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Problem: Given a infinitely small geodesic triangle $\triangle ABC$ on a smooth surface, denote the corresponding edges as $a,b,c.$ Prove: the area of $\triangle ABC$ (denote as $S$) and the Gauss curvature on $C$ ( denote as $K$) satisfies

$$c^2=a^2+b^2-2ab\cos\left(C-\frac{KS}{3}\right).$$

Since the triangle is infinitesimal, therefore it is naturally to use the first fundamental form and the Gauss-Bonnet theorem. However, I got stuck at how to handle the $\frac{KS}{3}$ in the equation. Since the coefficient is $\frac{1}{3}$, if using Gauss-Bonnet theorem, we would include new variables $\angle A,\angle B$. Could anyone help me out? Thanks a lot in advance!

LonnerT
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  • While I don't have a full proof of $C=\tfrac{KS}{3}+\arccos\frac{a^2+b^2-c^2}{2ab}$, note that $A,,B,,C$ average $\frac{\pi}{3}+\frac{KS}{3}$ instead of the Euclidean $\frac{\pi}{3}$, so the question is why each angle ends up exceeding Euclidean expectations by the same amount. – J.G. Dec 12 '20 at 09:50
  • Do you know the Taylor expansion of the metric in geodesic normal coordinates (centered at $p=C$), $$ds^2 = dr^2 + G,d\theta^2, \quad\text{with}\quad \sqrt G = r-K(p)r^3/6 + o(r^3)?$$ – Ted Shifrin Dec 12 '20 at 18:40
  • Having worked on this for a while, I am somewhat astonished never to have seen anything like this. Where did this come from? It doesn't quite seem to check out for the case of the sphere, but perhaps I'm not patient enough. – Ted Shifrin Dec 12 '20 at 22:10

2 Answers2

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Making certain reasonable approximations due to the assumption of infinitesimal triangles, this works on the sphere. For simplification of notation we will work on a unit sphere.

Assume $s=\frac{a+b+c}2$ is very small and $\frac{\min(a,b,c)}{\max(a,b,c)}\ge\lambda\gt0$ is not. That is, $\frac23s\le\max(a,b,c)\le s$ and $\min(a,b,c)\ge\frac23\lambda s$.

Let $\tilde{C}$ be the angle opposite side $c$ in a spherical triangle with sides $a,b,c$, and $C$ be the angle opposite side $c$ in a plane triangle with sides $a,b,c$. Then $$ \begin{align} \cos(\tilde C) &=\frac{\cos(c)-\cos(a)\cos(b)}{\sin(a)\sin(b)}\tag1\\ &=\frac{\left(1-\frac12c^2+\frac1{24}c^4\right)-\left(1-\frac12a^2+\frac1{24}a^4\right)\left(1-\frac12b^2+\frac1{24}b^4\right)+O\!\left(s^6\right)}{\left(a-\frac16a^3\right)\left(b-\frac16b^3\right)+O\!\left(s^6\right)}\tag2\\ &=\frac{a^2+b^2-c^2+\color{#C00}{\frac1{12}c^4-\frac1{12}a^4-\frac1{12}b^4-\frac12a^2b^2}}{2ab\color{#090}{-\frac13a^3b-\frac13ab^3}}+O\!\left(s^4\middle/\lambda^2\right)\tag3\\ &\approx\frac{a^2+b^2-c^2}{2ab}+\frac{\color{#C00}{\frac1{12}c^4-\frac1{12}a^4-\frac1{12}b^4-\frac16a^2b^2-\frac13a^2b^2}}{2ab}\\ &-\frac{a^2+b^2-c^2}{2ab}\frac{\color{#090}{-\frac13a^3b-\frac13ab^3}}{2ab}\tag4\\ &=\cos(C)+\frac{a^2+b^2-c^2}{12}\frac{a^2+b^2-c^2}{2ab}-\frac{ab}6\tag5\\ &=\cos(C)+\frac{(a-b)^2-c^2}{12}\frac{(a+b)^2-c^2}{2ab}\tag6\\ &=\cos(C)-\frac{16\Delta^2}{48\cdot\frac12ab}\tag7\\ &=\cos(C)-\tfrac\Delta3\sin(C)\tag8\\[6pt] &\approx\cos\left(C+\tfrac\Delta3\right)\tag9 \end{align} $$ Explanation:
$(1)$: Spherical Law of Cosines
$(2)$: $\cos(x)=1-\frac12x^2+\frac1{24}x^4+O\!\left(x^6\right)$
$\phantom{\text{(2):}}$ $\sin(x)=x-\frac16x^3+O\!\left(x^5\right)$
$(3)$: collect terms and move the error out of the fraction
$(4)$: $\frac{u+\color{#C00}{\mathrm{d}u}}{v+\color{#090}{\mathrm{d}v}}-\left(\frac uv+\frac{\color{#C00}{\mathrm{d}u}}v-\frac{u\,\color{#090}{\mathrm{d}v}}{v^2}\right)=\frac{(u\,\mathrm{d}v-v\,\mathrm{d}u)\,\mathrm{d}v}{v^2(v+\mathrm{d}v)}=O\!\left(s^4\middle/\lambda^2\right)$
$(5)$: apply the Law of Cosines to term $1$ and combine terms $2$ and $3$
$(6)$: factor
$(7)$: Heron's formula
$(8)$: $\Delta=\frac12ab\sin(C)$
$(9)$: $\cos(x+\mathrm{d}x)\approx\cos(x)-\sin(x)\,\mathrm{d}x$

Note that $$ \begin{align} \cos\left(C+\tfrac\Delta3\right) &=\cos(C)\overbrace{\ \cos\left(\tfrac\Delta3\right)\ }^{1-O\left(s^4\right)}-\sin(C)\overbrace{\ \sin\left(\tfrac\Delta3\right)\ }^{\frac\Delta3-O\left(s^6\right)}\tag{10a}\\[6pt] &=\cos(C)-\tfrac\Delta3\sin(C)+O\!\left(s^4\right)\tag{10b} \end{align} $$ so that the total error in the approximation $\tilde{C}\approx C+\frac\Delta3$ is on the order of $s^4$.

Comparing Heron's formula $$ \Delta=\sqrt{s(s-a)(s-b)(s-c)}\tag{11} $$ and L'Huillier's formula $$ \tan\left(\frac S4\right)=\sqrt{\tan\left(\frac{s}2\right)\tan\left(\frac{s-a}2\right)\tan\left(\frac{s-b}2\right)\tan\left(\frac{s-c}2\right)}\tag{12} $$ using $\tan(x)=x\!\left(1+O\!\left(x^3\right)\right)$, and noting that $K=1$ on the unit sphere, we get $$ KS=\Delta+O\!\left(s^4\right)\tag{13} $$

robjohn
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  • Nicely done. The factoring game is the same one that goes on in (one of) the proofs of Heron's formula. :) – Ted Shifrin Dec 12 '20 at 23:30
  • I'm a little confused in step (9) ,that $KS$ seems not to be infinitesimal due to Gauss-Bonnet theorem (the corollary that $\alpha_1+\alpha_2+\alpha_3=\pi+KS$). Do I misunderstand or there is a flaw? – LonnerT Dec 13 '20 at 15:23
  • We are dealing with infinitesimal triangles, so $KS$ is very small (it is the area of the image of the triangle under the Gauss map). – robjohn Dec 13 '20 at 17:25
  • So nice with the fourth order..Is the last cosine term valid for surfaces of positive and negative $K$ like $ \cos (C+ KS/3)?$ – Narasimham Dec 13 '20 at 21:18
  • @Narasimham: Since Gauss-Bonnet holds for negative $K$, I believe this should also, but I've only worked with positive curvatures in this answer, so don't quote me on that. – robjohn Dec 13 '20 at 21:36
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    In the hyperbolic case, as always, all the cos turn to cosh and the sin to sinh, and it should work out completely analogously. (Remember the sign changes in the Taylor polynomials.) I don't know why Narasimham removed the minus sign — does he mean $+|K|S/3$ in the case $K<0$? The sign is correct as it stands in the OP's post. – Ted Shifrin Dec 13 '20 at 21:47
  • @TedShifrin: won't the Law of Hyperbolic Cosines be $$\cosh(\gamma)=\cosh(\alpha)\cosh(\beta)-\sinh(\alpha)\sinh(\beta)\cos(C)$$ As I remember, we substitute $\sin(x)\mapsto i\sinh(x)$ and $\cos(x)\mapsto\cosh(x)$ and the $i$s resolve away. – robjohn Dec 13 '20 at 22:32
  • @robjohn: Yes, that's right. Oh, there's a minus sign different from the substitution I claimed. – Ted Shifrin Dec 13 '20 at 22:35
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Comment @ robjohn : Can we check for both signs of const $K?$

So for constant $K$ approximation we have $ \cos \tilde C$

$$= \cos (C+ \dfrac{KS}{3})= \cos (C\pm \dfrac{A+B+C-\pi}{3}) $$

K>0

For an equilateral spherical triangle $( a=b=c,A=B=C=\pi/2) $

$$\text{RHS}=\cos 2 \pi/3= -\dfrac12$$

$$ c^2= a^2+b^2+ab = 3 a^2 \to c= {\sqrt 3 a}, \text{ a contradiction }$$

K<0

For an equilateral hyperbolic spiky triangle $(A=B=C=\pi/6,a=b=c),$

$$ = \cos ( \dfrac{4C+A+B-\pi}{3}) = \cos (\dfrac{2C-A-B+\pi}{3}), $$

$$\text{RHS}=\cos \pi/3= \dfrac12$$

$$ c^2= a^2+b^2-ab = a^2 ,\; c=a.$$

Please comment what I am missing here in the K>0 approximation.

Narasimham
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  • In the $K\gt0$ approximation, your triangle covers an entire octant. Each of the sides is $\pi/2$. The approximation is for infinitesimal triangles, not ones this large. – robjohn Dec 16 '20 at 18:55
  • Thanks. For $K<0$ triangle where each vertex angle $ =\pi/6$ is not so small, however has a full tally.. – Narasimham Dec 16 '20 at 19:18