Evaluate the circular Wilson loop at finite N

The field theory has reduced to a Gaussian integral, but its Wilson insertion is an exponential of a matrix. The eigenvalues repel through their integration measure, so replacing them by independent Gaussian numbers would still give the wrong answer. We will evaluate the integral at every finite rank and then remove its central mode.

Use the Gaussian localization result and the normalization of observables. The physical target is the normalized fundamental half-BPS circle in massless four-dimensional $SU(N)$ N=4 SYM, $N\geq2$. Its flat parent has sixteen real Poincaré supercharges, in addition to conformal supercharges. The N=2 presentation and selected localization charge used to derive the matrix model do not change that theory. Here we evaluate the resulting finite-dimensional integral.

Keep the physical coupling and matrix variable fixed:

$$\lambda=g^2N,\qquad M=2\pi\ell a,\qquad \alpha=\frac{2N}{\lambda}>0. \tag{N15.1}$$

$M$ and $\lambda$ are dimensionless; $\ell$ is the sphere radius and $a$ is the Hermitian scalar of dimension one. $\operatorname{Tr}$ always means the ordinary fundamental trace. We first allow all Hermitian matrices, defining the explicitly labeled $U(N)$ benchmark

$$W_{U(N)}(\lambda)= \frac{\int dM\,e^{-\alpha\operatorname{Tr}M^2} \,N^{-1}\operatorname{Tr}e^M} {\int dM\,e^{-\alpha\operatorname{Tr}M^2}}. \tag{N15.2}$$

The $SU(N)$ answer instead integrates only traceless matrices. It will not be obtained by changing $N$ in (N15.2). The source’s matrix proposal is Drukker–Gross (2.25); its gauge-theory justification is the localization result Pestun (1.2)–(1.5), with the contour and mass conventions established in the prerequisite.

Keep the eigenvalue measure

Diagonalize $M=U\operatorname{diag}(m_1,\ldots,m_N)U^\dagger$. An infinitesimal change has a diagonal part $dm_i$ and an angular part $[\omega,\operatorname{diag}m]$, with $\omega=U^\dagger dU$. Consequently the metric on matrix space is

$$\operatorname{Tr}(dM)^2 =\sum_i dm_i^2+ 2\sum_{i<j}(m_i-m_j)^2|\omega_{ij}|^2.$$

Each off-diagonal complex angle has two real components. Their volume factor is $(m_i-m_j)^2$, giving the Vandermonde $\Delta(m)=\prod_{i<j}(m_i-m_j)$ and the eigenvalue measure $\Delta(m)^2\prod_i dm_i$. The angular volume and permutation factor cancel between numerator and denominator. We may integrate over all of $\mathbb R^N$, counting labeled eigenvalues in both.

Set $y_i=\sqrt\alpha\,m_i$ and $t=\alpha^{-1/2}=\sqrt{\lambda/(2N)}$. Symmetry among the eigenvalues gives

$$W_{U(N)} =\frac{\int_{\mathbb R^N}d^Ny\,\Delta(y)^2 e^{-\sum_i y_i^2}\,e^{t y_1}} {\int_{\mathbb R^N}d^Ny\,\Delta(y)^2 e^{-\sum_i y_i^2}}. \tag{N15.3}$$

The numerator originally contains $N^{-1}\sum_i e^{t y_i}$. Its $N$ equal integrals cancel that $1/N$ here; they do not remove the normalization of the loop.

Use orthogonality to integrate out the other eigenvalues

The physicists’ Hermite polynomials are defined by

$$H_j(y)=(-1)^j e^{y^2}\frac{d^j}{dy^j}e^{-y^2}, \qquad \int_{\mathbb R}dy\,e^{-y^2}H_j(y)H_k(y) =2^j j!\sqrt\pi\,\delta_{jk}. \tag{N15.4}$$

For example, $H_0=1$, $H_1=2y$, and $H_2=4y^2-2$. Integration by parts $j$ times proves the norm: lower-degree polynomials vanish, while the $j$th derivative of $H_j$ is $2^j j!$. The Gaussian removes every boundary term.

A determinant of monomials is a Vandermonde, up to an irrelevant sign. Replacing each degree-$j$ monomial by a monic polynomial of the same degree is a triangular column operation, so it leaves that determinant unchanged. We may therefore use $2^{-j}H_j$.

For a normalization check, define $Z_N^{\rm eig}(\alpha)=\int d^Nm\,\Delta(m)^2 e^{-\alpha\sum m_i^2}$, without a $1/N!$. The monic polynomials and their norms are

$$p_j(m)=\frac{H_j(\sqrt\alpha m)}{2^j\alpha^{j/2}}, \qquad h_j=\int_{\mathbb R}dm\,e^{-\alpha m^2}p_j(m)^2 =\frac{\sqrt\pi\,j!}{2^j\alpha^{j+1/2}}.$$

Expanding both determinants into permutations, orthogonality kills every term except matching permutations. The $N!$ surviving terms each give $\prod_j h_j$. Thus

$$Z_N^{\rm eig}(\alpha) =N!\prod_{j=0}^{N-1}h_j =N!\,\pi^{N/2}\,2^{-N(N-1)/2} \alpha^{-N^2/2}\prod_{j=0}^{N-1}j!. \tag{N15.5}$$

This is an eigenvalue normalization, not an absolute normalization of the four-dimensional sphere partition function.

Define normalized functions $\varphi_j(y)=e^{-y^2/2}H_j(y)/(2^j j!\sqrt\pi)^{1/2}$. The normalized joint density in the $y$ variables is then $\det[\varphi_{j-1}(y_i)]^2/N!$. Integrating out $y_2,\ldots,y_N$ matches the corresponding polynomial labels. For each remaining label $j$, there are $(N-1)!$ terms. The probability density of one eigenvalue is therefore

$$\rho_N(y)=\frac1N\sum_{j=0}^{N-1}\varphi_j(y)^2, \qquad \int_{\mathbb R}\rho_N(y)\,dy=1,$$

and

$$W_{U(N)}=\frac1N\sum_{j=0}^{N-1} \frac1{2^j j!\sqrt\pi} \int_{\mathbb R}dy\,e^{-y^2+t y}H_j(y)^2. \tag{N15.6}$$

This supplies the measure steps behind Drukker–Gross (A.1)–(A.5). The density integrates to one; the eigenvalue number density would instead be $N\rho_N$.

Evaluate the remaining integral

The generating function $e^{2uy-u^2}=\sum_{j\geq0}H_j(y)u^j/j!$ lets us integrate before extracting a coefficient. Completing the square gives

$$\frac1{\sqrt\pi}\int_{\mathbb R}dy\, e^{-y^2+t y}e^{2uy-u^2}e^{2vy-v^2} =e^{t^2/4}e^{2uv+t(u+v)}. \tag{N15.7}$$

The coefficient of $u^jv^j$ on the left is the desired integral divided by $(j!)^2\sqrt\pi$. On the right, choose $j-q$ powers from $e^{2uv}$ and $q$ powers from each of $e^{tu}$ and $e^{tv}$. After dividing by the norm in (N15.6), the result is

$$\frac{\int dy\,e^{-y^2+t y}H_j(y)^2}{2^j j!\sqrt\pi} =e^{t^2/4}\sum_{q=0}^{j}\binom jq\frac{(t^2/2)^q}{q!} =e^{t^2/4}L_j(-t^2/2). \tag{N15.8}$$

Here the last equality defines the ordinary Laguerre polynomial by a finite sum. No special-function background is required. The remaining sum over $j$ uses $\sum_{j=q}^{N-1}\binom jq=\binom N{q+1}$. Define

$$L_{N-1}^{\,1}(x) =\sum_{q=0}^{N-1}\binom N{q+1}\frac{(-x)^q}{q!}.$$

Substituting $t^2=\lambda/(2N)$ yields the exact finite-rank answer,

$$\boxed{ W_{U(N)}(\lambda) =\frac1N e^{\lambda/(8N)} L_{N-1}^{\,1}\!\left(-\frac{\lambda}{4N}\right). } \tag{N15.9}$$

This agrees with Drukker–Gross (A.6)–(A.8). The derivation used $\lambda>0$, where the Hermitian Gaussian converges. The resulting exponential times polynomial has a continuous limit at $\lambda=0$ and an analytic continuation; a negative real $\lambda$ does not itself define the same convergent Hermitian integral.

Since $L_{N-1}^{\,1}(0)=N$, the answer has $W(0)=1$. At $N=1$ it gives $W_{U(1)}=e^{\lambda/8}$, exactly the single real Gaussian average $\langle e^M\rangle$ with variance $\lambda/4$. The factor $1/N$ and the exponent’s coefficient are both tested by these limits.

Remove the central mode at fixed coupling

Write

$$M=X+c\mathbf1,\qquad \operatorname{Tr}X=0,\qquad c=\frac{\operatorname{Tr}M}{N}.$$

Then $\operatorname{Tr}M^2=\operatorname{Tr}X^2+Nc^2$. The flat matrix metric likewise separates into $\operatorname{Tr}(dX)^2+N\,dc^2$, so its constant Jacobian cancels in normalized averages. $X$ and $c$ are independent under this Gaussian measure. Moreover, $N^{-1}\operatorname{Tr}e^M=e^c\,N^{-1}\operatorname{Tr}e^X$. The central variance and its exponential average are

$$\langle c^2\rangle=\frac1{2\alpha N} =\frac{\lambda}{4N^2}, \qquad \langle e^c\rangle=e^{\lambda/(8N^2)}. \tag{N15.10}$$

Consequently, at the same $g$ and $\lambda=g^2N$,

$$\boxed{ W_{SU(N)}(\lambda) =e^{-\lambda/(8N^2)}W_{U(N)}(\lambda) =\frac1N e^{\lambda(N-1)/(8N^2)} L_{N-1}^{\,1}\!\left(-\frac{\lambda}{4N}\right). } \tag{N15.11}$$

This is the central Gaussian relation in Drukker–Gross (5.1). In generator language the center is $t^0=\mathbf1/\sqrt{2N}$, with $\operatorname{Tr}(t^0t^0)=1/2$. Its fundamental charge and variance are therefore fixed by this embedding, not by an independently chosen Abelian coupling.

The calculation separates a Lie-algebra integral for this sphere observable. It does not assert that every global sector or line operator of $U(N)=(SU(N)\times U(1))/\mathbb Z_N$ factorizes into freely chosen theories. The formal traceless $N=1$ integral is a single point with value one; the physical Abelian test is the $U(1)$ Gaussian.

As in the normalization prerequisite, source-only bulk factors cancel from this expectation value. Changing the defect’s scalar coupling or multiplying the line itself by a coupling-dependent finite factor changes the observable. The check $W(0)=1$ alone does not authorize such a redefinition. We keep the inherited supersymmetric loop prescription.

Worked example: a direct traceless two-color integral

For $SU(2)$ write $X=r\,\widehat n\cdot\boldsymbol\sigma$, with $r\geq0$ and $\widehat n$ a unit three-vector. Its eigenvalues are $\pm r$, so the normalized loop is $\cosh r$. The ordinary flat measure on the three matrix components is proportional to $r^2dr\,d\Omega$, and the action is $8r^2/\lambda$. Thus a calculation without Hermite polynomials gives

$$W_{SU(2)} =\frac{\int_0^\infty dr\,r^2e^{-\beta r^2}\cosh r} {\int_0^\infty dr\,r^2e^{-\beta r^2}}, \qquad \beta=\frac8\lambda.$$

Start from $\int_0^\infty dr\,e^{-\beta r^2}\cosh(pr) =\sqrt\pi\,e^{p^2/(4\beta)}/(2\sqrt\beta)$. Applying $-\partial_\beta$ inserts $r^2$. Setting $p=1$ gives

$$W_{SU(2)}=\left(1+\frac{\lambda}{16}\right)e^{\lambda/32}, \qquad W_{U(2)}=\left(1+\frac{\lambda}{16}\right)e^{\lambda/16}. \tag{N15.12}$$

The difference is precisely the central factor, not a different value of the coupling. At $\lambda=4$, the two values are approximately $1.4164355663$ and $1.6050317709$. For any traceless Hermitian $X$, convexity gives $N^{-1}\operatorname{Tr}e^X\geq e^{\operatorname{Tr}X/N}=1$, with equality only at $X=0$. Hence the $SU(N)$ answer is strictly greater than one for positive coupling and $N\geq2$, consistent with this example.

We now have a finite-rank result, without a weak-coupling or large-rank approximation. Taking a planar limit and then estimating strong coupling are separate operations, to be studied next.

The exact polynomial also lets us test approximations. Continue to the planar limit and strong coupling to derive the limiting density, resolve the exponential endpoint, and compare the two independent error budgets.

Exercises

1. Check the first two corrections using matrix contractions

For the unconstrained Gaussian, derive $\langle M_{ij}M_{kl}\rangle=\lambda\,\delta_{il}\delta_{jk}/(4N)$. Evaluate $N^{-1}\langle\operatorname{Tr}M^2\rangle$ and $N^{-1}\langle\operatorname{Tr}M^4\rangle$ by pairing entries. Obtain both $U(N)$ and $SU(N)$ loop expansions through order $\lambda^2$.

Hint 1

In $\operatorname{Tr}M^2$, a diagonal real entry appears once and an off-diagonal complex entry has weight twice. In the four-entry trace there are three pairings; count the independent color indices left by each.

Hint 2

The two noncrossing pairings each leave three color sums; the crossing pairing leaves one. For the traceless answer, use the already derived independent central Gaussian, or project the covariance.

Solution

The real diagonal variance is $1/(2\alpha)$. For $i<j$, the real and imaginary parts each have variance $1/(4\alpha)$, giving $\langle M_{ij}M_{ji}\rangle=1/(2\alpha)$. All other independent contractions vanish. With $v=1/(2\alpha)=\lambda/(4N)$,

$$\frac1N\langle\operatorname{Tr}M^2\rangle=vN=\frac\lambda4, \qquad \frac1N\langle\operatorname{Tr}M^4\rangle =\frac{v^2}{N}(2N^3+N) =\frac{\lambda^2}{16}\left(2+\frac1{N^2}\right).$$

Odd moments vanish by $M\mapsto-M$. The exponential supplies $1/2!$ and $1/4!$, so

$$\begin{aligned} W_{U(N)} &=1+\frac\lambda8+ \frac{\lambda^2}{384}\left(2+\frac1{N^2}\right)+O(\lambda^3),\\ W_{SU(N)} &=1+\frac\lambda8\left(1-\frac1{N^2}\right) +\frac{\lambda^2}{384} \left(2-\frac5{N^2}+\frac3{N^4}\right)+O(\lambda^3). \end{aligned}$$

The second line follows by multiplying the first by $1-\lambda/(8N^2)+\lambda^2/(128N^4)$. Equivalently, $\langle X_{ij}X_{kl}\rangle= v(\delta_{il}\delta_{jk}-\delta_{ij}\delta_{kl}/N)$ removes the trace. The first correction is $g^2C_F/4$, with $C_F=(N^2-1)/(2N)$ for $SU(N)$, checking the physical coupling normalization. At $N=2$ these coefficients agree with (N15.12).

2. Distinguish winding from a product of loops

Let $W_p=N^{-1}\langle\operatorname{Tr}e^{pM}\rangle$ for a positive integer winding $p$ with the same scalar coupling. Derive its finite-rank formula and central factor. Then, for $SU(2)$, compute $\langle w_1^2\rangle-\langle w_1\rangle^2$, where $w_1=\tfrac12\operatorname{Tr}e^X$ is the insertion before averaging. Why is this a different question from computing $W_2$?

Hint 1

Replace $t$ by $pt$ in (N15.7). In two colors, use eigenvalues $\pm r$ to relate $\cosh^2r$ and $\cosh 2r$.

Hint 2

The expectation of a product need not be the product of expectations. Keep the trace normalization in the identity $(\operatorname{Tr}e^X)^2=\operatorname{Tr}e^{2X}+2$.

Solution

The calculation depends on $p$ through $p^2\lambda$:

$$W_{p,U(N)} =\frac1N e^{p^2\lambda/(8N)} L_{N-1}^{\,1}\!\left(-\frac{p^2\lambda}{4N}\right), \qquad W_{p,SU(N)}=e^{-p^2\lambda/(8N^2)}W_{p,U(N)}.$$

This is a single traced multiply wound insertion, as in Drukker–Gross (4.1)–(4.2). It is not a product of separately traced insertions. For $SU(2)$, $w_1^2=(1+w_2)/2$, hence

$$\operatorname{Var}(w_1)= \frac12\left[1+\left(1+\frac\lambda4\right)e^{\lambda/8}\right] -\left(1+\frac\lambda{16}\right)^2e^{\lambda/16}.$$

This is nonnegative because it is the variance under a positive real Gaussian measure. Its expansion starts with $3\lambda^2/512$, not a term linear in $\lambda$. A connected two-insertion observable thus contains information beyond one winding. Even in $U(1)$, where the two insertions $e^{2M}$ and $(e^M)^2$ coincide, their expectation $e^{\lambda/2}$ differs from $\langle e^M\rangle^2=e^{\lambda/4}$.

3. Diagnose a missing measure factor

A calculation drops $\Delta(m)^2$ but retains the same coefficient $\alpha=2N/\lambda$, the normalized trace, and division by its own Gaussian partition function. What loop value does it obtain? Can agreement at $\lambda=0$, or a source-only bulk counterterm, repair the discrepancy? Make the comparison explicit at $N=3$.

Hint 1

The incorrect measure makes the $m_i$ independent real Gaussians. Every term in the normalized trace then has the same one-variable expectation.

Hint 2

Use $L_2^1(x)=3-3x+x^2/2$. A bulk multiplicative factor cancels in a normalized expectation; a new factor multiplying the insertion does not.

Solution

The incorrect answer is $W_{\rm independent}=e^{1/(4\alpha)}=e^{\lambda/(8N)}$. It has the right Abelian limit and equals one at zero coupling, but its linear correction is $\lambda/(8N)$ rather than $\lambda/8$. It lost the angular Jacobian of the actual matrix integral.

At $N=3$ the correct unconstrained answer is

$$W_{U(3)}= e^{\lambda/24}\left(1+\frac\lambda{12}+\frac{\lambda^2}{864}\right), \qquad W_{SU(3)}= e^{\lambda/36}\left(1+\frac\lambda{12}+\frac{\lambda^2}{864}\right).$$

A source-only bulk counterterm multiplies numerator and denominator by the same number, leaving either normalized result unchanged. One could artificially multiply the incorrect insertion by the missing polynomial, whose value at zero coupling is one, but that defines a different line normalization rather than repairs the missing Jacobian. The fixed supersymmetric operator and its measure must both be retained.

The Gaussian Wilson-loop laboratory tests this finite-rank answer with independent Hermitian matrices. Compare both gauge-group choices, retain their sampling covariance, and diagnose a deliberately incomplete matrix measure.

The same normalized answer determines a local response in From a circular loop to a defect observable. Derive the logarithmic coupling derivative before using it for a cusp or an accelerated probe.

What localization shares across three and four dimensions compares the two completed sphere calculations. Identify their different charge squares, surviving measures and observables, then calculate a physical coupling response that the deformation identity does not set to zero.

Paper guides

Connect this lesson to the original papers and their notation.

• Reading Pestun's four-sphere localization paper