and $T$ duality spontaneously, the BPS states do not form degenerate multiplets.

Let us now explore which states in the elementary string spectrum of the 4d heterotic string saturate the Bogomol’nyi bound. Absorbing the moduli $M^{(0)}$ in the definition of the Narain lattice, the BPS condition for purely electric states becomes $$({\rm Mass})^2 = {1\over 16 \rho_2^{(0)}} \hat{\alpha}^a
(I + L)_{ab} \hat{\alpha}^b =
{1\over 8\rho_2^{(0)}} (\hat{\alpha}_R)^2 ~,$$ where $\hat{\alpha} L \hat{\alpha} = \hat{\alpha}_R \cdot \hat{\alpha}_R -
\hat{\alpha}_L \cdot \hat{\alpha}_L$. ($\hat{\alpha}_L$ is 22-dimensional and $\hat{\alpha}_R$ is 6-dimensional. They correspond to the left-moving and right-moving internal momenta of the string.) Now we should compare the free string spectrum, which is given by $$({\rm Mass})^2 = {1\over 4\rho_2^{(0)}} \left[ {1\over 2}
(\hat{\alpha}_L)^2 + N_L - 1\right]
= {1\over 4\rho_2^{(0)}} \left[ {1\over 2} (\hat{\alpha}_R)^2 + N_R -
\delta\right] ~.$$ $N_L$ and $N_R$ represent left-moving and right-moving oscillator excitations. The parameter $\delta$ is 1/2 in the NS sector and $0$ in the R sector. Alternatively, it is simply 0 in the GS formulation. The factor of $(\rho_2^{(0)})^{-1}$ appears because the mass is computed with respect to the canonically normalized Einstein metric. It does not appear if one uses the string metric, which differs by a 