III Classical and Quantum Solitons

2Vortices

2.4 Bogomolny/self-dual vortices and Taubes’ theorem

As mentioned, we can think of the radial vortex solution as a collection of

vortices all superposed at the origin. Is it possible to have separated vortices all

over the plane? Naively, we would expect that the vortices exert forces on each

other, and so we don’t get a static solution. However, it turns out that in the

= 1 case, there do exist static solutions corresponding to vortices at arbitrary

locations on the plane.

This is not obvious, and the proof requires some serious analysis. We will

not do the analysis, which requires use of Sobolev spaces and PDE theory.

However, we will do all the non-hard-analysis part. In particular, we will obtain

Bogomolny bounds as we did in the sine-Gordon case, and reduce the problem

to finding solutions of a single scalar PDE, which can be understood with tools

from calculus of variations and elliptic PDEs.

Recall that for the sine-Gordon kinks, we needed to solve

= sin θ,

with boundary conditions

(

)

→

0 or 2

x → ±∞

. The only solutions we

found were

(x − X)

for any

X ∈ R

. This

is interpreted as the location of the kink. So the moduli

space of solutions is M = R.

We shall get a similar but more interesting description for the

= 1 vortices.

This time, the moduli space will be

, given by

complex parameters

describing the solutions.

Theorem

(Taubes’ theorem)

For

= 1, the space of (gauge equivalence classes

of) solutions of the Euler–Lagrange equations

δV

= 0 with winding number

is M

∼

To be precise, given

N ∈ N

and an unordered set of points

, ··· , Z

}

there exists a smooth solution

(

;

, ··· , Z

) and Φ(

;

, ··· , Z

) which

solves the Euler-Lagrange equations

δV

= 0, and also the so-called Bogomolny

equations

Φ + iD

Φ = 0, B =

(1 − |Φ|

Moreover, Φ has exactly

zeroes

, ··· , Z

counted with multiplicity, where

(using the complex coordinates z = x

+ ix

)

Φ(x; Z

, ··· , Z

) ∼ c

(z − Z

)

z → Z

, where

|{k

is the multiplicity and

is a nonzero

complex number.

This is the unique such solution up to gauge equivalence. Furthermore,

(A( ·, Z

, ··· , Z

), Φ( ·; Z

, ··· , Z

)) = πN (∗)

and

2π

B d

x = N = winding number.

Finally, this gives all finite energy solutions of the gauged Ginzburg–Landau

equations.

Note that it is not immediately clear from our description that the mod-

uli space is

. It looks more like

quotiented out by the action of the

permutation group

. However, the resulting quotient is still isomorphic to

. (However, it is important for various purposes to remember this quotient

structure, and to use holomorphic coordinates which are invariant under the

action of the permutation group — the elementary symmetric polynomials in

, . . . , Z

There is a lot to be said about this theorem. The equation (

∗

) tells us the

energy is just a function of the number of particles, and does not depend on

where they are. This means there is no force between the vortices. In situations

like this, it is said that the Bogomolny bound is saturated. The final statement

suggests that the topology is what is driving the existence of the vortices, as we

have already seen. The reader will find it useful to work out the corresponding

result in the case of negative winding number (in which case the holomorphicity

condition becomes anti-holomorphicity, and the sign of the magnetic field is

reversed in the Bogomolny equations).

Note that the Euler–Lagrange equations themselves are second-order equa-

tions. However, the Bogomolny equations are first order. In general, this is a

signature that suggests that interesting mathematical structures are present.

We’ll discuss three crucial ingredients in this theorem, but we will not

complete the proof, which involves more analysis than is a pre-requisite for this

course. The proof can be found in Chapter 3 of Jaffe and Taubes’s Vortices and

Monopoles.

Holomorphic structure

When there are Bogomolny equations, there is often some complex analysis

lurking behind. We can explicitly write the first Bogomolny equation as

Φ + iD

Φ =

∂Φ

∂x

+ i

∂Φ

∂x

− i(A

+ iA

)Φ = 0.

Recall that in complex analysis, holomorphic functions can be characterized as

complex-valued functions which are continuously differentiable (in the real sense)

and also satisfy the Cauchy–Riemann equations

∂f

∂¯z



∂f

∂x

+ i

∂f

∂x



= 0.

So we think of the first Bogomolny equation as the covariant Cauchy–Riemann

equations. It is possible to convert this into the standard Cauchy–Riemann

equations to deduce the local behaviour at Φ.

To do so, we write

Φ = e

Then

∂f

∂¯z

= e

−ω



∂Φ

∂¯z

−

∂ω

∂¯z



= e

−ω



i(A

+ iA

)

−

∂ω

∂¯z



Φ.

This is equal to 0 if ω satisfies

∂ω

∂¯z

= i

+ iA

So the question is — can we solve this? It turns out we can always solve this

equation, and there is an explicit formula for the solution. In general, if

smooth, then the equation

∂w

∂¯z

= β,

has a smooth solution in the disc {z : |z| < r}, given by

ω(z, ¯z) =

2πi

|w|<r

β(w)

w −z

dw ∧d ¯w.

A proof can be found in the book Griffiths and Harris on algebraic geometry,

on page 5. So we can write

Φ = e

where

is holomorphic. Since

is never zero, we can apply all our knowledge

of holomorphic functions to

, and deduce that Φ has isolated zeroes, where

Φ ∼ (z − Z

)

for some integer power n

The Bogomolny equations

We’ll now show that (

Φ) satisfies

(

Φ) =

πN ≥

0 iff it satisfies the

Bogomolny equations, i.e.

Φ + iD

Φ = 0, B =

(1 − |Φ|

We first consider the simpler case of the sine-Gordon equation. As in the

kinks, to find soliton solutions, we write the energy as

E =

∞

−∞



+ (1 − cos θ)



∞

−∞



+ 4 sin



∞

−∞



− 2 sin



+ 4θ

sin

∞

−∞



− 2 sin



dx +

∞

−∞

∂

∂x



−4 cos



∞

−∞



− 2 sin



dx +



−4 cos

θ(+∞)

+ 4 cos

θ(−∞)



We then use the kink asymptotic boundary conditions to obtain, say,

∞

) = 2

and

(

−∞

) = 0. So the boundary terms gives 8. Thus, subject to these boundary

conditions, we can write the sine-Gordon energy as

E =

∞

−∞



− 2 sin



dx + 8.

Thus, if we try to minimize the energy, then we know the minimum is at least 8,

and if we could solve the first-order equation

= 2

sin

, then the minimum

would be exactly 8. The solution we found does satisfy this first-order equation.

Moreover, the solutions are all of the form

θ(x) = θ

(x − X), θ

(x) = 4 arctan e

Thus, we have shown that the minimum energy is 8, and the minimizers are all

of this form, parameterized by X ∈ R.

We want to do something similar for the Ginzburg–Landau theory. In order

to make use of the discussion above of the winding number

, we will make

the same standing assumptions as used in that discussion, but it is possible

to generalize the conclusion of the following result to arbitrary finite energy

configurations with an appropriate formulation of the winding number.

Lemma. We have

(A, Φ) =



B −

(1 − |Φ|

)



+ 4|

∂

Φ|

x + πN,

where

∂

Φ =

Φ + iD

Φ).

It is clear that the desired result follows from this.

Proof. We complete the square and obtain

(A, Φ) =



B −

(1 − |Φ|

)



+ B(1 − |Φ|

) + |D

Φ|

+ |D

Φ|

We now dissect the terms one by one. We first use the definition of

∧

dA and integration by parts to obtain

(1 − |Φ|

) dA = −

d(1 − |Φ|

) ∧ A = 2

(Φ, DΦ) ∧ A.

Alternatively, we can explicitly write

B(1 − |Φ|

) d

x =

(∂

− ∂

)(1 − |Φ|

) d

∂

|Φ|

− A

∂

|Φ|

) d

= 2

(Φ, D

Φ) − A

(Φ, D

Φ).

Ultimately, we want to obtain something that looks like

∂

. We can write

this out as

Φ + iD

Φ, D

Φ + iD

Φ) = |D

Φ|

+ |D

Φ|

+ 2(D

Φ, iD

Φ).

We note that

Φ and Φ are always orthogonal, and

is always a real coefficient.

So we can write

Φ, iD

Φ) = (∂

Φ − iA

Φ, i∂

Φ + A

Φ)

= (∂

Φ, i∂

Φ) + A

(Φ, ∂

Φ) − A

(Φ, ∂

Φ).

We now use again the fact that (Φ

, i

Φ) = 0 to replace the usual derivatives with

the covariant derivatives. So we have

Φ, iD

Φ) = (∂

Φ, i∂

Φ) + A

(Φ, D

Φ) − A

(Φ, D

Φ).

This tells us we have



B(1 − |Φ|

) + |D

Φ|

+ |D

Φ|



x =



∂

Φ|

+ 2(∂

Φ, i∂

Φ)



It then remains to show that (∂

Φ, i∂

Φ) = j

(Φ). But we just write

(∂

+ i∂

, −∂

+ i∂

) = −(∂

, ∂

) + (∂

, ∂

)

= −j

(Φ)

= −det



∂



Then we are done.

Corollary.

For any (

Φ) with winding number

, we always have

(

Φ)

≥

πN, and those (A, Φ) that achieve this bound are exactly those that satisfy

∂

Φ = 0, B =

(1 − |Φ|

Reduction to scalar equation

The remaining part of Taubes’ theorem is to prove the existence of solutions to

these equations, and that they are classified by

unordered complex numbers.

This is the main analytic content of the theorem.

To do so, we reduce the two Bogomolny equations into a scalar equation.

Note that we have

Φ + iD

Φ = (∂

Φ + i∂

Φ) − i(A

+ iA

)Φ = 0.

So we can write

+ iA

= −i(∂

+ i∂

) log Φ.

Thus, once we’ve got Φ, we can get A

and A

The next step is to use gauge invariance. Under gauge invariance, we can fix

the phase of Φ to anything we want. We write

Φ = e

(u+iθ)

Then |Φ|

= e

We might think we can get rid of

completely. However, this is not quite

true, since the argument is not well-defined at a zero of Φ, and in general we

cannot get rid of θ by a smooth gauge transformation. But since

Φ ∼ c

(z − Z

)

near Z

, we expect we can make θ look like

θ = 2

j=1

arg(z − Z

We will assume we can indeed do so. Then we have

(∂

u + ∂

θ), A

= −

(∂

u − ∂

θ).

We have now solved for

using the first Bogomolny equation. We then use

this to work out

and obtain a scalar equation for

by the second Bogomolny

equation.

Theorem.

In the above situation, the Bogomolny equation

− |

) is

equivalent to the scalar equation for u

−∆u + (e

− 1) = −4π

j=1

This is known as Taubes’ equation.

Proof. We have

B = ∂

− ∂

= −

∂

u −

∂

u +

(∂

∂

− ∂

∂

)θ

= −

∆u +

(∂

∂

− ∂

∂

)θ.

We might think the second term vanishes identically, but that is not true. Our

has some singularities, and so that expression is not going to vanish at the

singularities. The precise statement is that (

∂

− ∂

∂

)

is a distribution

supported at the points Z

To figure out what it is, we have to integrate:

|z−Z

|≤ε

(∂

∂

− ∂

∂

)θ d

x =

|z−Z

|=ε

∂

θ dx

+ ∂

θ dx

|z−Z

|=ε

dθ = 4πn

where n

is the multiplicity of the zero. Thus, we deduce that

(∂

∂

− ∂

∂

)θ = 2π

But then we are done!

We can think of this

as a non-linear combination of fundamental solutions

to the Laplacian. Near the

functions, the

−

1 term doesn’t contribute much,

and the solution looks like the familiar fundamental solutions to the Laplacian

with logarithmic singularities. However, far away from the singularities,

−

forces u to tend to 0, instead of growing to infinity.

Taubes proved that this equation has a unique solution, which is smooth on

\{Z

}

, with logarithmic singularities at

, and such that

u →

0 as

|z| → ∞

Also, u < 0.

It is an exercise to check that the Bogomolny equations imply the second-order

Euler–Lagrange equations.

For example, differentiating the second Bogomolny equation and using the

first gives

∂

B = −(Φ, D

Φ) = (Φ, iD

Φ).

We can similarly do this for the sine-Gordon theory.