As I know it, T-duality essentially tells us that if we compactify a superstring theory on a circle of radius $R$, it is equivalent to a string theory compactified on a circle of radius $\tfrac{\alpha'}{R}$. So, take $R\to\infty$. Then $R\to\infty$ and $\tfrac{\alpha'}{R}\to 0$. Hence, this should tell us that a string theory in 10 (noncompact) dimensions is equivalent to one in 9D.

However, I know that superstring theory is only non-anomalous and free of ghosts in 10D, so where does this reasoning break down?

  • $\begingroup$ It doesn’t become a 9D theory. It remains a 10D theory. $\endgroup$ Jul 5, 2020 at 10:44
  • $\begingroup$ But if it's compactified on a circle of zero radius, shouldn't that be equivalent to removing a dimension? $\endgroup$
    – arow257
    Jul 5, 2020 at 18:50
  • $\begingroup$ It’s a limit of a dimension of very small size. That doesn’t mean the dimension disappears. Similarly if you increase the radius (even to infinity) you are not creating extradimensions, do you? $\endgroup$ Jul 5, 2020 at 18:58
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    $\begingroup$ But if you take the radius of a circle to zero, it just becomes a point. So, if I compactify an entire dimension down to just a point, it should be equivalent to not having that dimension $\mathbb{R}^{8,1}\times S^1_R\to \mathbb{R}^{8,1}\times\{0\}\simeq \mathbb{R}^{8,1}$. Taking the radius to infinity would just uncompactify the compactified dimension. $\endgroup$
    – arow257
    Jul 5, 2020 at 19:43
  • $\begingroup$ So as the theories with $R$ and $\alpha'/R$ are the same this means that in the $R\rightarrow \infty$ limit you also lose a dimension? $\endgroup$ Jul 6, 2020 at 12:31

1 Answer 1


Well, I think I finally understand why my argument is flawed. Consider the case $R\to 0$ for closed strings; all of the momentum states, i.e., states with KK mode $n\ne 0$, become infinitely massive. If we were studying field theory we would stop here, as this would be all that would happen—the surviving fields would simply be independent of the compact coordinate, and so we would have performed a dimension reduction down to 9D. However, in string theory, things are different: the pure winding states (i.e., $n = 0$, winding mode $w \ne 0$ states) form a continuum as $R \to 0$, since it is very "cheap" to wind around the small circle. Therefore, in the $R \to 0$ limit, an effective uncompactified dimension reappears.

Let us now consider the $R \to 0$ limit of the open string spectrum. Open strings do not have a conserved winding around the periodic dimension and so they have no quantum number comparable to $w$, so something different must happen, as compared to the closed string case. In fact, it is more like field theory: when $R \to 0$ the states with nonzero momentum go to infinite mass, but there is no new continuum of states coming from winding. So we are left with a theory in one dimension fewer. A puzzle arises when one remembers that theories with open strings have closed strings as well, so that in the $R \to 0$ limit the closed strings live in 10 spacetime dimensions but the open strings only in 9D. This is perfectly fine, though, since the interior of the open string is indistinguishable from the closed string and so should still be vibrating in 10 dimensions. The distinguished part of the open string is the endpoints, and these are restricted to a 9-dimensional hyperplane. In particular, this is a D9-brane.


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