Stone M., Goldbart P. Mathematics for Physics: A Guided Tour for Graduate Students
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15.2 Geometry of SU(2) 535
where a, b and c are now allowed to be complex, but with the restriction that S = I +A
be unitary. This requires A to be skew-hermitian, so a =−a
†
, and c =−b
†
, while b
(and hence c) remains symmetric. There are n
2
free real parameters in a, and n(n + 1)
in b,so
dim Sp(n) = (n
2
) + n(n + 1) = n(2n + 1),
as before.
Exercise 15.1: Show that
SO(2N ) ∩ Sp(2N , R)
∼
=
U(N ).
Hint: group the 2N basis vectors on which O(2N ) acts into pairs x
n
and y
n
, n = 1, ..., N .
Assemble these pairs into z
n
= x
n
+ iy
n
and ¯z = x
n
− iy
n
. Let ω be the linear map that
takes x
n
→ y
n
and y
n
→−x
n
. Show that the subset of SO(2N ) that commutes with ω
mixes z
i
’s only with z
i
’s and ¯z
i
’s only with ¯z
i
’s.
15.2 Geometry of SU(2)
To get a sense of Lie groups as geometric objects, we will study the simplest non-trivial
case, SU(2), in some detail.
A general 2-by-2 complex matrix can be parametrized as
U =
x
0
+ ix
3
ix
1
+ x
2
ix
1
− x
2
x
0
− ix
3
. (15.16)
The determinant of this matrix is unity, provided
(x
0
)
2
+ (x
1
)
2
+ (x
2
)
2
+ (x
3
)
2
= 1. (15.17)
When this condition is met, and if in addition the x
i
are real, the matrix is unitary:
U
†
= U
−1
. The group manifold of SU(2) can therefore be identified with the 3-sphere
S
3
. We will take as local coordinates x
1
, x
2
, x
3
. When we desire to know x
0
we will find
it from x
0
=
1 − (x
1
)
2
− (x
2
)
2
− (x
3
)
2
. This coordinate chart only labels the points
in the half of the 3-sphere having x
0
> 0, but this is typical of any non-trivial manifold.
A complete atlas of charts can be constructed if needed.
We can simplify our notation by using the Pauli sigma matrices
=σ
1
=
01
10
, =σ
2
=
0 −i
i 0
, =σ
3
=
10
0 −1
. (15.18)
These obey
[=σ
i
,=σ
j
]=2i
ijk
=σ
k
, and σ
i
=σ
j
+=σ
j
=σ
i
= 2δ
ij
I. (15.19)
536 15 Lie groups
In terms of them, we can write the general element of SU(2) as
g = U = x
0
I + ix
1
=σ
1
+ ix
2
=σ
2
+ ix
3
=σ
3
. (15.20)
Elements of the group in the neighbourhood of the identity differ from e ≡ I by real
linear combinations of the i=σ
i
. The three-dimensional vector space spanned by these
matrices is therefore the tangent space TG
e
at the identity element. For any Lie group,
this tangent space is called the Lie algebra, g = Lie G of the group. There will be a
similar set of matrices i
=
λ
i
for any matrix group. They are called the generators of the
Lie algebra, and satisfy commutation relations of the form
[i
=
λ
i
, i
=
λ
j
]=−f
k
ij
(i
=
λ
k
), (15.21)
or equivalently
[
=
λ
i
,
=
λ
j
]=if
k
ij
=
λ
k
. (15.22)
The f
k
ij
are called the structure constants of the algebra. The “i”’s associated with the
=
λ’s in this expression are conventional in physics texts because for quantum mechanics
application we usually desire the
=
λ
i
to be hermitian. They are usually absent in books
aimed at mathematicians.
Exercise 15.2: Let
=
λ
1
and
=
λ
2
be hermitian matrices. Show that if we define
=
λ
3
by the
relation [
=
λ
1
,
=
λ
2
]=i
=
λ
3
, then
=
λ
3
is also a hermitian matrix.
Exercise 15.3: For the group O(n) the matrices “i
=
λ” are real n-by-n skew symmetric
matrices A. Show that if A
1
and A
2
are real skew symmetric matrices, then so is [A
1
, A
2
].
Exercise 15.4: For the group Sp(2n, R) the i
=
λ matrices are of the form
A =
ab
c −a
T
where a is any real n-by-n matrix and b and c are symmetric (c
T
= c and b
T
= b) real
n-by-n matrices. Show that the commutator of any two matrices of this form is also of
this form.
15.2.1 Invariant vector fields
Consider a matrix group, and in it a group element I + i
=
λ
i
lying close to the identity
e ≡ I . Draw an arrow connecting I to I +i
=
λ
i
, and regard this arrow as a vector L
i
lying
in TG
e
. Next, map the infinitesimal element I + i
=
λ
i
to the neighbourhood an arbitrary
group element g by multiplying on the left to get g(I + i
=
λ
i
). By drawing an arrow
from g to g(I + i
=
λ
i
), we obtain a vector L
i
(g) lying in TG
g
. This vector at g is the
15.2 Geometry of SU(2) 537
push-forward of the vector at e by left multiplication by g. For example, consider SU(2)
with infinitesimal element I + i=σ
3
. We find
g(I + i=σ
3
) = (x
0
+ ix
1
=σ
1
+ ix
2
=σ
2
+ ix
3
=σ
3
)(I + i=σ
3
)
= (x
0
− x
3
) + i=σ
1
(x
1
− x
2
) + i=σ
2
(x
2
+ x
1
) + i=σ
3
(x
3
+ x
0
).
(15.23)
This computation can also be interpreted as showing that the multiplication of g ∈ SU(2)
on the right by (I +i=σ
3
) displaces the point g, changing its x
i
parameters by an amount
δ
⎛
⎜
⎜
⎝
x
0
x
1
x
2
x
3
⎞
⎟
⎟
⎠
=
⎛
⎜
⎜
⎝
−x
3
−x
2
x
1
x
0
⎞
⎟
⎟
⎠
. (15.24)
Knowing how the displacement looks in terms of the x
1
, x
2
, x
3
coordinate system let us
read off the ∂/∂x
µ
components of a vector L
3
lying in TG
g
:
L
3
=−x
2
∂
1
+ x
1
∂
2
+ x
0
∂
3
. (15.25)
Since g can be any point in the group, we have constructed a globally defined vector
field L
3
that acts on a function F(g) on the group manifold as
L
3
F(g) = lim
→0
1
[
F
(
g(I + i=σ
3
)
)
− F(g)
]
. (15.26)
Similarly, we obtain
L
1
= x
0
∂
1
− x
3
∂
2
+ x
2
∂
3
L
2
= x
3
∂
1
+ x
0
∂
2
− x
1
∂
3
. (15.27)
The vector fields L
i
are said to be left invariant because the push-forward of the
vector L
i
(g) lying in the tangent space at g by multiplication on the left by any g
produces a vector g
∗
[L
i
(g)] lying in the tangent space at g
g, and this pushed-forward
vector coincides with the L
i
(g
g) already there. We can express this statement tersely as
g
∗
L
i
= L
i
.
Using ∂
i
x
0
=−x
i
/x
0
, i = 1, 2, 3, we can compute the Lie brackets and find
[L
1
, L
2
]=−2L
3
. (15.28)
In general
[L
i
, L
j
]=−2
ijk
L
k
, (15.29)
538 15 Lie groups
which coincides with the matrix commutator of the i=σ
i
.
This construction works for all Lie groups. For each basis vector L
i
in the tangent
space at the identity e, we push it forward to the tangent space at g by left multiplication
by g, and so construct the global left-invariant vector field L
i
. The Lie bracket of these
vector fields will be
[L
i
, L
j
]=−f
k
ij
L
k
, (15.30)
where the coefficients f
k
ij
are guaranteed to be position independent because (see Exer-
cise 12.5) the operation of taking the Lie bracket of two vector fields commutes with
the operation of pushing-forward the vector fields. Consequently, the Lie bracket at any
point is just the image of the Lie bracket calculated at the identity. When the group is a
matrix group, this Lie bracket will coincide with the commutator of the i
=
λ
i
, that group’s
analogue of the i=σ
i
matrices.
The exponential map
Recall that given a vector field X ≡ X
µ
∂
µ
we define associated flow by solving the
equation
dx
µ
dt
= X
µ
(x(t)). (15.31)
If we do this for the left-invariant vector field L, with initial condition x(0) = e, we obtain
a t-dependent group element g(x(t)), which we denote by Exp (tL). The symbol “Exp ”
stands for the exponential map which takes elements of the Lie algebra to elements of the
Lie group. The reason for the name and notation is that for matrix groups this operation
corresponds to the usual exponentiation of matrices. Elements of the matrix Lie group
are therefore exponentials of matrices in the Lie algebra. To see this, suppose that L
i
is
the left-invariant vector field derived from i
=
λ
i
. Then the matrix
g(t) = exp(it
=
λ
i
) ≡ I + it
=
λ
i
−
1
2
t
2
=
λ
2
i
− i
1
3!
t
3
=
λ
3
i
+··· (15.32)
is an element of the group, and
g(t + ) = exp(it
=
λ
i
) exp(i
=
λ
i
) = g(t)
I + i
=
λ
i
+ O(
2
)
. (15.33)
From this we deduce that
d
dt
g(t) = lim
→0
1
[g(t)(I + i
=
λ
i
) − g(t)]
= L
i
g(t). (15.34)
Since exp(it
=
λ) = I when t = 0, we deduce that Exp (tL
i
) = exp(it
=
λ
i
).
15.2 Geometry of SU(2) 539
Right-invariant vector fields
We can also use multiplication on the right to push forward an infinitesimal group
element. For example:
(I + i=σ
3
)g = (I + i=σ
3
)(x
0
+ ix
1
=σ
1
+ ix
2
=σ
2
+ ix
3
=σ
3
)
= (x
0
− x
3
) + i=σ
1
(x
1
+ x
2
) + i=σ
2
(x
2
− x
1
) + i=σ
3
(x
3
+ x
0
)
(15.35)
This motion corresponds to the right-invariant vector field
R
3
= x
2
∂
1
− x
1
∂
2
+ x
0
∂
3
. (15.36)
Similarly, we obtain
R
1
= x
0
∂
1
+ x
3
∂
2
− x
2
∂
3
,
R
2
=−x
3
∂
1
+ x
0
∂
2
+ x
1
∂
3
, (15.37)
and find that
[R
1
, R
2
]=+2R
3
. (15.38)
In general,
[R
i
, R
j
]=+2
ijk
R
k
. (15.39)
For any Lie group, the Lie brackets of the right-invariant fields will be
[R
i
, R
j
]=+f
ij
k
R
k
(15.40)
whenever
[L
i
, L
j
]=−f
ij
k
L
k
(15.41)
are the Lie brackets of the left-invariant fields. The relative minus sign between the
bracket algebra of the left- and right-invariant vector fields has the same origin as the
relative sign between the commutators of space- and body-fixed rotations in classical
mechanics. Because multiplication from the left does not interfere with multiplication
from the right, the left and right invariant fields commute:
[L
i
, R
j
]=0. (15.42)
540 15 Lie groups
15.2.2 Maurer–Cartan forms
Suppose that g is an element of a group and dg denotes its exterior derivative. Then
the combination dg g
−1
is a Lie-algebra-valued 1-form. For example, starting from the
elements of SU(2)
g = x
0
+ ix
1
=σ
1
+ ix
2
=σ
2
+ ix
3
=σ
3
g
−1
= g
†
= x
0
− ix
1
=σ
1
− ix
2
=σ
2
− ix
3
=σ
3
(15.43)
we compute
dg = dx
0
+ idx
1
=σ
1
+ idx
2
=σ
2
+ idx
3
=σ
3
= (x
0
)
−1
(−x
1
dx
1
− x
2
dx
2
− x
3
dx
3
) + idx
1
=σ
1
+ idx
2
=σ
2
+ idx
3
=σ
3
. (15.44)
From this we find
dgg
−1
= i=σ
1
(x
0
+ (x
1
)
2
/x
0
)dx
1
+ (x
3
+ (x
1
x
2
)/x
0
)dx
2
+(−x
2
+ (x
1
x
3
)/x
0
)dx
3
+ i=σ
2
(−x
3
+ (x
2
x
1
)/x
0
)dx
1
+ (x
0
+ (x
2
)
2
/x
0
)dx
2
+(x
1
+ (x
2
x
3
)/x
0
)dx
3
+ i=σ
3
(x
2
+ (x
3
x
1
)/x
0
)dx
1
+ (−x
1
+ (x
3
x
2
)/x
0
)dx
2
+(x
0
+ (x
3
)
2
/x
0
)dx
3
.
(15.45)
Observe that the part proportional to the identity matrix has cancelled. The result of
inserting a vector X
i
∂
i
into dg g
−1
is therefore an element of the Lie algebra of SU(2).
This is what we mean when we say that dg g
−1
is Lie-algebra-valued.
For a general group, we define the (right-invariant) Maurer–Cartan forms ω
i
R
as being
the coefficient of the Lie algebra generator i
=
λ
i
. Thus, for SU(2), we have
dgg
−1
= ω
R
= (i=σ
i
)ω
i
R
. (15.46)
If we evaluate the 1-form ω
1
R
on the right-invariant vector field R
1
, we find
ω
1
R
(R
1
) = (x
0
+ (x
1
)
2
/x
0
)x
0
+ (x
3
+ (x
1
x
2
)/x
0
)x
3
+ (−x
2
+ (x
1
x
3
)/x
0
)(−x
2
)
= (x
0
)
2
+ (x
1
)
2
+ (x
2
)
2
+ (x
3
)
2
= 1. (15.47)
15.2 Geometry of SU(2) 541
Working similarly, we find
ω
1
R
(R
2
) = (x
0
+ (x
1
)
2
/x
0
)(−x
3
) + (x
3
+ (x
1
x
2
)/x
0
)x
0
+ (−x
2
+ (x
1
x
3
)/x
0
)x
1
= 0. (15.48)
In general, we discover that ω
i
R
(R
j
) = δ
i
j
. The Maurer–Cartan forms therefore constitute
the dual basis to the right-invariant vector fields.
We may similarly define the left-invariant Maurer–Cartan forms by
g
−1
dg = ω
L
= (i=σ
i
)ω
i
L
. (15.49)
These obey ω
i
L
(L
j
) = δ
i
j
, showing that the ω
i
L
are the dual basis to the left-invariant
vector fields.
Acting with the exterior derivative d on gg
−1
= I tells us that d(g
−1
) =−g
−1
dgg
−1
.
By exploiting this fact, together with the anti-derivation property
d(a ∧ b) = da ∧ b + (−1)
p
a ∧ db,
we may compute the exterior derivative of ω
R
. We find that
dω
R
= d(dgg
−1
) = (dgg
−1
) ∧ (dgg
−1
) = ω
R
∧ ω
R
. (15.50)
A matrix product is implicit here. If it were not, the product of the two identical 1-forms
on the right would automatically be zero. When we make this matrix structure explicit,
we see that
ω
R
∧ ω
R
= ω
i
R
∧ ω
j
R
(i=σ
i
)(i=σ
j
)
=
1
2
ω
i
R
∧ ω
j
R
[i=σ
i
, i=σ
j
]
=−
1
2
f
k
ij
(i=σ
k
)ω
i
R
∧ ω
j
R
, (15.51)
so
dω
k
R
=−
1
2
f
k
ij
ω
i
R
∧ ω
j
R
. (15.52)
These equations are known as the Maurer–Cartan relations for the right-invariant forms.
For the left-invariant forms we have
dω
L
= d(g
−1
dg) =−(g
−1
dg) ∧ (g
−1
dg) =−ω
L
∧ ω
L
, (15.53)
or
dω
k
L
=+
1
2
f
k
ij
ω
i
L
∧ ω
j
L
. (15.54)
542 15 Lie groups
The Maurer–Cartan relations appear in the physics literature when we quantize gauge
theories. They are one part of the BRST transformations of the Fadeev–Popov ghost
fields. We will provide a further discussion of these transformations in the next chapter.
15.2.3 Euler angles
In physics it is common to use Euler angles to parametrize the group SU(2). We can
write an arbitrary SU(2) matrix U as a product
U = exp{−iφ=σ
3
/2}exp{−iθ=σ
2
/2}exp{−iψ=σ
3
/2},
=
e
−iφ/2
0
0 e
iφ/2
cos θ/2 −sin θ/2
sin θ/2 cos θ/2
e
−iψ/2
0
0 e
iψ/2
,
=
e
−i(φ+ψ)/2
cos θ/2 −e
i(ψ −φ)/2
sin θ/2
e
i(φ−ψ)/2
sin θ/2 e
i(ψ +φ)/2
cos θ/2
. (15.55)
Comparing with the earlier expression for U in terms of the coordinates x
µ
, we obtain
the Euler-angle parametrization of the 3-sphere
x
0
= cos θ/2 cos(ψ + φ)/2,
x
1
= sin θ/2 sin(φ − ψ)/2,
x
2
=−sin θ/2 cos(φ −ψ)/2,
x
3
=−cos θ/2 sin(ψ + φ)/2. (15.56)
When the angles are taken in the range 0 ≤ φ<2π,0≤ θ<π,0≤ ψ<4π we cover
the entire 3-sphere exactly once.
Exercise 15.5: Show that the Hopf map, defined in Chapter 3, Hopf : S
3
→ S
2
is the
“forgetful” map (θ, φ, ψ) → (θ , φ), where θ and φ are spherical polar coordinates on
the 2-sphere.
Exercise 15.6: Show that
U
−1
dU =−
i
2
=σ
i
i
L
,
where
1
L
= sin ψ dθ − sin θ cos ψ dφ,
2
L
= cos ψ dθ + sin θ sin ψ dφ,
3
L
= dψ + cos θ dφ.
15.2 Geometry of SU(2) 543
Observe that these 1-forms are essentially the components
ω
X
= sin ψ
˙
θ − sin θ cos ψ
˙
φ,
ω
Y
= cos ψ
˙
θ + sin θ sin ψ
˙
φ,
ω
Z
=
˙
ψ + cos θ
˙
φ
of the angular velocity ω of a body with respect to the body-fixed XYZ axes in the
Euler-angle conventions of Exercise 11.17.
Similarly, show that
dUU
−1
=−
i
2
=σ
i
i
R
,
where
1
R
=−sin φ dθ + sin θ cos ψ dψ,
2
R
= cos φ dθ + sin θ sin ψ dψ,
3
R
= dφ + cos θ dψ .
Compute the components ω
x
, ω
y
, ω
z
of the same angular velocity vector ω, but now
taken with respect to the space-fixed xyz frame. Compare your answer with the
i
R
.
15.2.4 Volume and metric
The manifold of any Lie group has a natural metric, which is obtained by transporting
the Killing form (see Section 15.3.2) from the tangent space at the identity to any other
point g by either left or right multiplication by g. For a compact group, the resultant
left- and right-invariant metrics coincide. In the particular case of SU(2) this metric is
the usual metric on the 3-sphere.
Using the Euler angle expression for the x
µ
to compute the dx
µ
, we can express the
metric on the sphere as
ds
2
= (dx
0
)
2
+ (dx
1
)
2
+ (dx
2
)
2
+ (dx
3
)
2
,
=
1
4
dθ
2
+ cos
2
θ/2(dψ + dφ)
2
+ sin
2
θ/2(dψ − dφ)
2
,
=
1
4
dθ
2
+ dψ
2
+ dφ
2
+ 2 cos θ dφ dψ
. (15.57)
Here, to save space, we have used the traditional physics way of writing a metric. In
the more formal notation, where we think of the metric as being a bilinear function, we
would write the last line as
g( , ) =
1
4
(
dθ ⊗ dθ + dψ ⊗ dψ + dφ ⊗ dφ + cos θ(dφ ⊗dψ + dψ ⊗ dφ)
)
.
(15.58)
544 15 Lie groups
From (15.58) we find
g = det (g
µν
) =
1
4
3
!
!
!
!
!
!
10 0
0 1 cos θ
0 cos θ 1
!
!
!
!
!
!
=
1
64
(1 − cos
2
θ) =
1
64
sin
2
θ. (15.59)
The volume element,
√
gdθdφdψ , is therefore
d(Volume) =
1
8
sin θdθ dφdψ, (15.60)
and the total volume of the sphere is
Vo l(S
3
) =
1
8
π
0
sin θdθ
2π
0
dφ
4π
0
dψ = 2π
2
. (15.61)
This volume coincides, for d = 4, with the standard expression for the volume of S
d−1
,
the surface of the d-dimensional unit ball,
Vo l(S
d−1
) =
2π
d/2
(
d
2
)
. (15.62)
Exercise 15.7: Evaluate the Maurer–Cartan form ω
3
L
in terms of the Euler angle
parametrization, and hence show that
iω
3
L
=
1
2
tr (=σ
3
U
−1
dU ) =−
i
2
(dψ + cos θ dφ).
Now recall that the Hopf map takes the point on the 3-sphere with Euler angle coordinates
(θ, φ, ψ) to the point on the 2-sphere with spherical polar coordinates (θ , φ). Thus, if
we set A =−dψ − cos θ dφ, then we find
F ≡ dA = sin θ dθ dφ = Hopf
∗
d
7
Area S
2
8
.
Also observe that
A ∧ F =−sin θ dθ dφ dψ .
From this, show that the Hopf index of the Hopf map itself is equal to
1
16π
2
S
3
A ∧ F =−1.