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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, λ16\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 165=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,)\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,GL_{5,\mathbb{C}}. In fact, it’s not even a reductive group!

One can show that H=H 0V 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) 116{(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 March 27, 2010 2:47 PM

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Re: Pure Spinor Signature

The half-spinor representation S +S_+ in signature (1,9) consists of only two orbits: the zero spinor and everything else. So the analogue of your formula (3) in this signature is S (1,9){0}=Spin(1,9)/(Spin(7) 8),S_{(1,9)}\setminus\{0\} = \mathrm{Spin}(1,9)/(\mathrm{Spin}(7) \ltimes \mathbb{R}^8), which is a 16-dimensional orbit.
Posted by: Josť Figueroa-O'Farrill on March 28, 2010 9:58 AM | Permalink | Reply to this

Re: Pure Spinor Signature

… which sounds very bad for the pure-spinor formalism.

An 11-dimensional orbit was quite crucial for Nathan to get the central charge of the worldsheet theory correctly.

Posted by: Jacques Distler on March 28, 2010 10:15 AM | Permalink | PGP Sig | Reply to this

Re: Pure Spinor Signature

What makes you say that when scattering amplitudes are computed the spinor algebra is not Wick rotated? Last time I did perturbative calculations in QED (text book stuff) the spinor algebra *was* Wick rotated, or so I remember, and the anticommutation relations of the gamma’s had a delta_{\mu\nu} on the right hand side as opposed to g_{\mu\nu}.

Posted by: Trent on March 30, 2010 11:24 AM | Permalink | Reply to this

Dirac Algebra

Last time I did perturbative calculations in QED (text book stuff) the spinor algebra was Wick rotated

Well, then you probably got the wrong answer, because one frequently uses relations like η μ μ =(d2)\tensor{\eta}{_^\mu_\mu} = (d-2). In Euclidean signature, that would have been a “dd” on the RHS.

As a simple example (which occurs in just about every calculation of scattering amplitudes for fermions), what is Γ μΓ νΓ μ \Gamma^\mu \Gamma^\nu \Gamma_\mu (summed on μ\mu)? The answer is different in Euclidean and Minkowski signatures.

Posted by: Jacques Distler on March 30, 2010 12:05 PM | Permalink | PGP Sig | Reply to this

Re: Dirac Algebra

I don’t think η_{μμ}=d-2. It’s still d. How can you add the diagonal entries of η_{μν} with all the indices in the subscripts? The same can be said for Γ algebras.

Posted by: YT on March 30, 2010 12:32 PM | Permalink | Reply to this

Re: Dirac Algebra

What? \eta^\mu_\mu = d - 2 ??? If \eta_{\mu\nu} has both indices down it’s diag(-1,1,1,1) or diag(1,-1,-1,-1) but if you want to take the trace you need one index up and one index down: \eta^\mu_\nu which then equals \delta^\mu_\nu = diag(1,1,1,1). So I don’t agree, \eta^\mu_\mu does not equal d-2 even in Minkowski signature, it actually equals d for any signature.

Same is true for your \Gamma^\mu \Gamma^\nu \Gamma_\mu example. Since only upper and lower indices can be contracted, the sum will not know about the signature.

More nitpickingly: only a quadratic form has a signature, which has 2 indices either up or down, but a linear operator having an index up and another down does not have a signature. And you can only take the trace of a linear operator not a quadratic form. So any trace will be the same in all signatures.

Posted by: Trent on March 31, 2010 5:59 AM | Permalink | Reply to this

Re: Dirac Algebra

Sorry.

You’re both correct about the Dirac-matrix algebra, when one uses, η μν\eta^{\mu\nu} to raise and lower indices.

I should refrain from posting comments when I’m running off to catch a flight.

The difference between Minkowsi and Euclidean spinor algebra (in d=4d=4) occurs when one take the absolute square of the amplitude to compute a cross-section.

In both Euclidean and Minkowski signatures, the vector occurs in the tensor product of a left-handed and a right-handed spinor. In Minkowski signature, the complex-conjugate of a left-handed spinor is a right-handed spinor. So, when we square the amplitude and sum over polarizations, we get relations like

(1) su s(p)u¯ s(p)=p sv s(p)v¯ s(p)=p\begin{gathered} \sum_s u_s(p)\overline{u}_s(p) = \slash{p}\\ \sum_s v_s(p)\overline{v}_s(p) = \slash{p} \end{gathered}

(in the massless case) and su s(p)u¯ s(p)=p+m sv s(p)v¯ s(p)=pm \begin{gathered} \sum_s u_s(p)\overline{u}_s(p) = \slash{p}+m\\ \sum_s v_s(p)\overline{v}_s(p) = \slash{p}-m\\ \end{gathered} (in the massive case).

In Euclidean signature, the complex-conjugate of a left-handed spinor is still left-handed. So you never get a vector by multiplying a left-handed spinor by its complex-conjugate.

The story’s different in other dimensions, but it’s still true that the reality properties of spinors vary with signature, and it’s the Minkowski-signature spinors that we use in the analogue of (1).

Posted by: Jacques Distler on March 31, 2010 8:03 AM | Permalink | PGP Sig | Reply to this

Re: Pure Spinor Signature

A better example: in lattice gauge theory the fermions definitely obey Euclidean anti-commutation relations.

Posted by: Trent on March 30, 2010 11:26 AM | Permalink | Reply to this

Re: Pure Spinor Signature

My understanding is that the signature of spacetime is not really determined by the dynamic principle, at least not by the perturbative worldsheet action as it stands. The *prescription* of how to analytically continue certain momenta is an extra input needed to make sense of the theory.

The hope would be that a more complete understanding of string theory, for example a full non-perturbative action principle of M-theory, can shed light on this and replace prescriptions with results.

Out of cursiosity, what happens if one tries to naively define the spinor theory in Minkowski signature? What’s the (naive) critical dimension then? How hard is it to show that this theory does not exist (which I presume must be true, unless it’s somehow dual to what we know)?

Posted by: Rufus Anton on April 1, 2010 9:10 AM | Permalink | Reply to this

Re: Pure Spinor Signature

Come on, Jacques!

I agree that eta with suscript mu and superscript mu sums up to d, and not d-2 as you argue.

Concerning the spaces, you should always imagine the complexified spaces, and be very careful what the reality conditions say and how they’re interpreted.

I agree with everyone else that in the Euclidean space, we have to Wick-rotate all the spinors as well. How could we not?

You, Jacques, correctly say that in the Euclidean space, the complex conjugate spinor has the same chirality, while in the Minkowski space, it doesn’t.

But you’re missing one equally important point that cancels the sign flip back. If you Wick-rotate complex conjugation of a spinor at the same point (as another spinor) from the Minkowski space, you don’t just get the complex conjugate spinor at the Wick-rotated point.

Instead, the point gets modified, too. In particular, its time coordinate flips the sign. So the complex conjugation in the Minkowski space gets Wick-rotated to complex conjugation in the Euclidean space combined with the time-reversal - and this combination of operations does revert the chirality in the Euclidean space, much like it did in the Minkowski space.

The time reversal appears because the hermitean conjugate of the evolution, exp(i.H.t), is exp(-i.H.t).

Quite generally, if you can do some calculation in the Euclidean spacetime and obtain results that are physically consistent in the Minkowski spacetime, it’s clearly enough. So even if I didn’t tell you exactly where your reasoning went wrong, you should have known that all such reasoning is wrong. In most cases, it’s wrong to imagine things as “directly living” in the Minkowski space because the analytical properties are more transparent in the Euclidean space. After all, the Euclidean path integral is the more well-defined one, too.

Best wishes
Lubos

Posted by: Lubos Motl on April 2, 2010 8:13 AM | Permalink | Reply to this

Re: Pure Spinor Signature

Sorry, Lubos, I don’t know what spacetime points you’re talking about. I’m talking about the spinors that label external fermionic states. These are (in the usual formalism) momentum eigenstates. And, at least for Minkowski spinors, they obey completeness relations like the ones above.

The fact that left-handed and right-handed spinors in Euclidean signature are not related by complex-conjugation (as they are in Minkowski signature) has other consequences. When we write down the Lagrangian for the Euclidean field theory, we double the number of fermionic degrees of freedom. That is, we take Ψ¯\overline{\Psi} to be an independent field (not equal to Ψ Γ 0\Psi^\dagger \Gamma^0). This latter remark is, of course, intimately related to the first.

As to the general utility of Euclidean field theory, no one is questioning that. But the S-matrix (which is the central object of study, in perturbative string theory, which is what we happen to be discussing) is a distinctly Minkowskian concept, which has no analogue in Euclidean signature.

What it does have is an analytic continuation to complex momenta. When we study that analytic continuation, we don’t also analytically continue the polarizations (from representations of U(1)U(1) in the massless case, or of SU(2)SU(2) in the massive case, to representations of *\mathbb{C}^* or SL(2,)SL(2,\mathbb{C}), respectively). The above statements about the spinors labeling external states is just this remark, applied to spin-1/2.

Posted by: Jacques Distler on April 2, 2010 8:53 AM | Permalink | PGP Sig | Reply to this

Re: Pure Spinor Signature

A related remark.

As you’ve probably heard, Lubos, there has been a huge interest, in recent years, in applying twistor-space techniques to the study of scattering amplitudes in d=4d=4 super-Yang-Mills and supergravity.

These twistor methods are most straightforwardly applied in the split signature, where the Lorentz-group is Spin(2,2)SL(2,)×SL(2,)Spin(2,2)\simeq SL(2,\mathbb{R})\times SL(2,\mathbb{R}). But the polarizations of the external states are always still labeled by irreducible representations of Spin(2)U(1)Spin(2)\simeq U(1), not Spin(1,1)Spin(1,1)\simeq \mathbb{R} (let alone, by irreducible representations of the complexification Spin(2,)Spin(2,\mathbb{C})).

This is an important point, which becomes even more important in higher dimensions, where the irreducible representations of the little group are no longer 1-dimensional.

Posted by: Jacques Distler on April 2, 2010 9:57 AM | Permalink | PGP Sig | Reply to this

Re: Pure Spinor Signature

My previous long comment has disappeared. Too bad. A fraction of it is explained on my blog.

I agree with your twistor point - in fact, it’s precisely my point. Why don’t you criticize the twistor business because it is only naturally defined in an unphysical signature 2+2, and doesn’t look “simply” in 3+1? The perceived “problem” is completely isomorphic to Nathan’s case. It’s not a problem in either case. In some signature, things are just simpler.

Assuming for a while that this comment won’t disappear again, try to Wick-rotate real vector fields. What will you get? You will get the time reversal in the reality constraints just like in the case of spinors. There’s nothing special about the spinors. The reality condition A*(x,t)=B(x,t) gets translated to A*(x,-tE)=B(x,tE). It always gets a time reversal. For spinor, it’s also compensating the flipped reality condition.

If you think that the vector fields remain “real” in the Euclidean spacetime, it’s just an illusion. They’re complex fields with a more contrived reality condition - but the path integral over the “holomorphic” fields just behaves in the same way as the path integral over the real fields, so you don’t even distinguish them.

Cheers
LM

Posted by: Lubos Motl on April 2, 2010 10:17 AM | Permalink | Reply to this

Re: Pure Spinor Signature

Why don’t you criticize the twistor business because it is only naturally defined in an unphysical signature 2+2, and doesn’t look “simply” in 3+1? The perceived “problem” is completely isomorphic to Nathan’s case. It’s not a problem in either case. In some signature, things are just simpler.

Because you’re not paying attention.

In the twistor approach, the momenta are analytically-continued to the split signature (where the twistor transformation is “nice”). But the polarizations are still the polarizations appropriate to the Minkowski signature (a point which becomes even more obvious when you go to higher dimensions, where the irreducible representations of the little group are no longer 1-dimensional).

What we’re discussing here are precisely the generalizations of these considerations to 10 dimensions. There (as in 4d), we can (and do) analytically-continue in momenta. But (as in 4d), it is important that the polarization spinors be the ones appropriate to the Minkowski signature, not to some other signature (Euclidean, split, …).

Posted by: Jacques Distler on April 2, 2010 11:23 AM | Permalink | PGP Sig | Reply to this

Re: Pure Spinor Signature

Dear Jacques,

you can’t separate the momenta from the polarizations when you discuss reality conditions. Only whole well-defined degrees of freedom (or operators), such as components of a field at a particular point or a particular momentum, may be required to be real (or complex/Hermitean conjugates to others).

Polarizations are just bases of a smaller Hilbert space describing a particle with well-defined other quantum numbers such as the energy-momentum vector. Because they generate a (smaller) Hilbert space, they’re always bases of a complex space. It makes no sense whatsoever to claim that “polarizations are real” by themselves. Complex combinations of a particle’s polarizations are always physical states.

The only thing that is sometimes real or Hermitean is the actual field at points in the Minkowski spacetime - an operator. But when a field is real or Hermitean as a function of the Minkowski spacetime coordinate, it doesn’t mean that its correct Wick-rotation to 4+0 or 2+2-dimensional spacetime is also real. In fact, it’s not. The reality conditions change.

They also change if you go from a position basis to the momentum basis by Fourier transform, or by a twistor transform. The Fourier transform of a real function is not a real function, in general, and so on. You seem to be complaining about these facts but they’re facts about mathematics and they don’t imply the inconsistency of any theory in any way.

Best wishes
Lubos

Posted by: Lubos Motl on April 3, 2010 4:32 AM | Permalink | Reply to this

Re: Pure Spinor Signature

I’m sorry, Lubos.

You’ve descended to a level of handwavy vagueness where I have no idea what you’re talking about.

(I did, painful as it may be, take a look at your blog. The idea that one should define the Euclidean path integral to be an integral over complex-valued fields, obeying non-local constraints, is teh funny.)

I’ve explained what the issue is. Everyone (I suspect, including you, even though you’ll never admit it) seems to understand. Let’s just leave it there …

Posted by: Jacques Distler on April 3, 2010 9:45 AM | Permalink | PGP Sig | Reply to this

Re: Pure Spinor Signature

Dear Jacques,

what I wrote wasn’t any handwaving but elementary linear algebra relevant for quantum mechanics.

You haven’t found or described any “issue”. You have only presented an irrational criticism of the pure spinor formalism based on several rudimentary errors, such as the opinion that the trace of the metric tensor is d-2 rather than d, the assumption that the spinors shouldn’t be Wick-rotated, the misconception that the number of independent degrees of freedom doubles in Wick rotation, and the opinion that the mini-Hilbert space of “polarizations” differs after a Wick rotation.

I and others have explained you clearly why your assumptions are incorrect, which also happens to mean that your “big” conclusions about the pure spinor formalism are incorrect. There’s nothing wrong about rules to calculate that are naturally defined in terms of the Euclidean spacetime as long as the results can be Wick-rotated to the physical Minkowski signature.

Best wishes
Lubos

Posted by: Lubos Motl on April 4, 2010 2:15 AM | Permalink | Reply to this

Re: Pure Spinor Signature

One more comment about the quotients. The right way to “translate” SO(10)/U(5) to the Minkowski space must probably start from the complexification of the both groups, i.e. with SO(10,C)/GL(5,C). This probably doubles the dimension - the coset has complex dimension 22 or so.

In the coset, there exist various “real sections” that have real dimension 22, depending on the directions (and therefore signature). These sections don’t have to be cosets themselves but they can be the right Minkowskian counterpart of the Euclidean coset. The reality condition for the parameters/angles of an SO(9,1) transformation is not the same thing as the reality condition for a SO(9,1) spinor! But whatever the right “space” for the Minkowski signature is, it may be far less convenient to work with at the level of a CFT. So you shouldn’t have assumed that the right set in which the Minkowskian lambda take place is a coset of groups.

At any rate, it’s wrong to think that there is anything wrong with calculations done in the Euclidean signature of spacetime that are continued to the Minkowski states at the end.

Posted by: Lubos Motl on April 4, 2010 2:44 AM | Permalink | Reply to this

Re: Pure Spinor Signature

… the opinion that the trace of the metric tensor is d-2 rather than d …

A remark written in haste, and promptly corrected.

Like all humans, I do make mistakes, from time to time. But, unlike some people, I am man enough to admit them.

I’ll restate the bottom line point, again, for you. But – barring some miracle – I’m not sure further discussion of it will prove fruitful.

If you think the pure-spinor constraint can be “Wick-rotated to the physical Minkowski signature,” then be my guest, do so, and tell us what you get.

The right way to “translate” SO(10)/U(5) to the Minkowski space must probably start from the complexification of the both groups, i.e. with SO(10,C)/GL(5,C). This probably doubles the dimension - the coset has complex dimension 22 or so.

That’s certainly one way to start.

In the coset, there exist various “real sections” that have real dimension 22, depending on the directions (and therefore signature). These sections don’t have to be cosets themselves but they can be the right Minkowskian counterpart of the Euclidean coset.

Umh. OK. Let me ask the simplest possible question about the putative real section that you claim exists:

What is its signature?

The five cases listed in (3), each have different signature. What’s the signature of the real section corresponding to Spin(1,9)Spin(1,9)?

(Actually, things are not nearly so easy as you think. Spin(1,9)Spin(1,9) acts transitively on the space of solutions to the pure-spinor constraint, so your choice of real section is rather tightly-constrained by that requirement. I’d be a little surprised if demanding that some real form of Spin(10,)Spin(10,\mathbb{C}) acts transitively on the real section doesn’t just reduce you to the cases listed in (3).)

But whatever the right “space” for the Minkowski signature is, it may be far less convenient to work with at the level of a CFT.

That depends very much on what that space is. Since you think you know what it is, why don’t you tell us, so we can decide whether or not it’s something useful to calculate with.

Posted by: Jacques Distler on April 4, 2010 3:29 AM | Permalink | PGP Sig | Reply to this

Re: Pure Spinor Signature

Dear Jacques,

I wholeheartedly agree that this discussion wasn’t constructive in any way.

It seems that the very basic philosophy behind your questions is wrong. You seem to be looking for “the” new coset that makes the pure spinor better.

There’s nothing like that. The most useful way to formulate the pure spinor calculations is in terms of U(5)/SO(10). There’s nothing better. The proper Berkovits formalism may be analytically continued in many ways - but one would have to deal with the reality conditions and/or boundary conditions on the moduli space etc. It’s almost certain that there’s no “better” way to organize the calculation, and there doesn’t have to be anything like that.

When we calculate perturbative string theory in general, the worldsheets are also taken Euclidean: we’re integrating over the moduli spaces of Euclidean-signature Riemann surfaces, right? It’s the best way to formulate it - although not the only one. Why don’t you protest against string theory itself that it uses Euclidean-signature worldsheets in the loop calculations? The “right” worldsheets have Minkowskian signature, don’t they? Why don’t you criticize the twistors for SO(2,2) etc?

You just happened to pick Berkovits’ formalism for reasons that seem totally irrational and double-faced. Your criticism has no logic.

Cheers
LM

Posted by: Lubos Motl on April 9, 2010 3:00 AM | Permalink | Reply to this

Re: Pure Spinor Signature

Why don’t you protest against string theory itself that it uses Euclidean-signature worldsheets in the loop calculations?

The reason why I find discussions with you so frustrating is that, when you’ve completely run out of arguments, rather than admitting it, you start throwing out all sorts of irrelevant crap (or, in this case, recycling previously-dealt-with irrelevant crap), in the hopes of creating a distraction.

If you want to contribute something to the discussion, let me suggest that you do the following. Set up the computation of the 4-point amplitude (in IIB) for four self-dual RR 4-form gauge fields (ie, the guys whose field strength is a self-dual 5-form), at tree level and at one-loop, using Nathan’s pure-spinor formalism.

I won’t even ask you to compute the result. I think the act of precisely setting up the calculation will suffice to illuminate for you what the issue is.

Posted by: Jacques Distler on April 9, 2010 1:01 PM | Permalink | PGP Sig | Reply to this

Re: Pure Spinor Signature

Dear Jacques, I actually asked you a very serious question that goes to the very matter of your confusion. Why don’t you apply the same complaint against the conventional Euclidean signature of the worldsheets, even though the “real signature” of the world sheets of propagating strings is Minkowskian?

It may have been previously discussed but you have surely never provided any answer. You must know the reason: there’s no answer - simply because the Euclidean world sheets *are* more convenient to organize the integrals, just like U(5)/SO(10) is more convenient than any of its conceivable “non-compact” continuations. So even though the world sheets may be initially Minkowskian i.e. in light-cone gauge formulations, they’re usually continued and reduced to the Euclidean ones simply because the Euclidean calculations are more well-behaved. For example, their moduli spaces are compact so one doesn’t have to deal with new special boundary conditions at infinity.

These issues about the world sheet signature are not “irrelevant things”. They’re toy models for something much more general that you brutally misunderstand.

Your homework about the RR scattering is trivial and is covered in most reviews of the formalism by Nathan: supersymmetry is manifest in the pure spinor formalism so you surely don’t doubt that the calculation of R-R quanta is as straightforward as the calculation of NS-NS quanta’s scattering, do you?

You behave as dishonestly as Peter Woit, just your target and reach is much more moderate. Instead of attacking all of string theory by irrational pseudo-arguments, you have picked the pure spinor formalism.

Best wishes
LM

Posted by: Lubos Motl on April 10, 2010 3:17 AM | Permalink | Reply to this

Re: Pure Spinor Signature

Why don’t you apply the same complaint against the conventional Euclidean signature of the worldsheets …

Your homework about the RR scattering is trivial …

You behave as dishonestly as Peter Woit …

Lubos,

You are

  1. unable to read,
  2. lazy,
  3. abusive.

I could abide 2 out of the 3, but not all 3.

Your posting privileges here are at an end.

Posted by: Jacques Distler on April 11, 2010 12:11 PM | Permalink | PGP Sig | Reply to this

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