I am an engineering student and learned of Lie Group/Lie Algebra recently.

I can follow and understand all the formula derivation of Lie Algebra from Lie Group.

But I cannot grasp the meaning of "Lie Algebra being tangent space of the corresponding Lie Group".

Using $SO(2)$ as a simple example:

1) Why $\mathfrak s\mathfrak o(2)$ is tangent space of $SO(2)$?
2) How come $\mathfrak s\mathfrak o(2)$, as a tangent space, map bijectively to $SO(2)$ through $exp()$ and $log()$?

I don't have a strong math background and only want to have a simple intuition.


  • 1
    $\begingroup$ Please note that in the case of SO(2), the map you get from exp is not bijective as you say, only surjective (it is bijective to R — the universal cover of SO(2)) $\endgroup$
    – Weier
    Apr 8, 2020 at 12:37

2 Answers 2


Intuitively, you can think of the tangent space of a surface at a point as the space of all "directions" you can move from that point (with all velocities) while staying on the surface. That is, when you see common sketches of, e.g., spheres with a plane attached to a point, the plane is the set of all velocity vectors of paths thorough that point.

The paths themselves do not lie in that plane, but if you consider a very small portion of any path, it very nearly does. (That sounds rather vague, but you can make that precise -- it is basically the same situation as in a first-order Taylor expansion where a tangent line approximates a function close to a given point.) Of course, the same concept applies to higher-dimensional "surfaces", i.e. manifolds, which are harder to visualise.

Now for Lie groups: These are manifolds, and so you can look at tangent spaces at a point. Let's choose $SO(n)$ as an example and the identity element $\mathbb{1}$ as the point. Group elements $G$ are real $n\times n$ matrices defined by $$G^TG=\mathbb{1}\,.$$ (We can ignore the determinant condition for now.) The identity element obviously satisfies that, so in line with the tangent space intuition above, you can consider a "very small path" through the identity, i.e. a matrix $\mathbb1 +\alpha t$ with a real parameter $\alpha$ and an $n\times n$ matrix $t$. For $\alpha=0$, that is just the identity, and for small $\alpha$, in defines a "path" of matrices. Now, what is the condition that this path stays in $SO(n)$? Consider the defining relation, $$\left(\mathbb1 +\alpha t\right)^T\left(\mathbb1 +\alpha t\right)=\mathbb1 +\alpha\left( t+t^T\right) + \mathcal O(\alpha^2).$$ We see that for antisymmetric matrices $t$, the path stays in $SO(n)$ for very small $\alpha$. (Since we assume $\alpha$ to be small, we can neglect $\alpha^2$.) Hence, the antisymmetric matrices form the Lie algebra $\mathfrak{so}(n)$.

More generally, the Lie algebra gives the set of possible ways (directions) in which you can change the group elements while staying in the group, and that is just what the tangent space is for a surface.


The are several ways of defining the Lie algebra $\mathfrak g$ of a Lie group $G$. One of them is: it is the tangent space of $G$ at the identity element. Of course, this only defines it as a vector space; one still needs to define the Lie bracket in $\mathfrak g$.

In the case of $SO(2,\mathbb{R})$, you can see it as the circle$$S^1=\{z\in\mathbb{C}\,|\,\lvert z\rvert=1\}=\{\cos\theta+i\sin\theta\,|\,\theta\in\mathbb{R}\}.$$Then the identity element is $1\bigl(=\cos(0)+i\sin(0)\bigr)$ and it is natural to see its tangent space at $1$ as $\{i\theta\,|\,\theta\in\mathbb{R}\}$. But then$$(\forall\theta\in\mathbb{R}):\exp(i\theta)=\cos\theta+i\sin\theta\in S^1=SO(2,\mathbb{R}).$$


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