Why the matrix model becomes Gaussian

The surviving constant scalar is a matrix, but that alone does not make its probability distribution Gaussian. The physical action supplies one factor, ordinary fluctuations supply another, and the two poles can supply instanton contributions. We must evaluate all three before asserting an exact matrix model.

Start from the localization locus and its measure. The theory is massless four-dimensional $SU(N)$ N=4 SYM, $N\geq2$, on the round sphere of radius $\ell$. Its flat parent has sixteen real Poincaré and sixteen conformal supercharges. We retain its N=2 vector plus adjoint-hyper description and the selected odd charge $Q$. These choices do not reduce the full massless theory to an N=2 theory. A temporary hypermultiplet mass deformation will help us identify which cancellations require N=4 supersymmetry.

Our Hermitian absorbed fields satisfy $D=\nabla-i[\mathcal A,\cdot]$, $\operatorname{Tr}_N(t^at^b)=\delta^{ab}/2$, and the coupling $g$ is dimensionless. The scalar matrix $a$ has dimension one and is traceless.

The auxiliary field fixes the classical coefficient

On the smooth locus, $\mathcal A=\sigma=q_i=0$, $a(x)=a$ and $k_j=-w_j a$, with $\sum_jw_j^2=\ell^{-2}$. The relevant part of the physical off-shell action, after the same auxiliary contour continuation used for localization, is

$$S_{\rm cl}(a) =\frac1{g^2}\int_{S^4}\sqrt g\,d^4x\, \operatorname{Tr}_N\left[ \frac{2}{\ell^2}a^2+\sum_j k_j^2 \right] =\frac{8\pi^2\ell^2}{g^2}\operatorname{Tr}_N a^2. \tag{N14.1}$$

Indeed, the bracket is $3a^2/\ell^2$ and the sphere volume is $8\pi^2\ell^4/3$. The two terms have different origins: the first is the conformal scalar coupling; the second is the continued auxiliary action. Eliminating the auxiliary with its undeformed field equation before evaluating the localization locus would lose one third of this weight. Equation (N14.1) is the Hermitian translation of Pestun (2.33), (3.18)–(3.19).

The normalized fundamental equatorial loop uses the positive scalar coupling already fixed by its supersymmetry condition. Since the circle has length $2\pi\ell$ and all surviving matrices commute with themselves along it,

$$W_{\rm cl}(a)=\frac1N\operatorname{Tr}_N e^{2\pi\ell a}. \tag{N14.2}$$

The source writes $i\phi_0^{P,E}$ in the loop and $\phi_0^{P,E}=-ia$ on our contour, giving precisely $+a$ in the exponent. Replacing it by $ia$ would turn a real exponential into an oscillatory insertion and would describe a different observable.

What cancels in the normal fluctuations

The Gaussian ratio from the preceding lesson can be simplified by pairing modes related by $Q$. A mode with a partner contributes reciprocal factors; only the unpaired modes of the gauge-fixed complex remain. Their signed count at each rotation weight is its equivariant index. It is a count with signs and symmetry weights, not the total number of bosonic and fermionic components.

Here we use the precise index result of Pestun (4.43)–(4.46). Its derivation uses the local differential complexes at the two rotation-fixed poles. We will show how that result fixes the determinant; a proof of the full index theorem is not needed for this step.

Let $z$ temporarily record a rotation weight, rather than a spacetime coordinate. The northern vector contribution expands as $-(1+z^2)/(1-z)^2$, and the southern contribution has the corresponding expansion in $z^{-1}$. Since $(1-z)^{-2}=\sum_{j\geq0}(j+1)z^j$, their combined index, including the two constant gauge-sector entries, is

$$\begin{aligned} I_v(z) &=2-\left[1+2z+4z^2+6z^3+\cdots\right]\\ &\quad-\left[1+2z^{-1}+4z^{-2}+6z^{-3}+\cdots\right] =-2\sum_{n\geq1}n(z^n+z^{-n}). \end{aligned} \tag{N14.3}$$

The massless hypermultiplet has the opposite index, $I_h(z)=2\sum_{n\geq1}n(z^n+z^{-n})$. For example, its positive-weight expansion is $2z/(1-z)^2$. Thus the cancellation holds separately for every nonzero rotation weight. Because the hypermultiplet is adjoint, it also has the same gauge weights as the vector. A hypermultiplet in an arbitrary representation would not give this cancellation.

To see the eigenvalues rather than just their counts, temporarily introduce a real Euclidean mass parameter $m_E$, and write $x_\alpha=\ell\alpha(a)$, $\mu=\ell m_E$ for a positive root $\alpha$. Combining the two roots $\pm\alpha$ and the two rotation signs, the contribution at each positive integer $n$ is

$$\left[ \frac{(n^2+x_\alpha^2)^2} {\bigl(n^2+(x_\alpha+\mu)^2\bigr) \bigl(n^2+(x_\alpha-\mu)^2\bigr)} \right]^n . \tag{N14.4}$$

This is Pestun (4.47) written over positive roots. At $\mu=0$ it equals one before an infinite product is taken. A common regulator for the vector and hypermultiplet therefore gives $Z_{\rm 1-loop}'=1$, with the finite Vandermonde still outside that factor. The Cartan’s zero gauge weights do not create an additional eigenvalue Jacobian; their massless factors cancel in the same normalization.

For nonzero mass the product needs subtraction. A convenient source prescription uses the Barnes function,

$$\begin{aligned} H(x)&=G(1+ix)G(1-ix)\\ &=e^{(1+\gamma)x^2} \prod_{n=1}^{\infty} \left(1+\frac{x^2}{n^2}\right)^n e^{-x^2/n},\\ R(x,\mu)&=\frac{H(x)^2}{H(x+\mu)H(x-\mu)}. \end{aligned} \tag{N14.5}$$

Here $\gamma$ is Euler’s constant; the displayed product defines the needed combination of $G$. Our $H(x)$ is Pestun’s $H_P(ix)$. Terms independent of $a$, including mass-dependent normalization and zero-weight factors, cancel from normalized loop expectation values. Fixing an absolute sphere partition function additionally fixes the allowed background counterterms and measure conventions. None of this permits an arbitrary $a$-dependent modification of the answer. For the physical massless theory the cancellation is already present in (N14.4).

The mass in the instanton formula is shifted

There are three useful mass symbols. $m_P$ is the analytic hypermultiplet mass used in Pestun’s transformation laws, $m_E$ is real on the stated Euclidean contour, and $m_{\rm eq}$ is the parameter in the instanton fixed-point formula:

$$m_P=i m_E,\qquad m_{\rm eq}=m_P+\frac{\epsilon_1+\epsilon_2}{2},\qquad \epsilon_1=\epsilon_2=\ell^{-1}\quad\hbox{on the round sphere}. \tag{N14.6}$$

The massless N=4 limit is $m_P=m_E=0$, hence $m_{\rm eq}=\ell^{-1}$, not zero. This is a relation in the specified Euclidean source prescription, not a rule identifying a positive Lorentzian particle mass with an arbitrary complex parameter.

Why is there an additive shift? The rotation defining the local instanton problem is accompanied by an R-symmetry rotation that keeps $Q$ fixed. In its fluctuation complex the two hyper directions carry weights $\epsilon_3=-m_{\rm eq}$ and $\epsilon_4=m_{\rm eq}-\epsilon_+$, with $\epsilon_+=\epsilon_1+\epsilon_2$. Their R-symmetry combination is $-\epsilon_3-\epsilon_4=\epsilon_+$, as required to cancel the appropriate spacetime spin rotation. Their flavor combination is $\epsilon_4-\epsilon_3=2m_{\rm eq}-\epsilon_+$. The physical flavor doublet has eigenvalues $\pm m_P$, so the same Cartan parameter is $2m_P$. Equating these two descriptions derives (N14.6). These weights and their interpretation are established in Okuda–Pestun (4)–(7); Pestun v3 (5.11), footnote 9 incorporates the correction explicitly.

Why every nonempty instanton sector vanishes

We can see the cancellation concretely in the $U(N)$ instanton counting prescription. A partition $Y_\alpha$ is a diagram of boxes whose row lengths decrease; there is one diagram for each color. The total number of boxes $|\vec Y|$ is the instanton number in this local counting problem. This labeling and the fixed-point weight below are the sourced instanton-localization input, not a claim that ordinary smooth Abelian instantons exist on $S^4$.

For a box $s$ in $Y_\alpha$, let $h_{Y_\beta}(s)$ be the horizontal distance to the edge of $Y_\beta$ in its row, possibly negative when the box is outside $Y_\beta$. Let $v_{Y_\alpha}(s)$ be the vertical distance to the lower edge of its own column. With complex Coulomb inputs $\mathfrak a_\alpha$—on the sphere contour $\mathfrak a_\alpha=i a_\alpha$—define

$$E_{\alpha\beta}(s) =-\epsilon_1h_{Y_\beta}(s) +\epsilon_2\bigl(v_{Y_\alpha}(s)+1\bigr) +\mathfrak a_\beta-\mathfrak a_\alpha.$$

The regulated local factor is

$$Z_{\rm pole} =\sum_{\vec Y}\mathfrak q^{|\vec Y|} \prod_{\alpha,\beta=1}^N\prod_{s\in Y_\alpha} \frac{(E_{\alpha\beta}(s)-m_{\rm eq}) (\epsilon_+-E_{\alpha\beta}(s)-m_{\rm eq})} {E_{\alpha\beta}(s)(\epsilon_+-E_{\alpha\beta}(s))}. \tag{N14.7}$$

The denominator comes from tangent fluctuations of the instanton moduli; the adjoint hyper supplies the shifted numerator. This is Pestun (5.1)–(5.4), (5.11).

Every nonempty diagram has a removable corner: a box with neither a box to its right nor one below. For that box and $\beta=\alpha$, $h=v=0$ and $E_{\alpha\alpha}=\epsilon_2$. Its factor is

$$\frac{(\epsilon_2-m_{\rm eq})(\epsilon_1-m_{\rm eq})} {\epsilon_1\epsilon_2}. \tag{N14.8}$$

It vanishes at $m_{\rm eq}=\epsilon_1$ or $\epsilon_2$. Work first at generic distinct nonzero $\epsilon_i$ and generic Coulomb inputs so that the other fixed-point denominators are nonsingular. Every nonempty contribution then vanishes; the empty diagrams contribute one. The physical round limit has $m_{\rm eq}=\epsilon_1=\epsilon_2=\ell^{-1}$, so $Z_N=Z_S=1$ in this prescription.

The order of operations matters. Equal rotation frequencies can make individual fixed-point denominators zero. Define each coefficient by its regulated sum and analytic continuation; do not substitute equal frequencies into an unresolved $0/0$. The corner argument can be applied at $m_{\rm eq}=\epsilon_1$ before taking the round limit. This unequal-frequency intermediate step regulates the local pole formula; it is not a claim about a different global sphere background. It is consistent with the continued full coefficients.

For $SU(N)$, the conclusion is not obtained by declaring the $U(N)$ center irrelevant. The independent physical argument identifies right-handed supercharges with equivariant weights $\pm(m_{\rm eq}-\epsilon_1)$ and $\pm(m_{\rm eq}-\epsilon_2)$. The corresponding extra supersymmetries are restored when these weights vanish. They are broken by an anti-self-dual instanton configuration and generate fermionic zero modes; the opposite pole has the conjugate argument. With no insertion to absorb those modes, the polar contribution vanishes. Okuda–Pestun, pp. 5–6, gives this gauge-group-independent argument and the concrete diagram check. It applies to the partition function and the selected circular loop; it is not a statement that every observable in N=4 SYM has no instanton effects.

We keep the predecessor’s north/south labels and its translation $k_P=-k_{\rm site}$, $\theta_P=-\theta_{\rm site}$. Thus $\mathfrak q$ in (N14.7) is the appropriate local source fugacity, with the conjugate factor at the other pole. Since both nonempty-sector coefficients vanish here, neither factor leaves theta dependence. The moduli-space prescription must preserve the chosen supersymmetry and include the small-instanton resolution used in the source; a generic hard cutoff is not interchangeable with it.

A one-instanton test of the mass distinction

For $U(2)$ there are two one-box configurations. Set $d=\mathfrak a_1-\mathfrak a_2$. Each contribution contains (N14.8); adding their remaining cross-color factors gives

$$Z_1= \frac{2(m_{\rm eq}-\epsilon_1)(m_{\rm eq}-\epsilon_2) \bigl[m_{\rm eq}^2-d^2-m_{\rm eq}\epsilon_++\epsilon_+^2\bigr]} {\epsilon_1\epsilon_2(\epsilon_+^2-d^2)}. \tag{N14.9}$$

For example, the contribution with the box in the first color is (N14.8) times $(\epsilon_+-d-m_{\rm eq})(d-m_{\rm eq})/[(\epsilon_+-d)d]$; the second is obtained by $d\mapsto-d$. This derives (N14.9), which also agrees with Pestun (5.12).

At the physical round massless point it is zero. If instead $m_{\rm eq}=0$, every numerator in (N14.7) cancels its denominator and $Z_1=2$. More generally one obtains the colored-partition generating function $\prod_{j\geq1}(1-\mathfrak q^j)^{-N}$, not one. This is the different analytic mass choice $m_P=-\epsilon_+/2$. It is precisely the trap avoided by (N14.6).

Assemble the Gaussian model

The two independently justified cancellations remove the normal fluctuation and polar factors. They do not remove the finite eigenvalue Jacobian:

$$\langle W\rangle_{SU(N)} = \frac{\displaystyle\int_{\sum a_i=0}\! da\, \Delta(a)^2 e^{-\frac{8\pi^2\ell^2}{g^2}\sum_i a_i^2} \frac1N\sum_i e^{2\pi\ell a_i}} {\displaystyle\int_{\sum a_i=0}\! da\, \Delta(a)^2 e^{-\frac{8\pi^2\ell^2}{g^2}\sum_i a_i^2}}, \qquad \Delta(a)=\prod_{i<j}(a_i-a_j). \tag{N14.10}$$

The measure on the traceless hyperplane is induced by $\operatorname{Tr}(da)^2$; identical constants and Weyl factors cancel in the ratio. Introduce the dimensionless Hermitian matrix $M=2\pi\ell a$ and $\lambda=g^2N$. Then

$$\boxed{\displaystyle \langle W\rangle_{SU(N)} =\frac{\int_{\operatorname{Tr}M=0}dM\, e^{-\frac{2N}{\lambda}\operatorname{Tr}M^2} \frac1N\operatorname{Tr}e^M} {\int_{\operatorname{Tr}M=0}dM\, e^{-\frac{2N}{\lambda}\operatorname{Tr}M^2}}.} \tag{N14.11}$$

The explicit radius drops out of this normalized conformal observable. For $U(N)$ the same expression integrates all Hermitian matrices, including an independent central Gaussian. That central mode affects the loop and must not be discarded. Evaluating the finite-rank integral and its central-mode relation is the next calculation.

The Gaussian expression was proposed in Drukker–Gross (2.25). Their preceding discussion explicitly left an all-order interaction argument open. The sphere derivation supplies the additional localization and instanton input; it is not a deduction from a few perturbative diagrams.

Exercises

1. The wrong zero-mass limit

At $m_{\rm eq}=0$, enumerate the $U(2)$ diagrams through two boxes and compute $Z_{\rm pole}$ through $\mathfrak q^2$. Compare with the physical massless round-sphere answer. What value of $m_P$ does the first limit actually describe, and is it on the stated real-$m_E$ contour?

Hint 1

Each diagram has weight one at zero equivariant mass. Distinguish the color of a one-box diagram and the two possible shapes of a two-box diagram.

Hint 2

Use $m_P=m_{\rm eq}-\ell^{-1}$ and $m_P=i m_E$ on the round sphere.

Solution

There is one empty pair and two one-box pairs. At size two, both boxes can occupy either color with shape $(2)$ or $(1,1)$, giving four possibilities; one box in each color gives a fifth. Hence $Z_{\rm pole}=1+2\mathfrak q+5\mathfrak q^2+O(\mathfrak q^3)$. This also follows by expanding $(1-\mathfrak q)^{-2}(1-\mathfrak q^2)^{-2}$ to this order.

Zero equivariant mass means $m_P=-1/\ell$, or $m_E=i/\ell$. It is an analytic continuation away from the stated real-$m_E$ contour, not the physical N=4 massless point. The latter has $m_P=m_E=0$, $m_{\rm eq}=1/\ell$, and all positive-box coefficients vanish. A nontrivial partition-generating function obtained by setting the wrong parameter to zero does not contradict the Gaussian result.

2. A mass deformation really changes the integrand

For one positive-root pair, remove the $a$-independent normalization by defining $L(x,\mu)=\log[R(x,\mu)/R(0,\mu)]$ from (N14.5). Derive an absolutely convergent integral for $L$ using $\log(1+y^2/n^2)=2\int_0^\infty dt\,e^{-nt}(1-\cos yt)/t$. Show that $L>0$ for nonzero real $x,\mu$, and find its leading term when both are small.

Hint 1

In the combination $2\log H(x)-\log H(x+\mu)-\log H(x-\mu)+2\log H(\mu)$, all quadratic subtraction terms cancel.

Hint 2

Use $\sum_{n\geq1}ne^{-nt}=e^{-t}/(1-e^{-t})^2$ and $\cos((x+\mu)t)+\cos((x-\mu)t)=2\cos xt\cos\mu t$.

Solution

The four logarithms combine to

$$L(x,\mu)=4\int_0^\infty \frac{dt}{t}\frac{e^{-t}}{(1-e^{-t})^2} (1-\cos xt)(1-\cos\mu t). \tag{N14.12}$$

At the origin its integrand is $O(t)$; at infinity it is exponentially bounded. It is nonnegative and positive on open sets when $x\mu\ne0$, so the integral is strictly positive. The leading expansion is

$$L(x,\mu) =6\zeta(3)x^2\mu^2 -10\zeta(5)x^2\mu^2(x^2+\mu^2) +O\bigl((|x|+|\mu|)^8\bigr).$$

For the first term, expand the cosines and use $\int_0^\infty t^3 e^{-t}(1-e^{-t})^{-2}dt =6\sum_{n\geq1}n^{-3}$. The next term follows in the same way from the sixth-degree expansion. For example, $L(0.7,0.4)=0.2948669558615543181\ldots$, also obtained from the Barnes functions. The nonconstant $x$ dependence demonstrates an actual change of the eigenvalue weight; its higher powers cannot be absorbed into a single Gaussian coupling. At $\mu=0$ the integrand vanishes identically, as required.

3. Catch an omitted auxiliary contribution

Suppose someone keeps only $2a^2/\ell^2$ in (N14.1). Find the resulting wrong coefficient of $\operatorname{Tr}M^2$. Then compute the first nontrivial term in the normalized $SU(N)$ loop for the correct and wrong weights, using $M=M^at^a$ and $C_F=(N^2-1)/(2N)$.

Hint 1

The wrong classical action is two thirds of the correct action. With $\operatorname{Tr}(t^at^b)=\delta^{ab}/2$, each real matrix component has an ordinary Gaussian variance.

Hint 2

Expand $N^{-1}\operatorname{Tr}e^M$ through $M^2$. Odd moments vanish, and there are $N^2-1$ traceless components.

Solution

The wrong weight is $\exp[-4\operatorname{Tr}M^2/(3g^2)]$ instead of $\exp[-2\operatorname{Tr}M^2/g^2]$. For the correct weight, the component action is $\sum_a(M^a)^2/g^2$, so $\langle M^aM^b\rangle=g^2\delta^{ab}/2$. Consequently

$$\langle W\rangle =1+\frac{1}{2N}\langle\operatorname{Tr}M^2\rangle+O(g^4) =1+\frac{g^2(N^2-1)}{8N}+O(g^4) =1+\frac{g^2C_F}{4}+O(g^4).$$

The wrong component variance is $3g^2/4$, giving $1+3g^2C_F/8+O(g^4)$. The error is already visible in the first perturbative term of the four-dimensional circle. Matching only the field content and the word “Gaussian” would not catch it; the auxiliary contour and trace normalization fix the coefficient.

The remaining calculation is now finite-dimensional: evaluate the circular Wilson loop at finite N with the specified gauge group and normalization. The reason it is Gaussian has already been established, separately for the classical action, paired normal modes, and the correctly normalized polar sectors.

The Pestun paper companion maps these ingredients back to the corrected source. Its two-box example shows explicitly why a complete instanton coefficient must be formed before equal rotation frequencies are imposed.

Paper guides

Connect this lesson to the original papers and their notation.

• Reading Pestun's four-sphere localization paper