The free chiral determinant

The constant chiral multiplet in the localization lesson contributed $3$ to its Gaussian integral. What happens when we include every harmonic on the sphere? Bosonic and fermionic factors cancel extensively, but their remaining infinite product still needs a definition.

We will derive that product from the round-sphere chiral action, then evaluate it in a specified regularization. The theory is three-dimensional $\mathcal N=2$, with four real supercharges before Euclidean complexification, one free chiral multiplet, and no dynamical gauge field or superpotential. The background preserves two complex spinor directions in each family. These are the complexified supersymmetries, not eight additional real charges. Because the action is quadratic, its Gaussian, often called its one-loop determinant, is the entire free matter integral in the chosen prescription.

Specify the operators and the contour

Let $\ell>0$ be the sphere radius. Keep $H=-i/\ell$, $A=V=C=0$, and the paper’s curvature trace $R=-6/\ell^2$. In the global orthonormal frame,

$$\nabla_a\psi=E_a\psi+\frac{i}{2\ell}\gamma_a\psi, \qquad [E_a,E_b]=-\frac2\ell\epsilon_{abc}E_c, \qquad \gamma_a\gamma_b=\delta_{ab}+i\epsilon_{abc}\gamma_c.$$

The scalar, spinor and auxiliary fields have engineering dimensions $1/2,1,3/2$, and R-charges $r,r-1,r-2$. Their independent tilded partners have opposite R-charges. Here $r$ specifies a conserved R-current used to couple the theory to the background. Its free superconformal value is $1/2$; the calculation also allows other real $r$.

We can include a real mass without adding interactions. For a flavor charge $q_F$, freeze a flavor vector multiplet at

$$a_F=0,\qquad \sigma_F=\mu,\qquad D_F=-H\mu=\frac{i\mu}{\ell},\qquad \lambda_F=\widetilde\lambda_F=0 .$$

These values preserve the sphere supersymmetries. Define

$$M=z-q_F\mu,\qquad m=-M=q_F\mu-z,\qquad x=\ell m. \tag{O03.1}$$

The central charge $z$, $\mu$ and signed physical mass $m$ have mass dimension one; $x$ is dimensionless. One may take $z=0,q_F=1$, so $m=\mu$. Equivalently, the source paper describes a gauge-neutral real mass by $z=-m$. This is a background mass, not a complex coefficient in a superpotential. P, (7.11)

Substituting the sources into P, (6.32), the term $q_F(D_F+\mu H)$ vanishes. After integrating the scalar kinetic term by parts on the closed sphere,

$$\begin{aligned} S={}&\int\sqrt g\,d^3x\, \left(\widetilde\phi\,\mathcal K_B\phi+ \eta\,\mathcal K_F\psi-\widetilde F_\Phi F_\Phi\right),\\ \mathcal K_B={}&-\nabla^2+m^2+ \frac{2i(1-r)m}{\ell}+\frac{r(2-r)}{\ell^2},\\ \mathcal K_F={}&-i\gamma^\mu\nabla_\mu -\frac{r-\frac12}{\ell}-im,\qquad \eta=\widetilde\psi^T C,\qquad C=\begin{pmatrix}0&-1\\1&0\end{pmatrix}. \end{aligned} \tag{O03.2}$$

The independent row $\eta$ and column $\psi$ are Grassmann odd. The row has no extra phase in this formula. In particular, replacing it by a Hermitian conjugate or changing its phase without translating the measure changes the determinant convention.

Use the bosonic contour $\widetilde\phi=\phi^*$, $\widetilde F_\Phi=-F_\Phi^*$. For real $x$ and $0<r<2$, the real part of the scalar quadratic form is strictly positive. The auxiliary integral is also convergent. This is a sufficient domain for every finite Gaussian below; complex coefficients need not be positive real numbers.

Normalize each complex ordinary measure as $d^2c/\pi$, and order each odd pair so that $\int d\eta\,d\psi\,e^{-b\eta\psi}=b$. Thus an ordinary mode contributes $1/a$, an odd pair contributes $b$, and a normalized auxiliary mode contributes one. Radius powers cancel between the scalar and fermion determinants. We may therefore calculate with the dimensionless operators $\ell^2\mathcal K_B$ and $\ell\mathcal K_F$.

Build the harmonic blocks

A scalar harmonic of degree $n=0,1,\ldots$ is the restriction to $S^3$ of a homogeneous harmonic polynomial of degree $n$ on $\mathbb R^4$. The radial decomposition $\Delta_{\mathbb R^4}=\partial_\rho^2+3\rho^{-1}\partial_\rho+\rho^{-2}\Delta_{S^3,\mathrm{unit}}$ gives

$$-\ell^2\nabla^2Y_n=n(n+2)Y_n,\qquad d_n=\binom{n+3}{3}-\binom{n+1}{3}=(n+1)^2. \tag{O03.3}$$

The second binomial is zero for $n<2$. It subtracts the polynomials obtained by multiplying degree-$n-2$ polynomials by $\rho^2$.

For spinors, keeping only the eigenvalues of the scalar Laplacian would miss the spin connection. Define Hermitian angular-momentum generators $J_a=\ell E_a/(2i)$. Their commutator is $[J_a,J_b]=i\epsilon_{abc}J_c$, and (O03.3) gives $J^2=n(n+2)/4=j(j+1)$ with $j=n/2$. The degree-$n$ scalar space consists of $n+1$ copies of this spin-$j$ representation. The copy index is a spectator in the following calculation.

Let $S_a=\gamma_a/2$ be spin $1/2$. The dimensionless Dirac operator is

$$T=-i\ell\gamma^\mu\nabla_\mu =2\gamma\cdot J+\frac32 =2\bigl[(J+S)^2-J^2-S^2\bigr]+\frac32. \tag{O03.4}$$

Adding angular momenta gives $j+1/2$ and $j-1/2$; the second is absent when $j=0$. Substituting their Casimirs into (O03.4) gives the entire block:

Branch	Eigenvalue of $T$	Multiplicity
Total spin $j+1/2$	$n+3/2$	$(n+1)(n+2)$
Total spin $j-1/2$	$-n-1/2$	$n(n+1)$

The multiplicities sum to $2d_n$, the number of two-component spinor coefficients. Replacing $n$ by $k+1$ in the negative branch recovers the familiar spectrum $\pm(k+3/2)$, each with multiplicity $(k+1)(k+2)$. That relabeling is harmless for describing eigenvalues, but changes which blocks lie at a finite cutoff. The source’s conventional spectrum appears in W16, (3.20).

We retain degrees $0\le n\le N$ in every global-frame component of the multiplet, as in the preceding lesson. The commuting supersymmetry labels are constant in this frame; derivatives, gamma matrices and constant mass terms preserve each degree. The odd operation $Q=\delta_\zeta$, with $Q^2=0$ off shell, therefore preserves this finite space. This prescription is more specific than simply cutting off positive and negative Dirac eigenvalues at the same absolute value.

Cancel the paired factors before taking a limit

For one degree, define

$$A_n=n+2-r-ix,\qquad B_n=n+r+ix.$$

The scalar eigenvalue factorizes:

$$n(n+2)+x^2+2i(1-r)x+r(2-r)=A_nB_n.$$

The two eigenvalues of $\ell\mathcal K_F=T-(r-\tfrac12)-ix$ are $A_n$ and $-B_n$, with the multiplicities above. The complete Gaussian block is consequently

$$Z_n= \frac{A_n^{(n+1)(n+2)}(-B_n)^{n(n+1)}} {(A_nB_n)^{(n+1)^2}} =\left(\frac{A_n}{B_n}\right)^{n+1}. \tag{O03.5}$$

The sign disappears because $n(n+1)$ is even. Each scalar determinant has two first-order factors. Matching their powers to the fermion determinant leaves an excess of $n+1$ numerator factors and $n+1$ denominator factors. These are the net unpaired factors; cancellation does not mean that every second-order boson eigenvalue equals a first-order fermion eigenvalue.

At $r=1/2,x=0$, the constant block is $Z_0=3$, while $Z_1=25/9$. Including both gives $25/3$, not a plausible approximation to a finite full partition function yet. The remaining ultraviolet subtraction is essential.

Set

$$u=1-r-ix,\qquad K=N+1.$$

At this stage we have an ordinary finite product,

$$Z_{\le N}=\prod_{k=1}^{K}\left(\frac{k+u}{k-u}\right)^k. \tag{O03.6}$$

There has been no manipulation of divergent sums.

Give the product a finite definition

Initially work in the strip $-1<\operatorname{Re}u<1$. Choose the logarithms continuously from $u=0$, where every factor is one. At large $k$,

$$k\log\frac{k+u}{k-u} =2u+\frac{2u^3}{3k^2}+O(k^{-4}).$$

The divergence is $2Ku$. Subtracting it leaves a convergent sum, but does not by itself specify a finite normalization. We use the symmetric zeta prescription

$$\boxed{ \mathcal L(u)= \lim_{K\to\infty} \left[ \sum_{k=1}^{K}k\log\frac{k+u}{k-u}-(2K+1)u \right] =-u+\sum_{k=1}^\infty \left[k\log\frac{k+u}{k-u}-2u\right]. } \tag{O03.7}$$

The extra finite $-u$ has a reason. Introduce the spectral series $\mathscr D(p,u)=\sum_{k\ge1}k[(k-u)^{-p}-(k+u)^{-p}]$, initially for sufficiently large $\operatorname{Re}p$. Its derivative at $p=0$, after analytic continuation, is the desired logarithm. For $|u|<1$,

$$\mathscr D(p,u)=2pu\,\zeta_{\rm R}(p) +\frac{p(p+1)(p+2)}3u^3\zeta_{\rm R}(p+2)+\cdots .$$

Here $\zeta_{\rm R}(p)=\sum_{k\ge1}k^{-p}$ denotes the Riemann zeta function continued away from its defining half-plane. Since $\zeta_{\rm R}(0)=-1/2$, differentiating the first term gives $-u$. The remaining terms reproduce the convergent series in (O03.7); analytic continuation then extends the result throughout the strip. Deleting only $2Ku$ would define a different function, larger by $u$ in its logarithm.

There is a practical way to evaluate the answer without summing many modes. Differentiating the convergent series and using $\pi\cot\pi u=u^{-1}+2u\sum_{k\ge1}(u^2-k^2)^{-1}$ gives

$$\mathcal L'(u)=-1+2u^2\sum_{k=1}^\infty\frac1{k^2-u^2} =-\pi u\cot\pi u,\qquad \mathcal L(0)=0. \tag{O03.8}$$

The apparent singularity at zero is removable. Integrating this derivative along a path in the strip determines the complex logarithm, including its phase.

Only now introduce a compact name:

$$Z_\chi(r,x)=e^{\mathcal L(1-r-ix)} =s_1\!\left(i(1-r)+x\right). \tag{O03.9}$$

Here $s_1$ is the round-sphere double sine in this prescription. Its double-index product groups $m,n\ge0$ by $m+n+1=k$, giving exactly $k$ factors at each $k$. Thus the name packages the mode product already derived. In W16, (3.25)–(3.27), the dimensionless argument called $\widehat\sigma$ is $-x$ when matching this determinant to our chosen $H$, action and fermion-row conventions. Copying that symbol as $+x$ would complex-conjugate the answer for real $r,x$.

Evaluate the conformal free chiral

For $r=1/2,m=0$, integrate (O03.8):

$$\begin{aligned} \log Z_\chi&=-\int_0^{1/2}\pi v\cot(\pi v)\,dv\\ &=\int_0^{1/2}\log\sin(\pi v)\,dv =-\frac12\log2. \end{aligned}$$

The integration-by-parts boundary term vanishes at both ends. To obtain the last integral, let $I=\int_0^{\pi/2}\log\sin t\,dt=\int_0^{\pi/2}\log\cos t\,dt$. Adding the two and using $\sin t\cos t=\tfrac12\sin2t$ gives $2I=I-\tfrac\pi2\log2$. Therefore

$$Z_\chi\left(\frac12,0\right)=\frac1{\sqrt2}, \qquad F_\chi=-\log Z_\chi=\frac12\log2. \tag{O03.10}$$

This finite answer agrees with the round specialization of the independently represented free energy in P, (8.11):

$$F_\chi=-\int_0^\infty\frac{dt}{2t} \left(\frac1{\sinh t}-\frac1t\right).$$

Near zero its integrand tends to $1/12$; at infinity it decays as $1/(2t^2)$. Both ends are integrable. Direct quadrature and the regulated mode sum agree. Omitting the finite $-u$ in (O03.7) would instead multiply (O03.10) by $e^{1/2}$.

With $r=1/2,x=0.7$, integration of (O03.8) gives, in the same prescription, $Z_\chi\simeq0.2636355432+0.2001253745i$. A complex answer is compatible with the complex Euclidean quadratic action. Opposite masses provide a useful magnitude check in Exercise 2.

Know what continuation and phase do not establish

Within $0<r<2$, real $x$, all the finite Gaussian integrals above converge and have no zero modes. Outside this sufficient domain, convergence must be checked anew. In particular, meromorphic continuation of (O03.9) is not a substitute for integrating a bosonic zero mode.

The continued product has a pole of order $k$ at $u=k$ and a zero of order $k$ at $u=-k$, for positive integers $k$. At $x=0,r=0$, the constant scalar mode becomes massless while no negative-branch constant spinor exists: a simple pole remains. At $x=0,r=2$, the scalar constant mode and two constant fermion eigenvalues vanish. Their formal ratio has a simple zero, but the original ordinary Gaussian is divergent. Multiplying a divergent bosonic integral by a vanishing fermionic integral does not define it. The reflection identity derived in Exercise 2, together with $\mathcal L(\varepsilon)=-\varepsilon+O(\varepsilon^3)$, gives the limits $Z_\chi(r,0)\sim1/(2\pi r)$ near $r=0^+$ and $Z_\chi(2-\varepsilon,0)\sim2\pi\varepsilon$ near $\varepsilon=0^+$.

The prescription also matters for phases. For a unit-charge compact flavor source with $z=0$, one can add an integer background Chern–Simons counterterm. Its bosonic supersymmetric completion is

$$S_{\rm ct}=\frac{k_{\rm ct}}{4\pi} \int\sqrt g\,d^3x\, \left(i\epsilon^{\mu\nu\rho}a_{F\mu}\partial_\nu a_{F\rho} -2\sigma_FD_F\right),\qquad k_{\rm ct}\in\mathbb Z. \tag{O03.11}$$

The level is dimensionless. The integrality statement concerns a compact unit-charge $U(1)$ on spin manifolds, not just the topologically trivial connection used in this calculation. CT, (1.5)–(1.7) The completion agrees with P, (6.28) and CT, (4.5). On our source family, it changes $Z$ by $e^{i\pi k_{\rm ct}x^2}$; Exercise 3 derives the sign.

There is a further global qualification. A single charge-one chiral contains one charged Dirac fermion. Defining its determinant on arbitrary flavor bundles requires the parity-anomaly prescription: one cannot simultaneously impose a parity-symmetric fermion prescription and full background gauge invariance without the required anomaly completion. Integer counterterm changes do not erase the fractional information. CT, section 3.1 Formula (O03.9) specifies the symmetric determinant factor on this sphere family; it does not alone specify that global completion. When this factor enters a gauge-theory integral or a duality comparison, the regulator and background counterterms must be supplied consistently. We have not set all contact terms to zero.

Practice

1. Change the cutoff and find the extra modes

Set $r=1,x=0$. First retain scalar harmonic degrees $0\le n\le N$ in every multiplet component. Then consider a different prescription: keep the same scalar modes, but keep both Dirac eigenvalues $\pm(k+3/2)$ for $0\le k\le N$. Find the ratio of the two finite Gaussian answers. Explain why the difference cannot simply be discarded before regularization.

Hint 1

In the global-component labeling, the negative Dirac branch at index $k$ belongs to $n=k+1$.

Hint 2

At $r=1,x=0$, the fermion eigenvalues in degree $n$ are $n+1$ and $-(n+1)$. Count the negative branch added at $n=N+1$.

Solution

In the common-component cutoff, $A_n=B_n=n+1$, so every complete block contributes one. The other prescription retains the same positive branches but an extra negative branch in degree $N+1$. Its eigenvalue is $-(N+2)$, with multiplicity $(N+1)(N+2)$. The latter is even, so the ratio is

$$\frac{Z_{\rm changed}}{Z_{\rm common}} =(N+2)^{(N+1)(N+2)}.$$

Already at $N=0$, the ratio is four: two extra odd pairs each have eigenvalue $-2$. The common cutoff gives one.

The changed finite space is not preserved by the same component supersymmetry transformations. It can be the starting point for a different regulator with appropriate restoration and subtraction, but its endpoint factor is not equal to one and has no finite unregulated limit. An infinite relabeling that suppresses this boundary contribution has skipped part of the definition of the determinant.

2. Pair two opposite real masses

For two independent free chirals of R-charge $1/2$ and dimensionless masses $x,-x$, derive their product from (O03.8). Find the magnitude of either single-chiral determinant for real $x$. Does the calculation imply that either factor is real?

Hint 1

With $u=1/2-ix$, the other argument is $1-u$. Differentiate $\mathcal L(u)+\mathcal L(1-u)$ with respect to $u$.

Hint 2

Use $\cot(\pi-\pi u)=-\cot\pi u$, and fix the integration constant at $u=1/2$.

Solution

The derivative is

$$\frac{d}{du}\bigl[\mathcal L(u)+\mathcal L(1-u)\bigr] =-\pi u\cot\pi u-\pi(1-u)\cot\pi u =-\pi\cot\pi u.$$

At $u=1/2$ the sum is $-\log2$. Hence, in the strip connecting these points,

$$\mathcal L(u)+\mathcal L(1-u)=-\log(2\sin\pi u), \qquad Z_\chi\left(\frac12,x\right)Z_\chi\left(\frac12,-x\right) =\frac1{2\cosh\pi x}.$$

This reproduces the paired identity of W16, (3.28). The specified series also obeys $\mathcal L(\bar u)=\overline{\mathcal L(u)}$, so opposite-mass factors are complex conjugates and $\lvert Z_\chi(1/2,x)\rvert=(2\cosh\pi x)^{-1/2}$. At zero mass this gives $1/\sqrt2$, as required; for large $|x|$ the magnitude decays as $e^{-\pi|x|/2}$.

The product fixes the magnitude, not a zero phase for either factor. The complex numerical example above illustrates this distinction. The identity refers to the displayed prescription before additional counterterm factors are included.

3. Keep the background phase

Take $q_F=1,z=0$, so $\mu=x/\ell$, and add (O03.11). Using $\operatorname{Vol}(S^3_\ell)=2\pi^2\ell^3$, calculate $S_{\rm ct}$ and the change in $\partial_x^2\log Z$. What happens to $\operatorname{Re}F=-\log|Z|$ for real $x$? Why is replacing an allowed integer level by an arbitrary real number a different claim?

Hint 1

The gauge field vanishes, but the source auxiliary field does not: $D_F=i x/\ell^2$.

Hint 2

The Euclidean weight is $e^{-S}$. Apply this minus sign after evaluating the counterterm action.

Solution

Only the scalar–auxiliary term survives:

$$S_{\rm ct} =\frac{k_{\rm ct}}{4\pi} \left(-2\frac{x}{\ell}\frac{ix}{\ell^2}\right) (2\pi^2\ell^3) =-i\pi k_{\rm ct}x^2.$$

Thus $Z_{\rm new}=e^{i\pi k_{\rm ct}x^2}Z$ and $\partial_x^2\log Z_{\rm new}-\partial_x^2\log Z=2i\pi k_{\rm ct}$. All radius powers cancel. For real $x$, this is a unit-modulus phase, so $\operatorname{Re}F$ is unchanged by this particular counterterm while the imaginary response changes.

On the one-parameter sphere family an arbitrary real coefficient also defines an algebraic phase. That observation does not make it a legal change between globally defined compact-source theories. Requiring the counterterm on all spin manifolds and flavor bundles restricts the shift to an integer. The parity-anomaly completion and any other background terms must remain fixed while comparing prescriptions.

The determinant is now a usable building block: its spectral multiplicities, mass convention, normalization and domain are specified. A first gauge-theory matrix integral adds the gauge zero modes, their measure and classical weights, and the common counterterm prescription. The free calculation alone does not supply those data.

Paper guides

Connect this lesson to the original papers and their notation.

• Reading three-dimensional localization with matter