Polyakov action of a bosonic string propagating in Minkowskian spacetime is:

$$S[\gamma, X] = \frac{T}{2}\int \mathrm{d}^{2}\sigma{\sqrt{-\gamma}}\gamma^{ab}\partial _{a}X^{\mu}(\sigma)\partial_{b}X^{\nu }(\sigma)\tag{1}$$

Which exhibits gauge redundancy, i.e the action is invariant under Weyl symmetry $\gamma_{\alpha\beta}(\sigma, \tau) = \Lambda(\sigma, \tau)\eta_{\alpha\beta}$ and diffeomorphisms ${\displaystyle \sigma ^{\alpha }\rightarrow {\tilde {\sigma }}^{\alpha }\left(\sigma ,\tau \right)}$.

$$Z = \int\mathcal{D}[X]\mathcal{D}[\gamma]e^{\frac{i}{\hbar}S[X, \gamma]}\tag{2}$$

Therefore, to avoid oversummation in path integral, additional fields (ghosts) have to be introduced. However, can you show how topological twist of this $\sigma$-model can also be used besides the introduction of additional fields?


1 Answer 1


The usual B-twist that follows Witten's method, with a BRST Operator of $Q = G^+ + \tilde{G}^+$, leads to non-zero ghost number, so either on- or off-shell twisting leads to a theory with ghosts. What I shall briefly present is an approach where the BRST operator is changed.

See this answer for more on ghost numbers.

Instead of using supersymmetries and having a BRST parameter of $\Lambda = \epsilon^+ = \tilde{\epsilon}^+$ to construct the BRST Operator, we can identify the supersymmetries as the BRST and anti-BRST operators themselves. $$ \overline{s}X^{i*} = \psi^{i*}, \quad sX^{i*} = \xi^{i*}, \\ \overline{s}\psi^i = -\frac{i}{2} \partial_+ X^i, \quad s\psi^{i*} = -\frac{1}{2} F^{i*}, \\ \overline{s}\xi^{i*} = \frac{1}{2} F^{i*}, \quad s\xi^i = -\frac{i}{2} \partial_- X^i\\ \overline{s}F^i = i\partial_+ \xi^i, \quad sF^i = -i\partial_- \psi^i, $$

Where F and X are bosonic fields, ψ, and ξ are fermionic ones. F is the typical auxiliary field to close off the supersymmetry algebra off-shell.

Now we have a conservation condition of the s raising the ghost number by one, and $\mathbf{\bar{s}}$ lowering it by one.

As a result, Bosons have a 0 ghost number. See this answer for more on conservation.

We can now rewrite our action in terms of the gauge fixing term and the classical part: $$S = \mathbf{s}\left[-2ig_{ij^*} \partial_+ X^{j^*} \xi^i + 2g_{ij^*} \psi^{j^*} \left(F^i - 2\Gamma^i_{jk} \xi^j \psi^k\right)\right]. $$

The main point of interest now is that the classical part has no dependence on X, so there is only one gauging, namely $\delta X^{i^*} = \epsilon^{i^*}$, so there is no overcounting.

Now we can directly call upon the Goddard-Theorn No-Ghost theorem. If we have a positive energy representation of the Virasoro algebra with central charge 24, V with V-invariant bilinear form, then the image of $V\otimes \pi$, where $\pi$ is the irreducible module of the $\mathbf{R^{1,1}}$ Heisenberg Algebra attached to a non-zero vector in $\mathbf{R^{1,1}}$, under quantization, is isomorphic to the subspace V on which the Virasoro operator $L_0$ acts.

Thus, since $L_0$ annihilates unphysical states, it follows that in 26D we have no ghosts.

Connecting this to the topological twist construction for getting a term in our action not depending on X for double counting, allows us to have a quantization with no ghosts.

Here are a few papers for more reading:


No Ghost Theorem of BRST Quantized Bosonic String Theory

B-Twisted Topological Sigma Models


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.