General relativity, solution sheet 2.
General relativity, solution sheet 2.
General relativity, solution sheet 2.
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<strong>General</strong> <strong>relativity</strong>, <strong>solution</strong> <strong>sheet</strong> <strong>2.</strong><br />
HS 08<br />
1. Jacobi identity<br />
i)<br />
ii) For any α, β, γ,<br />
[[X, Y ], Z] + [[Y, Z], X] + [[Z, X], Y ]<br />
= [X, Y ]Z − Z[X, Y ] + [Y, Z]X − X[Y, Z] + [Z, X]Y − Y [Z, X]<br />
= XY Z − Y XZ − ZXY + ZY X + Y ZX − ZY X<br />
− XY Z + XZY + ZXY − XZY − Y ZX + Y XZ = 0 .<br />
0 = C µ αβ [Y µ, Y γ ] + C µ βγ [Y µ, Y α ] + C µ γα [Y µ, Y β ] = (C µ αβ Cν µγ + Cµ βγ Cν µα + Cµ γα Cν µβ )Y ν .<br />
Thus, the function C µ αβ Cν µγ +C µ βγ Cν µα+C µ γαC ν µβ must be identically zero by the linear<br />
independence of the basis vectors {Y ν } n ν=1 at each point p ∈ M.<br />
<strong>2.</strong> Lie groups and Lie brackets<br />
i) a) Let r(t) be a curve in O(n) such that r(0) = e, denote ṙ(0) ≡ s. Then<br />
0 = d dt r(t)T r(t)<br />
∣ = s T + s ,<br />
t=0<br />
so that Lie(O(n)) = {s ∈ gl(n, R) | s T = −s}.<br />
b) Similarly for a curve l(t) ∈ SO(1, 3) with l(0) = e and ˙l(0) ≡ j,<br />
0 = d dt l(t)T ηl(t)<br />
∣ = j T η + ηj .<br />
t=0<br />
Thus, Lie(SO(1, 3)) = {j ∈ gl(4, R) | ηjη −1 = −j T }.<br />
ii) Let m i (t) ∈ G (i = 1, 2), the two curves generated by x i , i.e. x i = ṁ i (0). The generator<br />
of the curve m 1 (α 1 t)m 2 (α 2 t) is<br />
d<br />
dt (m 1(α 1 t)m 2 (α 2 t))<br />
∣ = α 1 x 1 + α 2 x 2 ∈ Lie(G) . (3)<br />
t=0<br />
For any s, m 1 (t)m 2 (s)m 1 (t) −1 m 2 (s) −1 is a curve through e. Hence<br />
d<br />
dt (m 1(t)m 2 (s)m 1 (t) −1 m 2 (s) −1 )<br />
∣ = x 1 − m 2 (s)x 1 m −1<br />
2 (s) ∈ Lie(G) .<br />
t=0<br />
By the linearity just proven, the derivative of a Lie algebra valued function is again an<br />
element of the algebra, i.e.<br />
d<br />
ds (x 1 − m 2 (s)x 1 m −1<br />
2 (s)) ∣<br />
∣∣∣s=0<br />
= −x 2 x 1 + x 1 x 2 ∈ Lie(G) . (4)
Finally, (4) follows from (2,3) and the distributivity for n × n matrices.<br />
iii) It follows from ϕ ∗ [X, Y ] = [ϕ ∗ X, ϕ ∗ Y ], see lecture notes p.7, that<br />
(λ g ) ∗ [X, Y ] = [X, Y ]<br />
for any two left-invariant vector fields X, Y .<br />
iv) Applying (λ g ) ∗ to (1) gives<br />
[Y α , Y β ] = (λ g ) ∗ (C γ αβ Y γ) = (C γ αβ ◦ λ g)Y γ .<br />
Comparison with (1) yields, by the independence of the Y γ , C γ αβ ◦ λ g<br />
C γ αβ (gh) = Cγ αβ (h). Hence Cγ αβ<br />
is constant.<br />
= C γ αβ , i.e.<br />
v) Left-invariance means X gh = (λ g ) ∗ X h . For h = e this shows that X e ∈ T e (G) = Lie(G)<br />
determines X g at all g ∈ G.<br />
vi) Let us first recall that the relation X ↔ x between the (abstract) left-invariant vector<br />
field X and its matrix form x is<br />
(Xf)(e) = d dt f(m(t)) ∣<br />
∣∣∣t=0<br />
, (f ∈ F(G)) ,<br />
where ṁ(0) = x. Hence α 1 x 1 + α 2 x 2 corresponds to (see (3))<br />
d<br />
dt f(m 1(α 1 t)m 2 (α 2 t))<br />
∣ = d t=0<br />
dt f(m 1(α 1 t))<br />
∣ + d t=0<br />
dt f(m 2(α 2 t)) ∣<br />
= α 1 (X 1 f)(e) + α 2 (X 2 f)(e) ,<br />
by using the chain rule, i.e., to α 1 X 1 + α 2 X 2 . To establish the second correspondence<br />
consider the conjugation τ h : G → G, g ↦→ hgh −1 . One verifies λ g ◦ τ h = τ h ◦ λ h −1 gh,<br />
implying (λ g ) ∗ (τ h ) ∗ = (τ h ) ∗ (λ h −1 gh) ∗ . Hence (τ h∗ )X is left-invariant if X is; moreover<br />
(τ h∗ )X ↔ hxh −1 because<br />
d<br />
dt f(hm(t)h−1 )<br />
∣ = d t=0<br />
dt (f ◦ τ h)(m(t))<br />
∣ = (τ h∗ Xf)(e) (5)<br />
t=0<br />
by the definition of the tangent map. In particular<br />
(τ m2 (s)) ∗ X 1 ←→ m 2 (s)x 1 m 2 (s) −1<br />
and by the already established linearity of the correspondence it extends to the derivative<br />
w.r.t. s. On the l.h.s we have by (5)<br />
d<br />
ds τ m 2 (s)∗X 1 f(e)<br />
∣ = d d<br />
s=0<br />
ds dt f(m 2(s)m 1 (t)m 2 (s) −1 )<br />
∣<br />
s=t=0<br />
= d d f(m2 (s)m 1 (t)) + f(m 1 (t)m 2 (s)<br />
ds dt( −1 ) )∣ ∣∣s=t=0<br />
= d ds X 1(f ◦ λ m2 (s))(e)<br />
∣ − d s=0<br />
dt X 2(f ◦ λ m1 (t))(e) ∣<br />
= ((X 1 X 2 − X 2 X 1 )f)(e) ,<br />
∣<br />
t=0<br />
∣<br />
t=0
i.e.<br />
On the r.h.s. we have, see (4)<br />
d<br />
ds τ m(s)∗X 1<br />
∣<br />
∣∣∣s=0<br />
= [X 1 , X 2 ] .<br />
d<br />
ds m 2(s)x 1 m 2 (s) −1 ∣<br />
∣∣∣s=0<br />
= [x 1 , x 2 ] .