1
Solution to problem # 1
(a)
Observe that eφ and eθ from T S 2 are vector fields along the directions φ and θ, resp. To find their coordinates in R3 , one
should use Eq. (2.29) from the Altand’s book, since we have the mapping S 2 7→ R3 . Hence
eφ = ∂φ (x1 , x2 , x3 ) = (− sin θ sin φ, sin θ cos φ, 0),
eθ = ∂θ (x1 , x2 , x3 ) = (cos θ cos φ, cos θ sin φ, − sin θ).
(1)
(b)
In the same fashion as above,
1 + z̄ 2
2z̄
1 − z̄ 2
,
−i
,
ez = ∂z (x , x , x ) =
(1 + z z̄)2
(1 + z z̄)2 (1 + z z̄)2
1 − z2
1 + z2
2z
1
2
3
ez̄ = ∂z̄ (x , x , x ) =
,i
,
(1 + z z̄)2 (1 + z z̄)2 (1 + z z̄)2
1
2
3
(2)
(c) & (d)
One may start by relating spherical and stereographic coordinates using the mapping z(x) = (x1 + ix2 )/(1 − x3 ) to obtain
z(θ, φ) =
sin θeiφ
= cot(θ/2)eiφ ,
1 − cos θ
z̄(θ, φ) = cot(θ/2)e−iφ .
(3)
The inverse mapping C → S 2 is then
i
φ(z, z̄) = − ln (z/z̄) ,
2
θ(z, z̄) = 2 arccot(z z̄).
(4)
Next, by definition, dz(eφ ) = eφ (z) = ∂φ z, and etc. To be more formal, I’m using the material from page 539 of the
Altland’s book. Hence we may organize the matrix of transformation from stereographic to sperical basis,
!
eiϕ
ieiϕ cot θ2
∂φ z ∂θ z
cos(θ)−1
=
(5)
e−iϕ
∂φ z̄ ∂θ z̄
−ie−iϕ cot θ
2
For the inverse mapping one gets
∂z φ ∂z̄ φ
∂z θ ∂z̄ θ
=
− 21 ie−iϕ tan θ2
1 −iϕ
(cos(θ) − 1)
2e
cos(θ)−1
1 iϕ
θ
2 ie tan 2
1 iϕ
2 e (cos(θ) −
1)
Finally, one checks that matrices (5) and (6) are inverse to each other, as it should be.
(6)