Finite SU(2) Wilson integrals and transfer-sector well-definedness

For $V\in SU(2)$, the parametrization in [F2] gives $\frac{1}{2}{\mathrm{ReTr}}_{2}V\in [-1,1]$ and a real trace. Thus $\mathrm{\Phi }(V)=\beta (1-\frac{1}{2}{\mathrm{ReTr}}_{2}V)$ is continuous. Cyclicity of trace gives ${\mathrm{Tr}}_2(gV{g}^{-1})={\mathrm{Tr}}_{2}V$; unitarity gives ${\mathrm{Tr}}_2(V^{-1})=\overline{{\mathrm{Tr}}_{2}V}={\mathrm{Tr}}_{2}V$. Hence each plaquette potential is conjugation- and inverse-invariant as required by [F3].

Since ${\mathrm{ReTr}}_{2}V\in [-2,2]$, the normalized Wilson factor $1-\frac{1}{2}{\mathrm{ReTr}}_{2}V$ lies in $[0,2]$; hence each plaquette action term lies in $[0,2\beta ]$. Since ${\mathcal{P}}_{\mathrm{\Lambda }}$ is finite, $0\le S_{\mathrm{\Lambda }}\le 2\beta \mathrm{\mid }{\mathcal{P}}_{\mathrm{\Lambda }}\mathrm{\mid }$, and consequently $e^{-2\beta \mathrm{\mid }{\mathcal{P}}_{\mathrm{\Lambda }}\mathrm{\mid }}\le e^{-S_{\mathrm{\Lambda }}}\le 1$.

The actions $S_s,S_t$ are sums of nonnegative plaquette terms by step 1.1, so $0\le K(u^{\mathrm{\prime }},u)\le 1$. Matrix multiplication, inversion on $SU(2)$, and trace are continuous, so the temporal-action integrand is nonnegative and product-measurable. The slice and vertex product Haar measures are probabilities by [F1,F6], hence sigma-finite; Tonelli [F4] makes the $\eta$-integral $K$ measurable in $(u^{\mathrm{\prime }},u)$. Each slice Haar product is a probability by [F6], so $\int \mathrm{\mid }K{\mathrm{\mid }}^{2}d({\mu }_s\times {\mu }_s)\le 1$.

The compact Hausdorff group in [F2] has its normalized Haar probability by [F1]. Apply the two-factor product-measure theorem [F6] successively along the finite ordered link set; the rectangle formula at each stage gives total mass one, and [F3] identifies this iterated product with $d{\mu }_{\mathrm{\Lambda }}$. Integrating the bounds in step 2.1 gives $e^{-2\beta \mathrm{\mid }{\mathcal{P}}_{\mathrm{\Lambda }}\mathrm{\mid }}\le Z_{\mathrm{\Lambda }}\le 1$. For bounded $O$, $\mid \int O{e}^{-S_{\mathrm{\Lambda }}}d{\mu }_{\mathrm{\Lambda }}\mid \le \mathrm{\parallel }O{\mathrm{\parallel }}_{\mathrm{\infty }}{Z}_{\mathrm{\Lambda }}<\mathrm{\infty }$; division by the positive lower bound for $Z_{\mathrm{\Lambda }}$ defines its normalized expectation.

Apply [F5] to $X_s=SU(2{)}^{E_s}$ with ${\mathcal{A}}_s={⨂}_{e\in E_s}\mathcal{B}(SU(2))$ and its original product probability ${\mu }_s$ from [F6]; [A1] supplies Countable Choice, so ${\mathcal{H}}_{\mathrm{k}\mathrm{i}\mathrm{n}}$ is a complex Hilbert space with the pairing in [F5]. Let $\psi \in {\mathcal{H}}_{\mathrm{k}\mathrm{i}\mathrm{n}}$ and choose a measurable representative. Since $d{\mu }_s$ is a probability, [F7] applied to $\mathrm{\mid }\psi \mathrm{\mid }$ and the constant function $1$ of norm one gives $\mathrm{\parallel }\psi {\mathrm{\parallel }}_{L^1(d{\mu }_s)}\le \mathrm{\parallel }\psi {\mathrm{\parallel }}_{L^2(d{\mu }_s)}$. The kernel $K$ is measurable and satisfies $0\le K\le 1$ by step 2.2, so $\int K(u^{\mathrm{\prime }},u)\psi (u)d{\mu }_s(u)$ converges absolutely for every $u^{\mathrm{\prime }}$ and is independent of the representative of $\psi$. The product function $(u^{\mathrm{\prime }},u)\mapsto K(u^{\mathrm{\prime }},u)\psi (u)$ is measurable, and its absolute integral over $d{\mu }_s(u^{\mathrm{\prime }})d{\mu }_s(u)$ is at most $\mathrm{\parallel }\psi {\mathrm{\parallel }}_{L^1(d{\mu }_s)}$; Tonelli [F4] applied to the positive and negative parts of its real and imaginary parts shows that the section integral $T\psi$ is measurable. For each $u^{\mathrm{\prime }}$, both $K(u^{\mathrm{\prime }},\cdot {)}^{1\mathrm{/}2}$ and $K(u^{\mathrm{\prime }},\cdot {)}^{1\mathrm{/}2}\psi$ lie in $L^2(d{\mu }_s)$, so [F7] applied to $K(u^{\mathrm{\prime }},\cdot {)}^{1\mathrm{/}2}\psi$ and $K(u^{\mathrm{\prime }},\cdot {)}^{1\mathrm{/}2}$ gives $\mathrm{\mid }T\psi (u^{\mathrm{\prime }}){\mathrm{\mid }}^2\le (\int K(u^{\mathrm{\prime }},u)d{\mu }_s(u))(\int K(u^{\mathrm{\prime }},u)\mathrm{\mid }\psi (u){\mathrm{\mid }}^{2}d{\mu }_s(u))\le \int K(u^{\mathrm{\prime }},u)\mathrm{\mid }\psi (u){\mathrm{\mid }}^{2}d{\mu }_s(u)$, because the first factor is at most one. Integrating in $u^{\mathrm{\prime }}$ and applying Tonelli again yields $\mathrm{\parallel }T\psi {\mathrm{\parallel }}_{L^2(d{\mu }_s)}^2\le \int \mathrm{\mid }\psi (u){\mathrm{\mid }}^2(\int K(u^{\mathrm{\prime }},u)d{\mu }_s(u^{\mathrm{\prime }}))d{\mu }_s(u)\le \mathrm{\parallel }\psi {\mathrm{\parallel }}_{L^2(d{\mu }_s)}^2$, since $K\le 1$ and $d{\mu }_s$ is a probability. Linearity follows from linearity of the absolutely convergent section integral. Thus $T$ is a well-defined bounded linear operator with $\mathrm{\parallel }T\mathrm{\parallel }\le 1$.

Each spatial-link transformation is a left and right translation, so it preserves its Haar factor by [F1]; the rectangle formula and uniqueness in [F6] then show the product slice measure $d{\mu }_s$ is preserved. Under independent gauge transformations $h,h^{\mathrm{\prime }}$ of the incoming and outgoing slices, change temporal-link variables by ${\eta }_x\mapsto h_x{\eta }_x(h_x^{\mathrm{\prime }}{)}^{-1}$. Each such coordinate map also preserves Haar factors by [F1], and thus preserves $d{\mu }_V$ by [F6]. Substitution in $Q_{x,i}={\eta }_x{U}_{x,i}^{\mathrm{\prime }}{\eta }_{x+\widehat{i}}^{-1}{U}_{x,i}^{-1}$ cancels the neighboring $h^{\mathrm{\prime }}$ matrices and gives $Q_{x,i}\mapsto h_x{Q}_{x,i}{h}_x^{-1}$; each spatial plaquette is likewise conjugated at its base vertex. Trace invariance therefore preserves $S_t,S_s$, and the temporal integration gives $K(h^{\mathrm{\prime }}\cdot u^{\mathrm{\prime }},h\cdot u)=K(u^{\mathrm{\prime }},u)$ for every pair $(h^{\mathrm{\prime }},h)$. In particular, setting $h=1$ gives $K(h^{\mathrm{\prime }}\cdot u^{\mathrm{\prime }},u)=K(u^{\mathrm{\prime }},u)$. The pointwise integral definition established in step 3.2 now implies $(T\psi )(h^{\mathrm{\prime }}\cdot u^{\mathrm{\prime }})=(T\psi )(u^{\mathrm{\prime }})$ for every $u^{\mathrm{\prime }}$, hence as $L^2$ classes for each fixed $h^{\mathrm{\prime }}$, so $T\psi \in {\mathcal{H}}_{\mathrm{p}\mathrm{h}\mathrm{y}\mathrm{s}}$. The constant function has norm one because $d{\mu }_s$ is a probability and is fixed by every gauge transformation, so it is a nonzero physical vector.

Measure preservation from step 4.1 implies that pullback by $h^{-1}$ preserves measurable null sets, so $\mathcal{U}(h)$ is well-defined on almost-everywhere classes. It preserves the integral pairing from [F5] and has inverse $\mathcal{U}(h^{-1})$, hence is unitary on the complex Hilbert space established in step 3.2. Its fixed space is the kernel of the bounded operator $\mathcal{U}(h)-I$, hence is closed; the intersection over all $h$ is ${\mathcal{H}}_{\mathrm{p}\mathrm{h}\mathrm{y}\mathrm{s}}$, so ${\mathcal{H}}_{\mathrm{p}\mathrm{h}\mathrm{y}\mathrm{s}}$ is closed.

In the cyclic product of kernels, each time slice contributes one half of $S_s(u_t)$ from the incoming factor and one half from the outgoing factor, summing to the full spatial action. Each temporal-link integral contributes exactly $e^{-S_t(u_{t+1},u_t;{\eta }_t)}$. Therefore the product integrand before temporal-link integration is the full $e^{-S_{\mathrm{\Lambda }}}$; finite-product Tonelli [F4] gives the cyclic-kernel integral $Z_{\mathrm{\Lambda }}$. For a bounded slice observable $O(u_0)$, the absolute value of its inserted integrand is bounded by $\mathrm{\parallel }O{\mathrm{\parallel }}_{\mathrm{\infty }}{e}^{-S_{\mathrm{\Lambda }}}\le \mathrm{\parallel }O{\mathrm{\parallel }}_{\mathrm{\infty }}$ on the finite product probability space [F6], so it is integrable. Applying [F4] to the positive and negative parts of its real and imaginary components justifies exchanging the integrations; the result is the corresponding unnormalized finite-volume expectation.