 Solution

ID: past-exam-of-the-random-matrix-theory-course-of-the-university-of-oxford/2025/1/b/ii/solution

Write the four cyclic factors as

A = M_{ij}

B = M_{jk}

C = M_{k l}

and

D = M_{l i}

. The Gaussian moment pairing theorem gives three terms, corresponding to

(A B) (C D)

(A C) (B D)

and

(A D) (BC)

. We sum each term over

i, j, k, l

For

(A B) (C D)

, the entry covariance gives

i, j, k, l \sum E [A B] E [C D] = i, j, k, l \sum (δ_{ik} + δ_{ij} δ_{jk}) (δ_{ik} + δ_{k l} δ_{l i}) = n^{3} + 2 n^{2} + n .

Indeed the four products leave respectively three, two, two and one free indices. A cyclic relabelling gives exactly the same sum for

(A D) (BC)

For the crossing pairing

(A C) (B D)

i, j, k, l \sum E [A C] E [B D] = i, j, k, l \sum (δ_{ik} δ_{j l} + δ_{i l} δ_{jk}) (δ_{j l} δ_{ki} + δ_{ji} δ_{k l}) = n^{2} + 3 n .

The first product allows

i = k

and

j = l

freely, whereas each other product forces all four indices to coincide. Combining the three pairings and the trace normalization,

E [m_{4} (n)] = \frac{2 ( n ^{3} + 2 n ^{2} + n ) + ( n ^{2} + 3 n )}{n ^{3}} = 2 + \frac{5}{n} + \frac{5}{n ^{2}} ⟶ 2.

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