Paper 313 (Solitons, Instantons and Geometry)

, which is possible for a nonconstant finite-energy configuration. The oppositely scaling magnetic and potential terms therefore evade the usual Derrick obstruction and allow vortex solitons.

At infinity write

ϕ \to e^{i χ}

. The condition

D ϕ \to 0

gives

A \to d χ

, and the phase winding defines the Abelian Higgs vortex number

N = \frac{1}{2 π} \oint_{S_{\infty}^{1}} d χ \in Z .

(4)

By Stokes theorem,

\int_{R^{2}} F = \oint_{S_{\infty}^{1}} A = 2 π N .

(5)

Solved by gpt-5.6-sol high.

b

 0  0

i

 0  0

Solution

 0  0

Away from zeros of

ϕ

, write

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

(1)

The first Bogomolny vortex equation,

(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 .

(2)

If the zeros

z_{r}

have multiplicities

N_{r}

, the phase curl and the logarithmic singularities give, distributionally,

B = 2 π \sum_{r} N_{r} δ^{(2)} (z - z_{r}) - \frac{1}{2} Δ u .

(3)

Equating this with

Ω (1 - e^{u}) /2

yields the Taubes equation

Δ u + Ω (1 - e^{u}) = 4 π r \sum N_{r} δ^{(2)} (z - z_{r}) .

(4)

Solved by gpt-5.6-sol high.

ii

 0  0

Solution

 0  0

For a zero of order

N

at the origin and no other zeros,

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

(1)

Away from the zero, the flat Taubes equation is

Δ u = e^{u} - 1.

(2)

u

had a positive interior maximum, then the second-derivative test would give

Δ u \leq 0

there, while

e^{u} - 1 > 0

, a contradiction. The boundary values are

- \infty

at the zero and

0

at infinity, so the maximum principle gives

u \leq 0

. Hence

∣ ϕ ∣ = e^{u /2} \leq 1 .

(3)

Solved by gpt-5.6-sol high.

iii

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Solution

 0  0

The surface area is

Area (Σ) = \int_{R^{2}} \frac{a ^{2}}{( 1 + ∣ z ∣ ^{2} ) ^{2}} d x d y = 2 π a^{2} \int_{0}^{\infty} \frac{r d r}{( 1 + r ^{2} ) ^{2}} = π a^{2} .

(1)

Integrating the Taubes equation over the compact surface eliminates the Laplacian and gives

4 π N = \int_{Σ} Ω (1 - e^{u}) d^{2} x < Area (Σ)

(2)

for a nontrivial solution. For

N = 2

, the Bradlow bound therefore requires

π a^{2} > 8 π, a^{2} > 8 .

(3)

Equality is the dissolved-vortex limit with identically vanishing Higgs field and does not give the stipulated ordinary two-vortex solution.

Solved by gpt-5.6-sol high.

2

 0  0

a

 0  0

Solution

 0  0

For the curvature

F = d A + A \land A

, define the Second Chern form

C_{2} = \frac{1}{8 π ^{2}} Tr (F \land F) .

(1)

Using the graded cyclicity of the trace and

d^{2} = 0

d Tr (A \land d A) = Tr (d A \land d A),

(2)

while

d Tr (A \land A \land A) = 3 Tr (d A \land A \land A) .

(3)

Expanding

Tr (F \land F)

gives the same two terms with coefficient two on

d A \land A \land A

; the quartic term has vanishing trace by graded cyclicity. Therefore

C_{2} = d Y, Y = \frac{1}{8 π ^{2}} Tr (A \land d A + \frac{2}{3} A \land A \land A),

(4)

α = 2/3

. This is the Chern-Simons 3-form.

Solved by gpt-5.6-sol high.

b

 0  0

Solution

 0  0

Set

ω = d g g^{- 1}, A = g A g^{- 1} - ω .

(1)

The right Maurer-Cartan equation is

d ω = ω \land ω

. Direct substitution gives

F = g F g^{- 1},

(2)

so trace invariance proves

C_{2} = C_{2}

For the Chern-Simons 3-form, expansion and graded cyclicity give

Y (A) - Y (A) = \frac{1}{8 π ^{2}} d Tr (ω \land g A g^{- 1}) + \frac{1}{24 π ^{2}} Tr (ω^{3}) .

(3)

Since cyclicity also gives

Tr (ω \land g A g^{- 1}) = Tr (g^{- 1} d g \land A),

(4)

the required two-form may be chosen as

T = \frac{1}{8 π ^{2}} Tr (g^{- 1} d g \land A) .

(5)

Thus

Y ⟼ Y + d T + \frac{1}{24 π ^{2}} Tr [(d g g^{- 1})^{3}] .

(6)

Solved by gpt-5.6-sol high.

c

 0  0

Solution

 0  0

Finite Euclidean Yang-Mills action requires

F \to 0

sufficiently rapidly at spatial infinity. Consequently the connection approaches a pure gauge,

A ⟶ - d g g^{- 1}

(1)

under the convention of the question. Compactifying the asymptotic boundary identifies it with

S_{\infty}^{3}

, while

S U (2)

is itself topologically

S^{3}

. Thus

g ∣_{S_{\infty}^{3}}

has an integer topological degree.

C_{2} = d Y

and Stokes theorem, the instanton number is

k = \int_{R^{4}} C_{2} = \int_{S_{\infty}^{3}} Y (- d g g^{- 1}) = \frac{1}{24 π ^{2}} \int_{S_{\infty}^{3}} Tr [(d g g^{- 1})^{3}] = de g (g),

(2)

up to the common simultaneous choice of trace and orientation signs. Smoothness and finite action are imposed in the interior, and

F = \pm^{⋆} F

selects an instanton or anti-instanton representative of the topological sector.

Solved by gpt-5.6-sol high.

3

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a

 0  0

Solution

 0  0

For a matrix Lie group, the left-invariant Maurer-Cartan form is

ρ = g^{- 1} d g .

(1)

Differentiating

g^{- 1} g = I

gives

d (g^{- 1}) = - g^{- 1} (d g) g^{- 1}

. Hence

d ρ = d (g^{- 1}) \land d g = - g^{- 1} d g \land g^{- 1} d g = - ρ \land ρ,

(2)

and therefore

d ρ + ρ \land ρ = 0 .

(3)

Solved by gpt-5.6-sol high.

b

 0  0

Solution

 0  0

Substitute

ρ = σ^{α} T_{α}

into the Maurer-Cartan equation. Antisymmetry of the wedge product gives

ρ \land ρ = \frac{1}{2} σ^{α} \land σ^{β} [T_{α}, T_{β}] = \frac{1}{2} c^{γ}_{α β} σ^{α} \land σ^{β} T_{γ} .

(1)

Equating coefficients of

T_{γ}

yields

d σ^{γ} = - \frac{1}{2} α, β \sum c^{γ}_{α β} σ^{α} \land σ^{β},

(2)

f^{γ}_{α β} = - c^{γ}_{α β} /2

Solved by gpt-5.6-sol high.

c

 0  0

Solution

 0  0

Represent the real affine group by

g (a, b) = (e^{a} 0 b 1) .

(1)

Matrix multiplication reproduces

(a, b) (a^{'}, b^{'}) = (a + a^{'}, b + e^{a} b^{'}) .

(2)

The Maurer-Cartan form is

g^{- 1} d g = (d a 0 e^{- a} d b 0),

(3)

so a basis of left-invariant one-forms is

σ^{1} = d a, σ^{2} = e^{- a} d b .

(4)

The dual left-invariant vector fields are

E_{1} = \partial_{a}, E_{2} = e^{a} \partial_{b} .

(5)

Indeed

σ^{i} (E_{j}) = δ_{j}^{i}

, and left translation preserves the one-forms and vector fields.

Solved by gpt-5.6-sol high.

d

 0  0

Solution

 0  0

The metric is the left-invariant metric

g = d a^{2} + e^{- 2 a} d b^{2} .

(1)

Its right-invariant vector fields are

R_{1} = \partial_{a} + b \partial_{b}, R_{2} = \partial_{b} .

(2)

Their flows act by left translations, which preserve a left-invariant metric. Directly,

L_{R_{1}} g = 0, L_{R_{2}} g = 0,

(3)

so both are Killing vector fields and generate one-parameter isometry groups.

There is an additional Killing field. Put

y = e^{a} > 0

and

x = b

; then

g = \frac{d x ^{2} + d y ^{2}}{y ^{2}},

(4)

the hyperbolic plane of constant curvature

- 1

. Its isometry algebra is three-dimensional, whereas the space of right-invariant fields here is two-dimensional. For example, the third independent Killing field can be written

(b^{2} - e^{2 a}) \partial_{b} + 2 b \partial_{a},

(5)

which is not right invariant. Hence the answer is yes.

Solved by gpt-5.6-sol high.

Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 313 (Solitons, Instantons and Geometry)

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Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 313 (Solitons, Instantons and Geometry)

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