Show that the SU(2) group is a Lie group

Question

How can I prove that the SU(2) is a Lie group with the Pauli matrices as generators?

Answer 1

First study the group $U(2)$. This is the set of matrices $A$ satisfying $$A \overline{A}^t=1,$$ where the overline denotes complex conjugation of the entries and the $t$ is the transpose. The map $A \mapsto A \overline{A}^t$ is a differentiable map from the set of all $2$ by $2$ complex matrices to the set of all matrices $C$ satisfying $\overline{C}^t=C$. Its derivative at a matrix $A$ is the linear map $$B \mapsto A \overline{B}^t+B \overline{A}^t.$$ One verifies (do it first for $A=1$ where it is trivial and then derive the general case from this) that this is surjective at all points $A$ of $U(2)$, which is therefore an embedded real submanifold of $\mathbb{C}^4$ by the implicit function theorem. It is immediate that the group operations are smooth: they are given by rational functions of the entries of the matrix. This proves that $U(2)$ is a Lie group.

Now $SU(2)$ is the kernel of the Lie group homomorphism $$\mathrm{det}: U(2) \rightarrow S^1.$$ It is therefore an embedded subgroup.

I don't believe $SU(2)$ is generated as an abstract group by the Pauli matrices (at least, not by the Pauli matrices I know, which correspond via the isomorphism between the unit quaternions and $SU(2)$ to the quaternions $i$, $j$, and $k$). You can modify the Pauli matrices to obtain generators of the Lie algebra of $SU(2)$, though, and the one-parameter subgroups corresponding to these do generate $SU(2)$. If you clarify (give a precise definition of Pauli matrices) and it's still useful to you, I will give more details.