### Berkovits Update

I’ve spent the past few days in a lengthy email correspondence with Nathan Berkovits about some of the questions raised in my previous post about his “pure spinor” approach to the superstring.

The main progress has been in clarifying his construction of the bilocal operators, $\hat{b}(y,z)$, which play the role of anti-ghosts in his theory. Recall that I was more than a little worried about the apparent $z$-dependence of the amplitudes.

The idea (reminiscent of Friedan, Martinec and Shenker) is to augment the Hilbert Space of the $w_\alpha$ and $\lambda^\alpha$. In this “big Hilbert Space”, there is a dimension-zero conformal primary, $\xi_B(z)$, such that

$Z_B$ is $Q$-trivial in the big Hilbert Space, whereas it is $Q$-closed, but nontrivial, in the original “small Hilbert Space.” While $\xi_B$ is an operator in the big Hilbert Space,

lives in the small Hilbert space^{1}. Hence $\partial_z Z_B = \{ Q, \partial_z \xi_B \}$ really *is *$Q$-trivial.

The antighosts, $\hat{b}(y,z)$, can then be written as

where the normal-ordering is defined, as usual, by

The second and third terms, *separately*, are defined only in the big Hilbert Space. But the *difference* is an operator in the small Hilbert space.

$A(y)$ is an operator in the small Hilbert space, to whose construction, Nathan devotes several pages. It satisfies

Now,

which, modulo the usual worries about surface terms, integrates to zero.

In this “big Hilbert Space” formulation, $\hat{b}(y,z)$ is a sum of (products of) local operators. At least in the FMS construction, you can freely pass from the small Hilbert Space formulation to the large, by inserting an extra $\xi$ somewhere on $\Sigma$. Having done so, it seems to me that you can throw away the third term in expression for $\hat{b}(y,z)$. By exactly the arguments presented here, it yields a total derivative on the moduli space. We’d then have a simple *local* operator representing the antighost,

By construction, $\{ Q,b(y)\}=0$. At the cost of a single extra insertion, $\xi_B(z)$, we can eliminate Berkovits’s spurious dependence on $3g-3$ extra points, $z_i$.

What happens when you insert a $Q$-trivial state in the amplitude? Commuting the $Q$ past all the insertions, it gets hung-up on the lone insertion of $\xi_B$. This term would give zero in the FMS case (or here, if we use the $\hat{b}(y,z)$ insertions), because there’s nothing *else* available to absorb the zero mode of $\xi$. Instead, the only contributions would come from when $Q$ hits the $b$-insertions ($\hat{b}(y,z)$ insertions), giving a total derivative. Here, we’ve arranged that $Q$ annihilate the $b(y)$s. Instead, the term where it hits the lone $\xi_B$ is nonzero, because the zero mode can be absorbed by the $b(y)$ instead. The result is still a total derivative, as would be manifest, if we’d used the $\hat{b}(y,z)$ insertions instead.

Anyway, the details could certainly be worked out better (despite the superficial similarity with the FMS construction, there are some important differences, I think), but I feel a little more comfortable with the construction of $\hat{b}(y,z)$ than I did last week…

^{1} Just to be clear, it’s *not* obvious that you can actually do this. If the operator $B_{n m} \partial_z N^{n m} \delta (B N)$ has a single pole in its OPE with anything else in correlation function, then it *obviously* cannot be written as $\partial \xi_B$, for some conformal field $\xi_B$. If that should happen, I don’t think there’s a satisfactory definition of the $\hat{b}(y,z)$. Which makes the rest of the discussion here moot.