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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 (1,950 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 : $$m^2 = \frac{n^2}{R^2} + \frac{w^2R^2}{\alpha'^2} + \frac{2}{\alpha'}(N + \tilde N - 2)$$ $$0 = nw + N - \tilde N $$ $n$: momentum quantification number, $w$: winding number, $N,\tilde N$ : left/right levels We consider (16) bosonic left-movers, so there should be a relation between $n$ and $w$, something like $p_R=\frac{n}{R} -\frac{wR}{\alpha'} = 0$. 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 : $X \sim X +2\pi R$, $X$ being one of the 25 spatial dimensions, say $X^{25}$ The left and right momenta are :

$k_L =\frac{n}{R} +\frac{wR}{\alpha'} = 0$, $k_R =\frac{n}{R} - \frac{wR}{\alpha'} = 0$

The on-shell mass conditions are written :

$m^2 = k_L^2 + \frac{4}{\alpha'}(N - 1)$, $m^2 = k_R^2 + \frac{4}{\alpha'}(\tilde N - 1)$

From this we get :

$0 = k_L^2 - k_R^2 + \frac{4}{\alpha'}(N - \tilde N)$

Using a "dimensionless" momentum $l_{L,R} = k_{L,R}(\frac{\alpha'}{2})^{\frac{1}{2}}$, we get :

$0 = l_L^2 - l_R^2 + 2 (N - \tilde N)$

If we compactify 16 dimensions, we will have vectors $\vec l_L, \vec l_R$, with :

$0 = \vec l_L^2 - \vec l_R^2 + 2 (N - \tilde N)$

Now, in the heterotic string, we consider only left - movers, so $\vec l_R = \vec 0$, so we have :

$0 = \vec l_L^2 + 2 (N - \tilde N)$

If we consider a lattice $\Gamma$ made up with the $\vec l_L$, 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) :

$X_L(z_1) X_L(z_2) \sim -\frac{\alpha'}{2} \ln z_{12}$ and $X_R(z_1) X_R(z_2) \sim -\frac{\alpha'}{2} \ln \bar z_{12}$

Note that we have also :

$:e^{ik_LX_L(z)+ik_RX_R(\bar z)}: :e^{ik_L'X_L(0)+ik_R'X_R(\bar 0)}:~ \sim z^{\alpha'k_Lk_L'/2} (\bar z)^{\alpha'k_Rk_R'/2} ~:e^{i(k_L +k_L')X_L(0)+i(k_R + k_R')X_R(0)}:$

where the $z,\bar z$ term could be written $z^{l_Ll_L'} \bar z^{l_Rl_R'}$

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

$e^{2i\pi (l_Ll_L' - l_Rl_R')} = 1$, so $l_Ll_L' - l_Rl_R'$ is in $\mathbb Z$

This post imported from StackExchange Physics at 2014-03-09 09:11 (UCT), posted by SE-user Trimok
answered Jun 18, 2013 by Trimok (950 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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