Two Nonparallel One-Parameter Subgroups Generate SU(2) After Closure

Put $P_a=a\cdot \sigma$. By [F1], $P_a=P_a^{\ast }$, $P_a^2=I$ and $\mathrm{tr}{P}_a=0$. Its eigenvalues are $1,-1$ by [F2], so the power series gives $e^{tA}=I\cos t+i{P}_a\sin t$ and its eigenvalues are $e^{it},e^{-it}$. By [F4] it is unitary, and its determinant is one. The same holds for $B$. Thus the generated subgroup lies in $SU(2)$; its closure $H$ also lies there because the defining matrix equations for $SU(2)$ are closed. Continuity of multiplication and inversion makes $H$ a subgroup.

By [F1], $[A,B]=-2i(a\times b)\cdot \sigma$. Let $c=a\times b\mathrm{\ne }0$ and $s_0=\max (\mathrm{\parallel }A\mathrm{\parallel },\mathrm{\parallel }B\mathrm{\parallel })$. For $\mathrm{\mid }t\mathrm{\mid }\le 1$, multiplying the four exponential series $e^{tA}{e}^{tB}{e}^{-tA}{e}^{-tB}$ gives degree-one coefficient $A+B-A-B=0$ and degree-two coefficient $AB-BA=[A,B]$. Submultiplicativity bounds the sum of norms of all degree-$m$ terms by $(4s_0{)}^m\mathrm{/}m!$. Hence the terms of degree at least three have norm at most $K\mathrm{\mid }t{\mathrm{\mid }}^3$, where $K=(4s_0{)}^3{e}^{4s_0}\mathrm{/}6$. For fixed $r\ge 0$, put $t=\sqrt{r\mathrm{/}n}$. The commutator factor is $X_n=I+r[A,B]\mathrm{/}n+R_n$ with $\mathrm{\parallel }{R}_n\mathrm{\parallel }\le K{r}^{3\mathrm{/}2}{n}^{-3\mathrm{/}2}$. Set $Y_n=I+r[A,B]\mathrm{/}n$. For fixed $r$, the norms of $X_n^k$ and $Y_n^k$ are bounded uniformly over $0\le k\le n$, because each one-step norm is at most $1+C_r\mathrm{/}n$ for all sufficiently large $n$. The telescoping identity $X_n^n-Y_n^n={\sum }_{k=0}^{n-1}{X}_n^{n-1-k}(X_n-Y_n)Y_n^k$ bounds their difference by $O(n^{-1\mathrm{/}2})$. The binomial expansion of $Y_n^n$ converges to the exponential series for $e^{r[A,B]}$: its $k$th term is bounded by $(r\mathrm{\parallel }[A,B]\mathrm{\parallel }{)}^k\mathrm{/}k!$, a summable sequence. Swapping $A$ and $B$ gives $e^{-r[A,B]}$ for $r\ge 0$. Every approximant belongs to the generated subgroup, so closedness gives $e^{s[A,B]}\in H$ for every real $s$.

The vectors $a,b,c$ form a basis of ${\mathbb{R}}^3$, since $c$ is nonzero and orthogonal to both $a$ and $b$. By [F1], every traceless skew-Hermitian $2\times 2$ matrix $L$ is a real linear combination $xA+yB+z[A,B]$. For fixed matrices $S,T$, exponential-series expansion gives $e^{S\mathrm{/}n}{e}^{T\mathrm{/}n}=I+(S+T)\mathrm{/}n+O(n^{-2})$ and $e^{(S+T)\mathrm{/}n}=I+(S+T)\mathrm{/}n+O(n^{-2})$. The one-step difference is $O(n^{-2})$; the norms of intermediate powers up to $n$ are bounded by $e^{\mathrm{\parallel }S\mathrm{\parallel }+\mathrm{\parallel }T\mathrm{\parallel }+1}$ for sufficiently large $n$. Telescoping therefore gives $(e^{S\mathrm{/}n}{e}^{T\mathrm{/}n}{)}^n\to e^{S+T}$. The same estimate for three factors gives $(e^{xA\mathrm{/}n}{e}^{yB\mathrm{/}n}{e}^{z[A,B]\mathrm{/}n}{)}^n\to e^L$. Every factor lies in $H$ by step 1.1 and step 2.1, so closedness gives $e^L\in H$.

Every $U\in SU(2)$ is normal, so [F2] diagonalizes it. Its eigenvalues lie on the unit circle and their product is one, so they are $e^{i\theta },e^{-i\theta }$ for some real $\theta$. In that orthonormal eigenbasis, the matrix with diagonal entries $i\theta ,-i\theta$ is traceless and skew-Hermitian, and its exponential is $U$. Step 3.1 gives $U\in H$. Since $H\subseteq SU(2)$ by step 1.1, $H=SU(2)$.