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  Why do the mismatched 16 dimensions have to be compactified on an even lattice?

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The mismatched 16 dimensions between the left- (26 dimensional) and right- (10 dimensional) are compactified on even, unimodular lattices. I think I get the unimoduar part, at least intuitively, somewhat, but I don't understand why the lattice has to be even. From what I understand, an even lattice means that the vectors have even norm-squared. Why is that a necessary property for compactifying the 16 dimensions?

asked Jun 16, 2013 in Theoretical Physics by dimension10 (1,985 points) [ revision history ]
edited Apr 25, 2014 by dimension10
I suppose that one of the tracks is to consider mass-shell relations, considering closed-string and toroidal compactification (so leading to T-Duality), they could be written : m2=n2R2+w2R2α2+2α(N+˜N2)
0=nw+N˜N
n: momentum quantification number, w: winding number, N,˜N : left/right levels We consider (16) bosonic left-movers, so there should be a relation between n and w, something like pR=nRwRα=0. But I shamefully missed the final step...

This post imported from StackExchange Physics at 2014-03-09 09:11 (UCT), posted by SE-user Trimok

1 Answer

+ 1 like - 0 dislike

(Source : Polchinski)

Consider a toroidal compactification for a bosonic closed string. We make the identification : XX+2πR, X being one of the 25 spatial dimensions, say X25 The left and right momenta are :

kL=nR+wRα=0, kR=nRwRα=0

The on-shell mass conditions are written :

m2=k2L+4α(N1), m2=k2R+4α(˜N1)

From this we get :

0=k2Lk2R+4α(N˜N)

Using a "dimensionless" momentum lL,R=kL,R(α2)12, we get :

0=l2Ll2R+2(N˜N)

If we compactify 16 dimensions, we will have vectors lL,lR, with :

0=l2Ll2R+2(N˜N)

Now, in the heterotic string, we consider only left - movers, so lR=0, so we have :

0=l2L+2(N˜N)

If we consider a lattice Γ made up with the lL, we see that it must be a even lattice.


Note :

The expression of the dimensionless momentum may be justifyed by looking at the Operator Product Expansion (OPE) :

XL(z1)XL(z2)α2lnz12 and XR(z1)XR(z2)α2lnˉz12

Note that we have also :

:eikLXL(z)+ikRXR(ˉz)::eikLXL(0)+ikRXR(ˉ0): zαkLkL/2(ˉz)αkRkR/2 :ei(kL+kL)XL(0)+i(kR+kR)XR(0):

where the z,ˉz term could be written zlLlLˉzlRlR

In fact, single-valuedness of the last OPE under a circle means that :

e2iπ(lLlLlRlR)=1, so lLlLlRlR is in Z

This post imported from StackExchange Physics at 2014-03-09 09:11 (UCT), posted by SE-user Trimok
answered Jun 18, 2013 by Trimok (955 points) [ no revision ]
+1 Thanks a lot!

This post imported from StackExchange Physics at 2014-03-09 09:11 (UCT), posted by SE-user Dimensio1n0

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