# Propagators for the dual symmetric string and a familiar identity

If I could count all the moments so far when faced with a puzzle or question and it was appropriate to say, ‘it’s in my Polchinski!’ or ‘I should double-check my Polchinski!’. Use of the possessive here should be taken as an endearing reference to the role Joe Polchinski’s textbooks have played in one’s life. They’re like a trusty companion.

A great example comes from the other day. Actually, the story begins with Tseytlin’s first principle construction of the dual symmetric string, which serves as the basis of some work I am doing more generally in the doubled formalism. In one of the papers I have been reading that generalises from Tseytlin there are references made about the propagators in this dual symmetric formulation of string theory, followed by an assortment of assumptions including one about $z \rightarrow 0$ regularisation such that $\bar{\partial} z^{-1} = \pi \delta^{(2)}(z)$. When I first read this over, it wasn’t immediately obvious to me from where this identity originated; I was much more focused on the actual propagators and some of the important generalisations of the construction, so I sort of left it as something to be returned to. Then in the last week I was reminded of it again, so I went back quickly and I realised how silly it was of me to not immediately recognise why this identity is true. The equation from above just comes from eqn. (2.1.24) in Polchinski, $\partial \bar{\partial} \ln \mid z \mid^2 = 2\pi \delta^{(2)} (z,\bar{z})$. Furthermore, one may have also thought equivalently of eqn. (2.5.8) in the context of bc CFT, which was what I first recalled (it also happens to be the subject of problem 2.1 at the end of the chapter).

There are a few ways to verify $\partial \bar{\partial} \ln \mid z \mid^{2} = \partial \bar{z}^{-1} = \bar{\partial} z^{-1} = 2 \pi \delta^{2} (z, \bar{z})$. One direct way is to take the terms to the left of the first equality, noting that $\partial \bar{\partial} \ln \mid z \mid^{2} = 0$ if $z \neq 0$. What we want to do is integrate this over some region $R$ in the complex plane, using divergence theorem given in eqn. (2.1.9) which states $\int_R d^{2}z (\partial_z v^z + \partial_{\bar{z}} v^{\bar{z}}) = i \oint_{\partial R} (v^{z} d\bar{z} - v^{\bar{z}} dz)$, where the contour integral circles $R$ counterclockwise.

For the holomorphic case, using the test function $f(z)$,

$\int_{R} d^{2}z \ \partial \bar{\partial} \ \ln \mid z \mid^{2} \ f(z) \ (1)$

From derivative properties we see $\partial \bar{\partial} \ln \mid z \mid^{2} = \partial \bar{\partial} (\ln z + \ln \bar{z}) = \bar{\partial} z^{1}$. Taking this fact into account and then also finally invoking divergence theorem,

$= \int_{R} d^{2}z \ \bar{\partial} \ z^{-1} \ f(z)$
$=-i \ \oint_{\partial R} \ dz \ z^{-1}$
$= 2 \pi f(0) \ (2)$

Where we have used the fundamental result in complex analysis that the contour integral of $z^{-1}$ is $2\pi i$. The same procedure can also be used for the antiholomorphic case. Hence, $\int \ d^2z \ \partial \bar{\partial} \ln \mid z \mid^{2} \ f(z, \bar{z}) = 2\pi f(0,0)$, which therefore gives us $\partial \bar{\partial} \ln \mid z \mid^{2} = 2 \pi \delta^{2} (z, \bar{z})$.

As an aside, thinking of this reminds me of how I’ve been wanting to go back and update whatever notes I have so far uploaded to this blog as part of my ‘Reading Polchinski’ series, which I started writing in my first undergraduate year. I still like the idea of uploading my hundreds of pages of notes on Polchinski’s textbooks and formatting them into a pedagogical blog series, because there are so many subtleties and nuances that are fun to think through. I think I now also have a better sense of how I want to continue formatting the online version of the notes and communicate them, so after my thesis I intend to return to the project :).