Cartan Structure Equations

Cartan’s Structure Equations : Theory and Execution Sumanto Chanda S.N. Bose National Centre for Basic Sciences  JD Block, Sector-3, Salt Lake, Calcutta-700098, INDIA.

September 10, 2014

1

Basi Basics cs:: NonNon-coor coordi dinat nate e base basess & Conne Connect ctio ions ns

As you know from school, a vector can be expressed in the components in a particular basis as:

                 V  =  V  µ E µ  = V  V  i E  E i

(1)

The derivative of this vector can therefore be written as:

 µ  µ ∂ ν  E µ +  V  µ ∂ ν  ν V  =  ∂ ν  ν  V  E µ  =  ∂ ν  ν  V  ν  E µ

(2)

We remember that the  connection co-efficient / christoffel symbol  is defined as: λ ∂ ν  ν  E µ  = Γνµ E λ

(3)

Giving Giving the covariant the covariant derivative  from (2) as:

 µ λ ∂ ν  E µ +  V  µ Γνµ E λ = ν V  =  ∂ ν  ν  V 

µ

 µ  λ ∂ ν  E µ ν V  + Γ νλ V 

(4)

The covariant derivative in Euclidean (pseudo-Euclidean) space requires the  spin connection: connection : ∂ ν  E i  =  ω ν  j i E  E j  ν  E 

(5)

This gives us the alternative covariant derivative:

∂ ν  V  i + ων i j V  V  j E  E i ν V  = ∂ ν  ν V 

(6)

One must make an important note here about the spin-connection tensor ω. Since it is defined on a space with a diagonal Euclidean metric, it must be   anti-symmetric under anti-symmetric  under the permutation of its indices, since it purely represents rotations alone. This can be seen from demanding that the E i ⊗ E  E j   must vanish: covariant derivative of the inverse metric g −1 =  δ ij E 

  





 

δ  E  E i  = δx ν ων  j i E  E j  =  ω j i E  E j 

1

ω j i  =  ω ν  j i δx ν 

         

δ  g −1  =  δ  δ ij E i ⊗ E j   = 0

⇒

ωkj E k ⊗ E j   + E i ⊗ ω ki E k  = 0

⇒

ωij + ω ji = 0

⇒

⇒

       

δ ij ω k i E k ⊗ E j   + E i ⊗ ω k j E k  = 0

⇒

ω ij + ω ji Ei  ⊗ E j  = 0

ω i j = −ω j i

ω i j = −ω j i

∴

(7)

Remember the transformation rules:





V  i =  e i µ V  µ

E i  =  E i µ E µ

(8)

ei µ E i ν  =  δ µν 

ei µ E  j µ =  δ  ji

(9)

Thus, comparing (4) and (6) gives us:



µ

∂ ν V  µ + Γνλ V  λ = ∂ ν  ei λ V  λ + ων i j e j λ V  λ E i µ

⇒

1.1

          

∂ ν V  µ + Γµνλ V  λ E µ  = ∂ ν V  i + ων i j V  j E i µ E µ

⇒

Γµνλ V  λ = ∂ ν  ei λ + ων i j e j λ V  λ E i µ

∴

Γµνλ  = ∂ ν  ei λ + ων i j e j λ E i µ

(10)

1st Cartan Structure Equation

This definition allows us to move on to consider the Cartan structure equations. In the new coordinate system, we can define the 1-form co-ordinate differential as: dX i ≡ ei =  e i µ dxµ

(11)

We keep in mind that (10) is not truly a differential 1-form since it maps one differential 1form into another, and not a scalar function (0-form) into a 1-form  ie:-(df  =  ∂ µ f dxµ ). It’s 2-form differential is therefore  non-zero (ie:- dei =  d 2 X i  = 0) and is given by:





dei = ∂ ν ei µ − ∂ µ ei ν  dxν  ⊗ dxµ

=





  

∴

T i =  de i + ω i j ∧ e j

1 ∂ ν ei µ − ∂ µ ei ν  dxν  ∧ dxµ = (∂ µ ei ν ) dxµ ∧ dxν  2

Thus, we obtain the  1st Cartan structure equation  as:



dei = ei λ Γλµν  − ωµ i j e j ν  dxµ ∧ dxν 

=

1 i e λ Γλµν  − Γλνµ 2

=  T i − ω i j ∧ e j

dxµ ∧ dxν  − ωµ i j dxµ ∧ e j ν dxν 

2

(12)

1.2

2nd Cartan Structure Equation

The 2nd structure equation follows from the 1st in (12) as:



dT i =  d 2 ei + d ω i j ∧ e j

               

⇒



d ei λ Γλµν  dxµ ∧ dxν  =  dω i j ∧ e j − ω i j ∧ de j

LHS = dei λ Γλµν dxµ ∧ dxν  + ei λ dΓλµν  dxµ ∧ dxν  = =

∂ α ei λ Γλµν  +  ei λ ∂ α Γλµν  dxα ∧ dxµ ∧ dxν  i  j λ i λ α µ ν  ei γ Γγ  αλ − ωα  j e λ Γµν  +  e λ ∂ α Γµν  dx ∧ dx ∧ dx

= ei λ ∂ α Γλµν  + Γλαγ Γγ µν  − ωα i j e j λ Γλµν 

dxα ∧ dxµ ∧ dxν 

The Riemann curvature tensor us given by:

λ γ  Rλ ναµ  =  R αµ λ ν  = −Rαµν λ =  ∂ α Γλµν  − ∂ µ Γλαν  + Γλαγ Γγ  µν  − Γµγ Γαν 

 

(13)

Thus, we can use (13) to say:









1 λ γ  α µ ∂ α Γλµν  − ∂ µ Γλαν  + Γλαγ Γγ  µν  − Γµγ Γαν  dx ∧ dx 2 1 λ = R ναµ dxα ∧ dxµ 2 We can wrap up our demonstration with: α µ ∂ α Γλµν  + Γλαγ Γγ  µν  dx ∧ dx =

∴

and





1 i λ e λ R ναµ − ωα i j T  j µν  dxα ∧ dxµ ∧ dxν  LHS = 2 =  R i j ∧ e j − ω i j ∧ T  j RHS =  dω i j ∧ e j − ω i j ∧ T  j + ω i k ∧ ωk j ∧ e j

Thus, comparing LHS and RHS gives us the  2nd Cartan Structure equation  as: Ri j =  dω i j  +  ω i k ∧ ω k j

2

 

(14)

Execution of Cartan analysis

So far, we have covered the proof of the Cartan Structure equations. But it is one thing to convinced of the validity of a theory, and another to apply it in practice. Now, we will proceed to cover the steps of how to execute the Cartan analysis of any metric. Let us suppose that we have a metric, written as: ds2 =  g µν  dxµ ⊗ dxν 

(15)

It does not matter whether the metric is hyperbolic as it will usually be for general cases or elliptic as it is in special cases. The technique works for both cases. 3

2.1

Extracting the vierbeins

Our first step will be obtaining the 1st Cartan Structure equation, for which we will need to obtain the related vierbeins. This means that we are re-writing the co-ordinate system in way that the new metric appears to be flat and Minkowskian (Euclidean if you are using an elliptic metric). gµν  dxµ ⊗ dxν  ≡ ηab ea ⊗ eb ea =  e a µ dxµ

(16)

gµν  =  η ab ea µ eb ν 

 

(17)

The difficulty in executing this step will depend on the metric, your skill and familiarity with various metrics. The main difficulty will probably arise from the off-diagonal elements. For now, we will focus only on the simplest case; metrics with diagonal elements only. This is done by merely taking square roots of the diagonal elements. ds2 =  g µν  dxµ ⊗ dxν  =  η ab ea µ dxµ ⊗ eb ν dxν  =  η ab ea ⊗ eb ds2 ≡ gµµ dxµ ⊗ dxµ =  η aa

 

η aa gµµ dxµ ⊗

 

η aa gµµ dxµ =  η aa ea ⊗ ea

(18)

Hopefully, this step be executed without much hassle, and you should have a set of vierbeins. The next step will involve taking their exterior derivative.

2.2

Taking the exterior derivative

Depending on the co-ordinate dependence of the vierbeins, for an n-dimensional manifold you should get maximum up to n(n2−1) independent spin connection components. For your benefit, remember to write out your results as a n × n   matrix, and fill your entries as you work your way through each vierbein creating (12) to stay organized. Remember that we are considering only  torsion free cases, where T i = 0. Under this condition, (12) will become: dei = −ωi j ∧ e j

2.3

(19)

Writing the spin-connection matrix

Using (19), we fcan proceed to write all the non-zero components of the spin connection matrix ω , which we will use to obtain the curvature matrix. Often this will require cleverly coupling the wedge product terms with co-efficients to reproduce vierbein terms. i

de =  ∂ µ e

i

ν 

µ

ν 

            

dx ∧ dx =

∂ µ ei ν  dxµ e j ν 

∧ e j ν  dxν 

taking all j > i

 

(20)

ej

−ω i j

Overall, for an n-dimensional manifold, one needs to compute at most all the components that can fit on one side of the diagonal. This amounts to n(n2−1) independent components. So for a 4-dimensional space, we need at most 6 independent components. ω  =

 

0

ω0 1

ω02 ω12

0 −ω 0 1 0 −ω 2 −ω 1 2 0 0 1 −ω 3 −ω 3 −ω 2 3

ω03 ω13 ω23

0

 

 

(21)

The remaining half on the other side of the diagonal are the negative inverse triangle of elements in the upper half triangle. With this, the spin connection matrix is complete. The next step is the curvature matrix, which requires Cartan’s 2nd equation. 4

For Euclidean 4-dimensional spherical metrics There are circumstances when the metric one is dealing with is given in the form: 3

    2

2

2

2

ci (r) σi2

ds = c0 (r ) dr +

dσ i = −εi jk σ j ∧ σ k

(22)

i=1

This gives us the following vierbeins: e0 =  c 0 dr

ei =  c i σ i ,

(i  = 1, 2, 3)

(23)

One might be tempted to follow the same procedure as usual: de0 = 0

(24)

dei =  ∂ r ci dr ∧ σ i − ci εi jk σ j ∧ σ k ∂ r ci i ci k ci  j σ ∧ c0 dr + εi jk σ ∧ c j σ j + εi kj σ ∧ ck σ k =− c0 2c j 2ck

 

 

 

 

(25)

But here we need to take an extra precaution. The spin connections ω i j  could be written as: ω i j = −εi jk

ci k σ 2c j

(26)

but then it can easily be seen that this form is not antisymmetric under permutation of the indices i and j. This is why we need to use a more elaborate form of the spin connections. 2 ∂ r ci i i ci  j de = − σ ∧ c0 dr − ε  jk σ ∧ σ k c0 ci

 

i

εi jk

2c2i + (c j2 − c2k ) − (c j2 − c2k )  j c2i  j k i σ ∧ σ =  ε  jk σ ∧ σk ci 2ci c2i + c j2 − c2k k c2i + c2k − c j2  j i  j i σ ∧ c j σ − ε kj σ ∧ ck σ k = −ε  jk 2ci c j 2ci ck =

−εi jk

∂ r ci i dei = − σ ∧ e0 − c0

∴



c2i + c j2 − c2k

2ci c j

− εi jk

 

k

 j

i

σ ∧e −ε

c2i + c j2 − c2k

2ci c j

∂ r ci i ω 0  = σ c0 i

c2i + c2k − c j2 kj

2ci ck

σ k ∧ e j − εi kj

ω i j

=

−εi jk

 

σ j ∧ ek

c2i + c2k − c j2

2ci ck

c2i + c j2 − c2k

2ci c j

σk

σ j ∧ ek



 

(27)

(28)

Here we can clearly see that the spin connection components are anti-symmetric under permutation of i and j.

5

2.4

Writing the curvature matrix

Remembering the 2nd Cartan Structure equation (14), we can say that the curvatgure will comprise 2 matrices, formed by: R  =  dω  +  ω ∧ ω

dω  =

ω ∧ ω  =

   

0

dω 0 1

dω 0 2 dω 1 2

dω 0 3 dω 1 3 dω 2 3

−dω 0 1 0 0 −dω 2 −dω 1 2 0 0 1 −dω 3 −dω 3 −dω 2 3

0

ω01

ω02 ω12

0

ω03 ω13 ω23

−ω 0 1 0 0 −ω 2 −ω 1 2 0 0 1 −ω 3 −ω 3 −ω 2 3

0

 

∧

   

 

ω01

0

(29)

ω02 ω12

ω03 ω13 ω23

−ω0 1 0 0 −ω 2 −ω 1 2 0 0 1 −ω 3 −ω 3 −ω 2 3

0

 

 

(30)

Here the 2nd term is a matrix whose wedge product computation is performed in a manner similar to matrix multiplication, (ie. rows  ×  columns). Each individual matrix element is:

   ω×ω

i

 j

=

ω i k ∧ ω k j

 

(31)

k =i,j

Naturally, the inequality used to avoid zero value spin connection components in this case leaves us to deal with only n − 2 product terms at a time. This process as with the spin connections needs to be done only for n(n2−1) components. Furthermore, each and every matrix element is an individual 2-form by itself. Consequently, this means that if each component is written as  Rabcd , then they are anitsymmetric over c, d indices. Also, because the Rab  2-forms arise from ωab   1-forms which are anti-symmetric over a, b indices, the same symmetry property holds for Rab . Rabcd  = −Rbacd

Rabcd  = −Rabdc

 

(32)

Also, we have the symmetry relation: Rabcd  =  R cdab

 

(33)

This can be verified by using (32) and the 1st Bianchi identity for Riemann tensors: Rabcd  +  Racdb  +  Radbc = 0



 

Rabcd  = − Racdb +  Radbc  =  R cadb  +  Radcb



= − Rcdba  +  Rcbad − Rdcab  +  Rcadb





(34)



= 2Rcdab − Rcbad  +  Rcadb  = 2Rcdab  +  Rcdba  =  R cdab

Thanks to (32) and (33) one can also see that the identity applies for any index kept fixed, while the others cycle around it. 6