### Pure Spinor Signature

By some coincidence, I’ve had several discussions, recently, about Nathan Berkovits’s pure spinor formulation of the superstring. Which reminds me of something I’ve long puzzled over. Nathan invariably works in Euclidean signature, where the pure spinor constraint is

Here, $\lambda\in \mathbf{16}$, is a chiral spinor of $Spin(10)$, and $\gamma^\mu$ is a symmetric $16\times 16$ matrix, expressing the Clebsch-Gordon coefficient, $\mathop{Sym}^2(\mathbf{16})\supset \mathbf{10}$. In terms of $32\times 32$ gamma matrices,
$\Gamma^0 \Gamma^\mu = \begin{pmatrix}\gamma^\mu& 0 \\ 0& \tilde{\gamma}^\mu\end{pmatrix}$
Now, the $\mathbf{16}$ of $Spin(10)$ is a *complex* representation, so (1) naïvely looks like 10 complex equations in 16 complex variables. But, in reality, only 5 equations are independent, and the kernel of the pure spinor constraint is $16-5=11$ complex-dimensional (real dimension 22). In fact, it’s a complex cone over

This illuminates the above remark about the dimension of the kernel. Under the decomposition $\mathbf{16} = \mathbf{1}_{-5} + \mathbf{10}_{-1} + \mathbf{\overline{5}}_{3}$, the pure spinor constraint, (1) kills the $\mathbf{\overline{5}}_{3}$.

The dimension (22) is crucial to getting the critical central charge correctly, in Nathan’s formulation.

Thing is, we don’t live in Euclidean signature. While, in string theory and in field theory, we are quite happy to analytically-continue in momenta, when computing scattering amplitudes, we *don’t* Wick-rotate the spinor algebra. That’s always done in the correct, Minkowski, signature.

It turns out that there are analogues of (1),(2) for other signatures

There’s one for each real form of $A_4$. But, you’ll note, signature $(1,9)$ is notably absent. So it’s not so obvious (to me, at least) that the space of solutions to the constraint (1) has the desired dimension (22), when the signature is $(1,9)$.

Does anyone know how to see that it does (or, alternatively, what to do if it doesn’t)?

#### Update (4/16/2010): Non-Reductive

I had a private email exchange with Nathan, who explained the resolution of my conundrum.

The space of solutions to the pure spinor constraint (1), for the Minkowski signature is, of course, of the form of a complex cone over $X_{(1,9)}$. And (contra Lubos, below, who is, alas, a bit confused), $X_{(1,9)}$ is *necessarily* of the form $X_{(1,9)} = Spin(1,9)/H$, for some subgroup $H$. However, unlike the cases in (3), $H$ is not a real form of $GL_{5,\mathbb{C}}$. In fact, it’s not even a reductive group!

One can show that $H = H_0 \ltimes V$ where $H_0= Spin(1,1)\times U(4) \subset Spin(1,1)\times Spin(8) \subset Spin(1,9)$ such that the $\mathbf{16}$ decomposes, under $H_0$, as $\mathbf{16} = {(1_2 + 1_{-2} + 6_0)}^{+1} + {(4_1 +\overline{4}_{-1})}^{-1}$ (the superscript is the $Spin(1,1)$ weight). $V$ is the Abelian subgroup generated by the $M^{+j}$ generators of $so(1,9)$, where $j$ is an $SO(8)$ vector index. Under $H_0$, $V$ transforms as $V= {(4_{-1} +\overline{4}_{1})}^{+2}$ The pure spinor constraint kills the ${(1_2)}^{+1}+{(\overline{4}_{-1})}^{-1}\subset \mathbf{16}$, and $X_{(1,9)}$ indeed has the desired dimension.

I find this answer very striking, in how it differs from the cases in (3), In particular, it ought to have some rather important implications for the construction of amplitudes. Berkovits and Nekrasov develop a rather elaborate construction, involving Čech cohomology classes on the space $X_{(0,10)}=Spin(10)/U(5)$. Presumably, things change significantly, when dealing with $X_{(1,9)}=Spin(1,9)/(Spin(1,1)\times U(4))\ltimes V$.

Posted by distler at March 27, 2010 2:47 PM
## Re: Pure Spinor Signature