Exercises: Set 1
Q1. Matrices from U (2)/(U (1))2 can be parameterised
cos φ
sin φ
U=
−e−iα sin φ e−iα cos φ
where 0 < φ < π/2, 0 < α < 2π.
(a) Explain why the first entries in each column have been chosen to be real and
positive.
(b) Deduce that U † dU = −dU † U and thus conclude U † dU is equal to i times
an Hermitian matrix.
(c) Show that
2
sin φ dα
iγ
†
U dU = −i
−iγ̄
cos2 φ dα
where γ = dφ + i sin φ cos φ dα.
(d) From your answer to (c) deduce that (U † dU ) = sin φ cos φ dφ dα, clearly
stating the meaning you have given to (U † dU ).
Q2. According to Euler,

cos φ sin φ
− sin φ cos φ
0
0
an element of SO(3) can be decomposed



0
0
cos ψ
sin ψ 0
0 1
sin θ  − sin ψ cos ψ 0
0 0 cos θ
0
0
1
1 0 − sin θ cos θ
where 0 ≤ θ ≤ π, 0 ≤ φ, ψ < 2π.
(a) With the 3 columns denoted ~q1 , ~q2 , ~q3 , show that






cos θ cos ψ sin φ + cos φ sin ψ
sin θ sin φ
cos φ cos ψ − cos θ sin φ sin ψ
~q1 = − cos ψ sin φ − cos φ cos θ sin ψ  , ~q2 = cos φ cos θ cos ψ − sin φ sin ψ  , ~q3 = sin θ cos φ
− cos ψ sin θ
cos θ
sin θ sin ψ
(b) Explain why the invariant measure is equal to ~q1T d~q3 ∧ ~q2T d~q3 ∧ ~q1T d~q2 .
(c) Observe that ~q3 does not depend on ψ, and thus ~q1T d~q3 ∧ ~q2T d~q3 does not
contain dψ, implying that all dependence on dψ comes from ~q1T d~q2 . In particular,
only the term proportional to dψ is relevant to ~q1T d~q2 . Show that this term is equal
to


− cos θ sin ψ sin φ + cos φ cos ψ
− cos φ cos θ sin ψ − sin φ cos ψ  dψ
sin ψ sin θ
Take the dot product with ~q1T and simplify (using computer algebra) to down to
dψ.
(d) Noting that




cos θ sin φ
sin θ cos φ
d~q3 = cos θ cos φ dθ + − sin θ sin φ dφ,
− sin θ
0
with the help of computer algebra show
~q1T d~q3 = cos ψdθ + sin ψ sin θdφ
~q2T d~q3 = − sin ψdθ + cos ψ sin θdφ
(e) From the above working conclude ~q1T d~q3 ∧ ~q2T d~q3 ∧ ~q1T d~q2 = sin θdθdφdψ.
1
2
Q3. For U ∈ U (N ), the eigenvalue/ eigenvector decomposition reads U = V DV †
where D = diag (eiθ1 , . . . , eiθN ), and V is the matrix of eigenvectors.
(a) Show that
V † dU V = V † dV D + dD − DV dV † ,
and from this read off that the element in position (jk) of V † dU V is equal to
(eiθk − eiθj )~vj d~vk
while the element in position (jj) is ieiθj dθj .
(b) Use (a) to deduce that
Tr dU dU † =
N
X
(dθj )2 + 2
j=1
X
|eiθk − eiθj |2 ((~vj† d~vk )r )2 + ((~vj† d~vk )i )2
j<k
Q4. Show that for A, X, Y ∈ GLN (R), and with Y = AX,
(dY ) = (det A)N (dX)
Q5. (a) Let M ∈ GLN (R). In the singular value decomposition M = O1 DO2T ,
note that O2 can be interpreted as the matrix of eigenvectors of M T M and thus as
a member of O(N )/{−1, 1}.
(b) Use the fact that O1T dM O2 = O1T dO1 D + dD − DO2T dO2 , and the antisymmetry of OiT dOi to deduce that
Y
(dM ) = 2−N (O1T dO1 )(O2T dO2 )
(σj2 − σk2 )dσ1 · · · dσN
1≤j<k≤N
Q6. Define
Z
R
Z
dy1 · · ·
I(t) =
0
R
dyN δ t −
0
N
Y
Y
yl )
l=1
|yj2 − yk2 |
1≤j<k≤N
(a) Show that
Z
∞
ts−1 I(t) dt = 2−N RN s+N (N −1)
0
Z
1
Z
dx1 · · ·
0
1
dxN
0
N
Y
s/2−1
xl
l=1
Y
1≤j<k≤N
(b) By using the Selberg integral, deduce that for a suitable c
Z
N −1
Γ((s + j)/2)
RN (N −1) c+i∞ N s Y
R
ds
I(1) = 2−N
2πi
Γ((s
+ N + 1 + j)/2)
c−i∞
j=0
Q7. With s1 , s2 > 0, consider the decomposition
a11 a12
s1 r
cos θ
A=
=
Q,
Q=
a21 a22
0 s2
− sin θ
sin θ
cos θ
(a) Show that
ds1
dr + s1 dθ
dA Q =
−s2 dθ ds2 + rdθ
and thus (dA) = s2 ds1 ds2 drdθ.
R 2π R ∞
(b) Use (a) to show 0 dθ 0 ds2 δ(1 − s1 s2 ) (dA) = 2π dss12dr
T
1
|xk − xj |
3
Q8. (a) Let ~b1 , ~b2 be linearly independent vectors in R2 such that |~b1 | ≤ |~b2 |.
Show that the inequality 2|~b1 ·~b2 | ≤ |~b1 |2 is equivalent to the inequality ~b2 +n~b1 | ≥ |~b2
for all n ∈ Z.
(b) Suppose ~b1 , ~b2 are as in (a). Let ~u = n1~b1 + n2~b2 , n1 , n2 ∈ Z. Show that for
(n1 , n2 ) 6= (0, 0), |~u| ≥ |~b1 |, and for n1 6= 0, |~u| ≥ |~b2 |.
Q9. Consider the un-normalised measure 2πdr11 dr12 restricted to the region
2
2
2
r12
+ r22
≥ r11
and 2|r12 | ≤ r11 .
(a) Show that the volume of this region is equal to π/3.
(b) Show that the PDF of the distribution of the variable r11 is
12 s
− χs>1 (s2 − 1/s2 )1/2 , 0 < s < (4/3)1/4
π 2