Find the classical Coulomb branch of SU(2)

In the first Seiberg–Witten lesson, one constant scalar background gave masses to two gauge-field components. How many different vacua can we obtain this way? Counting every matrix with zero potential would count gauge copies as different answers. Choosing a real diagonal scalar would miss most of the family.

We will first solve the vacuum equation, then identify equivalent solutions. The result has one complex parameter. Its special point also tells us where the classical low-energy field content changes.

Solve the vacuum equation before taking a quotient

Keep the four-dimensional Lorentzian signature $(+---)$, pure $SU(2)$ gauge theory and eight real supercharges of the preceding lesson. N=1 notation makes four manifest. The canonical complex adjoint scalar is $\phi=\phi^at^a$, with

$$t^a=\frac{\sigma_a}{2},\qquad \operatorname{Tr}_2(t^at^b)=\frac12\delta^{ab},\qquad \mathscr D_\mu\phi=\partial_\mu\phi-ig[A_\mu,\phi].$$

Here $\sigma_a$ are the Pauli matrices, $\operatorname{Tr}_2$ is the ordinary fundamental trace, and $g>0$ is a fixed classical coupling. The scalar kinetic term is $2\operatorname{Tr}_2[(\mathscr D_\mu\phi)^\dagger\mathscr D^\mu\phi]$. We seek translation-invariant vacua on Minkowski space, with vanishing fermions and gauge field strength. There are no compact-space holonomies in this problem.

The auxiliary equations give $F_\Phi=0$ and $D=-g[\phi,\phi^\dagger]$. Consequently the scalar potential is

$$V(\phi)=g^2\operatorname{Tr}_2[\phi,\phi^\dagger]^2. \tag{SW02.1}$$

The commutator is Hermitian. If its real eigenvalues are $c_1,c_2$, its squared trace is $c_1^2+c_2^2$. Thus the zero-energy equation is precisely

$$[\phi,\phi^\dagger]=0. \tag{SW02.2}$$

This condition makes the raw $D$ auxiliary vanish, as well as the potential. Constant solutions with zero field strength preserve all eight supercharges. It is the pure-vector specialization of Tachikawa’s vacuum equations, section 2.2.

There is a useful elementary way to solve (SW02.2). Write $\phi=X+iY$, where $X,Y$ are traceless Hermitian matrices. Then

$$[\phi,\phi^\dagger]=-2i[X,Y].$$

Writing $X=\mathbf x\cdot\mathbf t$ and $Y=\mathbf y\cdot\mathbf t$ with real vectors gives

$$[\phi,\phi^\dagger]=2(\mathbf x\times\mathbf y)\cdot\mathbf t, \qquad V=2g^2|\mathbf x\times\mathbf y|^2 . \tag{SW02.3}$$

The two vectors must be parallel, with either allowed to vanish. Away from $\phi=0$ we can therefore write $\phi=b\,\mathbf n\cdot\mathbf t$, with $b$ complex and $\mathbf n$ a real unit vector. An $SU(2)$ conjugation rotates $\mathbf n$ to the third axis, leaving

$$\phi=b\,t^3= \begin{pmatrix}b/2&0\\0&-b/2\end{pmatrix}. \tag{SW02.4}$$

Equivalently, (SW02.2) says that $\phi$ is a normal matrix, so a unitary transformation diagonalizes it. Multiplying that transformation by an overall phase makes its determinant one without changing the conjugation. Tracelessness makes the two eigenvalues opposite. Both arguments establish diagonalization by an allowed gauge transformation, not merely by an arbitrary invertible matrix.

Identify the residual gauge copies

For $b\ne0$, the two eigenvalues in (SW02.4) are distinct. Any matrix commuting with this representative must preserve each eigenvector. Inside $SU(2)$ the commuting matrices are

$$T(\alpha)= \begin{pmatrix}e^{i\alpha/2}&0\\0&e^{-i\alpha/2}\end{pmatrix}, \qquad \alpha\sim\alpha+4\pi . \tag{SW02.5}$$

They form the unbroken $U(1)$ gauge group. The period is written in the original $SU(2)$ group; its center acts trivially on adjoint fields. This local scalar calculation does not change the specified global group to $SO(3)$.

Another allowed transformation exchanges the two eigenvectors:

$$w=i\sigma_2= \begin{pmatrix}0&1\\-1&0\end{pmatrix}, \qquad wt^3w^{-1}=-t^3. \tag{SW02.6}$$

Hence $b$ and $-b$ describe the same vacuum. Although $w^2=-\mathbf1$, that center element has trivial conjugation action, so the action on the diagonal coordinate has order two. This remaining identification is the Weyl group, $\mathbb Z_2$.

At a nonzero $b$, $w$ maps the chosen representative to a different representative of the same vacuum; it does not fix that matrix. The unbroken group is still (SW02.5). This distinction separates transformations that stabilize a representative from transformations that relate representatives.

There are no other identifications of diagonal values: conjugation preserves the unordered pair of eigenvalues $\{b/2,-b/2\}$. In particular, a general phase of $b$ cannot be gauged away. The classical scalar vacuum space is

$$\mathcal M_{\rm cl}=\mathbb C_b/(b\sim-b). \tag{SW02.7}$$

At the origin the stabilizer increases to all of $SU(2)$. Counting a generic orbit by subtracting three gauge directions would fail: its stabilizer already has dimension one. The nonzero vacuum configurations have four real parameters, two from $b$ and two from $\mathbf n$; their two-dimensional gauge orbit removes the latter, leaving two real vacuum parameters.

Normalize a coordinate that labels each vacuum once

The first Seiberg–Witten paper uses a rescaled scalar. Define

$$\phi_{\rm SW}=g\phi,\qquad a_{\rm cl}=gb,\qquad u=\operatorname{Tr}_2\phi_{\rm SW}^2 =g^2\operatorname{Tr}_2\phi^2. \tag{SW02.8}$$

On the diagonal representative this becomes

$$\phi_{\rm SW}=a_{\rm cl}t^3,\qquad u=\frac12a_{\rm cl}^2. \tag{SW02.9}$$

The dimensions are $[b]=[a_{\rm cl}]=1$ and $[u]=2$. The factor $1/2$ comes from the trace of $(t^3)^2$; it is not a convention that can be changed while retaining all other formulas. These are the scalar and invariant conventions in Seiberg–Witten, the discussion following (2.6).

Every complex $u$ has a square root. Its two choices give the same gauge orbit, and $u=0$ has just the zero vacuum. Conversely, two normal traceless matrices with the same $u$ have the same unordered eigenvalues and are unitarily gauge equivalent. Thus $u$ labels each classical vacuum once.

For holomorphic polynomials the quotient is equally explicit. A polynomial in $a_{\rm cl}$ invariant under $a_{\rm cl}\mapsto-a_{\rm cl}$ contains only even powers, so it is a polynomial in $u$. The quotient is a smooth complex plane in this invariant coordinate. This assertion about the coordinate and its polynomial functions does not assert smoothness of the physical kinetic metric at the origin.

Worked example: identify a vacuum without choosing its color axis

Let $b=m_0(1+2i)$, with $m_0>0$ a mass scale, and consider

$$\phi=\frac{b}{10} \begin{pmatrix}3&4\\4&-3\end{pmatrix}, \qquad U=\frac1{\sqrt5} \begin{pmatrix}2&-1\\1&2\end{pmatrix}.$$

The real matrix multiplying $b$ is Hermitian, so $\phi$ commutes with its conjugate. Direct multiplication gives $U^\dagger U=\mathbf1$, $\det U=1$ and $U^\dagger\phi U=bt^3$. Taking the invariant before or after this transformation gives

$$u=\frac{g^2m_0^2}{2}(-3+4i),\qquad |u|=\frac52g^2m_0^2.$$

The charged-vector threshold found previously is

$$M_W=\sqrt2\,g|b| =\sqrt2\,|a_{\rm cl}| =2\sqrt{|u|} =\sqrt{10}\,g m_0 . \tag{SW02.10}$$

The equality is classical. It is unchanged by either the color rotation or the Weyl sign. A changed complex phase of $b$ can leave this mass unchanged while changing $u$; a single mass therefore does not label the vacuum.

What remains light, and what happens at the origin?

The neutral components along $t^3$ give one photon, one complex scalar and two Weyl fermions. They form one N=2 vector multiplet with four bosonic and four fermionic propagating modes. The charged components have the threshold (SW02.10). At energies well below it, the remaining photon mediates a long-range Coulomb interaction, giving the Coulomb branch its name.

At the classical origin, $M_W$ vanishes and the full non-Abelian field content becomes light. The vacuum still has zero energy and preserves supersymmetry. What fails there is the description obtained by removing the charged fields.

The tree-level scalar kinetic term makes another part of this limit visible. In a local diagonal gauge, allowing a slowly varying $a_{\rm cl}$ gives

$$\mathcal L_{\rm scalar,kin} =\frac1{g^2}\partial_\mu a_{\rm cl}\, \partial^\mu\overline{a_{\rm cl}} .$$

Define the corresponding positive field-space line element by its kinetic coefficient. Since $du=a_{\rm cl}\,da_{\rm cl}$,

$$ds_{\rm cl}^2 =\frac1{g^2}|da_{\rm cl}|^2 =\frac{|du|^2}{2g^2|u|}, \qquad u\ne0 . \tag{SW02.11}$$

This is a metric on scalar values, not the spacetime metric. To interpret its apparent singularity, put $a_{\rm cl}=\rho e^{i\vartheta}$. The Weyl identification makes $\vartheta$ periodic with period $\pi$, and

$$ds_{\rm cl}^2=\frac1{g^2} \left(d\rho^2+\rho^2d\vartheta^2\right), \qquad \vartheta\sim\vartheta+\pi . \tag{SW02.12}$$

It is a flat cone away from its tip. A circle of proper radius $R=\rho/g$ has circumference $\pi R$, rather than $2\pi R$. The tip is at finite distance. Thus a smooth complex coordinate $u$ coexists with a singular classical quotient metric and an enhanced gauge stabilizer.

All these metric statements use the fixed classical coupling. They do not establish the quantum metric or the locations where the quantum theory develops light charged particles. Those require the quantum low-energy analysis.

The classical square root and the later electric period

On a local patch choose a branch and write

$$a_{\rm cl}=\sqrt{2u}. \tag{SW02.13}$$

For the positive real $u$ axis we choose the positive square root. No single-valued choice covers every loop around the punctured $u$ plane; the invariant $u$ remains well defined when the representative changes sign.

The quantum electric coordinate, conventionally denoted $a(u)$, is defined through the low-energy theory and later through a period integral. The classical calculation fixes its weak-coupling normalization:

$$a(u)\sim\sqrt{2u}, \qquad |u|\gg|\Lambda|^2 \quad\text{on the chosen weak-coupling branch}. \tag{SW02.14}$$

Here $\Lambda$ is the strong-interaction scale already introduced in the preceding lesson. The asymptotic relation is the one in Seiberg–Witten, (6.2). It is not an exact identity for the quantum function at every $u$. The quantum coordinate is the expectation value of the specified composite operator, $u=\langle\operatorname{Tr}_2\phi_{\rm SW}^2\rangle$, as explained in the paper’s concluding discussion of section 5. This retains a specified operator normalization; it does not identify a quantum fluctuating field with a constant diagonal matrix.

Tachikawa writes his classical scalar as $\Phi_T=\operatorname{diag}(a_T,-a_T)$ with a different kinetic normalization. The translation from the established canonical fields is $\Phi_T=\sqrt2g\phi$, hence $a_T=a_{\rm cl}/\sqrt2$. His $M_W=2|a_T|$ agrees with (SW02.10). A comparison of symbols called “$a$” without this dictionary would not. Tachikawa, (2.2.8)–(2.2.12)

Exercises

1. Does a vanishing invariant imply a vacuum?

Take $\phi=\kappa\begin{pmatrix}0&1\\0&0\end{pmatrix}$ with nonzero complex $\kappa$. Compute $u$, the commutator and the potential. Is this matrix a representative of the $u=0$ vacuum? Identify the missing hypothesis in an argument based only on its eigenvalues.

Hint 1

The square of this matrix vanishes, but its product with its Hermitian conjugate need not vanish.

Hint 2

The coordinate $u$ classified solutions of the vacuum equation. Check that equation before using the classification.

Solution

Here $\phi^2=0$, so $u=0$. However,

$$[\phi,\phi^\dagger] =|\kappa|^2 \begin{pmatrix}1&0\\0&-1\end{pmatrix}, \qquad V=2g^2|\kappa|^4>0 .$$

It is not a vacuum. It is also not unitarily gauge equivalent to zero: $\operatorname{Tr}_2(\phi^\dagger\phi)=|\kappa|^2$ is unchanged by unitary conjugation. The matrix has zero eigenvalues but is not normal and has no orthonormal eigenbasis. The diagonalization argument used (SW02.2); without it, the invariant does not classify all scalar configurations. At $\kappa=0$ the obstruction and the potential disappear, as they should.

2. Follow a representative around the invariant plane

For $r_0>0$ of mass dimension two, follow the closed path $u(t)=r_0e^{it}$, $0\le t\le2\pi$. Start at $a_{\rm cl}(0)=\sqrt{2r_0}$ and continue the square root continuously. Does the final representative describe another vacuum? Separately compare the vacua represented by $a_{\rm cl}$ and $ia_{\rm cl}$ at a nonzero starting point.

Hint 1

A continuous square root along this path has phase $t/2$. Resetting to a principal square root midway through the path changes the continuation.

Hint 2

Use the explicit matrix $w$ for a sign change. For multiplication by $i$, compare the invariant values instead.

Solution

The continuation is $a_{\rm cl}(t)=\sqrt{2r_0}e^{it/2}$, so its final value is $-a_{\rm cl}(0)$. The two endpoint representatives are related by $w$ and describe the same vacuum. The sign change records the two-to-one diagonal parametrization away from zero; it does not double the number of vacua.

In contrast, $a_{\rm cl}\mapsto ia_{\rm cl}$ sends $u\mapsto-u$. For $u\ne0$ these are different vacua, although their classical charged thresholds $2\sqrt{|u|}$ agree. A global symmetry can relate distinct vacua; that does not make them gauge copies. This exercise follows only the classical square root. Quantum electric–magnetic period transformations require additional data.

3. A finite-distance endpoint and an invalid low-energy truncation

Using the classical metric, calculate the distance along a radial path from $u=0$ to $u=r_0>0$ and the circumference of the circle $|u|=r_0$. For a fixed energy $E>0$, impose the threshold criterion $E/M_W\le\epsilon$ with $0<\epsilon\ll1$. Find the allowed range of $|u|$. Does finite distance to the origin justify retaining only the photon multiplet there?

Hint 1

Along a radial ray, $ds=dr/(g\sqrt{2r})$. Along the circle, $|du|=r_0\,dt$ with $t$ running through $2\pi$.

Hint 2

Use the threshold itself, $M_W=2\sqrt{|u|}$, to test the omitted fields. A field-space distance is not a mass gap.

Solution

The radial distance and circumference are

$$R=\int_0^{r_0}\frac{dr}{g\sqrt{2r}} =\frac{\sqrt{2r_0}}g,\qquad L=\int_0^{2\pi}\frac{r_0\,dt}{g\sqrt{2r_0}} =\frac{\pi\sqrt{2r_0}}g=\pi R.$$

Both are finite. The threshold condition instead requires

$$\frac{E}{2\sqrt{|u|}}\le\epsilon \quad\Longleftrightarrow\quad |u|\ge\frac{E^2}{4\epsilon^2}.$$

For example, $\epsilon=1/10$ requires $|u|\ge25E^2$. Approaching the classical origin at fixed nonzero $E$ violates this criterion. The charged fields become light and must be retained. The inequality is a classical scale-separation test; using perturbation theory in the quantum theory also requires weak coupling, and no claim about the quantum metric near the origin follows from the integral above.

We have obtained the classical vacuum family by solving a physical equation and removing gauge copies, with its normalization and light fields fixed. Next, construct a smooth monopole and derive its BPS energy bound. The boundary scalar direction identifies magnetic charge, while the energy square tests whether a field configuration saturates the algebraic bound.

Paper guides

Connect this lesson to the original papers and their notation.

• Reading the first Seiberg–Witten paper