Jump to ContentJump to Main Navigation
On the Topology and Future Stability of the Universe$

Hans Ringström

Print publication date: 2013

Print ISBN-13: 9780199680290

Published to Oxford Scholarship Online: September 2013

DOI: 10.1093/acprof:oso/9780199680290.001.0001

Show Summary Details
Page of

PRINTED FROM OXFORD SCHOLARSHIP ONLINE (www.oxfordscholarship.com). (c) Copyright Oxford University Press, 2017. All Rights Reserved. Under the terms of the licence agreement, an individual user may print out a PDF of a single chapter of a monograph in OSO for personal use (for details see http://www.oxfordscholarship.com/page/privacy-policy). Subscriber: null; date: 30 March 2017

(p.695) E The curvature of left invariant metrics

(p.695) E The curvature of left invariant metrics

Source:
On the Topology and Future Stability of the Universe
Publisher:
Oxford University Press

The purpose of the present appendix is to compute the scalar curvature of left invariant metrics on 3 -dimensional Lie groups. The reason we do so is that we need this information in Part VII of this book. Note, however, that as opposed to most of the chapters in Part VII, we shall here work with orthonormal bases of the Lie algebra with respect to the left invariant metrics under consideration.

E.1 Left invariant metrics on 3 -dimensional Lie groups

Let G be a 3 -dimensional Lie group and let { e i } , i = 1 , 2 , 3 , be a basis of the associated Lie algebra g . Let the structure constants associated with this basis be denoted by γ j k i . In other words,

[ e i , e j ] = γ i j k e k .

E.1.1 Decomposition of the structure constants

Define a symmetric matrix ν and a vector a by

ν i j = 1 2 γ k l ( i ϵ j ) k l , a k = 1 2 γ k i i ,

where ϵ i j k is antisymmetric in all of its indices and satisfies ϵ 123 = 1 . We shall also use ϵ i j k = ϵ i j k . Computing ϵ n m i ν i j using the standard identities concerning products of ϵ i j k ’s (cf., e.g., [129, Appendix B]) leads to the conclusion that

(E.1)
γ j k i = ϵ j k l ν l i + a j δ k i a k δ j i .

In order to prove that this decomposition is unique, assume that there is a symmetric matrix n and a vector b such that

γ j k i = ϵ j k l n l i + b j δ k i b k δ j i .

(p.696) Setting i = k and summing yields the conclusion that a = b . Thus

ϵ j k l ν l i = ϵ j k l n l i .

Contracting this identity with ϵ j k m yields ν = n . In other words, the decomposition (E.1) is unique.

E.1.2 Consequences of the Jacobi identity

It is of course not clear that every choice of ν and a is allowed in (E.1). In particular, the Jacobi identity has to hold. In order to express this identity in terms of ν and a, note that it can be written

(E.2)
γ i l m γ j k l + γ j l m γ k i l + γ k l m γ i j l = 0.

If we write the last term as γ k l m γ j i l , it becomes clear that the last two terms, taken together, form an expression which is antisymmetric in j and k. Since the first term on the left hand side of (E.2) is antisymmetric in j and k, it is clear that (E.2) is satisfied if and only if the contraction of the left hand side of (E.2) with ϵ j k p vanishes. In other words, (E.2) is equivalent to

ϵ j k p γ i l m γ j k l + 2 ϵ j k p γ j l m γ k i l = 0.

Compute

ϵ j k p γ j k l γ i l m = 2 ϵ i l k ν k m ν l p + 2 a i a j ϵ j m p + 4 a i ν m p 2 ν p l a l δ i m 2 a l ν m l δ i p , 2 ϵ j k p γ j l m γ k i l = 4 a i ν m p 2 ϵ i l k ν k m ν l p 2 a i a j ϵ j m p + 2 a l ν p l δ i m 2 a l ν m l δ i p .

In other words, the Jacobi identity is equivalent to

a l ν m l δ i p = 0.

Thus, (E.2) is equivalent to ν l m a m = 0 , or, in matrix notation,

(E.3)
ν a = 0.

E.1.3 Change of basis

In order to establish how ν and a transform under a change of basis of the Lie algebra, let

e i = A i i j e j

be another basis and let γ ˜ j k i be the associated structure constants. Then, if B = A 1 ,

γ ˜ i j k = A i i m A j j n γ m n l B l l k = ϵ m n q A i i m A j j n A r r q B s s r ν s l B l l k + A i i m a m δ j k A j j m a m δ i k = ( det A ) ϵ i j r B s s r ν s l B l l k + A i i m a m δ j k A j j m a m δ i k .

(p.697) In other words, if

(E.4)
ν ˜ r k = ( det A ) B s s r ν s l B l l k , a ˜ i = A i i j a j ,

then

γ ˜ i j k = ϵ i j l ν ˜ l k + a ˜ i δ j k a ˜ j δ i k .

Note that, in matrix notation, the relations (E.4) can be written

(E.5)
ν = 1 det A A t ν ˜ A , a ˜ = A a .

E.1.4 Preferred bases

Given a left invariant metric g on G, let { e i } , i = 1 , 2 , 3 , be an orthonormal basis of the Lie algebra. Let us consider the following two cases.

Non-unimodular Lie groups (Bianchi class B)

A Lie group is said to be non-unimodular if and only if a 0 . In that case, we may, without loss of generality, assume the basis to be such that a 1 0 and a 2 = a 3 = 0 (choose an appropriate orthonormal matrix A in (E.5)). If ν is the commutator matrix associated with this basis, we have, due to (E.3) and the symmetry of ν , ν i 1 = ν 1 i = 0 . Letting A be an appropriate orthonormal matrix such that A 1 1 i = A i i 1 = δ 1 i , we can arrange for the orthonormal basis to be such that, in addition, ν is diagonal. The reason for this is that ν is a symmetric matrix and that (E.5) holds. When considering left invariant Riemannian metrics on non-unimodular Lie groups, we shall (in the present appendix) restrict ourselves to orthonormal bases with the above properties. It is of particular interest to note that with respect to such a basis, we have

(E.6)
ν i j ν i j + 1 2 ( tr ν ) 2 = 1 2 ( ν 2 ν 3 ) 2 0 ,

where ν i are the diagonal components of the matrix with components ν i j . The reason the expression appearing on the left hand side of (E.6) is of interest is that it constitutes one term in an expression for the scalar curvature we shall derive below.

Unimodular Lie groups (Bianchi class A)

In the unimodular case, a = 0 , and we can diagonalise ν using an orthogonal matrix A. Due to arguments presented in [129, Chapter 19], it is possible to choose a basis such that the matrix ν falls into one (and only one) of the categories given in Table E.1; cf., in particular, [129, Lemma 19.8, p. 208]. Note that

(E.7)
ν i j ν i j + 1 2 ( tr ν ) 2 = 1 2 ν 1 2 + ( ν 2 ν 3 ) 2 2 ν 1 ( ν 2 + ν 3 ) .

By inspecting Table E.1, it becomes clear that this expression can only be positive in the case of Bianchi type IX, and it can only be zero in the case of Bianchi types I, VII0 and IX. A simply connected Lie group of Bianchi type IX is isomorphic to SU ( 2 ) .

E.1 Bianchi class A.

Type

ν 1

ν 2

ν 3

I

0

0

0

II

+

0

0

V I 0

0

+

VI I 0

0

+

+

VIII

+

+

IX

+

+

+

(p.698) E.2 Scalar curvature

Let G be a 3 -dimensional Lie group, let g be a left invariant metric on G, and let { e i } , i = 1 , 2 , 3 , be a basis of the Lie algebra which is orthonormal with respect to this metric. We shall assume the basis to be of one of the forms described in Subsection E.1.4. Let D be the Levi-Civita connection associated with the metric g and let Γ j k i be defined by

D e i e j = Γ i j k e k .

Note that, due to the Koszul formula,

(E.8)
Γ i j k = 1 2 γ j k i + γ k i j + γ i j k .

In order to compute the scalar curvature, note that

R e i e j e j , e m = D e i D e j e j D e j D e i e j D [ e i , e j ] e j , e m = Γ j j k Γ i k m Γ i j k Γ j k m γ i j k Γ k j m ,

where it is understood that we sum over k and j (since we use an orthonormal basis, indices are raised and lowered with the Kronecker δ ). As a consequence, the scalar curvature S is given by

(E.9)
S = Γ j j k Γ i k i Γ i j k Γ j k i γ i j k Γ k j i ,

where it is understood that we sum over all indices. Let us compute the terms appearing on the right hand side one by one. As a consequence of (E.8), we see that Γ j j k = 2 a k and that Γ i k i = 2 a k . Consequently

(E.10)
Γ j j k Γ i k i = 4 a k a k .

(p.699) Consider

Γ i j k Γ j k i = 1 4 γ j k i + γ k i j + γ i j k γ k i j + γ i j k + γ j k i = 1 4 γ j k i + γ k i j γ j i k γ k i j γ j i k + γ j k i .

Note that the first parenthesis is antisymmetric in j and k, but that the first two terms inside the second parenthesis, taken together, form a symmetric expression in the same indices. Consequently,

Γ i j k Γ j k i = 1 4 γ j k i + γ k i j γ j i k γ j k i = 1 4 γ j k i γ j k i + 1 2 γ k i j γ j k i .

However, it can be verified that

γ j k i γ j k i = 2 ν i j ν i j + 4 a k a k , γ k i j γ j k i = ν i j ν i j + ( tr ν ) 2 2 a k a k .

Thus

(E.11)
Γ i j k Γ j k i = ν i j ν i j + 1 2 ( tr ν ) 2 2 a k a k .

Compute

γ i j k Γ k j i = 1 2 γ i j k γ j i k + γ i k j + γ k j i = 1 2 γ i j k γ i j k + γ i j k γ i k j = 1 2 γ i j k γ i j k γ i j k γ k i j = 1 2 γ j k i γ j k i γ k i j γ j k i = 2 Γ i j k Γ j k i .

Combining this observation with (E.9), (E.10) and (E.11), we obtain

(E.12)
S = Γ j j k Γ i k i + Γ i j k Γ j k i = 6 a k a k ν i j ν i j + 1 2 ( tr ν ) 2 .

Due to the observations made in Subsection E.1.4, we see that this expression is non-positive except if the universal covering group of G is SU ( 2 ) , in which case the scalar curvature could be positive, negative or zero. Furthermore, we see that if we exclude Bianchi types I, VII0 and IX, the scalar curvature has to be strictly negative.