SU(2) one-plaquette matrix average at strong coupling

In $M_2(\mathbb{C})\cong {\mathbb{R}}^8$, the equations $U^{\ast }U=I_2$ and $\det U=1$ define a closed set, and $U^{\ast }U=I_2$ implies $\mathrm{\parallel }U{\mathrm{\parallel }}_F^2=\mathrm{Tr}(U^{\ast }U)=2$, so the set is bounded. By [F3] it is compact; [F2] gives the stated smooth Lie-group structure. It is noncommutative: $A=\mathrm{diag}(i,-i)$ and $B=\left(\begin{array}{cc}0& 1\\ -1& 0\end{array}\right)$ lie in $SU(2)$, while $AB=-BA$ and $AB\mathrm{\ne }0$. The set-theoretic premise inherited by the lattice definitions is ${\mathrm{A}\mathrm{C}}_{\omega }$ [A1].

The coordinate map $\mathrm{\Xi }:S^3\to SU(2)$, $x\mapsto U(x)$, is a bijection. Indeed, for $U=\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)\in SU(2)$, $U^{-1}=U^{\ast }=\left(\begin{array}{cc}\bar{a}& \bar{c}\\ \bar{b}& \bar{d}\end{array}\right)$ while the adjugate formula gives $U^{-1}=\left(\begin{array}{cc}d& -b\\ -c& a\end{array}\right)$, so $d=\bar{a}$ and $c=-\bar{b}$; unitarity gives $\mathrm{\mid }a{\mathrm{\mid }}^2+\mathrm{\mid }b{\mathrm{\mid }}^2=1$. Writing $a=x_0+i{x}_3$ and $b=x_2+i{x}_1$ recovers a unique $x\in S^3$, and direct multiplication shows every such $x$ gives an element of $SU(2)$. The map and its inverse are smooth coordinate maps. For round measure, use $x=q(\theta ,\phi ,\psi )=(\cos \theta ,\sin \theta \sin \phi \cos \psi ,\sin \theta \sin \phi \sin \psi ,\sin \theta \cos \phi )$ on $0<\theta ,\phi <\pi$, $0<\psi <2\pi$. Its three coordinate tangent vectors are pairwise orthogonal with squared lengths $1$, ${\sin }^2\theta$, and ${\sin }^2\theta {\sin }^2\phi$, so the induced surface-area density is ${\sin }^2\theta \sin \phi d\theta d\phi d\psi$. The chart boundary is a finite union of lower-dimensional smooth pieces and has zero three-dimensional surface area. Integration gives total area $({\int }_0^{\pi }{\sin }^2\theta d\theta )({\int }_0^{\pi }\sin \phi d\phi )({\int }_0^{2\pi }d\psi )=2{\pi }^2$. Normalizing therefore gives $d{\mu }_2=(2\mathrm{/}\pi ){\sin }^2\theta d\theta d{\omega }_2$ with $d{\omega }_2=(1\mathrm{/}(4\pi ))\sin \phi d\phi d\psi$, exactly the density stated in [F1], and total mass one. For $V\in SU(2)$, the maps $x\mapsto {\mathrm{\Xi }}^{-1}(V\mathrm{\Xi }(x))$ and $x\mapsto {\mathrm{\Xi }}^{-1}(\mathrm{\Xi }(x)V)$ are real-linear by [F6] and preserve $\mathrm{\mid }x\mathrm{\mid }$, since $\mathrm{\parallel }\mathrm{\Xi }(x){\mathrm{\parallel }}_F^2=2\mathrm{\mid }x{\mathrm{\mid }}^2$ and left and right multiplication by a unitary matrix preserve Frobenius norm. Their restrictions to $S^3$ are orthogonal transformations, which preserve round measure. The pushforward formula [F1] defines a Borel probability. The compact matrix subspace $SU(2)\subset {\mathbb{R}}^8$ is Hausdorff and locally compact, and its intersections with Euclidean balls of rational centers and positive rational radii form a countable base. Since ${\mu }_2$ has total mass one it is finite on compact sets; [F7] under [A1] therefore makes it inner regular on open sets and outer regular on Borel sets. Thus ${\mu }_2$ is a finite Radon probability, and the left and right invariance already proved make it bi-invariant in the declared supplier convention.

In the coordinates established in step 2.1, $\tau (U)=\cos \theta$. Because ${\mu }_2$ is a probability and $-\beta \le \beta \tau (U)\le \beta$, the continuous factor $e^{\beta \tau }$ is bounded between $e^{-\beta }$ and $e^{\beta }$; hence $e^{-\beta }\le I_0(\beta )\le e^{\beta }$ and $I_0(\beta )$ is finite and positive. The density $z_{\beta }^{-1}{e}^{\beta \tau }$ therefore defines the probability ${\nu }_{\beta }$ [F5]. The angular vector $n\in S^2$ has mean zero by the measure-preserving symmetry $n\mapsto -n$. The coordinate integrands are continuous and bounded on the finite product-coordinate probability domain, so they are $L^1$ by [F5]; Fubini separates the angular and radial integrals. The matrix coordinates then give $\int U{e}^{\beta \tau (U)}d{\mu }_2(U)=I_1(\beta )I_2$, proving $\int Ud{\nu }_{\beta }(U)=\lambda (\beta )I_2$.

The transformation $U\mapsto -U$ corresponds to the orthogonal map $x\mapsto -x$, so it preserves ${\mu }_2$ and sends $\tau$ to $-\tau$. Consequently $I_1(\beta )=\int \tau (U)\sinh (\beta \tau (U))d{\mu }_2(U)>0$ for $\beta >0$: the integrand is nonnegative everywhere and strictly positive on, for example, the coordinate region $0<\theta <\pi \mathrm{/}3$, which has positive measure by the density in step 2.1. Also $I_0-I_1=\int (1-\tau (U))e^{\beta \tau (U)}d{\mu }_2(U)>0$, since $\tau \le 1$ and equality holds only at the identity matrix, whose singleton has zero round measure by the coordinate density. Thus $0<\lambda <1$.

The measure from step 2.1 is invariant under every orthogonal transformation of ${\mathbb{R}}^4$. Coordinate sign changes give $\int x_{0}d{\mu }_2=0$ and permutation symmetry makes all four squared-coordinate integrals equal; integrating ${\sum }_{j=0}^3{x}_j^2=1$ therefore gives $\int x_0^{2}d{\mu }_2=1\mathrm{/}4$. Taylor's formula for $e^{\beta x_0}$ has a remainder uniform for $\mathrm{\mid }{x}_0\mathrm{\mid }\le 1$ on a neighborhood of $\beta =0$; its integrals are finite because ${\mu }_2$ is a probability. Odd moments vanish under $x_0\mapsto -x_0$. Thus $I_0(\beta )=1+{\beta }^2\mathrm{/}8+O({\beta }^4)$ and $I_1(\beta )=\beta \mathrm{/}4+O({\beta }^3)$, so division gives $\lambda (\beta )=\beta \mathrm{/}4+O({\beta }^3)$.