= Solution
A <SU(2) matrix> has the form $\begin{pmatrix}a&b\\-\overline b&\overline a\end{pmatrix}$ with $|a|^2+|b|^2=1$. Comparing this with the Pauli expansion gives
$$
a=u_0+iu_3,\qquad b=u_2+iu_1.
$$
Consequently all four Pauli coefficients must be real, and their norm condition is
$$
\boxed{u_0,u_1,u_2,u_3\in\mathbb R,\qquad u_0^2+u_1^2+u_2^2+u_3^2=1.}
$$
Conversely, the <Pauli matrix multiplication law> gives $AA^\dagger=(u_0^2+|\mathbf u|^2)I$ and $\det A=u_0^2+|\mathbf u|^2$, so these conditions imply unitarity and <determinant> one. For complex coefficients the <determinant> equation alone would not imply unitarity.
On either open hemisphere,
$$
\boxed{u_0=\pm\sqrt{1-|\mathbf u|^2},\qquad|\mathbf u|<1.}
$$
Both signs are necessary globally. The two hemispheres meet along $u_0=0$, and the full four-coefficient constraint is exactly the unit sphere in $\mathbb R^4$. The coefficient map and its inverse are smooth, establishing <SU(2) as the three-sphere> as a <manifold>, not merely a <dimension> count.
For two real coefficient quadruples, expanding the <matrix> product with the <Pauli matrix multiplication law> gives
$$
w_0=u_0v_0-\mathbf u\cdot\mathbf v,\qquad
\mathbf w=v_0\mathbf u+u_0\mathbf v-\mathbf u\times\mathbf v.
$$
For an infinitesimal second factor, $v_0=1+O(|\mathbf v|^2)$. Therefore
$$
du_i=u_0v_i+\epsilon_{jki}v_ju_k
=\boxed{v_j\mu_{ji},\qquad\mu_{ji}=u_0\delta_{ij}+u_k\epsilon_{jki}},
$$
while $du_0=-u_iv_i$. These variations are tangent to the unit sphere because $u_0du_0+u_idu_i=0$.
On a hemisphere, differentiation of the norm constraint gives $\partial_{u_i}u_0=-u_i/u_0$, and hence
$$
\partial_{u_i}A=-\frac{u_i}{u_0}I+i\sigma_i.
$$
Contracting with the <Pauli-coordinate left-invariant vector fields on SU(2)> and using $\mu_{ji}u_i=u_0u_j$ gives
$$
T_jA=i\mu_{ji}\partial_{u_i}A
=-iu_jI-u_0\sigma_j-u_k\epsilon_{jki}\sigma_i.
$$
The <Pauli matrix multiplication law> gives exactly the same expression for $-A\sigma_j$, so
$$
\boxed{T_jA=-A\sigma_j.}
$$
It follows that
$$
[T_i,T_j]A=A\sigma_i\sigma_j-A\sigma_j\sigma_i
=A[\sigma_i,\sigma_j]=2i\epsilon_{ijk}A\sigma_k
=-2i\epsilon_{ijk}T_kA.
$$
This uses the <Pauli matrix commutator identity>. The <commutator> of two first-order <vector fields> is again a <vector field>, and the <matrix> entries of $A$ span the four coordinate functions $u_0,u_1,u_2,u_3$. Equality of their actions on these entries determines the tangent field. Therefore
$$
\boxed{[T_i,T_j]=-2i\epsilon_{ijk}T_k.}
$$
The real tangent field underneath $T_j$ is generated by the flow $A\mapsto A e^{it\sigma_j}$; left multiplication transports that tangent consistently, explaining its left invariance. The rescaled generators $-T_j/2$ have the usual positive-sign angular-momentum <commutator>.
To determine the <Haar measure>, compute the metric induced from the unit three-sphere. Since $du_0=-u_i\,du_i/u_0$, its line element is
$$
ds^2=du_0^2+\sum_i du_i^2
=\left(\delta_{ij}+\frac{u_iu_j}{u_0^2}\right)du_i\,du_j.
$$
The rank-one <matrix determinant lemma> gives
$$
\det g=1+\frac{|\mathbf u|^2}{u_0^2}=\frac1{u_0^2}.
$$
Thus the <Pauli-coordinate Haar measure on SU(2)> is
$$
\boxed{d\rho(\mathbf u)=\sqrt{\det g}\,d^3u=\frac{d^3u}{|u_0|},}
$$
up to an overall normalization. To check its group invariance, fix one unit coefficient quadruple and multiply by it. The product formulas define a linear transformation of the other quadruple. The norm of a product is the product of norms, as follows either from $AA^\dagger$ or directly from the dot and cross products. Fixed multiplication by a unit quadruple is therefore an orthogonal transformation of $\mathbb R^4$. It preserves the round sphere metric and volume, on both the left and the right, proving that this is the invariant measure.
The same density can be checked directly against the <vector fields>. On a fixed hemisphere, $u_0/|u_0|$ is constant, and
$$
\partial_{u_i}\left(\frac{\mu_{ji}}{|u_0|}\right)
=\partial_{u_j}\left(\frac{u_0}{|u_0|}\right)
+\epsilon_{jki}\partial_{u_i}\left(\frac{u_k}{|u_0|}\right)=0.
$$
The second term vanishes because its contractions involve either $\delta_{ik}$ or $u_iu_k$, both symmetric in $i,k$. Thus the density has zero divergence along every invariant tangent direction.
The hemisphere chart becomes singular at the equator, but its volume is finite:
$$
\int_{|\mathbf u|<1}\frac{d^3u}{\sqrt{1-|\mathbf u|^2}}
=4\pi\int_0^1\frac{r^2\,dr}{\sqrt{1-r^2}}
=4\pi\int_0^{\pi/2}\sin^2\chi\,d\chi=\pi^2.
$$
Both hemispheres give total volume $2\pi^2$, so probability-normalized <Haar measure> is $d^3u/(2\pi^2|u_0|)$ on each hemisphere. Finally, the three-sphere is closed and bounded in $\mathbb R^4$, hence compact; its smooth identification with <SU(2)> proves that \b[$SU(2)$ is a <compact group>]. The finite integral reflects this compactness, and the equatorial divergence is only a coordinate artefact.
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