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A SU(2) matrix has the form $\left(\begin{array}{cc}a& b\\ -\overline{b}& \overline{a}\end{array}\right)$ with $|a{|}^2+|b{|}^2=1$. Comparing this with the Pauli expansion givesConsequently all four Pauli coefficients must be real, and their norm condition isConversely, the Pauli matrix multiplication law gives $A{A}^{†}=(u_0^2+|\mathbf{u}{|}^2)I$ 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,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 givesFor an infinitesimal second factor, $v_0=1+O(|\mathbf{v}{|}^2)$. Thereforewhile $d{u}_0=-u_i{v}_i$. These variations are tangent to the unit sphere because $u_{0}d{u}_0+u_{i}d{u}_i=0$.

On a hemisphere, differentiation of the norm constraint gives ${\partial }_{u_i}{u}_0=-u_i/u_0$, and henceContracting with the Pauli-coordinate left-invariant vector fields on SU(2) and using ${\mu }_{ji}{u}_i=u_0{u}_j$ givesThe Pauli matrix multiplication law gives exactly the same expression for $-A{\sigma }_j$, soIt follows thatThis 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. ThereforeThe 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 $d{u}_0=-u_{i}d{u}_i/u_0$, its line element isThe rank-one matrix determinant lemma givesThus the Pauli-coordinate Haar measure on SU(2) isup 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 $A{A}^{†}$ 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, andThe second term vanishes because its contractions involve either ${\delta }_{ik}$ or $u_i{u}_k$, 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:Both hemispheres give total volume $2{\pi }^2$, so probability-normalized Haar measure is $d^{3}u/(2{\pi }^2|u_0|)$ 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 $SU(2)$ is a compact group. The finite integral reflects this compactness, and the equatorial divergence is only a coordinate artefact.