Number of zero-modes on the sphere

Question

Is it true that a field of conformal dimension $h$ (integer or half integer) has $1-2h$ zero-modes on the sphere, if $1-2h \geq0$.

This seems to be right for different ghost fields :

• $c$ has weight $-1$ and 3 zero-modes
• $\gamma$ has weight $-1/2$ and 2 zero-modes
• $\xi$ has weight $0$ and 1 zero-modes
• $\eta$ has weight $1$ and 0 zero-modes

I'm not sure how to compute the number of zero-modes on the sphere... Is it the argument given at p.167 in volume 1 of Polchinski's book ?

Answer 1

Yes, OP is right, cf. Ref. 1 and 2. Consider a generalized $bc$ ghost system. (This can be generalized to any system with the pertinent mathematical properties used below.) Here $b$ and $c$ are primary fields of conformal weights $h_b$ and $h_c$, respectively, where $h_b+h_c=1$.

Counting globally defined, holomorphic zero-modes of a (generalized) $bc$ system is related to the Riemann-Roch theorem. Here applied to the Riemann sphere $S^2$, i.e. Euler characteristic $\chi=2$, or equivalently, genus zero.

We basically have to check that a zero-mode on the complex plane $\mathbb{C}$ can be holomorphically extended to the point $z=\infty$ on the Riemann sphere $S^2=\mathbb{C}\cup \{\infty\}$. To describe the neighborhood of $z=\infty$, we perform a coordinate transformation $\tilde{z}=-1/z$. In the new coordinates the ghost field reads

$$\tilde{c}~=~\left(\frac{\partial\tilde{z}}{\partial z}\right)^{-h_c} c~=~z^{2h_c} c.$$

In order that a zero-mode doesn't blow up at $z=\infty$, it must be a polynomial in $z$ of order at most $-2h_c$. Recall that an $n$'th order polynomial in $z$ has $n+1$ coefficients. In other words, the number of globally defined, holomorphic zero-modes for $c$ is

$$ \max(1-2h_c,0).$$

References:

1. J. Polchinski, String Theory Vol. 1, 1998; p.167.

2. M.B. Green, J.H. Schwarz and E. Witten, Superstring theory, Vol. 1, 1986; p. 161-163.