Lie algebra: intuition of "Lie Algebra is tangent space of corresponding Lie Group"?

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.

Thanks

• 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)) Apr 8, 2020 at 12:37

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 $$\mathfrak{so}(n)$$.
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$$(\forall\theta\in\mathbb{R}):\exp(i\theta)=\cos\theta+i\sin\theta\in S^1=SO(2,\mathbb{R}).$$