Matrix integrals, roots, weights, and eigenvalues

If an integrand depends only on a Hermitian matrix’s eigenvalues, why not integrate those eigenvalues as independent real variables? Diagonalization also changes the volume element. The missing factor vanishes when eigenvalues coincide, and counting all eigenvalue orderings introduces another factor. We will calculate both in a two-by-two example, including the absolute normalization.

The preparation is matrix-valued matter, representations, and traces. This is finite-dimensional mathematics applicable to both three- and four-dimensional field theories; no supersymmetry algebra or curved background is needed. Our matrix integral is a defined model, not yet a derivation of any field-theory observable. All matrix entries below are dimensionless. For a dimensionful matrix, introduce a reference scale before using these formulas.

First calculation: a Gaussian before diagonalization

Let $M$ range over all two-by-two Hermitian matrices. Write

$$M=\begin{pmatrix}u&x-iy\\x+iy&v\end{pmatrix}, \qquad u,v,x,y\in\mathbb R.$$

We use the ordinary fundamental trace $\operatorname{Tr}_2$ and fix the Euclidean metric on this four-dimensional vector space:

$$ds^2=\operatorname{Tr}_2(dM\,dM) =du^2+dv^2+2dx^2+2dy^2, \qquad dV=2\,du\,dv\,dx\,dy. \tag{T16.1}$$

The factor two is the square root of the metric determinant. It would be absent in a convention defining the measure as the bare product of entries. Neither convention is wrong; an absolute partition function must specify which it uses.

For $a>0$, define

$$Z_{U(2)}(a)=\int dV\,e^{-a\operatorname{Tr}_2M^2}.$$

Since $\operatorname{Tr}_2M^2=u^2+v^2+2x^2+2y^2$, four ordinary Gaussian integrals give

$$Z_{U(2)}(a) =2\left(\sqrt{\frac{\pi}{a}}\right)^2 \left(\sqrt{\frac{\pi}{2a}}\right)^2 =\left(\frac{\pi}{a}\right)^2. \tag{T16.2}$$

The label $U(2)$ means we retain every Hermitian direction, including the identity. We have not integrated over the group $U(2)$ itself or over unitary matrices.

Now set

$$M=cI_2+\boldsymbol v\cdot\boldsymbol\sigma,\qquad c=\frac{u+v}{2},\quad \boldsymbol v=(x,y,(u-v)/2),\quad r=|\boldsymbol v|.$$

Here $\boldsymbol\sigma$ are the usual Pauli matrices. Their multiplication rule gives $(\boldsymbol v\cdot\boldsymbol\sigma)^2=r^2I_2$, so the ordered eigenvalues are

$$\lambda_+=c+r,\qquad \lambda_-=c-r,\qquad r\geq0.$$

The metric and volume become

$$ds^2=2(dc^2+d\boldsymbol v^2),\qquad dV=4\,dc\,r^2dr\,d\Omega,$$

where $d\Omega$ is the ordinary solid-angle measure, with integral $4\pi$. The direction of $\boldsymbol v$ specifies the eigenspaces. It carries no information needed by a conjugation-invariant integrand $F(M)=F(UMU^{-1})$.

The Jacobian $\left|\partial(c,r)/\partial(\lambda_+,\lambda_-)\right|=1/2$ therefore gives

$$\int dV\,F(M) =2\pi\int_{\lambda_+\geq\lambda_-} d\lambda_+\,d\lambda_-\, (\lambda_+-\lambda_-)^2F(\lambda_+,\lambda_-).$$

An invariant $F$ is symmetric in its two eigenvalues. Extending to the whole plane counts each distinct pair twice:

$$\boxed{\displaystyle \int dV\,F(M)=\frac{2\pi}{2!} \int_{\mathbb R^2}d\lambda_1\,d\lambda_2\, (\lambda_1-\lambda_2)^2F(\lambda_1,\lambda_2).} \tag{T16.3}$$

The constant $2\pi$ is fixed by our metric and angular integral; $1/2!$ has the different role of removing duplicate orderings. If we keep only the ordered region, we must not also divide by $2!$.

Check the measure, not just the ratio of observables

Define $J_0=\int_{\mathbb R}e^{-a\lambda^2}d\lambda=\sqrt{\pi/a}$ and $J_2=\int_{\mathbb R}\lambda^2e^{-a\lambda^2}d\lambda=J_0/(2a)$. The odd cross term integrates to zero, giving

$$\int_{\mathbb R^2}(\lambda_1-\lambda_2)^2 e^{-a(\lambda_1^2+\lambda_2^2)}d\lambda_1d\lambda_2 =2J_0J_2=\frac{\pi}{a^2}.$$

Multiplying by $2\pi/2!$ recovers (T16.2). Omitting the permutation factor doubles the absolute answer. Omitting the squared difference instead gives the wrong power of $a$.

A normalized expectation is $\langle\mathcal O\rangle=Z^{-1}\int dV\,\mathcal Oe^{-a\operatorname{Tr}M^2}$. Differentiating the normalization supplies two further checks:

$$\langle\operatorname{Tr}_2M^2\rangle=\frac2a,\qquad \langle(\operatorname{Tr}_2M^2)^2\rangle=\frac6{a^2}.$$

These count four real Gaussian directions. Diagonalization leaves two integration variables, but the degree-two Jacobian remembers the other two directions. The suppression of a small eigenvalue gap is a volume effect; it does not assert that the original matrix entries are fermions.

Cartan variables, roots, and weights from matrices

A Cartan subalgebra is a maximal commuting set of Hermitian generators. For $U(N)$ we can choose all real diagonal matrices $H=\operatorname{diag}(\lambda_1,\ldots,\lambda_N)$. Its dimension, the rank, is $N$. For $SU(N)$ impose $\sum_i\lambda_i=0$, leaving rank $N-1$.

In a representation, simultaneous eigenvectors of the Cartan generators transform with definite weights. A weight $\rho$ is a linear function of $H$: if $v_\rho$ is its vector, $H v_\rho=\rho(H)v_\rho$. For the fundamental, the coordinate column $e_i$ has $\rho_i(H)=\lambda_i$. For the anti-fundamental, the corresponding weights are $-\lambda_i$.

For the adjoint, the action is a commutator. Let $E_{ij}$ have a single unit entry in position $(i,j)$. Direct multiplication yields

$$[H,E_{ij}]=(\lambda_i-\lambda_j)E_{ij}. \tag{T16.4}$$

The nonzero adjoint weights are called roots, $\alpha_{ij}(H)=\lambda_i-\lambda_j$ for $i\ne j$. We use the complex basis $E_{ij}$ to diagonalize this action even though the original integration contour consists of Hermitian matrices. Hermiticity pairs the two opposite-root coefficients. Diagonal adjoint directions have zero weight; they are not additional roots.

For $SU(2)$ take

$$H=\begin{pmatrix}t&0\\0&-t\end{pmatrix} =2t\,t^3,\qquad t^3=\frac{\sigma_3}{2}.$$

The generator norm stays $\operatorname{Tr}_2(t^at^b)=\delta^{ab}/2$, as in the prerequisite lesson. The eigenvalue $t$ is half the coefficient along $t^3$. The following table shows evaluations on this $H$:

Representation or subspace	Weight values	Number of complex directions
Fundamental	$t,-t$	$2$
Adjoint root spaces	$2t,-2t$	$2$
Adjoint Cartan direction	$0$	$1$
Central direction of a $U(2)$ adjoint	$0$	$1$

Thus the usual fundamental weight numbers $\pm1/2$ are their values on $t^3$, not on $H$. A root is a linear function; its numerical value depends on the chosen Cartan coordinates.

For example, a color-space operator $K=mI+H$ on the fundamental has $\det K=(m+t)(m-t)$. On the complexified $SU(2)$ adjoint, $K=m+\operatorname{ad}_H$ has determinant $m(m+2t)(m-2t)$. These are finite color determinants, not yet determinants over all spacetime modes.

Why every pair supplies a squared difference

For a matrix with distinct eigenvalues, write $M=U\Lambda U^\dagger$, with $\Lambda=\operatorname{diag}(\lambda_i)$. Define the anti-Hermitian infinitesimal angular matrix $\omega=U^\dagger dU$. Differentiation gives

$$U^\dagger dM U=d\Lambda+[\omega,\Lambda],\qquad ds^2=\sum_i d\lambda_i^2 +2\sum_{i<j}(\lambda_i-\lambda_j)^2|\omega_{ij}|^2. \tag{T16.5}$$

The diagonal and off-diagonal pieces are orthogonal in the trace. Each complex $\omega_{ij}$ contains two real coordinates. Scaling both lengths by $|\lambda_i-\lambda_j|$ scales their area by $(\lambda_i-\lambda_j)^2$. Consequently the eigenvalue-dependent part of the measure is

$$\Delta(\lambda)^2 =\prod_{i<j}(\lambda_i-\lambda_j)^2 =\prod_{\alpha>0}\alpha(H)^2.$$

Here choosing positive roots means retaining one root of each pair, for example $i<j$. This gives a nonnegative Jacobian. A product over all signed roots instead differs by $(-1)^{N(N-1)/2}$; it should not be mistaken for a positive volume without its sign convention.

Multiplying $U$ by a diagonal unitary matrix changes the phases of the eigenvectors without changing $M$. Those redundant phases are removed from the angular integration. Permuting the eigenvectors permutes $\lambda_i$; the residual permutation group $S_N$ is the Weyl group here. Integrating over the whole Cartan therefore uses $1/N!$, or one integrates over an ordered Weyl chamber without that factor. Overall angular constants depend on the metric and group-volume convention; (T16.3) fixes one explicitly. The squared Vandermonde in a Gaussian matrix integral appears in DG v2, (A.1)–(A.4); the residual Weyl quotient is explicit in KWY v4, (3.20).

At coincident eigenvalues the stabilizer grows: rotations inside a degenerate eigenspace leave $M$ unchanged. The angular coordinates then fail to be unique and the Jacobian vanishes. For a smooth integrable density, these lower-dimensional sets have measure zero. A singular integrand needs its own convergence analysis; one cannot dismiss a singularity just by pointing to the Vandermonde.

Removing the center with the induced metric

Write $M=cI_2+X$, with $\operatorname{Tr}_2X=0$. The orthonormal central coordinate is $s_0=\operatorname{Tr}_2M/\sqrt2=\sqrt2c$. Orthogonality gives

$$dV_{U(2)}=ds_0\,dV_{SU(2)},\qquad \operatorname{Tr}_2M^2=s_0^2+\operatorname{Tr}_2X^2.$$

For $X=\boldsymbol v\cdot\boldsymbol\sigma$ the induced three-dimensional measure is $dV_{SU(2)}=2\sqrt2\,d^3\boldsymbol v$. Its eigenvalue formula and Gaussian are

$$\begin{aligned} \int dV_{SU(2)}F(X) &=\frac{2\pi}{2!}\int_{\mathbb R}\sqrt2\,dt\,(2t)^2F(t,-t),\\ Z_{SU(2)}(a)&=\left(\frac{\pi}{a}\right)^{3/2},\\ Z_{U(2)}(a)&=\sqrt{\frac{\pi}{a}}\,Z_{SU(2)}(a). \end{aligned} \tag{T16.6}$$

The $\sqrt2\,dt$ is the induced Cartan length, since $\operatorname{Tr}(dH)^2=2dt^2$. Equivalently, for an integrand restricted to the traceless hyperplane,

$$\int dV_{SU(2)}F =\sqrt2\int dV_{U(2)}\,\delta(\operatorname{Tr}_2M)F.$$

Indeed, $\operatorname{Tr}_2M=\sqrt2s_0$, so the delta function alone contributes $1/\sqrt2$. Merely inserting $\delta(\lambda_1+\lambda_2)$ without this conversion changes the normalization. This separation concerns a Lie-algebra integral; it does not assert a global group identity $U(2)=SU(2)\times U(1)$.

Reading a field-theory matrix integral

Roots describe adjoint fluctuations; weights describe matter fluctuations. If a color-diagonal fluctuation operator has eigenvalue $f(\rho(H))$, its color determinant is $\prod_\rho f(\rho(H))$, including multiplicities. A separate product over spacetime modes can then produce functions such as hyperbolic sines. The zero-weight Cartan modes require their own treatment.

This explains how to read, but not yet derive, the structure of localization formulas. KWY v4, (3.28)–(3.32) combines the Jacobian with a fluctuation determinant: the polynomial root factors are already accounted for in the final hyperbolic expression. Its adjoint determinant notation there runs over roots, with the Cartan sector treated separately. Multiplying that final expression by another Vandermonde would double count.

For two independent $U(N)$ gauge nodes, there are two permutation quotients. A bifundamental has weights $\lambda_i-\mu_j$, one eigenvalue from each node. Accordingly, FG v3, (2.1) contains $(N!)^{-2}$, differences within each node in the numerator, and differences between nodes in the matter denominator. Its variables and factors of $2\pi$ are explicitly fixed in that formula; our angular constant is not an extra factor to append to it. These sourced formulas involve additional field-theory assumptions and, for oscillatory actions, an integration prescription.

Exercises

1. A trace observable detects the missing center

For the Gaussian above, calculate $W_b=\langle\operatorname{tr}_2e^{bM}\rangle_{U(2)}$, where $b$ is real and $\operatorname{tr}_2=\operatorname{Tr}_2/2$. Compare with the same observable for traceless $X$. Check the coefficient of $b^2$ against $\langle\operatorname{Tr}M^2\rangle$.

Hint 1

The central variable and traceless matrix are independent. For eigenvalues $\pm r$ of $X$, the normalized trace is $\cosh(br)$.

Hint 2

Differentiate the Gaussian generating integral $\int_{\mathbb R}e^{-2ar^2+br}dr$ twice with respect to $b$ to generate the radial factor $r^2$.

Solution

The central average is $\langle e^{bc}\rangle=e^{b^2/(8a)}$. The normalized traceless radial average is

$$\frac{\int_0^\infty r^2e^{-2ar^2}\cosh(br)dr} {\int_0^\infty r^2e^{-2ar^2}dr} =\left(1+\frac{b^2}{4a}\right)e^{b^2/(8a)}.$$

To obtain it, use $\int_{\mathbb R}e^{-2ar^2+br}dr=\sqrt{\pi/(2a)}e^{b^2/(8a)}$. Two $b$ derivatives give its $r^2$ moment; evenness supplies the half-line cosh integral. Therefore

$$W_b^{SU(2)}=\left(1+\frac{b^2}{4a}\right)e^{b^2/(8a)},\qquad W_b^{U(2)}=\left(1+\frac{b^2}{4a}\right)e^{b^2/(4a)}.$$

Both equal one at $b=0$. The unitary answer is $1+b^2/(2a)+O(b^4)$, agreeing with $b^2\langle\operatorname{Tr}_2M^2\rangle/4$. The special-unitary answer instead has coefficient $3/(8a)$, consistent with three real directions. Dropping the center changes a normalized observable, not only an overall partition-function constant.

These finite Gaussian results are consistent with the rank-two specialization of DG v2, (A.7). Identifying an exponential matrix insertion with a particular field-theory Wilson loop is a further physical step.

2. Four bifundamental weights

Let $H_L=\operatorname{diag}(c+t,c-t)$ and $H_R=\operatorname{diag}(d+s,d-s)$. The linear operator on a two-by-two bifundamental is $\mathcal K(Y)=mY+H_LY-YH_R$. Find its four eigenvalues and determinant. Which combination of the two centers acts on $Y$? Explain why imposing tracelessness on both nodes changes the example.

Hint 1

Apply the operator to each matrix unit $E_{ij}$, regarded as a map from the right color space to the left.

Hint 2

Group the two eigenvalues with the same left index. For a common shift of both centers, subtract the two contributions before taking the determinant.

Solution

Set $h=m+c-d$. The eigenvalues on $E_{11},E_{12},E_{21},E_{22}$ are

$$h+t-s,\quad h+t+s,\quad h-t-s,\quad h-t+s.$$

Consequently

$$\det\mathcal K=\big[(h+t)^2-s^2\big]\big[(h-t)^2-s^2\big].$$

As a trace check their sum is $4h$; after subtracting $m$, the sum of squared weights is $4[(c-d)^2+t^2+s^2]$. The common center cancels, whereas the relative center $c-d$ couples. Setting both centers to zero removes that relative direction as well; it is not merely a convenient rewriting of the same integral. There are four complex representation components, although the two special-unitary nodes together have Cartan rank two.

3. A non-Gaussian test of the same measure

For $a,g>0$ replace the Gaussian density by $\exp[-a\operatorname{Tr}M^2-g(\operatorname{Tr}M^2)^2]$. Derive its $U(2)$ partition function in two ways: use polar coordinates in the four orthonormal matrix coordinates, and use (T16.3) with polar coordinates in the eigenvalue plane. Reduce both to one integral, evaluate it, and recover the Gaussian as $g\to0^+$.

Hint 1

If $R^2=\operatorname{Tr}M^2$, the four-dimensional volume is $2\pi^2R^3dR$. In the eigenvalue plane set $(\lambda_1,\lambda_2)=R(\cos\theta,\sin\theta)$.

Hint 2

The angular integral of $(\cos\theta-\sin\theta)^2$ is $2\pi$. Set $u=R^2$ and differentiate $\int_0^\infty e^{-au-gu^2}du$ with respect to $a$.

Solution

Matrix polar coordinates give

$$Z(a,g)=2\pi^2\int_0^\infty R^3e^{-aR^2-gR^4}dR =\pi^2\int_0^\infty u\,e^{-au-gu^2}du.$$

The eigenvalue formula gives the same result: its prefactor is $\pi$, its angular integral is $2\pi$, and its radial integral is $R^3dR$. Completing the square yields

$$\begin{aligned} I(a,g)&=\int_0^\infty e^{-au-gu^2}du =\frac{\sqrt\pi}{2\sqrt g}\,e^{a^2/(4g)} \operatorname{erfc}\!\left(\frac{a}{2\sqrt g}\right),\\ Z(a,g)&=-\pi^2\partial_a I =\frac{\pi^2}{2g}\big[1-aI(a,g)\big]. \end{aligned}$$

Here $\operatorname{erfc}(x)=2\pi^{-1/2}\int_x^\infty e^{-t^2}dt$. For the limiting check it is safer to use the integral than subtract two nearly equal terms: dominated convergence gives $Z(a,0)=\pi^2/a^2$. The first derivative at $g=0^+$ gives $Z(a,g)=\pi^2a^{-2}[1-6g/a^2+O(g^2)]$, in agreement with the Gaussian fourth moment. This is a small-$g$ asymptotic statement, not a claim of a convergent Taylor series.

The positive quartic term changes the density, not the diagonalization Jacobian. It also illustrates why agreement of an absolute integral is more diagnostic than a ratio in which a missing constant could cancel.

You can now take a diagonal color background, list the numbers seen by each component, and identify which factors in its integral come from coordinates. The next ingredient in a field-theory calculation is the fluctuation spectrum that supplies the remaining determinant; the finite matrix Jacobian alone does not supply it.

For a large-rank integral, saddle points and eigenvalue densities supplies another next step. Derive the continuum force from this measure, normalize its density, and test how the resulting approximation differs from a finite-rank average.