Musings

The idea that there might be a topological version of the supersymmetric $\sigma$-model on a $G_2$ manifold dates back to Shatashvili and Vafa. They noticed that, in addition to the $N=1$ superconformal algebra (generated by $T(z)$ and $G(z)$), the theory has an extended chiral algebra, with the additional generators forming supermultiplets of spin 3/2 and 2. In (1,1)superspace (with $D= \partial_\theta + \theta\partial_z$), we have the $N=1$ supercurrent and stress tensor $$G+\theta T = - \frac{1}{2} g_{i j}(X) D X^i \partial_z X^j$$ as well as $$G_I + \theta K = \frac{i}{15} \phi^{(3)}_{i j k}(X) D X^i D X^j D X^k$$ and $$T_I + \theta M = \frac{1}{5} ( \phi^{(4)}_{i j k l} D X^i D X^j D X^k D X^l + \frac{1}{2} g_{i j}(X) D X^i \partial_z D X^j)$$ formed out of the covariantly-constant 3-form, $\phi^{(3)}$, and its Hodge dual, $\phi^{(4)}=*\phi^{(3)}$, associated to the existence of a $G_2$ structure. The key facts are

1. $G_I$ and $T_I$ form a second superconformal algebra, with central charge $c= 7/10$, i.e. there’s a hidden Tricritical Ising Model in this theory.
2. If we write $T=T_I+T_r$, then $T_I(z) T_r(w)=$ nonsingular, which is to say that, thought of as a conformal (as opposed to superconformal) theory, this model is the tensor product of a $c=7/10$ Tricritical Ising Model and a second theory with $c= 49/5$, whose stress tensor is $T_r$.

de Boer et al show that there’s a BPS bound on the conformal weight $$h = h_I + h_r \geq \frac{1+\sqrt{1+80 h_I}}{8}$$ which is saturated for the following conformal primaries in the NS sector: $|h_I,h_r\rangle=|0,0\rangle$, $|1/10, 2/5\rangle$, $|6/10,2/5\rangle$ and $|3/2,0\rangle$, whose Tricritical Ising components are just the primaries $\Phi_{n,1}$, $n=1,2,3,4$ in the Kač table.

Defining $P_n$ to be the projection (in the NS sector) onto the Virasoro representations corresponding to $\Phi_{n,1}$, the fusion rules of the Tricritical Ising Model allow us to decompose $$G(z) = G^\uparr(z) + G^\darr(z)$$ in the NS sector, where $$G^\uparr(z) = \sum_n P_{n+1} G(z) P_n,\qquad G^\darr(z) = \sum_n P_{n} G(z) P_{n=1}$$ de Boer et al define $$Q = G^\uparr_{-1/2}$$ which is nilpotent, by virtue of $$Q^2 = \sum_n P_{n+2} G_{-1/2}^2 P_n = \sum_n P_{n+2} L_{-1} P_n =0$$ where we used the above decomposition of $G(z)$ and the fact that $L_n$ has vanishing matrix elements between different Virasoro representations.

The spin field, which creates the ground state of the Ramond sector, has $h=7/16$, and lies entirely in the Tricritical Ising sector of the theory (it is $\Phi_{1,2}$ in the Kač table). We can decompose it into two conformal blocks $$\Phi_{1,2} = \Phi_{1,2}^+ + \Phi_{1,2}^-$$ defined by its action on the two Virasoro representations that comprise the R-sector ($\mathcal{H}_{1,2}$, with $h_I=7/16$ and $\mathcal{H}_{2,2}$, with $h_I=3/80$): $$\Phi_{1,2}^+ :\, \left\{ \array{ \mathcal{H}_{1,2}\to \mathcal{H}_{4,1}\\ \mathcal{H}_{2,2}\to \mathcal{H}_{3,1}}\right.\qquad \Phi_{1,2}^- :\, \left\{ \array{ \mathcal{H}_{1,2}\to \mathcal{H}_{1,1}\\ \mathcal{H}_{2,2}\to \mathcal{H}_{2,1}}\right.$$ Let $\mathcal{O}_{n,\alpha}$ be the operators corresponding to the “special” NS conformal primary states introduced above, $|h_I = \frac{(2n-3)(n-1)}{10}, h_r = \frac{(4-n)(n-1)}{5},\alpha\rangle$, where $\alpha$ is some discrete index labeling the possibly distinct operators with these conformal weights. ($\mathcal{O}_1=𝟙$ and $\mathcal{O}_4=G_I(z)$ presumably don’t need such a label if the “internal” $c=49/5$ theory is unitary.) The $\mathcal{O}_{n,\alpha}$ don’t commute with $Q$, but $$\mathcal{A}_{n,\alpha}(z)= \sum_m P_{n+m-1} \mathcal{O}_{n,\alpha}(z) P_m$$ do. The observables of the “topological” theory are defined as

where I’ve suppressed the right-movers, as in the rest of my summary. The claim is that precisely these amplitudes, for $n=1,\dots,4$, are independent of the insertion points, and constitute a 2D topological field theory.

de Boer et al also have a proposal for a higher-genus “topological string theory” generalization, but I have to say that I don’t really understand it. So, maybe I’d better stop here.