# From the definition of tangent space to the Lie algebra of a matrix group

I'm trying to understand how to go from the general definition of a tangent space to the explicit definition of the Lie algebra of a matrix group. Let $$G$$ be a Lie group, the Lie algebra is the tangent space at the identity

$$\mathfrak{g}=T_e G$$

We can see define the tangent space as follows: let $$\gamma: (-a,a)\to G$$ be a smooth curve such that $$\gamma(0)=e$$, we define an equivalence relation: $$\gamma \sim \tau$$ if for any chart $$\phi:U\to\mathbb{R}^n$$ such that $$e\in U$$ $$(\phi\circ\gamma)'(0)=(\phi\circ\tau)'(0)$$, we note by $$[\gamma]$$ the equivalence class of $$\gamma$$. For a curve $$\gamma$$ we can define a derivation

\begin{aligned}\gamma'(0):& C^\infty(G)\to\mathbb{R} \\ &f\mapsto(f\circ \gamma)'(0)\end{aligned}

then $$T_e G =\{[\gamma]'(0): \gamma:(-a,a)\to G \textrm{ smooth, }\gamma(0)=e\}$$ where by the derivative of the equivalence class I mean the one of a representative.

First question: do I have this definition correct?

Now suppose that $$G$$ is a matrix group, i.e. $$G\subset M_n(\mathbb{R})$$, the Lie algebra is usually defined as the space of the matrices

$$\Omega =\frac{d}{dt}R(t)|_{t=0}$$

where $$R(t):(-a,a)\to G$$ is a smooth curve. The derivative is well defined as $$R(t)$$ is a matrix for any $$t$$. To reconcile the two definitions my idea is to consider the functions

\begin{aligned}x^{ij}:&G\to \mathbb{R}\\&A\mapsto A_{ij}=\langle e_i, A e_j\rangle \end{aligned}

for some basis $$\{e_i\}$$, then $$R'(0)(x^{ij})=R'(0)_{ij}$$ and we automatically have a matrix associated to any derivation in $$T_eG$$. My question is, can we reconstruct the image of the derivation applied to any smooth function on $$G$$ via this matrix, i.e., is it true that

$$R'(0)(f)=\sum_{ij}\alpha_{ij} R'(0)(x^{ij})$$.

For some coefficient $$a_{ij}$$?. This seems to rely on the fact that in some way $$f(A)=\sum \alpha_{ij} x^{ij}$$, hence that $$f$$ is linear, so this doesn't seem like the right way to do it. How can I reconcile these two definitions?

You seem to be trying to identify $$M_n(R)$$ with $$\mathbb{R}^{n\times n}$$, which is a valid approach. The $$x^{ij}$$ are the coordinate projections.
We can write the curve $$R(t)= \pmatrix{R_{11}(t) & R_{12}(t) &\cdots & R_{1n}(t) \\ \vdots & \ddots &\\ R_{n1}(t) & R_{n2}(t) &\cdots & R_{nn}(t) \\} .$$
The only step you're missing is that $$f: G \to \mathbb{R}$$ now has to be identified as a function from a subset of $$\mathbb{R}^{n\times n}$$. Thus, we have $$f\circ R: I\subset \mathbb{R} \to \mathbb{R}.$$ Applying the chain rule we have, $$\frac{d}{dt} (f\circ R(t))\lvert_0 \ = \ df\vert_{R(0) }\circ \frac{d}{dt} R\vert_0.$$
Here $$df$$ is the derivative (linearisation/jacobian) of $$f$$ at $$R(0)$$. It is a linear map from the space of $$n\times n$$ real matrices to $$\mathbb{R}$$ (or from $$\mathbb{R}^{n\times n}$$ to $$\mathbb{R}$$).
Careful consideration of indices will show that your $$\alpha_{ij}$$ are in fact the directional derivatives of $$f$$ at $$R(0)$$.