Calculate the periods and effective coupling

The curve has supplied two contour integrals. We now turn them into functions that can be evaluated, differentiated, and continued between vacua. Two methods will meet: direct integration and an ordinary differential equation. Agreement between them tests more than either method alone.

Read the curve and its normalized differential first. We retain pure four-dimensional N=2 $SU(2)$ Yang–Mills, with eight real supercharges preserved by its Coulomb vacua and one generic low-energy Abelian vector multiplet. The answer concerns the exact two-derivative Wilsonian data. It does not determine the entire particle spectrum or justify integrating out a particle at a point where it becomes massless.

Evaluate the two oriented integrals

Write $s=\Lambda^2>0$ and $C=\sqrt2/(2\pi)$. Our curve and differential are

$$y^2=(x^2-s^2)(x-u),\qquad \lambda=C(x-u)\frac{dx}{y},\qquad \Pi=(a_D,a)^T. \tag{SW11.1}$$

Initially $u>s$ is real. The electric cycle goes from $-s$ to $s$ where $y$ is negative real and returns on the other sheet. The magnetic cycle goes from $s$ to $u$ where $y$ is positive imaginary and returns on the other sheet. These are the preceding lesson’s cycles, with geometric intersection $\gamma_e\cdot\gamma_D=+1$; the original paper uses the opposite sign convention for its intersection symbol. No contour is reversed here.

Both legs contribute, so

$$a=\frac{\sqrt2}{\pi}\int_{-s}^{s} \frac{\sqrt{u-x}}{\sqrt{s^2-x^2}}\,dx,\qquad a_D=\frac{i\sqrt2}{\pi}\int_s^u \sqrt{\frac{u-x}{x^2-s^2}}\,dx . \tag{SW11.2}$$

In particular $a>0$ and $a_D/i>0$. These are the scale-restored original period integrals, (6.22)–(6.23). The units are $[u]=[s]=2$ and $[a]=[a_D]=1$.

Define two complete elliptic integrals, which here mean the following ordinary definite integrals:

$$K(m)=\int_0^{\pi/2}\frac{d\theta}{\sqrt{1-m\sin^2\theta}}, \qquad E(m)=\int_0^{\pi/2}\sqrt{1-m\sin^2\theta}\,d\theta, \qquad m=\frac{2s}{u+s}. \tag{SW11.3}$$

We use the parameter $m$, not a modulus whose square is the argument. For $u>s$, $0<m<1$, and all square roots in (SW11.3) are positive.

For the electric integral put $x=-s+2s\sin^2\theta$. Then $dx/\sqrt{s^2-x^2}=2\,d\theta$ and $u-x=(u+s)(1-m\sin^2\theta)$. For the magnetic integral put $x=s+(u-s)\sin^2\theta$, and then replace $\theta$ by $\pi/2-\theta$. Its integral without $i\sqrt2/\pi$ becomes $2(u-s)\int_0^{\pi/2}\sin^2\theta\,d\theta/\sqrt{(u+s)[1-(1-m)\sin^2\theta]}$. The identity $K(q)-E(q)=q\int\sin^2\theta\,d\theta/\sqrt{1-q\sin^2\theta}$ therefore gives

$$\boxed{ a=\frac{2\sqrt{2(u+s)}}{\pi}E(m),\qquad a_D=\frac{2i\sqrt{2(u+s)}}{\pi} \big[K(1-m)-E(1-m)\big].} \tag{SW11.4}$$

This evaluates the contours in terms of explicitly defined real functions; it has not replaced the cycles by a new normalization.

Differentiate before forming the coupling

A prime below means $d/du$ at fixed $\Lambda$. Differentiating the defining integrals gives $dE/dm=(E-K)/(2m)$ and $dK/dm=E/[2m(1-m)]-K/(2m)$. The second relation follows by adding the integral of $d[\sin\theta\cos\theta/\sqrt{1-m\sin^2\theta}]/d\theta$, whose endpoint contribution vanishes. Using $m'=-m/(u+s)$ in (SW11.4) yields

$$a'=\frac{\sqrt2}{\pi\sqrt{u+s}}K(m),\qquad a_D'=\frac{i\sqrt2}{\pi\sqrt{u+s}}K(1-m),\qquad \boxed{\tau_{\rm eff}=\frac{a_D'}{a'} =i\frac{K(1-m)}{K(m)}.} \tag{SW11.5}$$

Both real integrals on the right are positive. Thus $\operatorname{Im}\tau_{\rm eff}>0$ on this real ray, as required for a positive gauge kinetic term. On a general smooth fiber, the preceding lesson’s bilinear identity gives the same positivity statement in a valid local frame.

The derivative ratio follows from $a_D=\partial\mathcal F/\partial a$ and $\tau_{\rm eff}=\partial^2\mathcal F/\partial a^2$. It is not $a_D/a$. For a worked value choose $u=3s$, so $m=1/2$:

Quantity	Value in the stated branch
$a/\Lambda$	$2.43201344685$
$a_D/\Lambda$	$0.906493919846\,i$
$\Lambda a'$	$0.417313420837$
$\tau_{\rm eff}$	$i$ exactly

Here $a_D/a\simeq0.372734\,i$ would give the wrong coupling. The equality of the two $K$ integrals fixes $\tau=i$ without numerical approximation.

Tachikawa’s geometric argument, (4.3.5)–(4.3.10), also computes a derivative ratio. Its curve, electric units and periods differ: its weak electric period is $\sqrt u$, and its displayed large-loop shear is $4$, rather than our $\sqrt{2u}$ and $2$. We use that discussion for the geometric method, not to import its coefficients.

Derive the equation satisfied by a period

A Picard–Fuchs equation is a differential equation for integrals of a differential over continuously transported cycles. We can derive the needed one directly, without assuming a course on such equations.

At fixed $x$, the chosen representative obeys $\partial_u\lambda=-C\,dx/(2y)$ and $\partial_u^2\lambda=-C\,dx/[4(x-u)y]$. Now calculate the following combination:

$$\begin{aligned} (u^2-s^2)\partial_u^2\lambda+\frac14\lambda &=\frac C4\frac{x^2-2ux+s^2}{(x-u)y}\,dx\\ &=\frac C2\,d_x\!\left(\frac{y}{x-u}\right). \end{aligned} \tag{SW11.6}$$

For the second equality, differentiate $y^2=(x^2-s^2)(x-u)$ with respect to $x$. The function $y/(x-u)$ is single-valued and meromorphic on the surface. Its exact differential integrates to zero around a closed cycle, even though a projection of that cycle uses two sheets. We transport and, where needed, deform the cycle away from poles before differentiating.

Consequently each period $p=a$ or $a_D$ satisfies

$$\boxed{(u^2-\Lambda^4)p''+\frac14p=0.} \tag{SW11.7}$$

This is the equation obtained by the different reconstruction in Bilal, (6.4), (6.11)–(6.13), with the scale restored. All terms have dimension one. The equation is singular precisely at the two finite degenerations; its large-$u$ behavior also matters.

At an ordinary point $u_0\ne\pm s$, a second-order linear equation needs two initial numbers, $p(u_0)$ and $p'(u_0)$. It determines a unique local solution. The electric and magnetic contours fix two different sets of these data. An equation alone, without those data, would not identify the physical periods.

For example, put $p=\sum_{n\ge0}c_n(u-u_0)^n$. Substitution gives the practical recurrence

$$(u_0^2-s^2)(n+2)(n+1)c_{n+2} +2u_0n(n+1)c_{n+1} +\left[n(n-1)+\frac14\right]c_n=0. \tag{SW11.8}$$

Starting with $c_0=p(u_0)$ and $c_1=p'(u_0)$ determines every higher coefficient. Evaluate within the distance to the nearest singularity, then use the resulting value and derivative as data for another disk. This continues the chosen solution, rather than resetting a square root or logarithm on each step.

There is also a conserved diagnostic. Define the Wronskian $W=a a_D'-a_D a'$. Equation (SW11.7) gives $W'=a a_D''-a_D a''=0$ because it has no first-derivative term. The normalization below fixes

$$W=\frac{2i}{\pi}. \tag{SW11.9}$$

A drifting numerical Wronskian signals an integration problem. A constant but incorrect value can instead signal incorrect initial normalization.

Recover weak coupling with its fixed scale

For real $u/s\gg1$, expand the electric integrand in (SW11.2). Its odd powers of $x$ integrate to zero. The normalized even moments are $\langle x^2\rangle=s^2/2$ and $\langle x^4\rangle=3s^4/8$, giving

$$a=\sqrt{2u}\left[ 1-\frac{s^2}{16u^2} -\frac{15s^4}{1024u^4} +O\!\left(\frac{s^6}{u^6}\right)\right]. \tag{SW11.10}$$

The logarithm in the magnetic period comes from an endpoint. To see its constant as well as its coefficient, write $K(1-\epsilon)=\int_0^\infty dt/\sqrt{(1+t^2)(1+\epsilon t^2)}$. Split the integral at $1\ll T\ll\epsilon^{-1/2}$. The small-$t$ part tends to $\log(2T)$ and the large-$t$ part to $\log[2/(\sqrt\epsilon T)]$. The sum is $\log(4/\sqrt\epsilon)$. Expanding the regular remainder gives $K(1-\epsilon)=\log(4/\sqrt\epsilon)+O(\epsilon|\log\epsilon|)$ and $E(1-\epsilon)\to1$.

Apply this with $\epsilon=m$ in (SW11.4). The leading result is $a_D=(i/\pi)\sqrt{2u}[\log(8u/s)-2+o(1)]$. To obtain the next correction without a delicate endpoint expansion, insert $a_D=(i/\pi)\sqrt{2u}[L-2+(cL+d)s^2/u^2+\cdots]$ into (SW11.7), where $L=\log(8u/s)$. The coefficients of $L$ and of the constant give $4c+1/4=0$ and $-4c+4d-1/2=0$. Hence

$$\begin{aligned} a_D&=\frac{i\sqrt{2u}}{\pi} \left[L-2-\frac{s^2}{16u^2}(L-1) +O\!\left(\frac{s^4}{u^4}L\right)\right],\\ \tau_{\rm eff}&=\frac{i}{\pi} \left[L-\frac{5s^2}{16u^2} +O\!\left(\frac{s^4}{u^4}\right)\right]. \end{aligned} \tag{SW11.11}$$

Differentiating both period expansions verifies the second line; the logarithmic correction cancels in the ratio. Their leading terms give $W=2i/\pi$. The constants $8$ and $-2$ belong to our already fixed curve scale and period frame. An earlier running-coupling formula that leaves a finite scale matching unspecified cannot fix these constants by itself.

The expansion parameter is $\Lambda^4/u^2$, not merely the inverse of a dimensionful $u$. For complex large $u$, these expansions use the square root and logarithm continued from the positive ray in a fixed sector avoiding their cuts. The displayed error estimates are asymptotic there; they are not estimates near $u=\pm s$.

Approach the monopole point in the same branch

Let $\delta=u-s>0$ with $\delta/s\ll1$. In the magnetic integral use $x=s+\delta t$:

$$a_D=\frac{i\sqrt2\,\delta}{\pi} \int_0^1\frac{\sqrt{(1-t)/t}}{\sqrt{2s+\delta t}}\,dt =\frac{i\delta}{2\Lambda} \left[1-\frac{\delta}{16s} +\frac{3\delta^2}{256s^2} +O\!\left(\frac{\delta^3}{s^3}\right)\right]. \tag{SW11.12}$$

For example $\int_0^1\sqrt{(1-t)/t}\,dt=\pi/2$, and its weighted averages of $t,t^2$ are $1/4,1/8$. These give every displayed coefficient. In particular $a_D'(s)=i/(2\Lambda)\ne0$: the vanishing period is a good local coordinate.

The electric period stays finite. At $\delta=0$ its integral gives $a(s)=4\Lambda/\pi$. In (SW11.5), $1-m=\delta/(2s+\delta)$, so the same endpoint logarithm now gives $a'=[\log(32s/\delta)+O((\delta/s)|\log(\delta/s)|)]/(2\pi\Lambda)$. Integrating from $\delta=0$ fixes the constant term:

$$\begin{aligned} a&=\frac{4\Lambda}{\pi} +\frac{\delta}{2\pi\Lambda} \left[\log\frac{32s}{\delta}+1\right] +O\!\left(\frac{\delta^2}{\Lambda^3}|\log(\delta/s)|\right),\\ \tau_{\rm eff}&\sim\frac{i\pi}{\log(32s/\delta)}, \qquad \tau_D=-\frac1{\tau_{\rm eff}} \sim\frac{i}{\pi}\log\frac{32s}{\delta}. \end{aligned} \tag{SW11.13}$$

The limits agree with the original local calculation, (6.27)–(6.30). The dual electric frame becomes weakly coupled. At the point itself the extra light monopole hypermultiplet must be retained, as explained in the strong-coupling lesson. A small electric $\operatorname{Im}\tau$ is not a negative kinetic term and is not permission to keep a photon-only action through its singularity.

Check integration against continuation

Numerical evaluation should compare independently formulated quantities. Removing square-root endpoints makes the contour quadrature well behaved: $a=(\sqrt2/\pi)\int_0^\pi\sqrt{u-s\cos\theta}\,d\theta$ and $a_D=(2i\sqrt2/\pi)(u-s)\int_0^{\pi/2}\cos^2\theta\,d\theta/\sqrt{2s+(u-s)\sin^2\theta}$. These formulas and their differentiated integrals provide initial data at, for example, $u=3s$.

Then continue those data with (SW11.8), independently of elliptic-integral evaluation. Compare at real points and at complex points reached within $\operatorname{Re}u>s$, where these interval integrals continue from their positive-real roots without ambiguity. Also compare $W$, $\operatorname{Im}\tau$, and the errors of the two asymptotic expansions as their small parameters decrease. Increasing arithmetic precision does not by itself control a truncated series: decrease the disk step or increase the retained order too.

Neither a finite collection of numerical points nor the ODE’s existence proves a stable-particle spectrum. The checks establish agreement of the normalized period calculations in their stated domain. Closed paths around punctures require the explicit continuation and loop conventions of the monodromy lesson.

Exercises

1. Invert the weak electric coordinate

Derive the $s^4/u^4$ coefficient in (SW11.10) directly from the contour integral. Then invert the result through order $s^4/a^6$ to find $u(a)$ on the positive weak branch. Why does $u=a^2/2$ fail beyond the classical approximation?

Hint 1

The fourth term of the binomial expansion of $\sqrt{1-z}$ is $-5z^4/128$. Average powers of $x=s\cos\theta$.

Hint 2

Square the period first and use an ansatz $u=a^2/2+A s^2/a^2+B s^4/a^6$. Every term must have dimension two.

Solution

The fourth moment is $3s^4/8$, giving $(-5/128)(3/8)=-15/1024$. Squaring the period gives $a^2=2u-s^2/(4u)-13s^4/(256u^3)+O(s^6/u^5)$. Substituting the proposed inverse, the $s^2/a^2$ coefficient requires $2A-1/2=0$, and the $s^4/a^6$ coefficient requires $2B+A-13/32=0$. Thus

$$u(a)=\frac{a^2}{2}+\frac{\Lambda^4}{4a^2} +\frac{5\Lambda^8}{64a^6} +O\!\left(\frac{\Lambda^{12}}{a^{10}}\right).$$

All terms have dimension two. The coordinate $u$ is a gauge-invariant vacuum expectation value, whereas $a$ is a quantum period. Their classical relation sets a normalization at infinity; it does not remove quantum corrections. This inversion is valid for $|a|\gg\Lambda$ on the declared branch.

2. Test the local coordinate and the scalar metric

Using (SW11.12), find $u$ as a series in $a_D$ through quadratic order. Then estimate $G(u)=\operatorname{Im}\tau_{\rm eff}|a'(u)|^2$ on the ray $\delta>0$ near the monopole point. Here $G$ omits any overall action normalization. Does its divergence place the singular point at infinite distance in this two-derivative metric?

Hint 1

Set $w=-2i\Lambda a_D$. Then $w=\delta-\delta^2/(16s)+\cdots$.

Hint 2

Use the logarithmic formulas for both $\tau$ and $a'$. A radial distance involves $\sqrt{G}\,d\delta$, not $G\,d\delta$.

Solution

Inverting gives $\delta=w+w^2/(16s)+O(w^3/s^2)$, hence $u=s-2i\Lambda a_D-a_D^2/4+O(a_D^3/\Lambda)$. In particular $du/da_D|_0=-2i\Lambda$ has the required dimension one and is nonzero.

With $L_\delta=\log(32s/\delta)$, $G\sim L_\delta/(4\pi\Lambda^2)>0$. The distance integral is proportional to $\int_0^\epsilon\sqrt{\log(32s/\delta)}\,d\delta$. Set $\delta=32s e^{-t}$: its tail is an exponentially convergent integral $\int^\infty e^{-t}\sqrt t\,dt$. The point is at finite distance although the photon-only metric coefficient diverges. This is a statement about the limiting effective metric, not a reason to omit the light hypermultiplet.

3. Diagnose a reassuring but incomplete numerical test

A program returns $\widehat a=a$ and $\widehat a_D=a_D+\kappa a$, where $\kappa$ is a real constant. It reports that both functions solve (SW11.7), their Wronskian is $2i/\pi$, and their derivative ratio has positive imaginary part. Are these tests enough to establish the specific contour answer? Find a discriminating boundary test. Contrast this error with $\widehat a_D=-a_D$.

Hint 1

The equation is linear. Compute the Wronskian and derivative ratio after adding a multiple of the other solution.

Hint 2

The assigned magnetic cycle shrinks at $u=s$, while the electric period approaches $4\Lambda/\pi$.

Solution

Linearity preserves the equation. The added terms cancel in $a\widehat a_D'-\widehat a_Da'$, and $\widehat\tau=\tau+\kappa$ has the same imaginary part. But $\widehat a_D(s)=4\kappa\Lambda/\pi$, so the assigned vanishing-period condition forces $\kappa=0$. An integral $\kappa$ could instead describe a different integral cycle basis, accompanied by the corresponding charge transformation. It is still not the contour specification of this problem. Arbitrary real $\kappa$ need not even preserve the integral cycle lattice.

Changing only the magnetic sign gives $\widehat W=-2i/\pi$ and $\operatorname{Im}\widehat\tau<0$ on the real ray. These diagnostics detect that orientation error immediately. Thus the ODE, positivity, Wronskian, and boundary normalization test different parts of the calculation; no one test replaces the rest.

What the periods now let us ask

We can calculate $a_D$, $a$ and the effective coupling at a smooth vacuum, and approach a degeneration in a controlled local coordinate. BPS masses and the stability of particles asks how $Z_\gamma=n_ma_D+n_ea$ constrains particle masses and decays. A computable central charge for an integer row is necessary information; deciding whether that row labels an existing stable BPS particle requires more.

For a different test of the same low-energy data, recover a prepotential coefficient by instanton counting. Evaluate the first equivariant contribution, translate its electric and magnetic normalization, and compare it with the weak-coupling periods derived here.

Paper guides

Connect this lesson to the original papers and their notation.

• Reading the first Seiberg–Witten paper