I'm studying about Lie groups and Lie algebras and the relations between them, and I've encountered the following formula to calculate the generators of a group:
$$ X_a = -i \frac{\partial}{\partial \alpha_a} D(\alpha) |_{\alpha=0} $$
Where $X_\alpha$ is the generator, a is a running index from 1 to N where N is the number of parameters and D is the representation of the group. I tried applying this formula to $SU(2)$, which has the form:
$$ D(a,b,c,d) = \begin{bmatrix}a+bi & -c+di\\\ c+di & a-bi\end{bmatrix} $$
With the requirement $a^2+b^2+c^2+d^2=1$, which means $d=\pm \sqrt{1-a^2-b^2-c^2}$ When I plug everything in the formula, instead of getting the 3 Pauli matrices up to some $i$, I get two of them and the identity matrix. What am I missing?
The formula is taken from page 44 of the following book: Georgi, Howard, Lie algebras in particle physics. From isospin to unified theories, Frontiers in Physics, 54. Reading, Massachusetts, etc.: The Benjamin/Cummings Publishing Company, Inc., Advanced Book Program. XXII, 255 p. (1982). ZBL0505.00036.:
It could be the case that you got lost in the identification of the identity $\begin{pmatrix}1&0\\0&1\end{pmatrix}$ of $SU(2)$ with the zero vector $(0,0,0)^T$ of the tangent space $\mathfrak{su}(2).$ The tangent space has the zero where the group has the identity matrix. These two points are where tangent space and manifold touch, i.e. they are the same geometric point.
Here is an example of how the calculation is done for $GL(n)$ and $SL(n):$ https://www.physicsforums.com/insights/pantheon-derivatives-part-iv/#B-%E2%80%93-Left-Invariant-Vector-Fields-and-GL-n and here are various calculations on $SU(2):$ https://www.physicsforums.com/insights/journey-manifold-su2mathbbc-part/#5-Tangent-Bundle