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April 26, 2004

Billiards at half-past E10

Posted by urs


Last week I gave a seminar talk on cosmological billiards and their relation to proposals that M-theory might be described by a 1+0 dimensional sigma-model on the group of E 10 /K(E 10 ). I had mentioned that already several times here at the Coffee table and we had some interesting discussion over at sci.physics.strings. But while preparing the talk it occured to me that the basic technical observation behind this conjecture is so simple and beautiful that it deserves a seperate entry. I’ll summarize pp. 65 of

T. Damour, M. Henneaux & H. Nicolai: Cosmological Billiards (2002).

So how would the equations of motion of geodesic motion on a Kac-Moody algebra group manifold look like, in general?

A Kac-Moody (KM) algebra is a generalization of an ordinary Lie algebra. It is determined by its rank r, an r×r Cartan matrix A=(a ij) and is generated from the 3 r Chevalley-Serre generators

(1){h i,e if i} i=1 ,,r

which have commutators

(2)[h i,h j]=0
(3)[e i,f j]=δ ijh j
(4)[h i,e j]=a ije j
(5)[h i,f j]=a ijf j.

(One should think of the SU(2) example where h=J 3 , e=J + and f=J .)

The elements of the algebra are obtained by forming multiple commutators of the e i and the f i.

(6)E α,s=[e i 1 ,[e i 2 ,[,[e i p1 ,e i p]]
(7)E α,s=[f i 1 ,[f i 2 ,[,[f i p1 ,f i p]].

Here, as always in Lie algebra theory, the α are the so called roots, i.e. the ‘quantum numbers’ with respect to the Cartan subalgebra generators h i:

(8)[h i,E α,s]=α iE α,s

and s=1 ,mult(α) is an additional index due to possible degeneracies of the αs. Not all of these elements are to be considered different, but instead one has to mod out by the Serre relations

(9)ad(e i) 1 a ij(e j)=0
(10)ad(f i) 1 a ij(f j)=0

which should be thought of as saying that the E α are nilpotent ‘matrices’ with entries above the diagonal, while the E α are nilpotent with entries below the diagonal.

(I’d be grateful if anyone could tell me why these Serre relations need to look the way they do…)

A set of simple roots α i generates, by linear combination, all of the roots and the Cartan matrix is equal to the normalized inner products of the simple roots

(11)a ij=2 α iα jα iα i,

where the product is taken with respect to the unique invariant metric (just as for ordinary Lie algebras).

The nature of the KM algebra depends crucially on the signature of the Cartan matrix A. We have three cases:

  • If A is positive definite, then the KM algebra is just an ordinary finte Lie algebra [T a,T b]=f ab cT c.
  • If A is semidefinite, then the KM algebra is an infinite affine Lie algebra, or equivalently a current algebra in 1+1 dimensions: [j m a,j n b]=mη abδ m+n,0 +f ab cj m+n c.
  • If, however, A is indefinite, we obtain an infinite KM algebra with exponential growth, which is, I am being told, relatively poorly understood in general. But this is the case of interest here!

A general element of the group obtained from such an algebra by formal exponentiation is of the form

(12)𝒱=exp(β ih i) =𝒜exp( α,sν α,sE α,s) =𝒩.

If we make the coefficients functions of a single parameter τ then we get the tangent vectors P to the trajectory in the group ‘manifold’ traced out by varying τ by writing:

(13)P:=1 2 (𝒱˙𝒱 1 +(𝒱˙𝒱 1 ) T).

This is exactly as for any old Lie group, the only difference being that we have here projected onto that part which is ‘symmetric’ with respect to the Chevalley involution which sends

(14)h i T=h i
(15)e i T=f i
(16)f i T=e i.

The algebra factored out this way is the maximal compact subalgebra of our KM algebra, so that we are really dealing with the remaining coset space (which, unless I am confused, should hence be a symmetric space).

Now the fun thing which I wanted to get at is this: If we define generalized momenta j α,s such that

(17)𝒩˙𝒩 1 = α,sj α,sE α,s,

then it is very easy to check, using the defining relations of the KM algebra, that

(18)P=β˙ ih i+1 2 α,sj α,sexp(α iβ i)(E α,s+E α,s).

Using the fact that the non-vanishing inner products of the algebra elements are

(19)h ih j=g ij
(20)E α,sE β,s=δ s,tδ α+β,0

one finally finds the Lagrangian describing geodesic motion of the coset space:

(21)PP=g ijβ˙ iβ˙ j+1 2 α,sexp(2 α iβ i)j α,s 2 .

(Here g is the invariant metric of the algebra.)

The point is that there is a free kinetic term in the Cartan subalgebra plus all the off-diagonal kinetic terms which all couple exponentially to the Cartan subalgebra coordinates.

It is obvious that for very large values of β the off-diagonal terms ‘freeze’ and leave behind effective potential walls which constrain the motion of the βs to lie within the Weyl chamber of the algebra, namely that poly wedge associated with the simple roots (all other roots generate potential walls which lie behind those of the simple roots.)

Anyone familiar with classical cosmology immediately recognizes the above Lagrangian as being precisely of the form as those mini/midi superspace Lagrangians that govern the dynamics of homogeneous modes of general relativity. There the β i are the logarithms of the spatial scale factors of the universe.

Indeed, it can be checked to low order that the Lagrangian of E 10 in the above sense reproduces that of 11d SUGRA when the latter is accordingly suitably expanded about homogeneous modes. That’s the content of

T. Damour, M. Henneaux & H. Nicolai: E 10 and the ‘small tension’ expansion of M Theory (2002).

But the crucial point is that there are many more degrees of freedom in the E 10 sigma model than can correspond to supergravity. There are indications that these can indeed be associated with brane degrees of freedom of M-theory:

Jeffrey Brown, Ori Ganor & C. Helfgott: M-theory and E 10 : Billiards, Branes, and Imaginary Roots,

which, unfortunately, I still have not read completely.

Posted at April 26, 2004 12:27 PM UTC

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Read the post Nicolai on E10 and Supergravity
Weblog: The n-Category Café
Excerpt: H. Nicolai on further progress in checking the hypothesis that the dynamics of supergravity is encoded in geodesic motion on a Kac-Moody group coset.
Tracked: November 29, 2006 6:33 PM

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