Construct the curve and normalize its differential

Before this lesson: Riemann surfaces from square roots: cuts and cycles.

A torus has two independent cycles, but topology does not tell us which integral is an electric mass parameter or what its units should be. We must choose a family of surfaces, a differential on each surface, and oriented cycles as one specification.

Read square-root surfaces, cuts and cycles first. Its dimensionless constructions now become a candidate description of pure four-dimensional N=2 $SU(2)$ Yang–Mills: eight real supercharges, all preserved by Coulomb vacua, with one generic low-energy Abelian vector multiplet. We determine two-derivative Coulomb-branch data, not arbitrary correlation functions or the full stable-particle spectrum.

State the reconstruction inputs

The field-theory information is a complex vacuum coordinate $u$ , two finite strong-coupling singularities exchanged by $u\mapsto-u$ , the monodromies already derived, and the weak-coupling condition $a\sim\sqrt{2u}$ . We retain the minimal two-finite-singularity pattern used in the original reconstruction.

Choose the definition of the strong scale so these finite points are $u=\pm\Lambda^2$ . For the initial real picture take $\Lambda>0$ . The proposed family is

E_u:\quad y^2=(x-\Lambda^2)(x+\Lambda^2)(x-u). \tag{SW10.1}

It has four projection branch points: $-\Lambda^2,\Lambda^2,u,\infty$ . The preceding lesson therefore gives genus one when they are distinct. The third finite point moves with the vacuum; the first two are fixed. This is a useful ansatz reproducing the required singularity pattern, not a theorem that supersymmetry alone selects every coefficient of every curve.

Put $s=\Lambda^2$ . The finite-root discriminant, the product of their squared pairwise differences, is

\Delta_x=4s^2(u^2-s^2)^2. \tag{SW10.2}

Thus, for fixed nonzero $\Lambda$ , finite degenerations occur at $u=\pm s$ . A loop of $u$ around either fixed root is a full winding of two labeled branch points, not merely their exchange. It produces the double twist discussed in the preceding lesson. The location at infinity and the precise period signs must also match the weak-coupling data.

The scale convention is part of the answer. The parameter $\Lambda$ in (SW10.1) need not equal a scale in an arbitrary subtraction scheme without a finite matching factor. The first paper sets its finite singularities to $\pm1$ ; here we restore dimensions:

[x]=[u]=2,\qquad [y]=3,\qquad X=x/\Lambda^2,\quad U=u/\Lambda^2,\quad Y=y/\Lambda^3. \tag{SW10.3}

Then $Y^2=(X^2-1)(X-U)$ . The curve is auxiliary; $x$ is not a spacetime coordinate.

Fix the sheet and the oriented periods

Start at a real base vacuum $u=P>s$ . Use cuts $[-s,s]$ and $[P,\infty]$ . In the upper half of the $x$ plane choose $y=\sqrt{x+s}\sqrt{x-s}\sqrt{x-P}$ , with each factor continued from positive real argument. In the lower half use the negative of that product of principal factors. The minus sign implements the stated cuts: the banks of $[-s,s]$ and $[P,\infty]$ lie on opposite sheets, while the gap $(s,P)$ has no cut.

The relevant upper-sheet boundary values are

Interval	Chosen value of the square root
$-s<x<s$	$y=-\sqrt{(s^2-x^2)(P-x)}$
$s<x<P$	$y=+i\sqrt{(x^2-s^2)(P-x)}$

Define $\gamma_e$ to go from $-s$ to $s$ on the first bank and return on the opposite sheet. Define $\gamma_D$ to go from $s$ to $P$ through the gap on the chosen sheet and return on the opposite sheet. The branch-point coordinates make these lifted contours smooth after a small deformation.

Compared with the generic diagram in the preceding lesson, $\gamma_e=-A$ and $\gamma_D=-B$ : that diagram’s upper-sheet arrows ran in the opposite directions. With its usual ordered-tangent intersection convention, $\gamma_e\cdot\gamma_D=+1$ . The original paper’s (6.7) writes the positive pairing in the opposite order, $\gamma_D\cdot\gamma_e=+1$ ; its intersection sign convention is therefore opposite to the geometric one used here. We fix the actual period signs by the explicit contours above, rather than altering a contour to reconcile two conventions for an intersection symbol.

For any proposed differential $\lambda$ , the physical ordering remains

a_D=\oint_{\gamma_D}\lambda,\qquad a=\oint_{\gamma_e}\lambda,\qquad \Pi=\begin{pmatrix}a_D\\a\end{pmatrix},\qquad Z_\gamma=(n_m,n_e)\Pi. \tag{SW10.4}

Charges transform inversely to changes of this period frame. No charge normalization has changed. In particular, the classical comparison is $u=\operatorname{Tr}_2\phi_{\rm SW}^2=a_{\rm cl}^2/2$ , with $\phi_{\rm SW}=g\phi_{\rm canonical}$ and $\operatorname{Tr}_2(t^at^b)=\delta^{ab}/2$ . It fixes the asymptotic electric period; it is not the exact identity $u=a(u)^2/2$ .

Find a differential with the right derivative

The differential must have dimension one, so its periods can enter the BPS mass $\sqrt2|Z_\gamma|$ . Two simple forms on this cubic are $dx/y$ and $x\,dx/y$ . The first is holomorphic everywhere, including infinity; the second has a double pole at infinity with zero residue.

Use the minimal ansatz

\lambda=\big[A(u)x+B(u)\big]\frac{dx}{y}. \tag{SW10.5}

Here $A$ is dimensionless and $B$ has dimension two. We ask that the derivative with respect to $u$ be a holomorphic differential. This condition makes the ratio of period derivatives a complex structure with positive imaginary part. It is the constructive condition in Seiberg–Witten, (6.15)–(6.20).

At fixed $x$ , $\partial_u(1/y)=1/[2(x-u)y]$ . Hence

\partial_u\lambda= \left[A'(u)x+B'(u)+\frac{A(u)x+B(u)}{2(x-u)}\right]\frac{dx}{y}. \tag{SW10.6}

At the moving branch point $x=u$ , the last term would create a pole unless $A(u)u+B(u)=0$ . At infinity, $A'(u)x\,dx/y$ would still have a double pole, so $A'=0$ . Calling the constant $C$ , we obtain

\lambda=C(x-u)\frac{dx}{y} =C\,\frac{y\,dx}{x^2-\Lambda^4}, \qquad \partial_u\lambda=-\frac C2\frac{dx}{y}. \tag{SW10.7}

This derives a representative within the declared minimal-pole ansatz. Adding the derivative of a single-valued meromorphic function does not change its closed periods. In a different representative the holomorphic derivative condition may hold only up to an exact differential, rather than literally as (SW10.7).

The local behavior checks the construction. At $x=u$ , write $x=u+\zeta^2$ ; then $\lambda$ is proportional to $\zeta^2\,d\zeta$ and has a double zero. At either fixed branch point it is regular. At infinity set $x=z^{-2}$ :

\lambda=-2C\,z^{-2} \frac{\sqrt{1-u z^2}}{\sqrt{1-\Lambda^4 z^4}}\,dz =\left(-2C z^{-2}+Cu+O(z^2)\right)dz. \tag{SW10.8}

There is a double pole but no $dz/z$ term, hence no residue. This is a meromorphic differential with controlled poles, not the holomorphic form $dx/y$ .

Normalize the electric mass parameter

The two legs of $\gamma_e$ contribute equally: both $dx$ and $y$ reverse sign on the return. Using the sheet table in (SW10.7),

a(P)=2C\int_{-s}^{s} \frac{\sqrt{P-x}}{\sqrt{s^2-x^2}}\,dx. \tag{SW10.9}

This is a complete cycle, so there is no additional hidden factor of two. For $P\gg s$ , the numerator is uniformly $\sqrt P[1+O(s/P)]$ . Since $\int_{-s}^{s}dx/\sqrt{s^2-x^2}=\pi$ , we find $a\sim2\pi C\sqrt P$ . Matching $a\sim\sqrt{2P}$ fixes

\boxed{\lambda_{\rm SW}= \frac{\sqrt2}{2\pi}(x-u)\frac{dx}{y}.} \tag{SW10.10}

This is the first paper’s (6.19), normalized by (6.21)–(6.24), with the mass scale restored.

The matching also provides a check beyond the leading power. The odd term in the expansion of $\sqrt{1-x/P}$ integrates to zero. The weighted average of $x^2$ is $s^2/2$ , so

\frac{a(P)}{\sqrt{2P}} =1-\frac{s^2}{16P^2} +O\!\left(\frac{s^4}{P^4}\right). \tag{SW10.11}

The small parameter is $\Lambda^4/P^2$ , as required by the weak-coupling expansion. The negative first correction follows from concavity of the square root as well as from the series. This short normalization check is not a calculation of the full period functions.

On $\gamma_D$ , the same choices make the period positive imaginary for real $P>s$ , and it vanishes as $P$ approaches $s$ . The integrand in its doubled interval is $iC\sqrt{(P-x)/(x^2-s^2)}\,dx$ . With $x=s+(P-s)t$ , the integral is proportional to $(P-s)/\sqrt s$ with a finite integral over $0<t<1$ . Thus the specified differential supplies the vanishing monopole period that topology alone could not guarantee. The full local expansion and period evaluation belong to the next calculation.

Why the kinetic coefficient stays positive

Let $\omega=dx/y$ and let its periods in the same oriented basis be $c_e=\oint_{\gamma_e}\omega$ and $c_D=\oint_{\gamma_D}\omega$ . Transport the cycles continuously as $u$ changes without crossing a degeneration. Differentiating their closed integrals in this local family gives

\frac{da_D}{du}=-\frac C2c_D,\qquad \frac{da}{du}=-\frac C2c_e,\qquad \tau_{\rm eff}=\frac{c_D}{c_e}. \tag{SW10.12}

The derivative ratio is essential. There is no replacement of it by $a_D/a$ .

Here is the positivity argument behind the derivative condition. Cut the torus along the oriented basis and choose a primitive of $\omega$ on the resulting polygon. Apply Stokes’ theorem to that primitive times $\overline\omega$ . Pairing opposite edges leaves their period jumps, giving

\frac i2\int_{E_u}\omega\wedge\overline\omega =\operatorname{Im}(\overline{c_e}\,c_D) =|c_e|^2\operatorname{Im}\!\left(\frac{c_D}{c_e}\right)>0. \tag{SW10.13}

In a local complex coordinate $z=X+iY$ , the integrand on the left is $|f(z)|^2\,dX\,dY$ when $\omega=f(z)\,dz$ . It is positive somewhere because $\omega$ is not zero. The identity also prevents $c_e=0$ for a primitive member of the basis. Therefore $\operatorname{Im}\tau_{\rm eff}>0$ at every smooth fiber in its valid local frame. At a degeneration the photon-only action still requires the light-field qualification of the earlier lessons.

Exercises

1. Check a scale change and a disappearing cut

Set $x=\Lambda^2X$ , $u=\Lambda^2U$ , $y=\Lambda^3Y$ . Find the overall scale of $\lambda_{\rm SW}$ and its periods. Then let $\Lambda\to0$ at fixed positive $u$ . The electric cut shrinks to a point: must the electric period vanish? Evaluate its limiting integral and identify which smoothness assumption fails at $\Lambda=0$ .

Hint 1

Count the powers of $\Lambda$ in $(x-u)dx/y$ . In the shrinking-cut integral, use $x=\Lambda^2\cos\theta$ before taking the limit.

Hint 2

The limiting polynomial has a repeated root. A contour can shrink while the limiting differential becomes singular there.

Solution

The differential is $\Lambda$ times $C(X-U)dX/\sqrt{(X^2-1)(X-U)}$ , so at fixed $U$ both periods scale as $\Lambda$ . This is not the same limit as fixed $u$ , in which $U\to\infty$ .

At fixed $u$ ,

a=2C\int_0^\pi\sqrt{u-\Lambda^2\cos\theta}\,d\theta \longrightarrow 2\pi C\sqrt u=\sqrt{2u}.

The limit is uniform in $\theta$ for sufficiently small $\Lambda$ . The electric period remains finite. At $\Lambda=0$ the curve is $y^2=x^2(x-u)$ , with a node at $x=y=0$ ; it is no longer the smooth genus-one fiber. The differential is singular on its normalization near that point. Applying a smooth-differential shrinking-cycle argument across this degeneration would be invalid.

2. Leading asymptotics are not enough

Consider the dimensionally allowed modification $\widehat\lambda=\lambda_{\rm SW}+\kappa\Lambda^2\,dx/y$ , where $\kappa\ne0$ is a dimensionless constant. Does matching only $a\sim\sqrt{2u}$ exclude it? Check instead its derivative near $x=u$ and its monopole period as real $u\downarrow s=\Lambda^2$ .

Hint 1

The added electric integral is of order $\Lambda^2/\sqrt u$ at large $u$ . For the derivative, differentiate $1/y$ while holding $x$ fixed.

Hint 2

On the monopole contour, $\oint dx/y=-2i\int_s^u dx/\sqrt{(x^2-s^2)(u-x)}$ . Set $x=s+(u-s)t$ and use $\int_0^1dt/\sqrt{t(1-t)}=\pi$ .

Solution

The leading electric period of the addition is $-2\pi\kappa\Lambda^2/\sqrt u$ on our chosen sheet. Its ratio to $\sqrt{2u}$ tends to zero, so leading asymptotic normalization alone does not exclude it.

However, $\partial_u\widehat\lambda=-C\,dx/(2y) +\kappa\Lambda^2\,dx/[2(x-u)y]$ . At $x=u+\zeta^2$ , the second term behaves as a nonzero multiple of $d\zeta/\zeta^2$ , and is not holomorphic.

More decisively, the specified monopole cycle gives

\oint_{\gamma_D}\frac{dx}{y} \longrightarrow-\frac{2\pi i}{\sqrt{2s}}, \qquad \widehat a_D(s)=-i\sqrt2\,\pi\kappa\Lambda\ne0.

The original period vanishes, but the added term does not. It therefore fails the same physical massless-monopole matching condition. A curve, one asymptotic power and a choice of units do not by themselves fix the whole physical differential.

3. Change the representative without changing the periods

Let $f=y/(x-u)$ , a single-valued meromorphic function on the surface, and set $\lambda'=\lambda_{\rm SW}+\eta\,df$ for constant dimensionless $\eta$ . Compute $df$ in the $x$ chart and show that every closed period avoiding its poles is unchanged. Explain why the pole in the resulting formula does not contradict the minimal ansatz used earlier.

Hint 1

Use $2y\,dy=P'(x)\,dx$ , where $P=(x^2-s^2)(x-u)$ .

Hint 2

The sign of $y$ is part of the surface point. Thus $f$ is a function on the surface even though it is not a single-valued function of $x$ alone.

Solution

Differentiating and substituting the curve equation gives

df=\left[\frac{P'(x)}{2y(x-u)} -\frac{y}{(x-u)^2}\right]dx =\frac{x^2-2ux+s^2}{2(x-u)y}\,dx.

Its mass dimension is one, as required. On a closed lifted contour, $\oint df=f_{\rm end}-f_{\rm start}=0$ , because the endpoint is the same surface point. The claim would fail for a projected path that ends on the opposite sheet.

The exact addition can introduce higher poles into a representative, but its residues and all closed periods vanish. The earlier calculation selected a convenient representative with minimal poles, not the only formula for the same period data. Also $\partial_u\lambda'=\partial_u\lambda_{\rm SW} +\eta\,d(\partial_u f)$ locally, so the holomorphic derivative condition survives modulo an exact form. Multiplying the entire differential by a new constant, in contrast, changes the periods and their physical mass normalization.

What has been fixed

The curve, scale convention, differential, base sheet and oriented cycles now form a calculable specification. Calculate the periods and effective coupling evaluates the period functions and derives the equation that allows their continuation away from the initial real vacuum. Those computations can test this specification; they must preserve its sheet choices and mass normalization.