I don't think this question has been asked before, however if it has, feel free to redirect me. I want to find the Lie Algebra su(2) of the Lie Group SU(2) via the Implicit function Theorem (i know how to do it using exp). This is how i do it (question at the end):
$$ SU(2) = \{ A \in Mat(2,C) | A^*A = I, \; \det(A) = 1 \} $$
Define the Function $$ \psi: \{ A \in Mat(2,C) | \det(A) = 1 \} \rightarrow \{ A \in Mat(2,C) | A^* = A \} $$ $$ A \mapsto A^*A - I $$
It is true that
- $\psi(I) = 0$
- $\psi^{-1}(\{0\}) = SU(2)$
Consider the derivative $$ D\psi(A)[B] = \frac{d}{dt} \psi(A + tB) \Bigg|_{t=0} = A B^* + BA^* $$ It follows that $$ D\psi(I) = B + B^* $$ The derivative is surjective at $I$ because for an arbitrary $C = C^*$ $$ D\psi(I)[\frac{1}{2}C] = \frac{1}{2}C + \frac{1}{2}C^* = C $$ The Implicit Function Theorem now states that $SU(2)$ is a smooth manifold whose tangent space at $I$ is given by $$ T_I SU(2) = \ker(D\psi(I)) = su(2) = \{B \in Mat(2,C) | B + B^* = 0, \; \det(B) = 1 \} $$ However, this is not what $su(2)$ should be. $su(2)$ should be given by $$ su(2) = \{B \in Mat(2,C) | B + B^* = 0, \; \textbf{trace}(B) = 0 \} $$ I tried to use the two properties $B + B^* = 0, \; \det(B) = 1$ to derive $\text{trace}(B) = 0$ but i did not succed. Now i have the feeling my mistake is rooted deeper. My question is: what is my mistake and how to do it right. Thank you in advance!
When you differentiate $\psi : SL(2, \mathbb C) \to H$ (where $H$ is the space of all Hermetian matrices), you have
$$ D\psi (I) : T_I SL(2, \mathbb C) \to T_0H$$ and from what you prove, $T_ISU(2)$ is the kernel of this map. So
\begin{align} T_ISU (2) &= \ker (D\psi (I) : T_I SL(2, \mathbb C) \to T_0H) \\ &= \{ B \in T_I SL(2, \mathbb C) : D\psi (I) (B) = 0 \} \\ &= \{ B \in T_I SL(2, \mathbb C) : B +B^* = 0 \} \\ \end{align}
Since $$T_I SL(2, \mathbb C) = \{ B \in Mat(2, \mathbb C) : \operatorname{trace} B = 0\},$$ one has
$$T_ISU (2) = \{ B \in Mat(2, \mathbb C) : \operatorname{trace} B = 0, B+B^* = 0\}.$$
I guess, at the end of the day, I don't see why you have $\det B = 1$ there.