There is a standard lore in superstring theory that 2-pt sphere amplitudes vanish. This is inline with the basic fact that string amplitudes give amputated Green functions with onshell asymptotic states (ref: Polchinski vol.1 p.104) -- these vanish in QFT in the case of 2 asymptotic states. After failing to make sense of this statement in string theory, and after failing to receive enlightening comments from my fellow string theorists I thought I'd put it out there to a broader audience, so here is the puzzle:

In string theory the vanishing of sphere 2-pt amplitudes is attributed to the fact that (in one of the many realisations) one divides by the infinite volume of $SL(2,\mathbb{C})$ (which I denote by ${\rm vol(CKG)}$) and not enough vertex operators to saturate this. In the Polyakov formalism (using Polchinski's definition of the Euclidean adjoint, $\overline{V(z,\bar{z})}$, p.202-203 in Polchinski vol.1) the quantity of interest is: $$ I_2 = \frac{1}{{\rm vol(CKG)}}\Big\langle\int d^2z\overline{V(z,\bar{z})} \int d^2wV(w,\bar{w})\Big\rangle_{S_2} $$ $V(z,\bar{z})$ denotes some physical and normal-ordered mass eigenstate vertex operator, e.g. $V(w,\bar{w})=ge^{ik\cdot x(w,\bar{w})}$ and $\overline{V(z,\bar{z})}=ge^{-ik'\cdot x(z,\bar{z})}$ with $k^2=k'^2=2$ (and $\alpha'=2$). Wick contractions may be carried out with: $$ \langle x^{\mu}(z,\bar{z})x^{\nu}(w,\bar{w})\rangle = -\eta^{\mu\nu}\ln |z-w|^2. $$ Carrying out contractions, if we look at each contribution (I’m dropping some real factors) there is the zero mode factor, $$ i(2\pi)^D\delta^D(k-k'), $$ (note the factor of $i$ from the $x^0$ zero mode), there are contractions in the numerator leading to: $$ \int \frac{d^2zd^2w}{|z-w|^4}, $$ which is independent of the choice of vertex operators (as the leading singularity is fixed by conformal invariance and the sub-leading singularities vanish in the correlator because of normal ordering), and there is the CKG volume, ${\rm vol(CKG)}$, $$ \int \frac{d^2zd^2wd^2u}{|z-w|^2|w-u|^2|u-z|^2}, $$ which appears in the denominator. Now both of these are infinite, so rather than $I_2=0$ (which would be the textbook answer), $$ I_2 = i(2\pi)^D\delta^D(k-k')\frac{\infty}{\infty}, $$ and one needs to regulate to find the actual limit (i.e. whether it is infinite, zero or finite, real or imaginary). I used a short-distance cutoff for simplicity (dim reg doesn't remove these infinities) and find: $$ \boxed{I_2 = i(2\pi)^D\delta^D(k-k')C} $$ where $C$ is a regularisation-dependent real and finite constant. This conclusion holds for all weight $(1,1)$ primaries, mass eigenstate (and also coherent state) vertex operators. I have gone through this carefully, but I cannot exclude the possibility that there is a mistake somewhere [update: I have presented one version of the calculation below]. One can also derive the same conclusion in the BRST formalism when the in and out vertex operators are related by duality (i.e. one vertex operator has ghost number 2 and the other ghost number 4 and the matter sector is again related by the Euclidean adjoint).

THE QUESTION: how can it be that $I_2$ is not zero given standard textbooks state $I_2=0$ (independently of the delta function).

some details:

The only real subtlety here is in calculating the regularised volume of the CKG, so maybe I should add a few comments on how I went about this. There are three UV and one IR divergence (the former are short distance so dim reg doesn't work whereas the latter won't need regulating, as we will see). To bring them into plain view consider ${\rm vol}$(CKG) given above and shift variables: $z\rightarrow z+w$, $w\rightarrow w+u$, leading to: \begin{equation} \begin{aligned} {\rm vol(CKG)} &=\int \frac{d^2zd^2wd^2u}{|z|^2|w|^2|z+w|^2}. \end{aligned} \end{equation} Our conventions are $d^2z\equiv idz\wedge d\bar{z}$, with every integral over the entire complex plane. Now change variables again, $z=re^{i\theta}$, $w=\rho e^{i\varphi}$ and subsequently shift $\theta\rightarrow \theta+\varphi$. We then integrate out $\varphi$ (trivially yielding $2\pi$) and $\theta$ using the standard integral: $$ \int\limits_0^{2\pi}d\theta \frac{1}{A+B\cos\theta} = \frac{2\pi}{\sqrt{(A-B)(A+B)}}, $$ with $A=r^2+\rho^2$ and $B=2r\rho$ (this is justified given the $A=B$ and $A=0$ regions are absent in the regularisation procedure below). The resulting expression reads: \begin{equation} \begin{aligned} {\rm vol(CKG)} &= \gamma\int \frac{dr d\rho}{r\rho}\frac{1}{|r^2-\rho^2|},\qquad \gamma\equiv (4\pi)^2\int d^2u \end{aligned} \end{equation} The quantity $\gamma$ is the IR divergent quantity allured to above and is uninteresting because the numerator in $I_2$ is (as we shall see) also proportional to $\gamma$, so $\gamma$ cancels out in $I_2$.

Both $r$ and $\rho$ integrals range from $0$ to $\infty$, so there are three distinct (codimension-1 in the $(r,\rho)$ plane) divergent regions:

  1. $r=0$ & $\rho>0$
  2. $\rho=0$ & $r>0$
  3. $r=\rho\geq 0$

We may regularise the integral by thickening these codimension-1 lines by a small amount $\epsilon$. The regularised integration domain is then: $$ r\geq \epsilon,\qquad \rho\geq\epsilon,\qquad {\rm and}\qquad |r-\rho|\geq\epsilon. $$ However, it is slightly easier (and presumably equivalent up to an immaterial $\epsilon$-independent multiplicative constant) to truncate the $r=\rho$ divergence by replacing $|r^2-\rho^2|\rightarrow|r^2-\rho^2|+\epsilon^2$. The required integrals are then easily evaluated: $$ \int_{\epsilon}^{\infty}\frac{dr}{r}\int_{\epsilon}^{\infty}\frac{d\rho}{\rho}\frac{1}{|r^2-\rho^2|+\epsilon^2} = \frac{\pi^2}{12\epsilon^2}, $$ leading to the final expression for the volume of $SL(2,\mathbb{C})$, $$ \boxed{{\rm vol(CKG)}=\frac{\pi^2}{12}\frac{\gamma}{\epsilon^2}} $$ For completeness I also give the corresponding result for the integral in the numerator of $I_2$ given above, $$ \boxed{\int \frac{d^2zd^2w}{|z-w|^4}=\frac{1}{8\pi}\frac{\gamma}{\epsilon^2}} $$ where again the same short-distance cutoff was used: we shift $z\rightarrow z+w$, change coordinates, $z=re^{i\theta}$, and integrate over $r\geq\epsilon$ and $2\pi>\theta\geq0$. Taking the ratio of the two boxed equations and taking the limit $\epsilon\rightarrow 0$ leads to the stated result for $I_2$, with $C=3/(2\pi^3)$.

  • $\begingroup$ Perhaps the textbooks that you mention give for granted that $k \neq k'$? $\endgroup$
    – Rexcirus
    Commented Feb 3, 2018 at 8:28
  • $\begingroup$ @Rexcirus: the problem is independent of the delta function; such a non-vanishing contribution would give wrong S-matrix elements. In particular, one might imagine that this calculation is really just a contribution to the trivial part of the S matrix, i.e. the $\mathbb{1}$ in $S=\mathbb{1}+iT$, but this cannot be true because of two reasons: $I_2$ is imaginary and secondly contains a delta function in all $D$ spacetime dimensions rather than $D-1$ as one would expect from the $\mathbb{1}$ in standard QFT. So I $\endgroup$ Commented Feb 3, 2018 at 13:04
  • $\begingroup$ .. So I believe there is a mistake in the above calculation and the only subtle part is calculating the volume of CKG (because there are multiple regions where the integrand diverges, reminiscent of overlapping divergences in higher loop Feynman diagrams), so the mistake must be there but I cannot find it .. $\endgroup$ Commented Feb 3, 2018 at 13:14

1 Answer 1



The cutoff approach discussed above is ambiguous. In fact, it can't even distinguish between $I_2=0$, or $I_2={\rm finite}$, or $I_2=i{\rm finite}$, or $I_2=\infty$. The question has now been answered: 2-pt amplitudes. The correct result is, remarkably, that of a free particle: $$ \boxed{\mathcal{A}_2 = 2k^0(2\pi)^{D-1}\delta^{D-1}(\mathbf{k}-\mathbf{k}')} $$ So the 2-point tree-level amplitude really does give the trivial contribution to the full connected S matrix, namely the $\mathbb{1}$ in $S=\mathbb{1}+iT$.

I will review this calculation here.

2-pt Amplitude

We consider a 2-pt open-string disc amplitude with any two external vertex operator insertions. (There is a similar procedure for closed strings.) We map the disc to the upper-half plane with holomorphic coordinate $z$. We also use the doubling trick, whereby anti-holomorphic quantities in the upper half plane are mapped to holomorphic quantities in the lower-half plane by the identification $z\sim z'$ when $z'=\bar{z}$.

The relevant 2-pt amplitude will be defined by: \begin{equation} \begin{aligned} \mathcal{A}_2 &\equiv \frac{\int dz_1dz_2\Big(\int Dx\,e^{-I(x)}V_a(z_1)V_b(z_2)\Big)}{\rm Vol\,G} \end{aligned}\qquad (1a) \end{equation} and define: $$ \boxed{ B(z_1,z_2)\equiv \int Dx\,e^{-I(x)}V_a(z_1)V_b(z_2) }\qquad (1b) $$ where $a,b$ denote collectively the various quantum numbers that characterise the vertex operators under consideration.


The residual symmetry group G is SL(2,$\mathbf{R}$)$/\mathbf{Z}_2$, under which an element $g\in {\rm G}$ maps a point $z\in H$ (the upper-half plane) to $z^g$ via, $$ g : z_i\longmapsto z_i^g=\frac{az_i+b}{cz_i+d},\qquad {\rm with}\qquad a,b,c,d\in \mathbf{R}, $$ where $a,b,c,d$ and their negatives map to the same $z_i^g$ which is why we mod out by $\mathbf{Z}_2$, and $ad-cb=1$. There are three generators (3 conformal Killing vectors). The 2-pt amplitude (1a) is G-invariant, in particular, $$ dz_1^gdz_2^g\,B(z_1^g,z_2^g)=dz_1dz_2\,B(z_1,z_2)\qquad (*) $$

Fadeev-Popov Gauge Fixing

We use Fadeev-Popov (FP) to gauge fix the residual symmetry G of (1a). Write, $$\boxed{ 1=\Delta(z_1,z_2,z_3)\int Dg\,\prod_{I=1}^3\delta\big(F_I(z_1^g,z_2^g,z_3^g)\big) }\qquad (2)$$ and since the measure is invariant under $g\rightarrow g'$, namely $Dg=Dg'$, it follows that $$ \Delta(z_1^g,z_2^g,z_3^g)=\Delta(z_1,z_2,z_3)\qquad (**) $$ Substitute (2) into (1), \begin{equation} \begin{aligned} \mathcal{A}_2 &= \frac{\int dz_1dz_2B(z_1,z_2)}{\rm Vol\,G}\\ &=\int \frac{Dg}{\rm Vol\,G}\int dz_1dz_2B(z_1,z_2)\Delta(z_1,z_2,z_3)\,\prod_{I=1}^3\delta\big(F_I(z_1^g,z_2^g,z_3^g)\big)\\ &=\int \frac{Dg}{\rm Vol\,G}\int dz_1^gdz_2^g\,B(z_1^g,z_2^g)\Delta(z_1^g,z_2^g,z_3^g)\,\prod_{I=1}^3\delta\big(F_I(z_1^g,z_2^g,z_3^g)\big)\\ \end{aligned} \end{equation} where in the last equality we made use of (*) and (**). Next change variables, $z_1^g,z_2^g\rightarrow z_1,z_2$ respectively and assume that $\mathcal{A}_2$ is independent of $z_3^g$ (so that we may replace $z_3^g$ by $z_3$), $$ \mathcal{A}_2=\int \frac{Dg}{\rm Vol\,G}\int dz_1dz_2\,B(z_1,z_2)\Delta(z_1,z_2,z_3)\,\prod_{I=1}^3\delta\big(F_I(z_1,z_2,z_3)\big) $$ (We will show below that $\mathcal{A}_2$ is indeed independent of $z_3$.) Since the integrand of the integral over group parameters, $g$, is independent of $g$ we can set, $$ \int \frac{Dg}{\rm Vol\,G}\equiv 1, $$ so that the gauge-fixed 2-pt amplitude takes the form: $$ \boxed{\mathcal{A}_2=\int dz_1dz_2\,B(z_1,z_2)\Delta(z_1,z_2,z_3)\,\prod_{I=1}^3\delta\big(F_I(z_1,z_2,z_3)\big)}\qquad(3) $$

Gauge-Fixing Functions

We now make a choice of gauge-fixing functions, $F_I(z_i)$, $I=1,2,3$. There is lots of freedom of choice, but we must pick functions that are not invariant under $G$. Since we have two vertex operator insertions we make the standard choice for $I=1,2$, whereas for $I=3$ we make a special choice, \begin{equation} \boxed{ \begin{aligned} F_1(z_1,z_2,z_3)&=z_1-z_1^0\\ F_2(z_1,z_2,z_3)&=z_2-z_2^0\\ F_3(z_1,z_2,z_3)&=x^0(z_3) \end{aligned}}\qquad (4) \end{equation} where $z_1^0$ and $z_2^0$ can be chosen to lie anywhere on the boundary of the worldsheet (the real axis, namely $\partial H$), and similarly for $z_3$ (so that $z_i=\bar{z}_i$ for $i=1,2,3$). (This choice is different from that in arXiV:1906.06051, where for $I=3$ the choice $F_3(z_1,z_2,z_3)=x^0(z_3,\bar{z}_3)$ was made, with $z_3\in H$ rather than $z_3\in \partial H$ as we do here.)

FP Determinant

To compute the Fadeev-Popov determinant, $\Delta$, we make use of (2) and (4), according to which, $$ \frac{1}{\Delta} = \int Dg\,\delta(z_1^g-z_1^0)\delta(z_2^g-z_2^0)\delta(x^0(z_3^g)) $$ But since $\Delta$ is a Jacobian we can compute it on the tangent space (recall that the Jacobian for a change of coordinates on the tangent space is equal to the Jacobian for a change of coordinates on the base space). Therefore, $$ \frac{1}{\Delta} = \int D(\delta g)\,\delta(\delta z_1)\delta(\delta z_2)\delta(\delta x^0(z_3)), $$ where $\delta$ denotes infinitesimal variations around the identity element, \begin{equation} \begin{aligned} \delta z_i &\equiv (z_i^g-z_i^0)-(z_1-z_1^0) \\ &\simeq \epsilon_{-1}+\epsilon_0 z_i+\epsilon_1 z_i^2+\dots\\ \delta x^0(z_3) &\equiv x^0(z_3^g)-x^0(z_3)\\ &\simeq (\epsilon_{-1}+\epsilon_0 z_3+\epsilon_1 z_3^2)\partial_{z_3}x^0(z_3)+\dots\\ D(\delta g) &= d\epsilon_{-1}\,d\epsilon_0\,d\epsilon_1, \end{aligned} \end{equation} so substituting these into the above and integrating out $\epsilon_{-1},\epsilon_0,\epsilon_1$ yields, $$ \boxed{\Delta(z_1,z_2,z_3) = z_{12}z_{13}z_{23}\partial_{z_3}x^0(z_3)}\qquad (5) $$

Gauged-Fixed 2-pt Amplitude

Now substitute (4) and (5) into the general expression for the gauge-fixed amplitude (3), \begin{equation} \begin{aligned} \mathcal{A}_2&=\int dz_1dz_2\,B(z_1,z_2)\Delta(z_1,z_2,z_3)\,\prod_{I=1}^3\delta\big(F_I(z_1,z_2,z_3)\big)\\ &= \int dz_1dz_2\,B(z_1,z_2)\Big(z_{12}z_{13}z_{23}\partial_{z_3}x^0(z_3)\Big)\, \delta(z_1-z_1^0)\delta(z_2-z_2^0)\delta(x^0(z_3))\\ &= \,B(z_1,z_2)\Big(z_{12}z_{13}z_{23}\partial_{z_3}x^0(z_3)\Big)\, \delta\big(x^0(z_3)\big) \end{aligned} \end{equation} where in the last equality we integrated out $z_1,z_2$ and renamed $z_1^0,z_2^0$ by $z_1,z_2$ for simplicity. We next substitute the definition of $B$ given in (1b) into the above, $$ \mathcal{A}_2=\int Dx\,e^{-I(x)}V_a(z_1)V_b(z_2)\,\partial x^0(z_3)\, \delta\big(x^0(z_3)\big)\,\big(z_{12}z_{13}z_{23}\big). $$

Up until now we the above analysis has been independent of basis for the external vertex operators. We now choose a momentum-eigenstate basis, whereby $V_a(z_1)={\rm Pol}_a\,(\partial^\# x)\,e^{ik_a\cdot x(z_1)}$ and similarly for $V_b(z_2)$. So extracting out the zero modes, $x^\mu(z_i)=x_0^\mu+\tilde{x}^\mu(z_i)$, $i=1,2$, in the above path integral the full amplitude takes the form, $$\boxed{ \mathcal{A}_2=\Bigg(i\int d^Dx_0\,e^{i(k_a+k_b)\cdot x_0} \delta\big(x^0(z_3)\big)\Bigg)\times \Big\langle V_a(z_1)V_b(z_2)\,\partial x^0(z_3)\Big\rangle_{\tilde{x}}\times \big(z_{12}z_{13}z_{23}\big)\times N}\qquad (6) $$ where the factor $i=\sqrt{-1}$ is from rotating $x_0^0$ back to Lorentzian signature (see Polchinski vol.1). The quantity $N$ is the normalisation of the path integral measure and will be computed below by unitarity.

Zero Modes

We now integrate out the zero modes, $x_0^\mu$, in (6), $$ \boxed{ i\int d^Dx_0\,e^{i(k_a+k_b)\cdot x_0} \delta\big(x^0(z_3)\big)=i(2\pi)^{D-1}\delta({\bf k}_a+{\bf k}_b)}\qquad (7) $$ where notice that the fluctuation term in the argument of the delta function does not contribute. Also, we have written $k^\mu=(k^0,{\bf k})$.

Fluctuations (Wick Contractions)

We now compute the fluctuations contributions in (6), namely, $$ \Big\langle V_a(z_1)V_b(z_2)\,\partial x^0(z_3)\Big\rangle_{\tilde{x}} $$ Since all insertions are on $\partial H$ and we are working in the Polyakov formulation all Wick contractions are carried out using: $$ \langle x^\mu(z_i)x^\nu(z_j)\rangle =-2\alpha'\ln z_{ij} $$

Suppose that $V_a(z_1)$, $V_b(z_2)$, are conformal primaries of weight $h=1$, which are the physical state conditions in the Polyakov formulation. Also $\partial x$ is a primary of weight $h=1$. So using a standard result from conformal field theory conformal invariance determines the correlation function up to an overall factor, $C$, $$ \Big\langle V_a(z_1)V_b(z_2)\,\partial x^0(z_3)\Big\rangle_{\tilde{x}} = \frac{C}{z_{12}z_{13}z_{23}}\qquad (8) $$

Notice that the right-hand side of (8) is analytic in $z_3$, so so must the left-hand side be analytic in $z_3$. Now make use of the fact that the momentum operator, $\mathbb{P}^\mu$, for open strings reads, $$ \mathbb{P}^\mu= \frac{1}{4\pi\alpha'}\oint dz\,\partial_zx^\mu(z), $$ and take into account that $\mathbb{P}V_a(z_1)=k_aV_a(z_1)$, according to which the contour integral (taken to encircle, e.g., $z_1$) of the left-hand side of (8) takes the form, \begin{equation} \begin{aligned} {\rm LHS}&=\frac{1}{4\pi\alpha'}\oint_{z_1} dz_3\,\Big\langle \partial x^0(z_3)V_a(z_1)V_b(z_2)\Big\rangle_{\tilde{x}}\\ &=k_a^0\Big\langle V_a(z_1)V_b(z_2)\Big\rangle_{\tilde{x}}\\ &=k_a^0\,\frac{g_0^2}{z_{12}^2}\delta_{a,b} \end{aligned} \end{equation} where we took into account that the OPE of two $h=1$ primaries are proportional to $1/z_{12}^2$, $g_0$ is the open string coupling constant, and used the standard normalisation (see Polchinski, vol. 1). (Strictly speaking $V_a$ should be the Euclidean adjoint of $V_b$ in order to guarantee positive norm.)

Hitting also the right-hand side of (8) with $\frac{1}{4\pi\alpha'}\oint dz_3$ and carrying out the contour integral around $z_1$ yields, \begin{equation} \begin{aligned} {\rm RHS}&=\frac{1}{4\pi\alpha'}\oint_{z_1} dz_3\,\frac{C}{z_{12}z_{13}z_{23}}\\ &=\frac{2\pi i}{4\pi\alpha'}\frac{C}{z_{12}^2}. \end{aligned} \end{equation}

So setting RHS $=$ LHS determines $C$, $$ C=-2i\alpha'g_o^2k_a^0\delta_{a,b} $$ and substituting this into (8) yields, $$ \boxed{\Big\langle V_a(z_1)V_b(z_2)\,\partial x^0(z_3)\Big\rangle_{\tilde{x}} = \frac{-2i\alpha'g_o^2k_a^0\delta_{a,b}}{z_{12}z_{13}z_{23}}}\qquad (9) $$

Full 2-pt Amplitude

We can now substitute the fluctuations contributions (9) into and the zero modes contribution (7) into the full expression for the 2-pt amplitude (6), \begin{equation} \begin{aligned} \mathcal{A}_2&=\Bigg(i\int d^Dx_0\,e^{i(k_a+k_b)\cdot x_0} \delta\big(x^0(z_3)\big)\Bigg)\times \Big\langle V_a(z_1)V_b(z_2)\,\partial x^0(z_3)\Big\rangle_{\tilde{x}}\times \big(z_{12}z_{13}z_{23}\big)\times N\\ &=\Big(i(2\pi)^{D-1}\delta^{D-1}\big({\bf k}_a+{\bf k}_b\big)\Big)\times \frac{-2i\alpha'g_o^2k_a^0\delta_{a,b}}{z_{12}z_{13}z_{23}}\times \big(z_{12}z_{13}z_{23}\big)\times N \end{aligned} \end{equation} and in particular, $$ \mathcal{A}_2(a,b)=\Big(2k_a^0(2\pi)^{D-1}\delta^{D-1}\big({\bf k}_a+{\bf k}_b\big)\delta_{a,b}\Big)\times \big(\alpha'g_o^2 N\big)\qquad (10) $$

Normalisation and Unitarity

The remaining objective is to compute the normalisation $N$. For this (as usual in string theory) we use unitarity. (This is the simplest possible unitarity calculation in string theory.) In particular, we expand the S matrix as usual, $S=\mathbb{1}+iT$, where $\mathbb{1}$ is the free piece and $T$ is the interaction piece. Unitarity is the statement $$ S^\dagger S=\mathbb{1}, $$ Applying this to the case of interest, we are to extract the interaction-free contribution of this unitarity relation, $$ \mathbb{1}^2=\mathbb{1}, $$ and adopting a Lorentz-covariant normalisation, we can write this in terms of $\mathcal{A}_2$, $$ \mathcal{A}_2(a,b) = \sum_c\int \frac{d^{D-1}{\bf k}_c}{(2\pi)^{D-1}}\frac{1}{2k_c^0}\,\mathcal{A}_2(a,c)\mathcal{A}_2(c,b)\qquad (11) $$ Substituting (10) into (11) yields precisely: $$ \boxed{N=\frac{1}{\alpha'g_o^2}}\qquad (12) $$ which is in fact identical to Polchinski's normalisation for all tree-level disc amplitudes, $C_{D_2}=N$.

Full Result

Summarising, we substitute (12) into the expression for the full 2-pt amplitude (10) and find that: $$ \boxed{\mathcal{A}_2(a,b)=2k_a^0(2\pi)^{D-1}\delta^{D-1}\big({\bf k}_a+{\bf k}_b\big)\delta_{a,b}} $$

So we see that indeed, as claimed in arXiV:1906.06051 all tree-level two-point amplitudes in string theory are equivalent to the standard field theory formula. (One can indeed show that the same result holds for closed strings.)

So we learn that the string path integral gives the full connected S matrix, not just the connected amputated Green's functions. (Of course the only distinction between the two is precisely in the tree-level 2-pt amplitude.)

It is astonishing how this simple (but subtle) result was missed for over 35 years.

  • 1
    $\begingroup$ I noticed that paper! Your question now serves as an introduction to the issues. Hopefully it will show up in arxiv trackback. $\endgroup$ Commented Jun 25, 2019 at 6:00

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