Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 307 / 4

Table of contents
- i 4
  - Solution i
- ii 4
  - Solution ii
- iii 4
  - Solution iii
- iv 4
  - Solution iv
- v 4
  - Solution v
- vi 4
  - Solution vi
- vii 4
  - Solution vii
- viii 4
  - Solution viii

i

Solution

The fermion in a chiral multiplet has R-charge

R [Φ] - 1

, while the gluino has charge one. Cancellation of the

Sp (N_{c})^{2} U (1)_{R}

anomaly gives

I (adj) + 2 N_{f} I (□) (R [Φ] - 1) = 0.

Using

I (adj) = 2 (N_{c} + 1)

and

I (□) = 1

yields

R [Φ] = 1 - \frac{N _{c} + 1}{N _{f}} = \frac{N _{f} - N _{c} - 1}{N _{f}} .

ii

Solution

The one-loop coefficient is

b_{0} = \frac{3}{2} 2 (N_{c} + 1) - \frac{1}{2} (2 N_{f}) = 3 (N_{c} + 1) - N_{f} .

Thus asymptotic freedom is lost at

N_{f} = 3 (N_{c} + 1),

and the theory is infrared free for larger

N_{f}

.

iii

Solution

In a four-dimensional supersymmetric conformal field theory, a chiral primary operator in four-dimensional N=1 supersymmetry obeys

Δ = 3 R /2

. The meson has R-charge

R [M] = 2 R [Φ]

, hence

Δ (M) = 3 R [Φ] = 3 (1 - \frac{N _{c} + 1}{N _{f}}) .

At

N_{f} = 3 (N_{c} + 1)

this gives

Δ (M) = 2 .

iv

Solution

A free scalar has scaling dimension one. Setting the meson dimension to one gives

3 (1 - \frac{N _{c} + 1}{N _{f}}) = 1, N_{f} = \frac{3}{2} (N_{c} + 1) .

The expected interacting supersymmetric conformal window is therefore

\frac{3}{2} (N_{c} + 1) < N_{f} < 3 (N_{c} + 1) .

v

Solution

The magnetic gauge group is

Sp (N_{c})

with

N_{c} = N_{f} - N_{c} - 2

. Repeating the gauge-anomaly cancellation gives

R [q] = 1 - \frac{N _{c} + 1}{N _{f}} = \frac{N _{c} + 1}{N _{f}} .

vi

Solution

The magnetic superpotential

W \sim qMq

must have R-charge two, so

R [M] = 2 - 2 R [q] = 2 (1 - \frac{N _{c} + 1}{N _{f}}) .

Therefore

Δ (M) = \frac{3}{2} R [M] = 3 (1 - \frac{N _{c} + 1}{N _{f}}),

exactly matching the R-charge and dimension of the electric meson

Φ^{i} Φ^{j}

.

vii

Solution

For the magnetic theory,

b_{0}^{mag} = 3 (N_{c} + 1) - N_{f} = 2 N_{f} - 3 (N_{c} + 1) .

It ceases to be asymptotically free when

N_{f} \leq \frac{3}{2} (N_{c} + 1) .

This is precisely the lower edge of the electric conformal window. Inside the window both descriptions flow to the same interacting infrared fixed point; below it the magnetic variables provide a weakly coupled infrared description.

viii

Solution

The electric quarks give

2 N_{c}

copies of the

S U (2 N_{f})

fundamental, so

A_{el} [S U (2 N_{f})^{3}] = 2 N_{c} .

In the magnetic theory the

2 N_{c}

color components of

q

transform as flavor antifundamentals and contribute

- 2 N_{c}

. The singlet

M^{ij}

is the two-index antisymmetric representation of

S U (2 N_{f})

, whose cubic anomaly coefficient is

2 N_{f} - 4

. Hence

A_{mag} [S U (2 N_{f})^{3}] = - 2 (N_{f} - N_{c} - 2) + (2 N_{f} - 4) = 2 N_{c} = A_{el} .

This is a direct anomaly check of Seiberg duality.

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