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Polyakov action reduces to the Nambu–Goto action on-shell

Proved
PolyakovAction.polyakov_eq_nambuGoto

by Lucas · Oct 1, 2026 · Mathlib 0df444a (Lean v4.33.1)

mathematical-physicsstring-theory

Let TTT be the string tension, gμνg_{\mu\nu}gμν​ a target metric, XXX the embedding of the worldsheet, habh_{ab}hab​ the worldsheet metric, and Gab=gμν(X)∂aXμ∂bXνG_{ab}=g_{\mu\nu}(X)\partial_aX^\mu\partial_bX^\nuGab​=gμν​(X)∂a​Xμ∂b​Xν the induced metric. Suppose that on a measurable worldsheet region UUU the metric equations of motion hold,

Tab=T(Gab−12habhcdGcd)=0,T_{ab}=T\Big(G_{ab}-\tfrac12h_{ab}h^{cd}G_{cd}\Big)=0,Tab​=T(Gab​−21​hab​hcdGcd​)=0,

and that hcdGcd≥0h^{cd}G_{cd}\ge0hcdGcd​≥0 on UUU. Then the Polyakov action equals the Nambu–Goto action:

T2∫Ud2σ −h habGab=T∫Ud2σ −G.\frac T2\int_U d^2\sigma\,\sqrt{-h}\,h^{ab}G_{ab}=T\int_U d^2\sigma\,\sqrt{-G}.2T​∫U​d2σ−h​habGab​=T∫U​d2σ−G​.

This is the classical equivalence of the two string actions ("Relation with Nambu–Goto action" in the source): solving the auxiliary metric's equation of motion, −h=2−G/(hcdGcd)\sqrt{-h}=2\sqrt{-G}/(h^{cd}G_{cd})−h​=2−G​/(hcdGcd​), and substituting back gives the Nambu–Goto action.

Formalization Note. The positivity hcdGcd≥0h^{cd}G_{cd}\ge0hcdGcd​≥0 is implicit in the source's formula −h=2−G/(hcdGcd)\sqrt{-h}=2\sqrt{-G}/(h^{cd}G_{cd})−h​=2−G​/(hcdGcd​) (both square roots are non-negative); without it the two actions differ by a sign. No signature condition on hhh is needed for the identity.

Preamble
import Mathlib
import Definitions.Def_PolyakovAction_Defs

open MeasureTheory Filter Asymptotics
open scoped Matrix Topology
Formal statement
namespace PolyakovAction

theorem polyakov_eq_nambuGoto {D : ℕ} (T : ℝ) (g : Spacetime D → Matrix (Fin D) (Fin D) ℝ)
    (h : Worldsheet → Matrix (Fin 2) (Fin 2) ℝ) (X : Worldsheet → Spacetime D)
    (U : Set Worldsheet) (hU : MeasurableSet U)
    (heom : ∀ σ ∈ U, stressEnergyTensor T g h X σ = 0)
    (hpos : ∀ σ ∈ U, 0 ≤ ∑ c, ∑ d, (h σ)⁻¹ c d * inducedMetric g X σ c d) :
    polyakovAction T g h X U = nambuGotoAction T g X U := by sorry

end PolyakovAction
Source
Wikipedia, "Polyakov action", revision oldid=1292212311 (https://en.wikipedia.org/w/index.php?title=Polyakov_action&oldid=1292212311)
Read-back

What the Lean code literally says, in plain math · Aristotle (Harmonic)

Non-blind read-back — not independent testimony. This read-back was written by the same agent (Aristotle, by Harmonic) that drafted the Lean statements of this proposal, at the account owner's explicit instruction. Its author knew the source material and the intended meaning of the code while writing it, so it is not a blind audit and must not be mistaken for independent testimony. Reviewers should compare it against the Lean code themselves.

Throughout, σ=(σ0,σ1)∈R2\sigma=(\sigma^0,\sigma^1)\in\mathbb R^2σ=(σ0,σ1)∈R2 is a worldsheet point (index 000 is τ\tauτ, index 111 is σ\sigmaσ), X:R2→RDX:\mathbb R^2\to\mathbb R^DX:R2→RD is the embedding, ∂aXμ(σ)\partial_a X^\mu(\sigma)∂a​Xμ(σ) is the μ\muμ-th component of the Fréchet derivative of XXX at σ\sigmaσ applied to the aaa-th standard basis vector (it is 000 wherever XXX is not differentiable), Gab(σ)=∑μ,νgμν(X(σ)) ∂aXμ ∂bXνG_{ab}(\sigma)=\sum_{\mu,\nu} g_{\mu\nu}(X(\sigma))\,\partial_aX^\mu\,\partial_bX^\nuGab​(σ)=∑μ,ν​gμν​(X(σ))∂a​Xμ∂b​Xν is the induced metric, habh^{ab}hab denotes the entries of the matrix inverse (hab)−1(h_{ab})^{-1}(hab​)−1 (Mathlib's inverse, which is the zero matrix when det⁡h=0\det h=0deth=0), and ⋅\sqrt{\cdot}⋅​ is the real square root, which returns 000 on negative inputs. The Polyakov density is LP=−det⁡h ∑a,bhabGab\mathcal L_P=\sqrt{-\det h}\,\sum_{a,b}h^{ab}G_{ab}LP​=−deth​∑a,b​habGab​, the Polyakov action on a set UUU is SP[U]=T2∫ULP d2σS_P[U]=\frac T2\int_U\mathcal L_P\,d^2\sigmaSP​[U]=2T​∫U​LP​d2σ (Lebesgue measure; Bochner integral, equal to 000 if the integrand is not integrable on UUU), and the Nambu–Goto action is SNG[U]=T∫U−det⁡G d2σS_{NG}[U]=T\int_U\sqrt{-\det G}\,d^2\sigmaSNG​[U]=T∫U​−detG​d2σ. The stress–energy tensor is Tab=T(Gab−12hab∑c,dhcdGcd)T_{ab}=T\big(G_{ab}-\tfrac12 h_{ab}\sum_{c,d}h^{cd}G_{cd}\big)Tab​=T(Gab​−21​hab​∑c,d​hcdGcd​), and η(n)=diag(1,−1,…,−1)\eta^{(n)}=\mathrm{diag}(1,-1,\dots,-1)η(n)=diag(1,−1,…,−1) is the n×nn\times nn×n Minkowski matrix.

Statement. For every natural number DDD, real TTT, target metric g:RD→RD×Dg:\mathbb R^D\to\mathbb R^{D\times D}g:RD→RD×D (arbitrary), worldsheet metric h:R2→R2×2h:\mathbb R^2\to\mathbb R^{2\times2}h:R2→R2×2 (arbitrary, no symmetry, invertibility or signature assumed), map X:R2→RDX:\mathbb R^2\to\mathbb R^DX:R2→RD (no differentiability assumed) and set U⊆R2U\subseteq\mathbb R^2U⊆R2, assume

  1. UUU is Lebesgue (Borel) measurable;
  2. for every σ∈U\sigma\in Uσ∈U the matrix Tab(σ)T_{ab}(\sigma)Tab​(σ) is the zero matrix;
  3. for every σ∈U\sigma\in Uσ∈U, ∑c,dhcd(σ)Gcd(σ)≥0\sum_{c,d}h^{cd}(\sigma)G_{cd}(\sigma)\ge0∑c,d​hcd(σ)Gcd​(σ)≥0.

Then

T2∫U−det⁡h ∑a,bhabGab d2σ  =  T∫U−det⁡G d2σ.\frac T2\int_U\sqrt{-\det h}\,\sum_{a,b}h^{ab}G_{ab}\,d^2\sigma \;=\; T\int_U\sqrt{-\det G}\,d^2\sigma .2T​∫U​−deth​a,b∑​habGab​d2σ=T∫U​−detG​d2σ.

Edge cases: if T=0T=0T=0, hypothesis 2 holds trivially and both sides are 000. Both integrals use the convention that a non-integrable integrand has integral 000; no integrability is assumed.

Human review
  • Endorsed by Shuze Chen · Oct 8, 2026

    Confirmed by the moderator at approval.

  • Endorsed by Lucas · Oct 8, 2026

    Confirmed by the mission captain (proposal self-audit).

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