Skip to the Main Content

Note:These pages make extensive use of the latest XHTML and CSS Standards. They ought to look great in any standards-compliant modern browser. Unfortunately, they will probably look horrible in older browsers, like Netscape 4.x and IE 4.x. Moreover, many posts use MathML, which is, currently only supported in Mozilla. My best suggestion (and you will thank me when surfing an ever-increasing number of sites on the web which have been crafted to use the new standards) is to upgrade to the latest version of your browser. If that's not possible, consider moving to the Standards-compliant and open-source Mozilla browser.

March 27, 2010

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

(1)λ tγ μλ=0\lambda^t \gamma^\mu \lambda = 0

Here, λ∈16\lambda\in \mathbf{16}, is a chiral spinor of Spin(10)Spin(10), and γ μ\gamma^\mu is a symmetric 16×1616\times 16 matrix, expressing the Clebsch-Gordon coefficient, Sym 2(16)⊃10\mathop{Sym}^2(\mathbf{16})\supset \mathbf{10}. In terms of 32×3232\times 32 gamma matrices, Γ 0Γ μ=(γ μ 0 0 γ˜ μ) \Gamma^0 \Gamma^\mu = \begin{pmatrix}\gamma^\mu& 0 \\ 0& \tilde{\gamma}^\mu\end{pmatrix} Now, the 16\mathbf{16} of Spin(10)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=1116-5=11 complex-dimensional (real dimension 22). In fact, it’s a complex cone over

(2)X (0,10)=Spin(10)/U(5)X_{(0,10)} = Spin(10)/U(5)

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

(3)X (0,10)=Spin(10)/U(5) X (2,8)=Spin(2,8)/U(1,4) X (4,6)=Spin(4,6)/U(2,3) X (5,5)=Spin(5,5)/GL(5,ℝ)\begin{gathered} X_{(0,10)}=Spin(10)/U(5)\\ X_{(2,8)}=Spin(2,8)/U(1,4)\\ X_{(4,6)}=Spin(4,6)/U(2,3)\\ X_{(5,5)}= Spin(5,5)/GL(5,\mathbb{R}) \end{gathered}

There’s one for each real form of A 4A_4. But, you’ll note, signature (1,9)(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)(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)X_{(1,9)}. And (contra Lubos, below, who is, alas, a bit confused), X (1,9)X_{(1,9)} is necessarily of the form X (1,9)=Spin(1,9)/HX_{(1,9)} = Spin(1,9)/H, for some subgroup HH. However, unlike the cases in (3), HH is not a real form of GL 5,ℂGL_{5,\mathbb{C}}. In fact, it’s not even a reductive group!

One can show that H=H 0⋉V H = H_0 \ltimes V where H 0=Spin(1,1)×U(4)⊂Spin(1,1)×Spin(8)⊂Spin(1,9)H_0= Spin(1,1)\times U(4) \subset Spin(1,1)\times Spin(8) \subset Spin(1,9) such that the 16\mathbf{16} decomposes, under H 0H_0, as 16=(1 2+1 −2+6 0) +1+(4 1+4¯ −1) −1 \mathbf{16} = {(1_2 + 1_{-2} + 6_0)}^{+1} + {(4_1 +\overline{4}_{-1})}^{-1} (the superscript is the Spin(1,1)Spin(1,1) weight). VV is the Abelian subgroup generated by the M +jM^{+j} generators of so(1,9)so(1,9), where jj is an SO(8)SO(8) vector index. Under H 0H_0, VV transforms as V=(4 −1+4¯ 1) +2 V= {(4_{-1} +\overline{4}_{1})}^{+2} The pure spinor constraint kills the (1 2) +1+(4¯ −1) −1⊂16{(1_2)}^{+1}+{(\overline{4}_{-1})}^{-1}\subset \mathbf{16}, and X (1,9)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)X_{(0,10)}=Spin(10)/U(5). Presumably, things change significantly, when dealing with X (1,9)=Spin(1,9)/(Spin(1,1)×U(4))⋉VX_{(1,9)}=Spin(1,9)/(Spin(1,1)\times U(4))\ltimes V.

Posted by distler at 2:47 PM | Permalink | Followups (23)

March 3, 2010

Coupling to Supergravity

There seems to be a certain amount of confusion about the claims of Seiberg and Komargodski in their latest paper. I have to say that I was confused, and there’s at least one recent paper arguing (more-or-less correctly) against claims that I don’t think they’re making.

So here’s my attempt to clear things up.

Consider an 𝒩=1\mathcal{N}=1 supersymmmetric nonlinear σ\sigma-model in d=4d=4, ie a Wess-Zumino model with target space, MM, a Kähler maninfold of complex dimension, nn. When can one couple such a theory to supergravity? A naïve reading of their paper might lead one to think that the possibilities are

  1. One can couple the theory to minimal supergravity if and only if the Kähler form, ω\omega, on MM, is exact.
  2. One can couple the theory to “new minimal” supergravity if and only if the theory has an exact U(1) RU(1)_R symmetry. In this case, ω\omega could be cohomologically nontrivial.
  3. If ω\omega is cohomologically nontrivial, and the theory does not have a U(1) RU(1)_R symmetry, then the only possibility is to couple to non-minimal “16|16” supergravity.

One might think that, but one would be wrong. As Bagger and Witten showed, nearly 30 years ago, coupling to minimal supergravity does not require the Kähler form to be exact. Rather, [ω][\omega] must be an even integral class.

Continue reading “Coupling to Supergravity” ...
Posted by distler at 8:53 PM | Permalink | Post a Comment