 '22 at 15:05
  • why does this approach work for matrix groups then, like why specifically differentiating the constraint – Morten Nissov Oct 06 '22 at 15:26
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    If the group is not embedded in some other space, like R^n, then there are no constraints to differentiate... – Mariano Suárez-Álvarez Oct 06 '22 at 16:27
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    Highly related: https://math.stackexchange.com/q/3111143/96384. Also possibly helpful: https://math.stackexchange.com/q/3037396/96384, https://math.stackexchange.com/q/3766220/96384. – Torsten Schoeneberg Oct 06 '22 at 17:41
  • I mean a space doesn't need to be embedded in R^n to have constraints on its surface right? – Morten Nissov Oct 06 '22 at 17:56
  • I guess the core of my question is why differentiating the constraint works, to me it would seem more natural to differentiate an element on the surface, as this should lead towards the velocity/tangent space at this point – Morten Nissov Oct 06 '22 at 17:57
  • @TorstenSchoeneberg thanks, ill look at these – Morten Nissov Oct 06 '22 at 17:57
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    @MortenNissov I think you're misunderstanding. These are not constraints on its surface but constraints identifying it as a subset within a larger space. In this case, we are finding $SO(3)$ within the nice vector space of 3x3 matrices. All the members of our subset satisfy the property $XX^T = I$ so any curve in that space does as well. Now pick a curve passing through the identity and differentiate it there. The condition on those tangent vectors is exactly the derivative of the condition for the group as Daniel's answer shows in this example – Callum Oct 06 '22 at 22:08

2 Answers2

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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).

Daniel Teixeira
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    this just applies te same technique as I demonstrated in my question on another group. This is not what I am asking – Morten Nissov Jan 08 '23 at 23:15
  • The difference between my answer and Equation 9 in the paper you mentioned is that there they use any tangent space to describe the Lie algebra, instead of the tangent space to the identity. So you wouldn't have $\gamma(0) = I$ and you would be stuck with $\gamma'(0)\gamma(0) + \gamma(0)\gamma'(0) = 0$ which is Equation 9. (in the paper they use $\mathcal X$ instead of $\gamma$) – Daniel Teixeira Jan 09 '23 at 15:27
  • This works because the map "multiplication by $g$", i.e. $h\mapsto h\cdot g$, is such that 1. it is invertible, 2. it sends the identity $I$ to $g$, so it's an isomorphism between (the Lie algebra) $T_I\mathcal M$ and (the Lie algebra) $T_g\mathcal M$. Is this what you asked for? – Daniel Teixeira Jan 09 '23 at 15:30
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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.

castillon
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