Download Lie algebras and Lie groups, Homework 3 solutions

Lie algebras and Lie groups, Homework 3 solutions
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1. Representations
In all the cases we show that ϕ([x, y]) = [ϕ(x), ϕ(y)] where ϕ is the representation. We simply write RV (x)v = x · v, and respectively for w ∈ W .
a) ϕ⊕ ≡ RV ⊕ RW , then
[ϕ⊕ (x), ϕ⊕ (y)](v ⊕ v) = (ϕ⊕ (x) · ϕ⊕ (y) − ϕ⊕ (y) · ϕ⊕ (x))(v ⊕ w)
= x · y · v ⊕ x · y · w − (y · x · v ⊕ y · x · w)
= (x · y · v − y · x · v) ⊕ (x · y · w − y · x · w)
= [x, y] · v ⊕ [x, y] · w = ϕ⊕ ([x, y])(v ⊕ w)
b) ϕ⊗ ≡ RV ⊗ RW , then
[ϕ⊗ (x), ϕ⊗ (y)](v ⊗ w) = ϕ⊗ (x)(y · v ⊗ w + v ⊗ y · w)−
− ϕ⊗ (y)(x · v ⊗ w + v ⊗ x · w)
= x · y · v ⊗ w + y · v ⊗ x · w + x · v ⊗ y · w + v ⊗ x · y · w−
−y · x · v ⊗ w − y · v ⊗ x · w − x · v ⊗ y · w − v ⊗ y · x · w
= [x, y] · v ⊗ w + v ⊗ [x, y] · w
= ϕ⊕ ([x, y])(v ⊗ w)
c) Again ϕ⊗ ≡ RV ⊗ RW . The map defined on the sheet is not even linear
ϕ⊗ (x + y)(v ⊗ w) = (x + y) · v ⊗ (x + y) · w
=x·v⊗x·w+x·v⊗y·w+y·v⊗x·w+y·v⊗y·w
6= x · v ⊗ x · w + y · v ⊗ y · w = (ϕ⊗ (x) + ϕ⊗ (y))(v ⊗ w) ,
or even simpler
ϕ⊗ (ax)(v ⊗ w) = ax · v ⊗ ax · w = a2 (x · v ⊗ x · w) = a2 ϕ⊗ (x)(v ⊗ w)
6= aϕ⊗ (x)(v ⊗ w) if a2 6= a ,
and, thus, it is no Lie-algebra representation.
2. The complex Lie algebra so(3)
a) Let us start with sl(2, C) whose definition in terms of matrices can be
given by {A ∈ gl(2, C) | Tr A = 0}. It is 3-dim. and a suitable basis is given
by
1 0
0 1
0 0
h=
, e=
, f=
.
0 −1
0 0
1 0
It is now easy to derive
[h, e] = 2e ,
[h, f ] = −2f ,
[e, f ] = h .
Next, let us investigate so(3, C) that can be defined as the matrix algebra
{A ∈ gl(3, C) | AT = −A}. An obvious basis is given by


0 0 0
X = 0 0 −1 ,
0 1 0


0 0 1
Y =  0 0 0 ,
−1 0 0


0 −1 0
Z = 1 0 0  ,
0 0 0
with easy to derive commutation relations
[X, Y ] = Z ,
[Y, Z] = X ,
[Z, X] = Y .
Let us now define a basis change by H = −2iZ , E = X − iY , and F =
−X − iY . In this basis the commutation relations are given by
[H, E] = −2i[Z, X − iY ] = −2iY + 2X = 2E
[H, F ] = −2i[Z, −X − iY ] = 2iY + 2X = −2F
[E, F ] = [X − iY, −X − iY ] = −i[X, Y ] + i[Y, X] = −2iZ = H ,
which shows that sl(2, C) and so(3, C) are isomorph as Lie-algebras.
b) The complexified Lie-algebra su(2)C can be regarded as a Lie algebra over
the field C, where scalar multiplication of a ∈ C with g ⊗b ∈ su(2)C is defined
via
a(g ⊗ b) ≡ g ⊗ (ab) .
(1)
A possible basis is given by σ̃ i = σ i ⊗ 1, where σ i , i = 1, 2, 3 is a basis of
su(2) with commutation relations
[σ i , σ j ] = ij k σ k .
These are the commutation relations of the Pauli matrices multiplied with
√i , and are exactly the same as for X, Y, Z in a). Hence, su(2)C is isomorphic
2
to so(3, C) which is isomorphic to sl(2, C).
3. Lie-algebra automorphisms of sl(n)
a) For Ad(A) to be a automorphism, we have to show that (i) for any x ∈
sl(n) also Ad(A)(x) = A−1 xA ∈ sl(n), (ii) that there exist a z ∈ sl(n)
s.t. Ad(A)(z) = x, and (iii) that Ad(A)([x, y]) = [Ad(A)(x), Ad(A)(y)]
for any two x, y ∈ sl(n). The first condition is ensured by the invariance
of Tr under cyclic permutation
Tr(A−1 xA) = Tr(AA−1 x) = Tr(x) = 0 .
The second one by defining z = AxA−1 which is in sl(n) due to the
same fact. For the last condition we can use that the Lie-bracket in
matrix representation is given by the commutator:
Ad(A)([x, y]) = A−1 (xy − yx)A = A−1 xAA−1 yA − A−1 yAA−1 xA
= Ad(A)(x)Ad(A)(y) − Ad(a)(y)Ad(A)(x)
= [Ad(A)(x), Ad(A)(y)]
b) f (x) := −xT . f is linear and bijective (easy task) and
f ([x, y]) = −(xy − yx)T = −(y T xT − xT y T )
= (−xT )(−y T ) − (−y T )(−xT ) = [f (x), f (y)] ,
which shows that it is a Lie-algebra automorphism.
However, for n ≥ 3 this is no inner automorphism: Consider e.g. the
element x = diag(1, ..., 1, 1 − n) ∈ sl(n). An inner automorphism must
not change the eigenvalues (it is just a basis change) but f (x) = −xT =
diag(−1, ..., −1, n − 1) has different eigenvalues than x for n > 2 and
hence f is no inner automorphism.
4. The Lie groups U (1), T n and Rn
a) Any g ∈ U (1) has a 1d matrix representation g = eiϕ where ϕ ∈ (0, 2π]
parametrises a curve in U (1). A generator of the Lie algebra then is
1 ∂eiϕ = 1,
i ∂ϕ ϕ=0
(2)
which shows that the Lie algebra u(1) is simply R ≡ span{1}.
b) A possible choice of curves to construct basis vectors in the tangent
space of the unit element in T n are
γk (ϕ) = (1, . . . , 1, |{z}
eiϕ , 1, . . . , 1)t ,
k = 1, . . . , n .
(3)
kth pos.
The n corresponding basis elements are given by
1 ∂γk (ϕ) = (0, . . . , 0, |{z}
1 , 0, . . . , 0)t ,
ek =
i ∂ϕ ϕ=0
(4)
kth pos.
and the Lie algebra is given by Rn = span{ek | k = 1, . . . , n} with the
trivial Lie bracket.
c) A possible choice of curves to construct basis vectors in the tangent
space of the unit element in Vn are
γk (ϕ) = (0, . . . , 0,
ϕ , 0, . . . , 0)t ,
|{z}
k = 1, . . . , n .
(5)
kth pos.
The n corresponding basis elements are given by
∂γk (ϕ) = (0, . . . , 0, |{z}
1 , 0, . . . , 0)t ,
ek =
∂ϕ ϕ=0
(6)
kth pos.
and the Lie algebra is again given by Rn = span{ek | k = 1, . . . , n} with
the trivial Lie bracket. The exponential map is simply the identity map.
d) The matrices Φ(v) are invertible since

1
 ..

det(Φ(v)) = det  .
0
0

. . . 0 v1
. . . .. .. 
. .
=1
. . . 1 vn 
... 0 1
(7)
In components we can write Φ(v)ij = δij + vi δn+1,j when we define
vn+1 := 0. We then can calculate
(Φ(v)Φ(w))ij =
X
=
X
Φ(v)ik Φ(w)kj
k
k
δik δkj + vi δn+1,k δkj + δik wk δn+1,j + vi δn+1,k wk δn+1,j
| {z }
→0
= δij + (vi + wi )δn+1,j = Φ(v + w)ij ,
which shows that Φ is a group homomorphism. By construction its
bijective.