I am trying to understand some general statements made in the lecture notes by Vafa entitled "Lectures on Strings and Dualities" concerning toroidal compactifications (arXiv:hep-th/9702201).

Question 1: First, consider the heterotic string (D = 10) bosonic string compactified on a torus $T^d$. It is stated on page 12 that the space of inequivalent choices of the left and right- moving momenta $(P_L, P_R)$ is given by the coset

$$\frac{SO(d,d)}{SO(d) \times SO(d) \times O(d,d;\mathbb{Z})}$$

I understand that this is basically $\frac{SO(d,d)}{SO(d) \times SO(d)}$ further modded out by the discrete T-duality group $O(d,d;\mathbb{Z})$ which corresponds to Lorentz boosts with integer coefficients. However, in the main text, it is stated that it is $O(d) \times O(d)$ transformations that do not change the string states. So strictly speaking, shouldn't we be modding $SO(d,d)$ by $O(d) \times O(d)$ instead of $SO(d) \times SO(d)$?

Question 2: Now consider $\mathcal{N} = 2$ theories on $T^d$.

The following statement appears on page 13 about the type-IIA and type-IIB theories:

For more general compactification on $T^d$ the part of the T-duality group which does not exchange the two theories is $SO(d,d;\mathbb{Z})$; the elements of T-duality which are in $O(d,d;\mathbb{Z})$ but not in $SO(d,d; \mathbb{Z})$ will exchange IIA and IIB and thus are not symmetries of either one.

So does this mean that the T-duality group consists of transformations that do take IIA to IIB (and vice-versa) but also transformations that do not take IIA to IIB and that the former correspond to $O(d,d;\mathbb{Z}) \setminus SO(d,d;\mathbb{Z})$? So when textbooks make statements like $IIA \leftrightarrow IIB$ under T-duality, they really mean under this set theoretic subtraction?

Question 3: This question, somewhat embarrasingly, is more about notation than physics. On page 13 of the same lecture notes, the following paragraph appears:

In compactifying type II strings on tori, the scalars parametrized by the coset $SO(d,d)/SO(d) \times SO(d)$ correspond to choices of the metric of the torus ($d(d+1)/2$ degrees of freedom) and the antisymmetric field $B_{ij}$ on the torus ($d(d-)1/2$ degrees of freedom).

Here, does the author mean $\frac{SO(d,d)}{SO(d) \times SO(d)}$ or does he mean $(SO(d,d)/SO(d)) \times SO(d)$, i.e. $\frac{SO(d,d)}{SO(d)} \times SO(d)$? The two are vastly different things.

In fact, often he refers to things like $SO(5,5)/SO(5) \times SO(5)$ or $SO(5,5)/SO(5) \times SO(5) \times SO(5,5;\mathbb{Z})$. What is the order in which these are to be read/interpreted?

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    $\begingroup$ In response to question 2: $O(d,d;\Bbb Z)\setminus SO(d,d;\Bbb Z)$ does not make much sense (because you haven't specified in what sense you consider $SO$ as a subset of $O$, i.e. how it is embedded). The relevant notions here are subgroups, not subsets; $O\cong \Bbb Z_2\rtimes SO$ (isometries of spheres are generated by rotations plus a reflection), and your question is properly formulated in terms of these subgroups. I guess that it translates to asking whether $\det=-1$ elements are the things that exchange IIA and IIB, while $\det=1$ elements preserve it. $\endgroup$
    – Danu
    Jul 18, 2016 at 16:51
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    $\begingroup$ In response to your last question, it seems that, in the given context, $SO/SO\times SO$ should be interpreted as $SO/(SO\times SO)$. $\endgroup$
    – Danu
    Jul 18, 2016 at 16:54
  • $\begingroup$ Thank you @Danu for your reply. Isn't it natural to think of the orthogonal group as containing the special orthogonal group as a subgroup? Isn't that also the sense implied in this paper when he says "the elements of T-duality which are in $O(d,d;\mathbb{ℤ})$ but not in $SO(d,d;\mathbb{ℤ})$ will exchange IIA and IIB "? $\endgroup$ Jul 18, 2016 at 18:23
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    $\begingroup$ The sense implied in the paper is the one I outlined above---from the decomposition I gave you you can see that there are two copies of SO inside O; it is not natural to take one of them out, because taking complements is not a natural thing to do in the context of a group. The elements of O that are not elements of SO are exactly those with $\det=-1$, and this is a more natural way to express your intent. $\endgroup$
    – Danu
    Jul 18, 2016 at 19:14

1 Answer 1


Regarding Question 1, you are talking about bosonic strings, rather than heterotic, since for heterotic strings left- and right- moving modes are compactified on $T^d$ and $T^{16+d}$ (or vice versa). Now, for bosonic strings on $T^d$ the moduli space should be $$M=\frac{O(d,d;\mathbb{R})}{O(d;\mathbb{R})\times O(d;\mathbb{R})\times O(d,d;\mathbb{Z})}$$ because the evenness condition is invariant under $O(d,d;\mathbb{R})$ transformations, and the mass formula $$m^2=2(p_L^2+p_R^2)+...$$ is invariant under $O(d;\mathbb{R})\times O(d;\mathbb{R})$ transformations (see Becker, Becker, Schwarz textbook, pages 270 and 283).

Question 2. As already answered by Danu, in type II superstrings, the transformations of $O(d,d;\mathbb{Z})$ with determinant $-1$ takes you between IIA and IIB (since they change relative chirality of fermions), whereas $det=+1$ transformations are symmetries within IIA and IIB separately (they don't change chiralities). So the moduli spaces of both type II theories (compactified on a torus) are the same.

As for Quation 3, the quotient by a product is the right one.

PS You probably already know the answers, but I'll leave it here anyway.


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