Utrecht University

Gerard ’t Hooft CERN Black Hole Institute

What will space-time look like at scalesmuch smaller than the Planck length?No physical degrees of freedom anymore

Gravity may become topological

It may well make sense to describe space-timethere as if made of locally flat pieces gluedtogether (as in “dynamical triangulation” or “Regge calculus”)

The dynamical degrees of freedom are then line-like defectsThe dynamical degrees of freedom are thenpointlike defects[??]

This theory will have a clear vacuum state:flat Minkowski space-time

“Matter fields” are identified with the defects.

There are no gravitons: all curvature comes fromthe defects, therefore:

Gravity = matter. Furthermore:

The vacuum has 0

What are the dynamical rules? What is the “matter Lagrangian”?

First do this in

A gravitating particle in 2 + 1 dimensions:

A moving particle in 2 + 1 dimensions:

x x

xx

A′A

A′

A

xx v t

'

'

x x

t t

The defect angleis the energy

A many-particle universe in 2+1 dimensions can now be constructed by tesselation

(Staruszkiewicz, G.’t H, Deser, Jackiw, Kadar)

There is no localcurvature; the onlyphysical variablesare N particle coordinates, withN finite.

2 + 1 dimensional cosmology is finite and interesting

BANG

CRUNCH

Quantization is difficult.

How to generalize this to 3 +1 (or more) dimensions ?

straight strings are linelike grav. defects

0

0

Calculation deficit angle:

2

33 00

2

2

11 22

21

( )

( ) ( )

( ) ,

8

T t x

t t

x r

R r x

r r G

Deficit angle →22 (1 cos ) 8 G

In units such that :11 22 1R R

A static universe would contain a large number of such strings

But what happens when we makethem move ?

One might have thought:

But this cannot be right !

New s

tring

??

Holonomies on curves around strings: Q : member of Poincaré group:

Lorentz trf. plus translationC

(3,1) (2, )SO SL C

Static string: pure rotation,

Lorentz boost:

Moving string:

(2)Q U SU †Q Q V

1Q V U V

Tr real , Tr 2U U

Tr real , Tr 2Q Q All strings must have holonomies that are constrained by

C1Q1

2Q

1Q1

1 2Q Q 12Q

12 1Q Q

1 1 1 1 12 1 1 2 1 2 1 2( )C Q Q QQ Q Q QQ

In general, Tr (C) = a + ib can be anything. Only if the angle is exactly 90° can the newly formed object be a string.What if the angle is not 90°?

12 1

2 12

Can this happen ?

a

c

1 11 2 1 2 a cC Q Q QQ Q Q

This is a delicate exercise in mathematical physicsAnswer: sometimes yes, but sometimes no !

bd

1Q 2Q

4Q 3Q

aQ

bQcQ

dQ

1 1 1 13 1 2 1 4 3 1 3

13

2 3 4

2 1

1 ,,

, .,b c b d aa

Q QQ Q Q Q Q Q

Q Q Q Q Q

QQ Q

Q Q Q

Q

Q

1

Notation and conventions:

1Q 2Q

4Q 3Q

aQ

bQcQdQ

The Lorentz group elements have 6degrees of freedom. Im(Q) = 0 is one constraint for each of the 4 internal lines →

We have a 2 dimensional space ofsolutions.

Strategy: write in SL(2,C) 1 1 2 2

3 3 4 4a

a ib a ibQ

a ib a ib

Im(Tr(Q)) = 0 are four linear constraints on the coefficients a, b .

|Re(Tr(Q))| < 2 gives a 4 - dimensional hypercube.

Two quadratic conditions remain: Re (det( )) = 1Im (det( )) = 0

aQaQ

In some cases, the overlap is found to be empty !- relativistic velocities - sharp angles,

- and others ...

In those cases, one might expect

1

2

N

where we expect free parameters.

We were unable to verify wether such solutions always exist

6( 1) (3 2) 3 4N N N

1. When two strings collide, new string segments form.2. When a string segment shrinks to zero length, a similar rule will give newly formed string segments.3. The choice of a solution out of a 2- or more dimensional manifold, constitutes the

matter equations of motion.4. However, there are no independent gravity degrees of freedom. “Gravitons” are composed of matter.5. it has not been checked whether the string constants can always be chosen positive. This is probably not the case. Hence there may be negative energy.

Assuming that some solutions always exist, we arriveat a dynamical, Lorentz-invariant model.

6. The model cannot be quantized in the usualfashion: a) because the matter - strings tend to generate a continuous spectrum of string constants; b) because there is no time reversal (or PCT) symmetry; c) because it appears that strings break up, but do not often rejoin.

7. However, there may be interesting ways to arrive at more interesting schemes. For instance:

crystalline gravity

❖ Replace 4d space-time by a discrete, rectangular lattice:

This means that the Poincaré group is replaced byone of its discrete subgroups

Actually, there are infinitely many discrete subgroups of thePoincaré group

One discrete subgroup: (3,1, )SO

Compose SO(3) rotations over 90° and powers of

0 1 1 1

1 0 1 1

1 1 0 1

1 1 1 2

B

But, there are (probably infinitely) many more scarcediscrete subgroups. Take in SL(2,C)

with

and

are rotated by 90° along z and y axis.Then, generically,

diverge as stronger and stronger boosts.

xB

2, , ,

1 ; | |, | |, | |, | | 1

,y zB B

1 2 3 4...n n n nx y z xB B B B

No double counting!

❖ add defect lines :

A

A′

identify

x

❖ postulate how the defect lines evolve:

1

2

N

Besides defect angles, surplus angles will probably beinevitable.

❖ quantize... ❖ pre - quantize...

Define the basis of a Hilbert space as spanned byontological states / equivalence classes ..

Diagonalize the Hamiltonian (evolution operator) tofind the energy eigenstates

... and do the renormalization group transformationsto obtain and effective field theory with gravity

At this point, this theory is still in its infancy.

Do the two quadratic curves have points within our hypercube ?

1. Choose the Lorentz frame where string #1 is static and in the z direction:

1

11 1

0

0

i

i

eQ

e

Im(Tr( )) 0 ; Im(Tr( )) 0 ; Det( ) 1a d aQ Q Q

2 1 2 1 2 2 22 1 1 2

1 1 2 2

1; 1 ( )

( ) 1a

ia b ibQ a b b

b ib ia

In the frame string a is static as well.

We must impose: because

is the boost giving the velocity of the joint.

1 1

1 0 1

1

0

0 1/zV

1Q 2Q

4Q 3Q

aQ

bQcQdQ

2. next, go to the frame where string #3 is static and in the z direction:

3

33 3

0

0

i

i

eQ

e

There, one must have:

1 2 1 2

3 1 2 1 2( )c

p ip q iqQ

q iq p ip

If one would choose and , then One would expect this to fix all coefficientsa, b, p and q.However, something else happens ... !

1 3

1Q 2Q

4Q 3Q

aQ

bQcQdQ

We found that for all initial string configurationsthere are four coefficients, A, B, C, D, depending onlyon the external string holonomies, such that

13

1

A B

C D

So one cannot choose freely ! Instead, the coefficients a, b, ... are still underdetermined, with one free parameter.

3

3. This leaves us the option to choose instead.One finds (from the symmetry of the system):

2

24

2

D C

B A

That determines a, b, p and quniquely !

1

3

2

4

If and only if all μ ’s are positive, the strings and the joints between the strings ar all moving with subluminal velocities. This is necessary for consistency of the scheme. In

13

1

A B

C D

we must demand that this is consistent. Consequently, the case

must be excluded. But we found that this will happen if strings move fast !

( , , , ) ( , , , ) or ( , , , )A B C D