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A Lie group is a group which is a finite-dimensional smooth Hausdorff second-countable manifold, with smooth multiplication and inversion. For the special unitary group, consider the real vector spaceThe diagonal entries of a skew-Hermitian matrix supply $n$ real parameters, and the entries above the diagonal supply $2\left(\genfrac{}{}{0ex}{}{n}{2}\right)$; the trace-zero condition removes one real parameter. Thus ${\dim }_{\mathbb{R}}\mathfrak{su}(n)=n^2-1$.

The matrix facts used to construct charts are these: the matrix exponential is analytic, its derivative at zero is the identity, and it has an analytic inverse matrix logarithm on a sufficiently small neighborhood of the identity. On those neighborhoods $\log (e^X)=X$, $e^{\log U}=U$, and $\det (e^X)=e^{\mathrm{tr}X}$. For a unitary matrix in that neighborhood, the spectral theorem writes $U=W\mathrm{diag}(e^{i{\theta }_j})W^{\ast }$ with principal angles close to zero, and $\log U=W\mathrm{diag}(i{\theta }_j)W^{\ast }$ is skew-Hermitian.

Choose a norm ball of radius $\epsilon <\pi /n$ small enough to lie in the exponential's inverse domain. If $U$ is unitary, $\det U=1$ and $\Vert \log U\Vert <\epsilon$ in operator norm, thenso its trace is zero. Conversely $X\in \mathfrak{su}(n)$ gives $(e^X{)}^{\ast }{e}^X=I$ and $\det (e^X)=1$. Thus the logarithm identifies an open neighborhood $V$ of $I$ in $SU(n)$, with its matrix subspace topology, with the open ball $B_{\epsilon }\subset \mathfrak{su}(n)$. Choosing a real linear identification $\mathfrak{su}(n)\cong {\mathbb{R}}^{n^2-1}$ turns this into a chart.

For every $A\in SU(n)$ take the translated chartTheir overlap maps are $X\mapsto \log (B^{-1}A{e}^X)$ wherever both charts are defined, so they are real analytic. They cover the group; the matrix topology is Hausdorff and second-countable. This constructs the real manifold directly, rather than merely counting equations. Matrix multiplication is polynomial in real/imaginary entries and inversion here is $U\mapsto U^{\ast }$; in the logarithm charts both are smooth. The logarithm charts for the special unitary group therefore proveFor $n=1$ it is the one-point group of dimension zero.

For $n=2$, let the first column be $(a,b)$, of unit norm. Every unit vector orthogonal to it is a scalar of modulus one times $(-\bar{b},\bar{a})$. The determinant of the resulting matrix is that scalar, so determinant one fixes it to $1$. Henceis bijective, with inverse given by taking the first column. The forward map is real polynomial in the four coordinates; composing with the local logarithm charts verifies smoothness into the constructed manifold. The inverse is a smooth matrix-coordinate projection and, in the usual embedded sphere charts, is smooth into $S^3$. Thus $\boxed{SU(2)\cong S^3\text{ by a diffeomorphism}}$, the explicit realization of SU(2) as the three-sphere.