general relativity - Does gravity require strings?

OK, before I ask my question, let me frame it with a few (uncontroversial?) statements:

1. 
   

   The low-field-limit plane-wave solution to Einstein's equations is helicity-2.

   
   



2. 
   

   In the early days of string theory, when people were factorizing the Veneziano amplitude and its generalizations in order to find a spectrum of physical states, they could never get rid of the massless spin-2 state (i.e., closed loop) that kept cropping up, although they tried mightily to do so.

   
   


3. 
   

   In string theory, one can only formulate a physical theory of interacting massless spin-2 states (i.e., closed loops) if the gauge symmetry of those states can be deformed into the diffeomorphism symmetry of GR.

   
   


4. 
   

   Conformal invariance of the 2-D worldsheet action of a closed loop in string theory leads via state-operator correspondence to gravitational-wave solutions of GR.

   
   

Points (3) and (4) were shamelessly lifted from Lubos Motl's blog. All this goes to say is that string theory "postdicts" gravity, as Ed Witten put it in a talk I saw back in the 90s. In other words, it seems that as soon as one starts writing down a quantum theory of interacting strings, gravity drops out of it like magic and there's nothing you can do about it.

This brings me to my question. If a closed loop is intrinsically spin-2 and by its very nature must satisfy the symmetry requirements of general relativity, why can't one work backwards from these seemingly stringent symmetry requirements and demonstrate that a massless spin-2 point particle can NEVER satisfy them? Is it possible that only an extended object can serve as the carrier of a force which manifests itself as the curvature of spacetime?

I would think that such a demonstration would show that not only does string theory postdict gravity, but gravity requires strings.

Answer

First, a correction to (3). It is not really incorrect but the diffeomorphisms are required for any spin-2 particles in any relativistic theory, not just string theory. It's because one needs to get rid of the negative-normed time-like modes of the spin-0 particles, excitations of $g_{0i}$, roughly speaking (where $g$ is a name of the spin-2 field, not necessarily a priori the metric tensor). A gauge symmetry is needed to do so. Because it is a symmetry, it must have a conserved current. Because it has Lorentz indices and other choices would be too constraining, this conserved quantity has to coincide with the stress-energy tensor of the remaining matter. The stress-energy tensor is the density and flux of the energy and momentum which is by Noether's theorem associated with spacetime translations. By densitizing it, we clearly get diffeomorphisms. The only semi-consistent modification of the procedure above would be to make the symmetry spontaneously broken in some way, obtaining massive gravitons.

Second, a closed string is not "just" a spin-2 particle. A closed string may carry any integral value of the spin, depending on the state (and for a closed superstring, also all half-integral values are allowed).

Third, your attempt to eliminate ordinary spin-2 particles in this simple way doesn't work because the closed string in the graviton spin-2 vibration mode is the graviton. They have indistinguishable properties at low energies. Indeed, the consistent completion of the graviton's interactions at high energies requires the graviton to be completed to the whole "string multiplet" with the corresponding symmetry structure but this statement has not been rigorously proved and your observations are not enough to prove it because the spin-2 massless strings and gravitons are indistinguishable from the low-energy physics viewpoint.