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?
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asked Jun 16, 2013 at 11:21
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1$\begingroup$ 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... $\endgroup$– TrimokCommented Jun 17, 2013 at 18:38
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1 Answer
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(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 $\mathbb Z$
