Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 313 / 2 / a

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- Solution a

Solution

Away from a zero of the Higgs field, write

ϕ = e^{u /2 + i χ}, u = lo g ∣ ϕ ∣^{2} .

Separating real and imaginary parts of

(D_{x} + i D_{y}) ϕ = 0

gives

A_{x} = \partial_{x} χ + \frac{1}{2} \partial_{y} u, A_{y} = \partial_{y} χ - \frac{1}{2} \partial_{x} u .

Thus

B = - Δ u /2

away from zeros. Combining this with

B = (1 - e^{u}) /2

gives

Δ u + 1 - e^{u} = 0.

If

ϕ

has a zero of multiplicity

N

at the origin, its vortex number is the winding number

N

and

u = 2 N lo g r + O (1)

. Since

Δ lo g r = 2 π δ^{(2)} (x)

as a distributional identity, the complete Taubes equation is

Δ u + 1 - e^{u} = 4 π N δ^{(2)} (x) .

The boundary conditions for a single

N

u (r) = 2 N lo g r + O (1) (r \to 0), u (r) \to 0 (r \to \infty) .

The latter is the finite-energy condition

∣ ϕ ∣ \to 1

.

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