Derive the ABJM matrix integral

The sphere action has two matrix-valued gauge multiplets and four matrix-valued matter multiplets. Localization replaces their field integrals by an integral over two lists of real numbers. The main work is to determine what survives, which fluctuations have already been integrated out, and how often the remaining configurations are counted.

Use the complete sphere action, the first gauge-theory matrix integral, and roots, weights and eigenvalue measures as preparation. We will derive an absolute partition function in a specified normalization, rather than suppress factors that would cancel only in an expectation value.

Our theory is three-dimensional compact $U(N)_k\times U(N)_{-k}$, with equal finite $N\geq1$, integer $k>0$, ordinary fundamental traces, and no extra common-center quotient. There are no real masses or FI terms. The full generic theory has twelve real Poincaré supercharges, enhanced to sixteen at levels one and two; its $\mathcal N=2$ presentation displays four. The calculation uses one localizing supercharge in the four-generator rigid sphere subalgebra. Gauge rank is $N$ per node.

Find the locus before substituting the action

Keep the radius $\ell$, with $A=V=C=0$, $H=-i/\ell$ and the paper’s $R=-6/\ell^2$. Add a positive supersymmetric Yang–Mills deformation for each gauge factor. Its positive trace is a sum of the two ordinary traces, even though the physical Chern–Simons action has opposite levels. Its coefficient is a deformation parameter, not a new defining Yang–Mills coupling of ABJM.

In the inherited variables its bosonic zeros obey

$$f_{\mu\nu}=0,\qquad \mathcal D_\mu\sigma=0,\qquad D+\sigma H=0, \tag{AB12.1}$$

and similarly at the other node. The positivity argument, ghost treatment and matter deformation are the localization inputs from KWY §§3.1–3.4, with the auxiliary continuation already fixed in our gauge-integral prerequisite.

Every smooth flat connection on the simply connected sphere is gauge equivalent to zero. The $U(N)$ bundles here are also trivial; there is no magnetic-flux sum without defects. The remaining configurations can be written

$$\begin{aligned} a=\widehat a&=0,& \sigma&=\ell^{-1}\operatorname{diag}(s_1,\ldots,s_N),& D&=i\sigma/\ell,\\ &&\widehat\sigma&=\ell^{-1}\operatorname{diag}(t_1,\ldots,t_N),& \widehat D&=i\widehat\sigma/\ell . \end{aligned} \tag{AB12.2}$$

Here $s_i,t_j\in\mathbb R$ are dimensionless. They are eigenvalues of Lie-algebra scalars, not periodic holonomies. The auxiliary fluctuation runs along $D=i\sigma/\ell+ih$, with $h$ Hermitian, so $-\operatorname{Tr}(D+\sigma H)^2=\operatorname{Tr}h^2$.

The matter deformation at R-charge $1/2$ has no scalar zero modes on this real scalar contour. Thus the matter fields and their auxiliaries vanish on the locus. The original quartic superpotential has no quadratic term there. Its interactions therefore do not enter the Gaussian fluctuation determinant; their coefficient and the R assignments still define which superconformal theory we are computing. This is not permission to change the superpotential while declaring the whole theory unchanged.

In particular, do not first use the undeformed algebraic equation $\sigma=(2\pi/k)\mu_L$ and then put matter to zero. That would incorrectly remove the Cartan integral. The auxiliary equation in the deformed functional integral contains the deformation, and its limiting zero is (AB12.1).

Calculate the classical phase directly

The connection three-forms vanish at (AB12.2). The nonzero contribution from the first node is its $-2\sigma D$ term:

$$\begin{aligned} S_{{\rm cl},L} &=\frac{k}{4\pi}(2\pi^2\ell^3) \operatorname{Tr}\!\left(-\frac{2i}{\ell}\sigma^2\right) =-i\pi k\sum_i s_i^2,\\ S_{{\rm cl},R}&=+i\pi k\sum_j t_j^2. \end{aligned} \tag{AB12.3}$$

Consequently the weight $e^{-S_E}$ is

$$e^{-S_{\rm cl}}= \exp\!\left[i\pi k\left(\sum_i s_i^2-\sum_j t_j^2\right)\right]. \tag{AB12.4}$$

All powers of $\ell$ have canceled. There is no extra $H\sigma^2$ term and no real Gaussian from the Yang–Mills deformation, which vanishes on the locus.

KWY absorb $k/(4\pi)$ into their “Tr”; their classical exponent (3.7) has the opposite sign at matched real scalar and level. To compare their formula with ours, take $k_{\rm KWY}=-k$, or exchange their two node eigenvalue lists. This translation concerns the localized phase. It does not identify every off-shell auxiliary or fermion in the two conventions.

Normalize the gauge measure

An infinitesimal off-diagonal gauge transformation changes a diagonal scalar by $\delta\sigma_{ij}\propto(s_i-s_j)\omega_{ij}$. Its two real directions give a squared difference. Thus diagonalization supplies

$$\Delta(s)^2=\prod_{i<j}(s_i-s_j)^2.$$

There are $N!$ ways of ordering a generic list of eigenvalues at each node. Integrating over both full real Cartan spaces therefore requires $1/(N!)^2$. The two permutations are independent. Exchanging the nodes is not an additional gauge identification.

The nonconstant gauge, ghost and gaugino modes complete this ordinary Vandermonde into a hyperbolic one. It is useful to separate the shape from its normalization. For one positive root, $x=s_i-s_j$, the paired nonzero-mode product has relative factor

$$\prod_{n=1}^\infty\left(1+\frac{x^2}{n^2}\right)^2 =\left(\frac{\sinh\pi x}{\pi x}\right)^2. \tag{AB12.5}$$

The remaining constant is fixed together with the gauge-volume and Gaussian normalization. In the standard normalization of KWY (1.1), (3.20) and (3.68), the completed factor per positive root is

$$x^2(2\pi)^2 \left(\frac{\sinh\pi x}{\pi x}\right)^2 =[2\sinh\pi x]^2. \tag{AB12.6}$$

The constant has a spectral origin: the corresponding paired integer-mode constant is $\prod_{n\geq1}^{\zeta}n^4=(2\pi)^2$. Here the superscript means zeta regularization: $\prod^\zeta n=\exp[-\zeta_{\rm R}'(0)]=\sqrt{2\pi}$, where $\zeta_{\rm R}(z)$ continues $\sum_{n\geq1}n^{-z}$. One may allocate constants differently between a determinant and the gauge-volume measure; the completed normalized measure (AB12.6) is what enters the answer.

This is not the unnormalized Euclidean volume of all Hermitian matrix entries in the matrix-tools lesson. There the angular volume remains in the integral; here gauge orbits are divided out. We use the inherited unit-charge Cartan measure $\prod_i ds_i$, extended independently to both nodes. No finite-matrix angular factor is to be multiplied into it afterward. The center remains a genuine integration direction.

In particular the two vector factors are

$$Z_{\rm vec}(s)Z_{\rm vec}(t)= \prod_{i<j}[2\sinh\pi(s_i-s_j)]^2 \prod_{i<j}[2\sinh\pi(t_i-t_j)]^2. \tag{AB12.7}$$

Zero adjoint weights are not roots. Inserting them into a hyperbolic product would set the answer to zero. Their constant scalar modes have already become the integration variables.

Count the bifundamental weights

For the matrix entry $(A_a)_{ij}$, $a=1,2$, the vector scalar acts by $\sigma A_a-A_a\widehat\sigma$. Its dimensionless mass is $s_i-t_j$. The entry $(B_a)_{ji}$ has the opposite mass. Both scalars have R-charge $1/2$.

For one conjugate pair the fixed determinant prescription gives

$$Z_\chi\!\left(\frac12,x\right) Z_\chi\!\left(\frac12,-x\right) =\frac1{2\cosh\pi x}. \tag{AB12.8}$$

For example, using the prerequisite’s $Z_\chi(r,x)=\exp[\mathcal L(1-r-ix)]$, differentiation of the logarithm of the left side gives $-\pi\tanh\pi x$. Its value at $x=0$ is $1/2$, fixing the constant as well as the derivative.

There are two conjugate pairs at each of the $N^2$ color pairs, so

$$Z_{\rm mat}(s,t)= \prod_{i,j=1}^N[2\cosh\pi(s_i-t_j)]^{-2}. \tag{AB12.9}$$

A single chiral does not supply a whole inverse hyperbolic cosine. Counting all four chirals as four conjugate pairs would double the exponent. The opposite-weight determinant phases cancel in the symmetric prescription used here; we append no extra background Chern–Simons phase. The integer $k$ is the level of this supersymmetric regulated theory, not a bosonic level to which another shift should now be added.

Assemble the integral and translate its variables

Combining (AB12.4), (AB12.7), (AB12.9) and the gauge quotient gives

$$\boxed{\begin{aligned} Z(N,k)&=\frac1{(N!)^2} \int_{\mathbb R^{2N}}\prod_{i=1}^N ds_i\,dt_i\, e^{\,i\pi k\sum_i(s_i^2-t_i^2)}\\ &\qquad\times\frac{ \prod_{i<j}[2\sinh\pi(s_i-s_j)]^2 [2\sinh\pi(t_i-t_j)]^2 }{ \prod_{i,j}[2\cosh\pi(s_i-t_j)]^2 }.\end{aligned}} \tag{AB12.10}$$

The contour and limiting prescription are specified below. This is an absolute partition function in the stated scheme, not an expectation value normalized to one. The empty-rank convention is $Z(0,k)=1$.

A commonly used set of eigenvalues is $\mu_i=2\pi s_i$, $\nu_i=2\pi t_i$. Then

$$\begin{aligned} Z(N,k) ={}&\frac1{(N!)^2} \int\prod_{i=1}^N\frac{d\mu_i}{2\pi}\frac{d\nu_i}{2\pi}\, e^{\,\frac{ik}{4\pi}\sum_i(\mu_i^2-\nu_i^2)}\\ &\quad\times \frac{ \prod_{i<j}[2\sinh((\mu_i-\mu_j)/2)]^2 [2\sinh((\nu_i-\nu_j)/2)]^2 }{ \prod_{i,j}[2\cosh((\mu_i-\nu_j)/2)]^2 }. \end{aligned} \tag{AB12.11}$$

Each $1/(2\pi)$ comes from a variable change. There is no new integration over a periodic angle.

The weight assignments and hyperbolic powers agree with KWY (4.2)–(4.4) after the phase translation above. For absolute normalization retain the Weyl factor explicit in their general (3.68), even though it is suppressed in the specialized (4.4). Their product over both signs of each root differs from positive-root squares by a sign at each node; the two signs cancel at equal rank. Their subsequent (4.5) explicitly drops constants and should not be used to normalize $Z$. The v4 introduction’s footnote 1 records the restored factors of two in the determinants.

Give the oscillatory expression a meaning

On the real contour the hyperbolic factors have no poles, but the integral is not absolutely convergent. A simultaneous real translation $s_i\mapsto s_i+c$, $t_j\mapsto t_j+c$ leaves its absolute value unchanged. Even the rank-one integrand has an undamped common-center direction.

A useful Abel prescription is

$$Z(N,k)=\lim_{\epsilon\downarrow0}Z_\epsilon(N,k). \tag{AB12.12}$$

Here $Z_\epsilon$ is (AB12.10) with the extra factor $e^{-\epsilon\sum_i(s_i^2+t_i^2)}$ in its integrand. At each $\epsilon>0$ the Gaussian dominates the at-most-exponential hyperbolic growth, so ordinary changes of variables and iterated integration are justified. The limit is the oscillatory prescription, not a proof of absolute convergence. Equivalently one approaches the two quadratic phases with the damping appropriate to their opposite signs. This regulator is for the remaining finite integral; it is not the vanishing value of the supersymmetric Yang–Mills deformation. Nor should one independently rotate both eigenvalue contours without checking the intervening hyperbolic poles.

The two node exchange maps the integrand to its complex conjugate and preserves this regulator. Hence $Z_\epsilon(N,k)$ is real and even in $k$. Reality does not make the integrand a positive probability density.

Worked check at finite regulator

At rank one the root products and Weyl factors are absent. Set $x=s-t$, $y=(s+t)/2$, whose absolute Jacobian is one. Then $s^2-t^2=2xy$ and $s^2+t^2=2y^2+x^2/2$. The regulated $y$ integral is an ordinary Fourier transform of a Gaussian:

$$Z_\epsilon(1,k)= \sqrt{\frac{\pi}{2\epsilon}} \int_{\mathbb R} \frac{\exp[-(\epsilon^2+\pi^2k^2)x^2/(2\epsilon)]} {4\cosh^2\pi x}\,dx. \tag{AB12.13}$$

This is a positive, convergent one-dimensional check on an initially complex two-dimensional integral. At $k=1,\epsilon=1$ it is approximately $0.1484589215$; reducing the regulator is a separate limit, not justification for replacing an undamped integral by an ordinary Gaussian. The first exact partition-function checks takes that limit and obtains the exact rank-one partition function.

There is also a useful normalization bound at finite regulator. Put $C_\epsilon=\epsilon^2+\pi^2k^2$. Since $0\leq1-\operatorname{sech}^2u=\tanh^2u\leq u^2$,

$$0\leq \frac{\pi}{4\sqrt{C_\epsilon}}-Z_\epsilon(1,k) \leq\frac{\pi^3\epsilon}{4C_\epsilon^{3/2}}. \tag{AB12.14}$$

Both coefficients follow from zeroth and second Gaussian moments. A missing factor four in the matter determinant or a missing Jacobian fails this finite-regulator test before any delicate limit is taken.

Exercises

1. Resolve a two-eigenvalue collision

For $N=2$, set $s=(u+\delta,u-\delta)$ and $t=(v+\eta,v-\eta)$. Derive the complete transformed measure and phase, and the leading hyperbolic density as $\delta,\eta\to0$ at fixed $u-v$. Explain why no additional ordinary Vandermonde should be inserted.

Hint 1

Each two-eigenvalue change has absolute Jacobian two. The original Weyl factor is $1/4$.

Hint 2

There is one positive root at each node and four color pairs in the matter product. Use $\sinh w=w+O(w^3)$ only after counting those factors.

Solution

The Jacobian four cancels $1/(2!)^2$, so the measure is $du\,dv\,d\delta\,d\eta$ over all four real variables. The phase is

$$e^{\,2\pi ik(u^2+\delta^2-v^2-\eta^2)}.$$

The exact hyperbolic ratio is

$$\frac{[2\sinh(2\pi\delta)]^2[2\sinh(2\pi\eta)]^2} {\displaystyle\prod_{a,b=\pm1} [2\cosh\pi(u-v+a\delta-b\eta)]^2}.$$

Its leading term is $256\pi^4\delta^2\eta^2/[2\cosh\pi(u-v)]^8$. The two independent eigenvalue collisions each have a quadratic zero; the matter denominator is smooth and nonzero. The ordinary Vandermonde has already combined with the nonconstant determinant in (AB12.6). Adding it again would incorrectly make each collision quartic. Alternatively one could restrict both gaps to be nonnegative and restore the corresponding ordered-domain measure, but must not retain both prescriptions at once.

2. Does the common center disappear?

Under $s_i\mapsto s_i+c$, $t_i\mapsto t_i+c$, calculate the change of the classical exponent. Does matter neutrality under the diagonal central gauge transformation allow both trace directions to be deleted? Also prove reality of the regulated answer without assuming a positive measure.

Hint 1

The $Nc^2$ terms cancel, but the terms linear in $c$ generally do not.

Hint 2

A change of integration variables is not a new gauge quotient. For reality, exchange the two whole lists rather than discarding either one.

Solution

The exponent changes by $2\pi ikc(\sum_i s_i-\sum_j t_j)$. The common center couples to the relative trace through the Chern–Simons quadratic form. Matter neutrality does not remove this coupling or its integral. Imposing both trace-zero conditions would change $U(N)\times U(N)$ to a different theory and requires a separately specified global group and measure.

At rank one both imposed conditions would simply remove both integrations and leave the matter factor $1/4$, with no information about $k$. The actual rank-one expression retains the two scalar integrals in (AB12.13).

For $\epsilon>0$ the hyperbolic ratio and damping are real and symmetric under $(s,t)\leftrightarrow(t,s)$, while the phase changes sign. Absolute convergence therefore gives $\overline{Z_\epsilon}=Z_\epsilon$ and $Z_\epsilon(-k)=Z_\epsilon(k)$. The same statements survive whenever the chosen Abel limit exists. They prove neither positivity nor invariance under arbitrary changes of global gauge data.

3. Test the complex Gaussian normalization

As an independent integration test, replace the rank-one matter factor by $e^{-b(s-t)^2}$, where $b\geq0$. This defines a test integral, not a changed supersymmetric matter theory. For $\epsilon>0$, evaluate

$$I_{\epsilon,b}(k)= \int_{\mathbb R^2}ds\,dt\, e^{-\epsilon(s^2+t^2)+i\pi k(s^2-t^2)-b(s-t)^2}.$$

Check it by two different Gaussian reductions. What happens at $k=0$ if the regulator is removed?

Hint 1

The complex symmetric quadratic matrix has diagonal entries $\epsilon+b-i\pi k$, $\epsilon+b+i\pi k$ and off-diagonal entries $-b$.

Hint 2

Its real part is positive definite. Alternatively use $x=s-t$, $y=(s+t)/2$ as in the worked example.

Solution

The determinant is $(\epsilon+b)^2+\pi^2k^2-b^2 =\epsilon^2+2b\epsilon+\pi^2k^2$. Continuous continuation from the positive real Gaussian fixes the square-root branch, so

$$I_{\epsilon,b}(k)= \frac{\pi}{\sqrt{\epsilon^2+2b\epsilon+\pi^2k^2}}.$$

In the second method, integrate $e^{-2\epsilon y^2+2\pi ikxy}$ first. The remaining $x$ coefficient is $b+\epsilon/2+\pi^2k^2/(2\epsilon)$. Multiplying the two Gaussian normalizations gives the same answer, including its absolute factor.

At $k=0$ the result diverges like $\pi/\sqrt{2b\epsilon}$ for $b>0$, or $\pi/\epsilon$ for $b=0$. The first divergence is the unsuppressed common center; the second additionally lacks any relative-direction damping. Thus a convergent relative factor alone does not regularize the whole integral. The nonzero physical level and its oscillatory prescription matter.

The matrix integral now contains no hidden color multiplicity or measure factor. Its first use is an exact finite-rank partition-function check; operator insertions and large-rank approximations require additional arguments of their own.

The original localization companion follows the measure and insertion through the source, including its displayed phase conventions.

Paper guides

Connect this lesson to the original papers and their notation.

• Reading three-dimensional localization with matter