supersymmetry - T-Duality between Type HE String theory and Type HO string theory

My question is regarding T-Duality between the 2 Type H string theories.

I know that the Type II String theories are T-dual to each other because T-Duality changes the sign of the Gamma Matrix so $$\operatorname{T}:{{\cal P}}_{{\mathop{\rm GSO}\nolimits} }^ - \leftrightarrow {{\cal P}}_{{\mathop{\rm GSO}\nolimits} }^ + $$

Since the Type IIB String theory employs the same GSO Projections on the left and right movers while Since the Type IIA String theory employs the different GSO Projections on the left and right movers, the 2 theories are T-Dual to each other.

However, when one considers T-Dualities on the Type HE string theory and the Type HO String theories, why are they T-Dual to each other?

I presume that maybe T-Duality switches $\operatorname{Spin}(32)/\mathbb Z^2$ and $E(8)\times E(8)$, so therefore, the theories to T-Dual to each other. But if so, then why? Why does T-Duality switch $\operatorname{Spin}(32)/\mathbb Z^2$ and $E(8)\times E(8)$?

Thanks in advance.

Answer

One shouldn't imagine the T-duality between the two heterotic strings to be a $Z_2$ group, like in the case of type II string theories' T-duality. In type II string theory, there is only one relevant scalar field, the radius of the circle producing T-duality, and it gets reverted $R\to 1/R$ under T-duality.

In the heterotic case, it's more complicated because more scalar fields participate in the T-duality. Instead of a $Z_2$ map acting on one scalar field, one must correctly adjust the moduli, especially the Wilson lines generically breaking the 10D gauge group to $U(1)^{16}$, and find an identification between the points of the moduli space of the two heterotic string theories: there is one theory at the end.

A fundamental reason why the T-duality holds is that one may define the heterotic string theories in the bosonic representation, using 16-dimensional lattice $\Gamma^{16}$ which is the weight lattice of $Spin(32)/Z_2$, and $\Gamma^{8}\oplus \Gamma^8$ which is the root lattice of $E_8\times E_8$.

One may also describe the compactification on a circle in terms of lattices. It corresponds to adding ($\oplus$) the lattice $\Gamma^{1,1}$ of the indefinite signature to the original lattice. The extra 1+1 dimensions correspond to the compactified left-moving and right-moving boson (of the circle), respectively.

Now, the key mathematical fact is that the 17+1-dimensional even self-dual lattice exists and is unique which really means $$\Gamma^{16}\oplus \Gamma^{1,1} = \Gamma^{8}\oplus \Gamma^8\oplus \Gamma^{1,1}$$ Even self-dual lattices in $p+q$ dimensions (signature) exist whenever $p-q$ is a multiple of eight and if both $p$ and $q$ are nonzero, the lattice is unique.

It's unique up to an isometry – a Lorentz transformation of a sort – and that's how the identity above should be understood, too. So there is a way to linearly redefine the 17+1 bosons on the heterotic string world sheet so that a basis that is natural for the $E_8\times E_8$ heterotic string gets transformed to the $Spin(32)/Z_2$ string or vice versa. The compactified boson has to be nontrivially included in the transformation – the 17+1-dimensional Lorentz transformation that makes the T-duality manifest mixes the 16 chiral bosons with the 1+1 boson from the compactified circle.

A different derivation of the equivalence may be found e.g. in Polchinski's book. One may start with one of the heterotic strings and carefully adjust the Wilson lines to see that at a special point, the symmetry broken to $U(1)^{17+1}$ is enhanced once again to the other gauge group.