In his biography of Gauss, G. Waldo Dunnington describes the Gauss-Bolyai episode and their correspondence. In particular, he describes the contents of one letter from Gauss to Janos-Bolyai:

In the above mentioned letter Gauss gave as a sample of his own research a proof that in non-euclidean geometry the area of a triangle is proportional to the deviation of the sum of the angles from 180 degrees... In the same letter Gauss urged Johann to busy himself with the corresponding problem for space, namely, to determine the cubic content of the tetrahedron (space bounded by four planes). The papers of Johann contain several processes which can serve as a solution, among them the method which Gauss had in mind and which he indicated in one of his notebooks at the time of sending his letter to Wolfgang on March 6, 1832.

In addition, Stillwell's book "Mathematics and Its History" (p.379) emphasizes that Gauss "did have many of the results of non-euclidean geometry by this time, including the answer to the volume problem he raised to test his young rival (see Gauss (1832 - Cubierung der Tetraeder)". Looking at the same letter, I also saw that Gauss wrote that unlike the case of 2-dimensional content of simplexes (triangles), where the area is proportional to the angular deficit, in the case of 3-dimensional content of tetrahedrons such a simple formula does not exist.

A translation of Gauss's note is:

In the tetrahedron $1234$, whose faces $124$ and $134$ are orthogonal. Its volume is $$\Delta,$$ then it holds that: $$\partial \Delta = -24\cdot\partial341,$$ and if the face angles at vertex 3 are constant, then the following also holds: $$\alpha^2\cdot \mathbb{cotg}^2341 - \beta^2\cdot \mathbb{tanh}^2(l_{24}) = 1$$ where $$\alpha = \mathbb{cotg}431,\quad \beta = \mathbb{cotg}234.$$

For the tetrahedron under the conditions described in Gauss's note, Gauss's formula therefore connects the length of the side $24$ with the angle $341$; literally speaking, it says the length of the side $24$ is an inverse hyperbolic function of a trigonometric function of the angle $341$. In his commentary on Gauss's note, Paul Stackel says Gauss missed a factor of $1/2$ in his expression for the volume differential.

As this is Gauss's only result dealing with calculations of volume in hyperbolic space $H^3$ (and not only in the hyperbolic plane), this is a very significant point in his work on non-euclidean geometry that has not recieved enough attention by historians of mathematics (except Stackel's and Stillwell's remarks).

Therefore, my questions are:

  • What is the meaning of Gauss's results? how to derive them?
  • How to derive a volume function $\Delta(l_{24})$ for the tetrahedron satisfying Gauss's conditions by the analytic procedure outlined by Gauss?
  • $\begingroup$ Bolay? do you mean Bolyai? $\endgroup$ Dec 14, 2017 at 11:43
  • 1
    $\begingroup$ I corrected the spelling of his name. $\endgroup$
    – user2554
    Dec 14, 2017 at 14:53

2 Answers 2


I am not sure of the notation, but I assume this can be derived from the Schlafli formula for the volume of a tetrahedron (so this seems to indicate that Gauss knew Schlafli's formula three quarters of a century prior to Schlafli):

$$ dV = -\frac12 \sum_{ij}l_{ij} d \alpha_{ij},$$ where $l$ is the length of the edge, and $\alpha$ is the dihedral angle of the edge. Notice, in particular, that if the dihedral angle at an edge does not change, then that edge does not contribute to the sum. Further note that if you look at the link of the vertex $1$ (wlog), this is a spherical triangle, whose angles are the dihedral angles $12, 13, 14$ while its sides are the face angles of the three adjacent faces. By the Gauss-Bonnet theorem (note the first author), the variation of the area of the face is equal to minus the sum of the variations of the angles.

Put this all together, and you should get Gauss' formula. As for Schlafli's formula, there are many nice proofs, a simple geometric one by Vinberg (which appeared in the Geometry of Spaces of Constant Curvature survey), and a very pretty analytic one by Hellmuth Kneser, which appeared in Deutsche Mathematik, and thus is hard to find, but there is a more recent exposition a paper of Feng Luo https://arxiv.org/abs/math/0412208.

  • $\begingroup$ I voted your answer because it really helped me understand things about Gauss's formula (now i know it's form is similar to the one given by Schlafli, and that helps me a lot). Thanks!, but i still need further explanation. In particular, as far as i know, Gauss meant hyperbolic geometry when he talked about "non-euclidean geometry", but when you mentioned spherical triangle it made suspect that perhaps we are not talking about the same thing. $\endgroup$
    – user2554
    Dec 15, 2017 at 7:55
  • $\begingroup$ In addition, since i don't understand well the rimannnian theory of 3-manifolds, i have a basic misunderstanding: in three-dimensional hyperbolic spaces with constant space curvature k, the Gauss-Bonnet theorem can be applied to figures with volume, but does it mean it can be applied to figures with finite area but zero volume? i ask because by analogy with the Gauss-Bonnet theorem for 2-manifolds, it cannot by applied to cross sections (sections of one dimension lower). $\endgroup$
    – user2554
    Dec 15, 2017 at 8:04
  • $\begingroup$ @user2554 The link of a vertex of a polyhedron in any geometry is a spherical triangle. As for Gauss-Bonnet, in its original form (due to Gauss) it is stated for triangles. In any case, area as angle defect is a fundamental fact of hyperbolic geometry. $\endgroup$
    – Igor Rivin
    Dec 15, 2017 at 16:51

Gauss's procedure leads to Bolyai's result on the volume of orthoscheme tetrahedron, as I'll show here. However, Gauss's result is a little bit more limited than Bolyai, since Gauss refers to an orthoscheme tetrahedron of which 4 of the 12 face angles of the tetrahedron are right (each face is an hyperbolic right triangle), while Bolyai refers to a slightly more general tetrahedron whose only 3 face angles ar right.

In order to help visualize the relations, I added here a pic of Gauss's note.

enter image description here

Preliminary discussion:

To see the connection between Schlafli formula and the first formula in Gauss's fragment, one needs to understand that Gauss thinks of the tetrahedron 1234 in such a way that the faces 124 and 134 are perpendicular and the edges 24 and 13 meet the intersection line 14 also at right angles. Therefore, the dihedral angles at sides 12 and 14 are constant right angles and don't contribute to the sum in Schlafli formula. In addition, Gauss defines the tetrahedron in such a way that the angles at vertex 3 are constant (so that an "observer" in hyperbolic space which is located at vertex 3 sees the rest of the vertexes at constant lines of sight). Since the three face angles at vertex 3 correspond to the length of sides of a spherical triangle, and the dihedral angles at sides 31,32,34 correspond to the angles of this spherical triangle, one gets that constancy of face angles at vertex 3 implies constancy of the dihedral angles 31,32,34.

Therefore, only the dihedral angle of the side 24 changes. The dihedral angle 24 is equal to face angle 341 since two face angles at vertex 4 are right so the third face angle 341 (which is one side of a spherical triangle) is equal to the opposite angle - which is dihedral angle 24. This leads directly to the first formula in Gauss's fragment (apart from a missing factor of $\frac {1}{2}$).

Derivation of explicit volume formula from Gauss's formula:

For the sake of consistency, we denote the angles 431, 234, and 214 as $\alpha$, $\beta$ and $\gamma$, respectively. Now lets look at the link of vertex 3 of the tetrahedron: it is a spherical triangle whose two edges lengths are $\alpha$ and $\beta$ and one angle is $\gamma$ (it is the dihedral angle of edge 31 and it is also equal to $\gamma$). In addition the sides $\alpha$$, \beta$ of this spherical triangle are orthogonal to each other. Therefore, by a combination of the spherical sine theorem and the spherical pythagoras theorem, we get:

$$ \frac{{\mathbb{sin}(\mathbb{arccos}(\mathbb{cos}\alpha\cdot \mathbb{cos}\beta))}}{{\mathbb{sin} (\frac{\pi}{2}})} = \frac {{\mathbb{sin}\beta}}{{\mathbb{sin}\gamma}}$$, or:

$$(1) \mathbb{sin}\gamma = \frac {{\mathbb{sin}\beta}}{{\sqrt{{1 - (\mathbb{cos}\alpha \cdot \mathbb{cos}\beta)^2}}}}$$

Now, denote the length of side $24$ as $l_{24} = x$ and the angle $341$ as $\varphi$. Since $\varphi$ is related to $x$ by the equation $c^2_1 \mathbb{cot}^2\varphi - c^2_2\mathbb{tanh}^2x = 1$ (here $c_1 = \mathbb{cot}\alpha,c_2 = \mathbb{cot}\beta$), one can write:

$$\varphi = \mathbb{arccot}(\frac{\sqrt{1+c^2_2\mathbb{tanh}^2x}}{c_1})$$

Gauss's procedure for the calculation of the volume, which uses the relation $\partial \Delta = -\frac{1}{2}x d\varphi$, leads to the following integral:

$$\Delta = -\frac{1}{2}\int x d\varphi = -\frac{1}{2}\int x \frac{d\varphi}{dx}dx$$

so one can compute the derivative of $\varphi$ with respect to $x$ by an application of the chain rule:

$$-\frac{d\varphi}{dx} = \frac{1}{1+\frac{1+c^2_2\mathbb{tanh}^2x}{c^2_1}}\frac{c^2_2\mathbb{tanh}x\cdot \frac{1}{\mathbb{cosh}^2x}}{c_1\sqrt{1+c^2_2\mathbb{tanh}^2x}}$$

Now make a very long algebraic simplification:

$$ -\frac{d\varphi}{dx} = \frac{\mathbb{sinh}x (c^2_2/c_1)}{\mathbb{cosh}^2x(1+\frac{1+c^2_2\mathbb{tanh}^2x}{c^2_1})\mathbb{cosh}x\cdot c_2\sqrt{\frac{1}{c^2_2}+\mathbb{tanh}^2x}} = \frac{\mathbb{sinh}x (c_2/c_1)}{\mathbb{cosh}^2x(1+\frac{1+c^2_2\mathbb{tanh}^2x}{c^2_1})\sqrt{\frac{\mathbb{cosh}^2x}{c^2_2}+\mathbb{sinh}^2x}} = \frac{\mathbb{sinh}x (c_1/c_2)}{(c_1/c_2)^2(\mathbb{cosh}^2x(1+\frac{1+c^2_2\mathbb{tanh}^2x}{c^2_1}))\sqrt{\frac{\mathbb{cosh}^2x}{c^2_2}+\mathbb{sinh}^2x}}$$

The expression under the square root is $$\sqrt{\frac{\mathbb{cosh}^2x}{c^2_2}+(\mathbb{cosh}^2x-1)} = \sqrt{\mathbb{cosh}^2x(\frac{1}{c^2_2}+1)-1} = \sqrt{{\frac{{\mathbb{cosh}^2x}}{{\mathbb{cos}^2\beta}} - 1}}$$

while the expression in the left side of the denominator is equal to:

$$(c_1/c_2)^2\mathbb{cosh}^2x(1+\frac{1}{c^2_1})+\mathbb{sinh}^2x = \mathbb{cosh}^2x(1+\frac{c^2_1+1}{c^2_2}) -1 = \mathbb{cosh}^2x(1+\frac{1}{\mathbb{sin}^2\alpha \mathbb{cot}^2\beta})-1$$

Recalling that $\frac{c_1}{c_2} = \frac {\mathbb{tan}\beta}{\mathbb{tan} \alpha}$, the resulting expression for the integral is:

$$\Delta = \frac {{\mathbb{tan}\beta}}{{2 \mathbb{tan} \alpha}}\int_{0}^{c}\frac {{x \mathbb{sinh}(x) dx}}{{(\mathbb{cosh}^2x(1 + \frac {{1}}{{\mathbb{sin}^2\alpha \mathbb{cot}^2\beta}})-1)\sqrt{{\frac{{\mathbb{cosh}^2x}}{{\mathbb{cos}^2\beta}} - 1}} }}$$

Now, the left factor of the denominator $\mathbb{cosh}^2(x)(1 + \frac{{1}}{{\mathbb{sin}^2\alpha \ \mathbb{cot}^2\beta}})-1$, is exactly equal to $\mathbb{cosh}^2(x)\cdot \frac{{1}}{{\mathbb{cos}^2\gamma}}-1$, because subtitution of $\mathbb{cos}\gamma = \sqrt {1 - \frac {{\mathbb{sin}^2\beta}}{{1-(\mathbb{cos}\alpha\cdot \mathbb{cos}\beta)^2}}}$ (this substitution is true because of relation (1)) in this expression gives the previous one.

Concluding remarks:

  • As can be seen from this presentation - Hyperbolic Volumes and Symmetry, the Bolyai's volume integral is written in my notation in this way (see Theorem 5, p. 12, at this presentation) :

$$Vol(T) = \frac {{\mathbb{tan}\beta}}{{2 \mathbb{tan} \alpha}}\int_{0}^{c}\frac {{x \mathbb{sinh}(x) dx}}{{(\frac {{\mathbb{cosh}^2(x)}}{{\mathbb{cos}^2\gamma}} - 1)\sqrt{{\frac{{\mathbb{cosh}^2x}}{{\mathbb{cos}^2\beta}} - 1}} }}$$

and in the case treated here Bolyai's integral coincides with the result of Gauss's procedure. Important Note: the differences in notation between the Bolyai integral in the presentation and Gauss's integral are just due the different symbols of the angles 431, 234, and 214 - $\alpha,\beta,\gamma$ in the presentation correspond to $\gamma, \alpha ,\beta$ in my notation.

However, for the case treated by Gauss, his formulas are absolutely correct. He should also be given credit for the identification of the calculation of the orthoscheme tetrahedron as the basis for volume formulas of general tetrahedrons (without right angles). In one of his letters, he refered to those calculations of volumes as "die jungle" - I guess he refered to the extremely complicated integrals that arise in the attempts to the decompose the general tetrahedron into orthoscemes (this problem was only solved very recently).

  • Paul Stackel, the mathematician who edited Janos Bolyai's geometric works, had the following things to say about Bolyai's derivation of his integral formula:

It is most remarkable that the method that Gauss used for cubing the tetrahedron, is exactly the same as that of Johann. This is shown in a note from March 1832, from Gauss's estate, which is printed in the works (vol. VIII, p. 228); Gauss has exactly the same special tetrahedron (only 3142 instead $abc\delta$ means) and exactly the same decomposition by planes perpendicular to ab (31).

This quotation is taken from p. 113 of the book "Wolfgang und Johann Bolyai geometrische Untersuchungen" (here is a link: https://archive.org/details/wolfgangundjohan01stuoft/page/112/mode/2up), which was edited and translated to german by the Stackel.

  • It's still necessary to understand how Gauss arrived at the formula $\partial \Delta = -\frac{{1}}{{2}}(24)d(341)$ (he missed the factor $\frac {{1}}{{2}}$ at the first attempt); the second formula from his note can be derived with relative ease. In his commentary on Gauss's note, Stackel derives it in the following way:

The tetrahedron $1234$, whose volume is called $\Delta$, may now experience an infinitely small increase in volume $1 2 4 1' 2' 4' = \partial \Delta$, by lengthening the edge $31$ by the infinitely small amount $11' = d(13)$ and through $1'$ a perpendicular to $31'$ laying in the plane $1'2'4'$, which intersects the edges $32$ and $34$ in $2'$ and $4'$, respectively. The angles at the corner $3$, and thus also the sizes $\alpha$ and $\beta$, remain unchanged. The angle $(341)$ changes into the angle $$(34'1') = (3 4 1)+d(3 4 1)$$ namely, like the consideration of the quadrilateral $1 1' 4' 4$ with the infinitely small base line $1 1' = d(1 3)$ and right angles at $1$ and $1'$ recognized immediately: $$d(3 4 1) = \mathbb{sinh}(14)\cdot d(13)$$ The increase in volume $1 2 4 1' 2' 4'$ is bounded laterally by the triangles $1 2 4$ and $1' 2' 4'$, whose planes are both perpendicular to $1 1'$, and hence (see p.233 of this volume): $$\partial \Delta = -\frac{1}{2}d(13)\cdot(24)\mathbb{sinh}(14)$$ hence:$$\partial \Delta = -\frac {1}{2}(24)\cdot d(3 4 1) $$ and that is, apart from the missing factor $\frac{1}{2}$, Gauss's formula.

Since Stackel refers to Gauss's second fragment on volume determinations in non-euclidean geometry (p. 233 of he same volume), which was written in 1840 and was found next to Gauss's copy of one of Lobachevski's publications, I think understanding Gauss's second fragment may help understanding Gauss's reasoning.


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