Electric–magnetic duality as a change of variables

An electric photon can be an inconvenient variable near a vacuum where a magnetically charged particle is light. How can a magnetic description retain the same scalar geometry and mass formula? The answer begins with an ordinary Maxwell calculation, before any strong-coupling solution is known.

Read the low-energy prepotential first. We remain in pure four-dimensional N=2 $SU(2)$ theory, with eight real supercharges and four manifest in N=1 superspace. On a regular Coulomb-branch patch the light fields form one Abelian vector multiplet. The two-derivative description applies below the masses of the omitted charged fields. Here there are no explicit electric or magnetic current insertions.

First make the field strength independent

Use signature $(+---)$, $\epsilon_{0123}=+1$, and hence $\epsilon^{0123}=-1$. For two antisymmetric tensors define

$$(*f)^{\mu\nu}=\frac12\epsilon^{\mu\nu\rho\sigma}f_{\rho\sigma}, \qquad f\mathbin{\cdot}h=f_{\mu\nu}h^{\mu\nu}, \qquad **f=-f. \tag{SW06.1}$$

Indices on $*f$ are lowered with the same metric when necessary. Thus $(*f)^2=-f^2$ and $f\cdot *h=h\cdot *f$. Write $\tau=s+it$, with $s,t$ real and $t>0$. The inherited photon Lagrangian is

$$\mathcal L_\gamma(f) =\frac{-t f^2+s f\cdot *f}{16\pi}, \qquad \tau=\frac{\theta}{2\pi}+\frac{4\pi i}{g_{\rm eff}^2}. \tag{SW06.2}$$

Initially $f=dv$. The connection $v$ has the coupling absorbed into it: at the microscopic tree-level matching point $v=gA^3$ and $a=gb$, where the canonically normalized adjoint scalar is $b\,t^3$ and $\operatorname{Tr}_2(t^3t^3)=1/2$. Equation (SW06.2) is already an Abelian component expression; no additional trace factor belongs in it.

Temporarily allow all six components of $f$ to vary independently. Introduce a second connection $v_D$, with $f_D=dv_D$, and set

$$\boxed{\displaystyle \mathcal L_{\rm first} =\frac{-t f^2+s f\cdot *f-2f\cdot *f_D}{16\pi}.} \tag{SW06.3}$$

The coefficient fixes the normalization of $v_D$. We will check it against the supersymmetric action rather than import a Maxwell convention with a different electric unit. All terms have mass dimension four: $[v]=[v_D]=1$, $[f]=[f_D]=2$, and $[\tau]=0$.

Take compactly supported variations, or boundary conditions that remove the following surface term. Integration by parts gives

$$-\frac1{8\pi}\int d^4x\,f\cdot *f_D =\frac1{4\pi}\int d^4x\,v_{D\nu}\partial_\mu(*f)^{\mu\nu} \quad\text{up to a boundary term}. \tag{SW06.4}$$

Varying $v_D$ therefore imposes the Bianchi identity $\partial_\mu(*f)^{\mu\nu}=0$, equivalently $df=0$. On a contractible spacetime patch a closed two-form is $dv$. Substituting this solution recovers the original local action, since the mixed term is then a boundary term. A closed form need not be globally exact; we will return to that limitation.

The normalization also has a charge test. In the first paper’s units a magnetic unit has flux $\int_{S^2}\mathbf B\cdot d\mathbf S=4\pi$, where $B_i=\epsilon_{ijk}f_{jk}/2$ and $\epsilon_{123}=+1$. Equation (SW06.4) couples $v_{D0}$ to $\nabla\cdot\mathbf B/(4\pi)$, giving that unit dual electric charge one. This uses the charge convention of SW1, (3.13); it does not yet sum over global flux sectors.

Instead eliminate $f$. Since the quadratic terms each contribute twice on variation,

$$-t f+s*f-*f_D=0, \qquad f_D=sf+t*f, \qquad f=\frac{s f_D-t*f_D}{s^2+t^2}. \tag{SW06.5}$$

The middle equality follows by applying $*$, including its minus sign on the second application. Let $d=s^2+t^2$. Then

$$f^2=\frac{(s^2-t^2)f_D^2-2st\,f_D\cdot *f_D}{d^2}, \qquad f\cdot *f_D=\frac{s f_D\cdot *f_D+t f_D^2}{d}. \tag{SW06.6}$$

At the stationary point the quadratic part of (SW06.3) equals $f\cdot *f_D/(16\pi)$; its mixed part is minus twice that. Consequently

$$\mathcal L_D =\frac{-\frac{t}{d}f_D^2-\frac{s}{d}f_D\cdot *f_D}{16\pi}, \qquad \boxed{\displaystyle\tau_D=-\frac1\tau.} \tag{SW06.7}$$

Substitution into $\mathcal L_\gamma$ alone would miss the mixed term and give the wrong answer. This is a Legendre transformation of the quadratic action, not a relabeling inside its original formula.

The algebra remains valid when $s$ and $t$ depend on the scalar and hence on spacetime. The original Maxwell equation is $\partial_\mu(t f^{\mu\nu}-s(*f)^{\mu\nu})=0$. By (SW06.5) it becomes the Bianchi identity for $f_D$. Conversely the original Bianchi identity becomes the dual Maxwell equation. Derivatives act on $s$ and $t$ too. Although (SW06.5) relates field strengths locally on solutions, finding $v_D$ requires solving $dv_D=f_D$: it is not a local off-shell substitution $v_D=v_D(v)$.

The same transformation for the whole multiplet

The supersymmetric construction in Seiberg–Witten, (3.15)–(3.21) treats the chiral spinor $W_\alpha$ as independent and imposes its Bianchi constraint with a real vector superfield. In the normalization of the preceding lesson the gauge part becomes

$$\frac1{8\pi}\operatorname{Im}\int d^2\theta\, \left[\tau(A)W^\alpha W_\alpha-2W_D^\alpha W_\alpha\right]. \tag{SW06.8}$$

Its field-strength component is precisely (SW06.3). Indeed the mixed chiral product contributes $-f\cdot f_D/2+i f\cdot *f_D/2$; taking the imaginary part of its real coefficient $-2$ gives the required mixed term. The independent chiral-spinor equation is $W_D=\tau(A)W$. Completing the square in (SW06.8) leaves $-W_D^2/\tau(A)$, with the same overall factor. This argument uses the constrained-superfield formulation; it does not assert that multiplying an arbitrary off-shell field strength by $\tau(A)$ automatically preserves its Bianchi identity.

Now transform the chiral scalar coordinate. Put

$$b=a_D=\mathcal F'(a),\qquad \mathcal F_D(b)=\mathcal F(a)-ab. \tag{SW06.9}$$

The inverse $a=a(b)$ exists locally because $db/da=\tau$ and $\operatorname{Im}\tau>0$. Differentiating the whole expression, including that inverse, gives

$$d\mathcal F_D=b\,da-b\,da-a\,db=-a\,db, \qquad b_D=\mathcal F_D'(b)=-a, \qquad\mathcal F_D''(b)=-\frac1{\tau(a(b))}. \tag{SW06.10}$$

This is the holomorphic Legendre transform explained in Bilal, (4.1)–(4.8). Its scalar term also agrees: $\operatorname{Im}(\bar b b_D)=\operatorname{Im}(\bar a a_D)$. In differential form,

$$\frac{\operatorname{Im}\tau_D}{4\pi}|db|^2 =\frac{t}{4\pi|\tau|^2}|\tau\,da|^2 =\frac{t}{4\pi}|da|^2. \tag{SW06.11}$$

Thus the photon and scalar coupling transformations agree, and the positive scalar metric survives. At zero theta angle, $\tau=4\pi i/g_{\rm eff}^2$ gives $g_D=4\pi/g_{\rm eff}$. At nonzero theta it is the complex $\tau$ that inverts, not the real coupling alone.

A field and charge test

For $f_{03}=E$, $f_{12}=B$, our orientation gives $(*f)_{03}=B$ and $(*f)_{12}=-E$. Choose $\tau=1+2i$ and, in a common field-strength unit, $E=1$, $B=3$. Then

$$E_D=E+2B=7,\qquad B_D=B-2E=1,\qquad \tau_D=\frac{-1+2i}{5}. \tag{SW06.12}$$

The original density is $-11/(4\pi)$, the mixed first-order term is $11/(2\pi)$, and their sum is $11/(4\pi)$, exactly the dual density. Constant fields here test the pointwise algebra; they are not assumed to obey falloff conditions on all of Minkowski space.

To test the scalar and charge signs, choose the illustrative prepotential $\mathcal F=(1+2i)a^2/2$. This constant-coupling model is not the exact pure-$SU(2)$ prepotential. The period column and its new version obey

$$\Pi=\begin{pmatrix}a_D\\a\end{pmatrix},\qquad \Pi'=\begin{pmatrix}b_D\\b\end{pmatrix} =S\Pi,\qquad S=\begin{pmatrix}0&-1\\1&0\end{pmatrix}. \tag{SW06.13}$$

This defines our sign for $S$. The generator printed in SW1 (3.11) and Bilal (4.10) is $-S$; it gives the same coupling inversion but the opposite signs for both new periods. Mixing those two choices in a charge formula would be an error.

A charge is a row $\gamma=(n_m,n_e)$, with central charge $Z_\gamma=\gamma\Pi=n_ma_D+n_ea$ and BPS inequality $M\geq\sqrt2|Z_\gamma|$. Invariance requires the inverse change of charge labels:

$$\gamma'=\gamma S^{-1}=(-n_e,n_m),\qquad \gamma'\Pi'=\gamma\Pi. \tag{SW06.14}$$

For $a=(1+i)m_0$, $m_0>0$, we have $a_D=(-1+3i)m_0$. The label $(2,-1)$ gives $Z=(-3+5i)m_0$ in either frame: in the dual frame its label is $(1,2)$ and $Z=b_D+2b$. The bound is $\sqrt{68}\,m_0$. This computation establishes a frame-invariant bound for that label, not the existence of a particle saturating it. The central-charge statement and the separate question of a BPS spectrum are distinguished in SW1, section 4.

What the local calculation does not integrate over

The Lorentzian calculation above is a classical local equivalence. A Euclidean Gaussian clarifies both its contour and its quantum limit. Continue $x^0=-ix^4$, $v_0=iv_4$, and use $\epsilon_{1234}=+1$. Then $f_L\cdot *_Lf_L=i f_E\cdot *_Ef_E$, and the continued first-order density is

$$\mathcal L_{{\rm first},E} =\frac{t f^2-is f\cdot *_Ef+2i f\cdot *_Ef_D}{16\pi}, \qquad *_E^2=1. \tag{SW06.15}$$

On the real $f$ contour its real quadratic part is positive. For real $f_D$ the stationary value $f=(s f_D-it*_Ef_D)/d$ is generally complex. At a finite Gaussian cutoff the contour can be shifted to this saddle: the integrand is entire and decays along the translated real directions. The resulting dual density is $[t f_D^2+is f_D\cdot *_Ef_D]/(16\pi d)$, as required by (SW06.7).

There is also a prefactor. For one Euclidean cell of four-volume $\Delta V$, define dimensionless components $\mathsf f=\sqrt{\Delta V}\,f$ and similarly $\mathsf f_D$. Set $\mathsf f_{i4}=\sqrt{4\pi}\,x_i$ and $\mathsf f_{jk}=\sqrt{4\pi}\,\epsilon_{ijk}y_i$, with $\epsilon_{123}=+1$. Write the similarly scaled dual components as $u_i,v_i$. One of the three identical two-variable integrals is

$$\begin{aligned} I(u,v)&=\int_{\mathbb R^2}dx\,dy\, e^{-\frac t2(x^2+y^2)+isxy-i(xv+yu)}\\ &=\frac{2\pi}{\sqrt d} \exp\!\left[-\frac{t(u^2+v^2)+2isuv}{2d}\right]. \end{aligned} \tag{SW06.16}$$

The rotation $(x,y)\mapsto((x+y)/\sqrt2,(x-y)/\sqrt2)$ reduces this to two convergent one-dimensional Gaussians. Its exponent checks the dual action and its prefactor shows why an unqualified equality of normalized partition functions would be premature. A continuum theory additionally needs gauge fixing, a regulator, boundary data, sums over flux sectors, and the associated measure and local terms.

Likewise $df=0$ alone does not impose quantized fluxes on a spacetime with two-cycles. The compact gauge group and its allowed electric and magnetic probes fix those data. A real matrix of determinant one can preserve the local scalar metric without preserving the charge lattice. Even an integral change of basis must be interpreted with the chosen global form and line operators; it need not act as a symmetry of a fixed global theory. For the distinction between probes, dynamical particles and bundle sums, see Aharony–Seiberg–Tachikawa, section 1.1. These qualifications do not obstruct the local variable change. They prevent it from being mistaken for a proof of microscopic $SU(2)$ S-duality or a construction of every stable charged state.

Exercises

1. Do not freeze a varying coupling inside a derivative

In the Maxwell sector, prescribe $s=x$, $t=2+x>0$ on $x>-2$, where $x=x^1/L$ is dimensionless and $L>0$. Let $f_{01}=E(x)$, $f_{23}=B=2$, with all other independent components zero. Fix the electric displacement constant by $tE-sB=1$. Find $E$, show that both original equations hold, and compute the dual magnetic component. Does this construct a complete solution of the dynamical scalar theory?

Hint 1

The Bianchi identity differentiates $f_{23}$ in the $x^1$ direction. The nontrivial Maxwell equation instead differentiates $tE-sB$.

Hint 2

The $(23)$ component of $f_D=sf+t*f$ is $sB-tE$. Do not discard derivatives of $s$ or $t$.

Solution

The prescribed displacement gives $E=(1+2x)/(2+x)$. The two-form is closed: $B$ is constant and differentiating $E(x)\,dx^0\wedge dx^1$ produces a repeated $dx^1$. Maxwell’s only nontrivial equation is $\partial_1(tE-sB)=\partial_1(1)=0$. In the dual description, $E_D=xE+2(2+x)$ and $B_D=2x-(2+x)E=-1$. Thus its Bianchi identity holds. Its Maxwell equation follows by the inverse transformation, or by substituting the displayed fields into the dual coefficients $s_D=-s/d$, $t_D=t/d$.

If one incorrectly replaces the Maxwell equation by $t\,\partial_1 E=0$, this solution appears to fail: $dE/dx=3/(2+x)^2\ne0$. The source of the discrepancy is the omitted coefficient derivatives. We prescribed a coupling profile only to test electromagnetic dualization. A full scalar-photon solution must also satisfy the scalar equation and its boundary conditions; that has not been shown.

2. A nonconstant Legendre transform

Let $\mathcal F(a)=ca^2/2+\kappa a^3/(6M)$, with $c=iT$, $T>0$, $M>0$, and real $\kappa$. Work sufficiently near $a=0$ that $\operatorname{Im}\tau>0$ and $|\kappa a/M|\ll T$. Invert $b=\mathcal F'(a)$ through order $b^2$, find $\mathcal F_D$ through order $b^3$, and check the transformed coupling and scalar metric.

Hint 1

Insert $a=\alpha b+\beta b^2+O(b^3)$ into $b=ca+\kappa a^2/(2M)$ and compare powers.

Hint 2

Differentiate $\mathcal F_D=\mathcal F-ab$. Its derivative must equal $-a(b)$; this checks the cubic sign independently of substitution.

Solution

The coefficients give $\alpha=1/c$ and $\beta=-\kappa/(2Mc^3)$. Therefore

$$a(b)=\frac bc-\frac{\kappa b^2}{2Mc^3}+O(b^3),\qquad \mathcal F_D(b)=-\frac{b^2}{2c}+\frac{\kappa b^3}{6Mc^3}+O(b^4).$$

Terms omitted here carry the appropriate powers of $M$ and the dimensionless constants; the expansion is about $b=0$ with $c,\kappa,M$ fixed. Differentiating twice gives $\tau_D=-1/c+\kappa b/(Mc^3)+O(b^2)$. Expanding $-1/[c+\kappa a(b)/M]$ gives exactly the same result. Both prepotentials have dimension two and both periods dimension one.

The exact local identity $db=\tau\,da$ and $\operatorname{Im}(-1/\tau)=\operatorname{Im}\tau/|\tau|^2$ prove equality of the metrics, without truncating. At $a=b=0$, the dual coefficient is $1/T$ and $|db|^2=T^2|da|^2$, giving the original coefficient $T$. Positivity is a local assumption here, not a claim about this cubic model on the whole complex plane.

3. A good local transformation can fail a global test

For real $\lambda$, consider $\Pi'=T_\lambda\Pi$, where $T_\lambda=\begin{pmatrix}1&\lambda\\0&1\end{pmatrix}$. Find the coupling, prepotential and charge transformation. Test the integer particle-charge lattice containing $(1,0)$ and $(0,1)$ when $\lambda=1/2$. Explain what changes, and what is still unproved, if $\lambda=1$.

Hint 1

The electric coordinate stays $a$. Differentiate the new magnetic period $a_D+\lambda a$.

Hint 2

The charge row transforms with $T_\lambda^{-1}$. Test it on a magnetic unit before drawing any conclusion from the positive scalar metric.

Solution

Here $\tau'=\tau+\lambda$ and the new prepotential is $\widehat{\mathcal F}=\mathcal F+\lambda a^2/2$. The scalar metric is unchanged because $\lambda$ is real; the photon action changes by $\lambda f\cdot *f/(16\pi)$, a theta term. The inverse charge map is $(n_m,n_e)\mapsto(n_m,n_e-\lambda n_m)$. For $\lambda=1/2$, $(1,0)$ becomes $(1,-1/2)$, outside the declared integer particle lattice. Local metric preservation was insufficient.

For $\lambda=1$, both the matrix and its inverse preserve that integer lattice and the antisymmetric pairing $n_m\widetilde n_e-n_e\widetilde n_m$. This is a necessary lattice test, not the full global argument: the allowed line operators can contain additional probe information, and theta periodicity must be evaluated with the actual bundle sum. Nor does a lattice element guarantee a stable particle or a short multiplet. A frame change transports labels of states that exist; it does not populate the spectrum.

You can now replace the electric variables by magnetic ones without changing the local scalar metric or any specified central charge. Continue with what becomes massless at strong coupling to determine how a light charged state changes the local effective action.

Paper guides

Connect this lesson to the original papers and their notation.

• Reading the first Seiberg–Witten paper