Seiberg–Witten periods and couplings

A plausible complex number is not yet a period. Its sign remembers an oriented cycle, its units remember a scale, and its derivative determines a physical coupling. This lab makes those choices testable. We will evaluate the same periods by direct quadrature and by transporting initial data with a differential equation, then deliberately report the wrong ratio and watch a simple test reject it.

The required preparation is periods and the effective coupling. The final instanton-coefficient section additionally requires the one-instanton calculation. You can complete the main program and the first two exercises before that extension.

Fix the theory, scale and sheet

We study flat four-dimensional N=2 pure $SU(2)$ Yang–Mills: eight real supercharges, manifest as four in N=1 notation, no hypermultiplets, and an unbroken eight-charge Coulomb vacuum with generic low-energy gauge group $U(1)$. There is no curved background in the main calculation. The answer describes the two-derivative Wilsonian action; it does not enumerate stable BPS particles.

Use the first-paper convention with $u=\langle\operatorname{Tr}\phi^2\rangle$, ordinary fundamental trace, and classical eigenvalues chosen so that $a\sim\sqrt{2u}$. For real positive $\Lambda$,

$$y^2=(x^2-\Lambda^4)(x-u),\qquad \lambda_{\rm SW}=\frac{\sqrt2}{2\pi}\frac{(x-u)\,dx}{y},\qquad \Pi=\begin{pmatrix}a_D\\a\end{pmatrix}. \tag{LAB13.1}$$

Initially $u>\Lambda^2$. The electric cycle follows $[-\Lambda^2,\Lambda^2]$ with $y$ negative real and returns on the other sheet. The magnetic cycle follows $[\Lambda^2,u]$ with $y$ positive imaginary and returns on the other sheet. Thus $a>0$ and $a_D/i>0$. Both legs are included. These are the original period integrals, (6.22)–(6.23), with the branch choice made explicit.

Let $m_0$ be a fixed mass unit. The program accepts the dimensionless scale $\Lambda/m_0$ and coordinate $z=u/\Lambda^2$, and calculates $A=a/\Lambda$, $D=a_D/\Lambda$. Dimensions are $[u]=[x]=2$, $[y]=3$ and $[a]=[a_D]=1$. Derivatives denoted by primes below are with respect to $z$. Consequently $da/du=A'/\Lambda$, not $\Lambda A'$.

We restrict the program to $\operatorname{Re}z\ge1.0001$, $|z|\le100$. The straight path from $z_0=3$ lies wholly in $\operatorname{Re}z>1$. The square roots in the following integrals have positive real part there, so continuation from the real ray is unambiguous. This restriction deliberately excludes the singularities and loops around them; those require a separately specified continuation path.

Two computations with different failure modes

Removing square-root endpoints from the contour integrals gives smooth finite intervals:

$$\begin{aligned} A(z)&=\frac{\sqrt2}{\pi}\int_0^\pi\sqrt{z-\cos t}\,dt,\\ D(z)&=\frac{2i\sqrt2(z-1)}{\pi} \int_0^{\pi/2}\frac{\cos^2t}{\sqrt{2+(z-1)\sin^2t}}\,dt. \end{aligned} \tag{LAB13.2}$$

The electric substitution is $x=\Lambda^2\cos t$, with the reversed limits absorbed. For the magnetic contour use $x=\Lambda^2+(u-\Lambda^2)\sin^2t$. These substitutions remove endpoint singularities rather than cutting off the contour short of its endpoints.

The derivative data are

$$\begin{aligned} A'&=\frac{\sqrt2}{2\pi}\int_0^\pi\frac{dt}{\sqrt{z-\cos t}},\\ D'&=\frac{i\sqrt2}{\pi}\int_0^{\pi/2} \frac{dt}{\sqrt{2+(z-1)\sin^2t}},\qquad \tau_{\rm eff}=\frac{D'}{A'}. \end{aligned} \tag{LAB13.3}$$

Differentiating the second integral in (LAB13.2) initially gives an expression with both $\cos^2t$ and a denominator cubed. Integration of the derivative of $\sin t\cos t/\sqrt{2+(z-1)\sin^2t}$, whose boundary values vanish, reduces it to (LAB13.3). Differentiating an integral and simply omitting its moving factor $z-1$ would fail this check.

The second computation uses the derived period equation, independently recognizable in Bilal, (6.4), (6.11):

$$(z^2-1)p''+\frac14p=0. \tag{LAB13.4}$$

For a local series $p(z_0+h)=\sum_{n=0}^{q}c_nh^n$, set $c_0=p(z_0)$ and $c_1=p'(z_0)$. Matching the coefficient of $h^n$ gives

$$c_{n+2}=-\frac{2z_0n(n+1)c_{n+1}+ \left[n(n-1)+\frac14\right]c_n} {(z_0^2-1)(n+2)(n+1)}. \tag{LAB13.5}$$

The program evaluates this polynomial and its derivative, moves to the next disk, and repeats. Each step is at most a fraction $\eta$ of the distance to the nearest puncture $z=\pm1$. Only the starting values at $3$ come from quadrature; no contour integral or elliptic function is called during a continuation step. The default comparisons use $(q,\eta)=(40,0.2)$ and $(60,0.1)$. Changing both order and step size separates series truncation from arithmetic precision.

A third check evaluates the explicit elliptic functions from the lesson, with parameter $m=2/(z+1)$, not modulus $\sqrt m$. The derivative ratio is $iK(1-m)/K(m)$. At $z=3$, $m=1/2$, so $\tau=i$ exactly. In contrast, $D/A\simeq0.372734i$ there. This is why the diagnostic can catch a wrong coupling even when the periods themselves are correct.

Read the errors before trusting the result

The equation has no $p'$ term. Differentiating $W=AD'-DA'$ and using (LAB13.4) therefore gives $W'=0$. The contour normalization fixes

$$W=\frac{2i}{\pi},\qquad \operatorname{Im}\tau_{\rm eff}>0. \tag{LAB13.6}$$

The Wronskian checks relative normalization as well as numerical drift. Positivity checks a physical local-frame condition. Neither alone proves that the initial cycles were the intended ones: adding an integer multiple of $A$ to $D$ preserves $W$ and $\operatorname{Im}\tau$ while shifting $\operatorname{Re}\tau$. The base-point data and specified cycles still matter.

For comparing computed numbers $x,y$, the program reports $|x-y|/\max(1,|x|,|y|)$. All tested equalities must be below $10^{-25}$; the absolute Wronskian error has its own field. It repeats quadrature at twenty additional decimal digits. These are finite convergence diagnostics, not interval-certified error bounds. Near $z=1$, more disks and a sharper endpoint profile are expected. A high precision setting does not repair a wrong branch, normalization or coupling definition.

Weak coupling supplies a different, asymptotic check. Expanding the electric integrand and using the average $\langle\cos^2t\rangle=1/2$ gives

$$\frac{A}{\sqrt{2z}}=1-\frac1{16z^2} -\frac{15}{1024z^4}+O(z^{-6}). \tag{LAB13.7}$$

Thus $z^2(A/\sqrt{2z}-1)$ approaches $-1/16$, while the remainder after subtracting the first correction decreases by roughly sixteen when $z$ doubles. This error is omitted asymptotic terms, not quadrature noise.

Run the portable calculation

Download the complete lab (ZIP) with the program, pinned dependencies, reference output and run instructions.

Save the Python program, pinned requirements, default output and output with the instanton extension in one directory. Python 3.10 or later and mpmath 1.3.0 suffice. The program uses no site files, online service or random input.

python -m venv .venv
.venv/Scripts/python.exe -m pip install -r requirements.txt
.venv/Scripts/python.exe sw_periods.py
.venv/Scripts/python.exe sw_periods.py --real 5 --imag -0.7 --scale 2
.venv/Scripts/python.exe sw_periods.py --swap-period-ratio

On macOS or Linux use .venv/bin/python. Input decimal strings specify $z$ and $\Lambda/m_0$; --dps 40 through --dps 100 changes working precision. The default is $z=2+0.5i$, $\Lambda/m_0=1$, with fifty digits.

Default quantity	Result, rounded
$a/m_0$	$1.9863556644+0.2602606142i$
$a_D/m_0$	$-0.2250108091+0.4783388996i$
$\tau_{\rm eff}$	$-0.0912969471+0.8702468746i$
$a_D/a$	$-0.0803466352+0.2513396635i$
Maximum scaled quadrature/ODE difference, order 40	$5.88\times10^{-31}$ or smaller
Same comparison, order 60	At the fifty-digit rounding scale

The reference JSON retains fourteen significant digits of values and diagnostic errors; the last error digits can depend on arithmetic details. The negative mode intentionally fills reported_coupling with the period ratio. Its base-point test becomes false and intentional_ratio_error_detected becomes true. The correctly computed derivative ratio remains separately visible. A program that prints many successful arithmetic checks while reporting the wrong observable has not solved the physical problem.

Compare a first instanton coefficient

This extension uses the extra preparation SW16. Run sw_periods.py --instanton. The parent theory is unchanged; the microscopic computation temporarily introduces the Euclidean rotational parameters $\epsilon_1,\epsilon_2$ and its chosen localization supercharge. It is not a second flat-space vacuum with a different SUSY count.

Nekrasov, (1.3)–(1.5), introduces the instanton generating function. To evaluate its first coefficient we use the explicit Euler orientation and imaginary equivariant coordinate of Nakajima–Yoshioka, Theorem 3.2 and §4.2, with the full translation derived in SW16:

$$\alpha=\frac{a}{\sqrt2},\quad \beta=-i\alpha,\quad q_I=\frac{\Lambda^4}{4},\quad \mathcal F_{\rm site,inst}=\frac{\mathcal F_{I,\rm inst}}{2\pi i}. \tag{LAB13.8}$$

Here $\beta$ is the code’s b and $q_I$ is an instanton fugacity of mass dimension four, not the Taylor order $q$. The magnetic source period is $\alpha_D=\sqrt2a_D$ and its period matrix is $2\tau_{\rm eff}$; changing the electric coordinate alone would miss this factor. These are the matched curve/differential conventions, not an integral relabeling of the original charge lattice.

Put $E=\epsilon_1+\epsilon_2$. The two one-box fixed points have tangent weights $(\epsilon_1,\epsilon_2,2\beta,E-2\beta)$ and $(\epsilon_1,\epsilon_2,-2\beta,E+2\beta)$. Summing their inverse products gives

$$Z_1=\frac{2}{\epsilon_1\epsilon_2(E^2-4\beta^2)},\qquad \lim_{\epsilon_1,\epsilon_2\to0}\epsilon_1\epsilon_2Z_1 =\frac1{2\alpha^2}. \tag{LAB13.9}$$

The logarithm is expanded in $q_I$ before removing the regulator, as in NY, §5. At first order its coefficient is $Z_1$. At fixed nonzero fugacity, taking the logarithm of a one-instanton truncation and then removing the regulator is a different, nonuniform limit.

Using (LAB13.8) and differentiating with respect to the site’s $a$ gives

$$\mathcal F_{\rm site,inst}^{(1)}=\frac{\Lambda^4}{8\pi i a^2},\qquad a_{D,\rm inst}^{(1)}=\frac{i\Lambda^4}{4\pi a^3}. \tag{LAB13.10}$$

Compare this with the independently integrated periods through

$$C(a)=\frac{a^3}{i\Lambda^4} \left[a_D-\frac{ia}{\pi} \left(\log\frac{4a^2}{\Lambda^2}-2\right)\right] \longrightarrow\frac1{4\pi}. \tag{LAB13.11}$$

Use the positive weak ray and its real logarithm. The constant $4$ and subtraction $-2$ follow from the previously matched period expansion; they are not fitted to this comparison. The program evaluates the coefficient along $\epsilon_1=h$, $\epsilon_2=2h$ at fixed $\alpha=1$, and the period diagnostic at $z=20,40,80$:

$z$	$C(a)$	Difference from $1/(4\pi)$
$20$	$0.07960081445$	$2.33429\times10^{-5}$
$40$	$0.07958330180$	$5.83025\times10^{-6}$
$80$	$0.07957892877$	$1.45722\times10^{-6}$

The shrinking difference is a higher-instanton/asymptotic remainder. It should not be forced below the equality tolerance at finite $z$. The comparison tests this coefficient and its dictionary, not an all-order instanton-counting theorem.

Exercises

1. A normalization that the coupling misses

Suppose the return leg is accidentally omitted from both period contours and their derivatives. Predict the coupling, base-point test and Wronskian. Which test catches the error? What happens to the BPS mass bound for a fixed charge row?

Hint 1

Each closed-cycle integral contains two equal contributions. Apply the same factor to values and derivatives.

Hint 2

The coupling is a ratio; the Wronskian is bilinear. Use $Z_\gamma=n_ma_D+n_ea$.

Solution

All four numbers are halved. The derivative ratio and $\tau(3)=i$ test are unchanged, but $W$ becomes $i/(2\pi)$, a quarter of the required $2i/\pi$. The absolute normalization test catches the missing sheet. For fixed $(n_m,n_e)$, $Z_\gamma$ and the bound $\sqrt2|Z_\gamma|$ are halved. Relabeling the same integer charges does not repair the normalization. This bound still does not establish that a stable particle of each charge exists.

2. Change the scale or change the vacuum

At fixed $z$, multiply $\Lambda/m_0$ by two. Predict all four dimensionful output entries and $\tau$. Then explain why holding the dimensionful $u$ fixed instead is a different experiment. Finally, at the default complex point, does conjugating $z$ simply conjugate both periods?

Hint 1

Use $a=\Lambda A(z)$ and $da/du=A'(z)/\Lambda$. Holding $u$ fixed sends $z$ to $z/4$.

Hint 2

In the chosen patch $A$ is real on the positive ray, while $D$ contains an explicit factor of $i$.

Solution

At fixed $z$, $a/m_0$ and $a_D/m_0$ double; $m_0da/du$ and $m_0da_D/du$ halve. Their ratio is unchanged, and the Wronskian in dimensionful variables is unchanged. At fixed $u$, the coordinate becomes $z/4$, changing the elliptic parameter and the coupling; it may even leave this program’s domain.

The real coefficients and positive-root continuation give $A(\bar z)=\overline{A(z)}$ but $D(\bar z)=-\overline{D(z)}$, and the same rule for derivatives. Consequently $\tau(\bar z)=-\overline{\tau(z)}$ retains positive imaginary part. Conjugating both periods without the magnetic minus sign would select the wrong orientation.

3. Separate regulator error from instanton error

After SW16, use $\epsilon_1=h$, $\epsilon_2=2h$ to derive the leading error in $\alpha^2\epsilon_1\epsilon_2Z_1$. If both $h$ and $1/z$ are halved, what different quantities approach their limits, and why is neither convergence an arithmetic-precision estimate?

Hint 1

Insert $E=3h$ and $\beta=-i\alpha$ in (LAB13.9) before expanding.

Hint 2

The next instanton contribution to $a_D$ scales as $\Lambda^8/a^7$. Multiply it by the prefactor in $C(a)$.

Solution

The regulated coefficient is $2\alpha^2/(4\alpha^2+9h^2)=1/2-9h^2/(8\alpha^2)+O(h^4/\alpha^4)$. Halving $h$ quarters its leading deviation from $1/2$; this is a residual rotational deformation. Meanwhile $C(a)-1/(4\pi)=O(\Lambda^4/a^4)=O(z^{-2})$ on the weak ray, so doubling $z$ quarters that leading deviation. This second remainder comes from higher instanton terms in the exact periods. Increasing decimal precision at fixed $h,z$ resolves the same nonzero deviations more accurately; it removes neither. The two limits concern different calculations, and each must retain its own control parameter.