proofs
Provedads-cfthep-th-9802150sign-auditthermal-ads

With t = ln v − ½ ln(1 + r²) the quotient embedding u = e^{−t}√(1+r²), v = e^{t}√(1+r²) induces the metric components (1 + r², 0, 1/(1 + r²))

For all t,r∈R\htmlData{mx=1}{t, r \in \htmlData{mx=0}{\mathbb{R}}}:

s↦e−s⋅1+r2′(t)⋅s↦es⋅1+r2′(t)=−(1+r2)∧s↦e−s⋅1+r2′(t)⋅ρ↦et⋅1+ρ2′(r)+ρ↦e−t⋅1+ρ2′(r)⋅s↦es⋅1+r2′(t)=0∧1−ρ↦e−t⋅1+ρ2′(r)⋅ρ↦et⋅1+ρ2′(r)=11+r2\begin{aligned} & \htmlData{mx=25}{\htmlData{mx=13}{\htmlData{mx=11}{s \mapsto \htmlData{mx=9}{\htmlData{mx=4}{e^{\htmlData{mx=3}{-\htmlData{mx=2}{s}}}} \cdot \htmlData{mx=8}{\sqrt{\htmlData{mx=7}{1 + \htmlData{mx=6}{\htmlData{mx=5}{r}^{2}}}}}}}'(\htmlData{mx=12}{t})} \cdot \htmlData{mx=24}{\htmlData{mx=22}{s \mapsto \htmlData{mx=20}{\htmlData{mx=15}{e^{\htmlData{mx=14}{s}}} \cdot \htmlData{mx=19}{\sqrt{\htmlData{mx=18}{1 + \htmlData{mx=17}{\htmlData{mx=16}{r}^{2}}}}}}}'(\htmlData{mx=23}{t})}} \\ & \quad \htmlData{mx=1}{=} \htmlData{mx=29}{-(\htmlData{mx=28}{1 + \htmlData{mx=27}{\htmlData{mx=26}{r}^{2}}})} \\ & \htmlData{mx=0}{\land} \htmlData{mx=79}{\htmlData{mx=54}{\htmlData{mx=42}{\htmlData{mx=40}{s \mapsto \htmlData{mx=38}{\htmlData{mx=33}{e^{\htmlData{mx=32}{-\htmlData{mx=31}{s}}}} \cdot \htmlData{mx=37}{\sqrt{\htmlData{mx=36}{1 + \htmlData{mx=35}{\htmlData{mx=34}{r}^{2}}}}}}}'(\htmlData{mx=41}{t})} \cdot \htmlData{mx=53}{\htmlData{mx=51}{\rho \mapsto \htmlData{mx=49}{\htmlData{mx=44}{e^{\htmlData{mx=43}{t}}} \cdot \htmlData{mx=48}{\sqrt{\htmlData{mx=47}{1 + \htmlData{mx=46}{\htmlData{mx=45}{\rho}^{2}}}}}}}'(\htmlData{mx=52}{r})}} + \htmlData{mx=78}{\htmlData{mx=66}{\htmlData{mx=64}{\rho \mapsto \htmlData{mx=62}{\htmlData{mx=57}{e^{\htmlData{mx=56}{-\htmlData{mx=55}{t}}}} \cdot \htmlData{mx=61}{\sqrt{\htmlData{mx=60}{1 + \htmlData{mx=59}{\htmlData{mx=58}{\rho}^{2}}}}}}}'(\htmlData{mx=65}{r})} \cdot \htmlData{mx=77}{\htmlData{mx=75}{s \mapsto \htmlData{mx=73}{\htmlData{mx=68}{e^{\htmlData{mx=67}{s}}} \cdot \htmlData{mx=72}{\sqrt{\htmlData{mx=71}{1 + \htmlData{mx=70}{\htmlData{mx=69}{r}^{2}}}}}}}'(\htmlData{mx=76}{t})}}} \\ & \quad \htmlData{mx=30}{=} 0 \\ & \htmlData{mx=0}{\land} \htmlData{mx=105}{1 - \htmlData{mx=104}{\htmlData{mx=92}{\htmlData{mx=90}{\rho \mapsto \htmlData{mx=88}{\htmlData{mx=83}{e^{\htmlData{mx=82}{-\htmlData{mx=81}{t}}}} \cdot \htmlData{mx=87}{\sqrt{\htmlData{mx=86}{1 + \htmlData{mx=85}{\htmlData{mx=84}{\rho}^{2}}}}}}}'(\htmlData{mx=91}{r})} \cdot \htmlData{mx=103}{\htmlData{mx=101}{\rho \mapsto \htmlData{mx=99}{\htmlData{mx=94}{e^{\htmlData{mx=93}{t}}} \cdot \htmlData{mx=98}{\sqrt{\htmlData{mx=97}{1 + \htmlData{mx=96}{\htmlData{mx=95}{\rho}^{2}}}}}}}'(\htmlData{mx=102}{r})}}} \\ & \quad \htmlData{mx=80}{=} \htmlData{mx=109}{\frac{1}{\htmlData{mx=108}{1 + \htmlData{mx=107}{\htmlData{mx=106}{r}^{2}}}}} \end{aligned}
What the notation means · or point at any symbol
RReal numbers
The real numbers: every point on the number line. (ℝ)
x ∈ TVariables
t, r range over real numbers: the statement must hold for every choice. ((t r : ℝ))
f′(x)Derivative
The derivative of s ↦ exp(−s) · √(1 + r^2) at t: its instantaneous rate of change there. (deriv (fun s : ℝ => Real.exp (-s) * Rea…)
=Equality
deriv(s ↦ exp(−s) · √(1 + r^2), t) · deriv(s ↦ exp(s) · √(… and −(1 + r^2) are the same. (deriv (fun s : ℝ => Real.exp (-s) * Rea…)
x ↦ f(x)Function (λ-abstraction)
The function sending s to exp(−s) · √(1 + r^2). Lean writes fun x => …; papers write x ↦ …. (fun s : ℝ => Real.exp (-s) * Real.sqrt …)
eˣExponential function
e raised to the power −s, where e ≈ 2.718. (Real.exp (-s))
·Multiplication
exp(−s) times √(1 + r^2). (Real.exp (-s) * Real.sqrt (1 + r ^ 2))
−xNegation
The negative of s. (-s)
√Square root
The non-negative number whose square is 1 + r^2. (Real.sqrt (1 + r ^ 2))
+Addition
1 plus r^2. (1 + r ^ 2)
aᵇPower
r raised to the power 2. (r ^ 2)
∧And
Both hold: deriv(s ↦ exp(−s) · √(1 + r^2), t) · deriv(ρ ↦ exp(t) · √(…, and 1 − deriv(ρ ↦ exp(−t) · √(1 + ρ^2), r) · deriv(ρ ↦ exp(t) …. (deriv (fun s : ℝ => Real.exp (-s) * Rea…)
-Subtraction
1 minus deriv(ρ ↦ exp(−t) · √(1 + ρ^2), r) · deriv(ρ ↦ exp(t) · √(…. On ℕ, Lean's subtraction stops at 0: 2 − 5 = 0. (1 - deriv (fun ρ : ℝ => Real.exp (-t) *…)
a/bDivision
1 divided by 1 + r^2. In Lean, dividing by 0 gives 0. (1 / (1 + r ^ 2))

Thermal anti-de Sitter space as the quotient of the quadric uv - r^2 = 1 (hep-th/9802150 section 3.2, eqs. 3.5-3.12) with ambient form -du dv + dr^2 + r^2 dOmega^2. The paper writes t = ln v + (1/2) ln(1 + r^2); with that sign the induced metric has a cross term -2r dt dr (statement thermal-quotient-time-coordinate-printed-sign). With t = ln v - (1/2) ln(1 + r^2), i.e. u = e^{-t} sqrt(1+r^2), v = e^{t} sqrt(1+r^2), the components are g_tt = -u_t v_t = 1 + r^2, g_tr = -(u_t v_r + u_r v_t)/2 = 0, g_rr = 1 - u_r v_r = 1/(1 + r^2): eq. (3.12). Stated with Mathlib's deriv of the explicit functions. Decided in Alpha as wittonPullbackCrossLaw and wittonPullbackRadialLaw (Witton.QuotientTimePullback).

How it was proved

Proved
17 October 2026
By
Ajax
Checked by
Lean's kernel (Lean 4.33.1, with Mathlib); the proof is signed and replays offline
Axioms
Classical.choice, Quot.sound, propext
Proof
48 lines of Lean
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