I'm leaning Lie theory in robotics. My question comes from a paper: A micro Lie theory for state estimation in robotics by Joan Sola et al. at page 4:

For multiplicative groups this yields the new constraint $\mathcal{X}^{-1} \dot{\mathcal{X}} + \dot{\mathcal{X}^{-1}} \mathcal{X} = 0$, which applies to the elements tangent at $\mathcal{X}$ (the term $\dot{\mathcal{X}^{-1}}$ is the derivative of the inverse). The elements of the Lie algebra are therefore of the form $$ \mathbf{v}^\wedge = \mathcal{X}^{-1} \dot{\mathcal{X}} = -\dot{\mathcal{X}^{-1}} \mathcal{X} \,. \tag{$9$} $$

(Original screenshot)

To my understanding, the symbol $\mathbf{v}^{\wedge}$ in equation (9) should more precisely be $^\mathcal{X}\mathbf{v}^{\wedge}\in T_{\mathcal{X}}\mathcal{M}$, which is an element in the local tangent space of $\mathcal{X}$. But I didn’t get how the form of $\mathbf{v}^{\wedge}$ was derived from the new constraint.

I have 2 questions:

  1. Could anyone tell me in detail what is going on in the bold sentence?

  2. Why isn’t $\mathbf{v}^{\wedge}$ equal to $\dot{\mathcal{X}}=\partial\mathcal{X}/\partial t$, which I thought was just the tangent element at $\mathcal{X}$ in $T_{\mathcal{X}}\mathcal{M}$?

Thank you.

  • $\begingroup$ This looks to me as the differentiation of $x\cdot x^{-1}=1$ in the group. This gives us equation nine in ... where? The Lie algebra elements are the tangents at the group element $1.$ So $v^{\wedge }$ is a condition at $1$ which turns into the $0\in T_1M$ per identification. This condition is independent of $\mathcal{X}$ because $xx^{-1}=1$ holds for all group elements, and we actually consider $T_1M$ and not $T_{\mathcal{X}}M.$ All we have here is a constraint transported from group to tangent space (at $M\ni 1 \longleftrightarrow 0\in T_1M).$ $\endgroup$ Commented Dec 20, 2022 at 21:57

2 Answers 2


I had the same confusion as the asker, but now I agree with the above explanation. Here is my understanding.

Premise: all tangent spaces described in local coordinate are the same. I do not understand the detailed math proof, but we can find this in Figure 5 of this paper.

If this premise is agreed, then we can explain equation(9) like this: The Lie algebra is just a special tangent space at the identity point, whose description in either global or local coordinate is the same. Recall our goal is to find the tangent space at the identity point. If the premise above holds true, therefore, we can turn our goal into finding the tangent space at any point $\mathcal{X}$ but described in local coordinate. This can be derived via the following two steps. First, the tangent space can be calculated as $\dot{\mathcal{X}}=\partial \mathcal{X} / \partial t$. However, it is described in global coordinate, that is ${}^{\mathcal{E}} \dot{\mathcal{X}}$, where $\mathcal{E}$ is the origin. Second, the tangent space can be regarded as a regular space and can be transformed into the local coordinate by $\mathcal{X}^{-1}$ (recall $\mathcal{X}$ can be the transformation from local to global). Then, the Lie algebra can be written as ${}^{\mathcal{X}}\mathbf{v}^{\wedge}={}^{\mathcal{E}}\mathcal{X}^{-1} \cdot {}^{\mathcal{E}}\dot{\mathcal{X}}={}^{\mathcal{E}}\mathbf{v}^{\wedge}=$Lie algebra.


I'm not totally familiar with the nomenclature used here, but I think your statement

To my understanding, the symbol $\mathbf{v}^{\wedge}$ in eq(9) should more precisely be $^\mathcal{X}\mathbf{v}^{\wedge}\in T_{\mathcal{X}}\mathcal{M}$, which is an element in the local tangent space of $\mathcal{X}$.

is right.

I don't fully understand what you mean with your first question, but I think that your confusion in the second questions comes from not taking into account the corresponding reference frames. In "II-D. The exponential map", you can find equation (12):

equation (12)

If I have interpreted it correctly, we can rewrite this equation 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 $\mathcal{X}$: $T_{\mathcal{X}}\mathcal{M}$.



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