Haar measure and gauge integration

A lattice gauge field assigns a group element $U_\ell$ to each oriented link. To integrate over these fields, we need an integration rule that does not favor a particular group element or coordinate system. For a compact group, normalized Haar measure supplies it.

Invariance is the defining property

Prerequisites: elementary matrix representations and ordinary integration.
Outcomes: write a normalized group expectation; verify lattice gauge invariance; explain why a finite integral is not a continuum construction.

For a compact group $G$, write its normalized Haar measure as $dU$, with $\int_GdU=1$. It is invariant under left and right multiplication:

$$\int_G f(VU)\,dU=\int_G f(UV)\,dU =\int_G f(U)\,dU.$$

Here $V$ is any fixed element of $G$. Compactness permits a finite normalization. Inversion also preserves this normalized measure.

For $U(1)$, take $U=e^{i\theta}$ and

$$\int_{U(1)} f(U)\,dU =\frac1{2\pi}\int_0^{2\pi}f(e^{i\theta})\,d\theta.$$

The elementary identity

$$\int_{U(1)}U^n\,dU=\delta_{n0}$$

already shows how group integration projects onto invariant combinations.

Why this helps on a lattice

For a finite set of independent links $E$, define

$$Z=\int_{G^{|E|}} e^{-S(U)}\prod_{\ell\in E}dU_\ell.$$

For a continuous real action on this compact configuration space, $Z$ is finite and positive. This is an ordinary finite-dimensional integral. The difficulty of quantum Yang–Mills theory has not disappeared: it has been moved into the limits of small spacing and large physical volume.

At a vertex, a gauge transformation acts by

$$U_{xy}\mapsto g_xU_{xy}g_y^{-1}.$$

Haar invariance preserves the product measure. A plaquette holonomy transforms by conjugation at its starting vertex, so its trace is unchanged. These two observations explain why the Wilson lattice formulation can preserve gauge invariance without first choosing a gauge.

A useful matrix identity

For the fundamental representation of $SU(N)$, $N\geq2$,

$$\int dU\,U_{ij}\overline{U_{kl}} =\frac1N\delta_{ik}\delta_{jl}.$$

The invariant tensors fix the form of the answer; normalization fixes its coefficient. Summing $i=k$ and using unitarity gives the factor $1/N$.

This identity is a first example of representation orthogonality. More complicated versions organize strong-coupling and character expansions. The averaging removes combinations that fail to form a singlet.

For $SU(2)$, a group element can be written

$$U=a_0I+i\sum_{j=1}^3a_j\sigma_j,\qquad a_0^2+a_1^2+a_2^2+a_3^2=1.$$

Normalized Haar measure corresponds to the uniform surface measure on this three-sphere. Uniformly sampling each coefficient in a box would produce the wrong group measure; rotationally invariant sampling followed by normalization is the appropriate geometric idea.

What the integral does not establish

A finite lattice measure is an excellent starting point, but its existence is not yet a continuum construction. Gauge invariance alone also does not imply reflection positivity or a mass gap. Those are separate structural and dynamical questions.

Exercise. For one $U(1)$ variable with action $S(\theta)=-\beta\cos\theta$, write the normalized expectation of $\cos\theta$ and express it as a derivative of the partition function.

Solution. With $Z(\beta)=\int_0^{2\pi} e^{\beta\cos\theta}\,d\theta/(2\pi)$, one has $\langle\cos\theta\rangle=\partial_\beta\log Z$. The normalized Haar measure remains present in both numerator and denominator.

Continue

The probability bridge spells out normalization and observables. Two-dimensional Yang–Mills uses character orthogonality in an exact heat-kernel calculation with explicit geometry and coupling normalization.

Read the lattice-definition lesson. The original lattice formulation is Kenneth Wilson, Confinement of Quarks. For the place of lattice approximation in the construction problem, see section 6.5 of Jaffe and Witten.