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According to p. 68 of Paul Stackel's essay "Gauss as geometer" (which deals with "complex quantities with more than two units") , Gauss calculated the coordinates of the vertices of the dodecahedron (regular polyhedron with 12 faces and 20 vertices) and the icosahedron (regular polyhedron with 20 faces and 12 vertices). After some unsuccessful efforts to derive these coordinates by myself, I gradually found this calculation of Gauss to be very interesting. After some searches I found this calculation, which was not published in the collected works, but appears in pp. 166-167 of Gauss's handbuch 6. Here is a translation of the first few sentences in this fragment:

The coordinates for the vertices of the regular icosahedron (circumscribed by a sphere of radius $\sqrt{a^2+b^2}$) are:$$0,\pm a, \pm b$$ $$\pm b, 0, \pm a$$ $$\pm a, \pm b, 0,$$ where $a=(\frac{1+\sqrt{5}}{2})b$.

($\frac{1+\sqrt{5}}{2}$ is the golden ratio $\phi$).

Under one conventional choice of a coordinate system, Gauss gives the correct modern results for the coordinates of the vertices (they agree with the data given in wolfram mathworld article "Regular Icosahedron"). Gauss then writes $\rho^5=1$ and expresses the coordinates as a kind of "tricomplex number" (with basis $1,i,k$) using this fifth root of unity:

Even more delicate, set the radius as $\sqrt{5}$, then the 12 vertices are defined by the third coordinate $k$ through $$\pm(2\rho^n +k)$$ and $$\pm\sqrt{5}k = \pm(\rho-\rho^2-\rho^3+\rho^4)k$$

(here $n = 0,1,2,3,4$ so there are $10 = 2\times 5$ vertices of the first kind (lateral vertices) and 2 vertices of the second kind (axial vertices)). Note that since $\rho$ is a root of unity of odd order, $-\rho^n$ is not a fifth root of unity, so the ten lateral vertices consist of two pentagons, laying on parallel planes, and one pentagon is rotated by an angle $\pi/5$ with respect to the second. This means that if we take, for example, the vertex $2\rho+k$, than its 5 adjacent vertices are: $\sqrt{5}k, 2\rho^2+k,2+k, -2\rho^3-k,-2\rho^4-k$.

Regarding the coordinates of the vertices of the regular dodecahedron, Gauss writes the following:

The 20 vertices of the dodecahedron; radius of circumscribing sphere:$\sqrt{15}$, will be most accurately represented by $(P,Q)$, where $P$ is a complex number that represents the first two coordinates, and $Q$ the third coordinate, by:$$P=2(\rho-\rho^4)\rho^n, Q = \rho-\rho^2+\rho^3-\rho^4$$ for which $n$ can take the values $0,1,2,3,4$ and $\rho$ is substituted by all its conjugate values.

I guess that Gauss meant that $\rho$ should be substituted by $\rho,\rho^2,\rho^3,\rho^4$ in both $P,Q$ (this interpretation is consistent with what is written in the deleted lines above it), while $n$ should take the values $0,1,2,3,4$. This yields $20=5\times 4$ triples, so it amounts to a total of 20 vertices. I am not sure I translated the passage on dodecahedron accurately, but I checked it and the radius of the circumscribing sphere for these coordinates is really $\sqrt{15}$.

As far as I know, Euclid's Elements contains several metrical results on the platonic solids, such as the ratio of edge of regular polyhedrons to the radius of its circumscribing sphere. However, since the cartesian coordinates system was invented only in early 17th century, I am not sure ancient geometers tried to calculate something like coordinates of vertices (but maybe I am wrong). I don't know who was the first to calculate these coordinates.

Some thinking on the way Gauss expressed the coordinates of the icosahedron and dodecahedron made me suspect something more interesting lays beneath this calculation. It appears to me to be interesting because Gauss did not have to express them using fifth roots of unity, as one could simply write the coordinates as real numbers. This made me wonder if there is a point of view according to which these cumbersome expressions for coordinates (involving powers of $\rho$) appear naturally?

After reading in several sources, including Felix Klein's "Lectures on the Icosahedron", I suspected that there might be an implicit connection of Gauss's calculation with Hamilton's "Icosian calculus" and the "Icosahedral group". What made me suspecting this is that $\rho$ is the generator of the cyclic group of order 5, and one of the generators of the Icosian group (which has two generators $x,y$) has the property $y^5 = I$. Gauss casted his calculations using $\rho$ in a way that resembles the second chapter of Klein's book, and I wondered if there is a way to arrive at the exact coordinates of the vertices of these polyhedra from purely group theoretic principles?

So I will be glad if anyone familiar with group theory and non-commutative algebras will offer a computational procedure which naturally results in an expressions involving certain sums of fifth roots of unity. Such a procedure is a curiosity to me not only because of the math in it, but because it is an interesting point in history of mathematics that might shed additional light on Gauss's understanding of non-commutative algebra. After all, Paul Stackel mentioned Gauss's calculations on the icosahedron and dodecahedron in the same section in which he mentioned Gauss's anticipation of the quaternions. It is also possible that Gauss's use of complex numbers here was just a result of his fascination with those numbers and not a result of anticipation of non-commutative algebra (and that is why I want to check it).

One possible direction: Gauss's statements on the vertices of regular dodecahedron can be restated more explicitly: the vertices can be divided into two groups of 10 vertices, with coordinates:

$$P = \pm 4i(\mathbb{sin}(\frac{2\pi}{5}))\rho^n, Q = \pm 2i(\mathbb{sin}(\frac{2\pi}{5})-\mathbb{sin}(\frac{\pi}{5}))$$

(here the signs of $P,Q$ are not independent so there are $10=2\times 5$ vertices in this group) and:

$$P = \pm 4i(\mathbb{sin}(\frac{\pi}{5}))\rho^n, Q = \pm 2i(\mathbb{sin}(\frac{2\pi}{5})+\mathbb{sin}(\frac{\pi}{5}))$$

so the radius of circumscribing sphere is:

$$(\sqrt{15})^2 = |P|^2+|Q|^2 = (4\mathbb{sin}(\frac{2\pi}{5}))^2+(2(\mathbb{sin}(\frac{2\pi}{5})-\mathbb{sin}(\frac{\pi}{5})))^2 = (4\mathbb{sin}(\frac{\pi}{5}))^2+(2(\mathbb{sin}(\frac{2\pi}{5})+\mathbb{sin}(\frac{\pi}{5})))^2$$

In a similar vein, Gauss's statements on the regular icosahedron lead to several equivalent trigonometric expressions for the edge length of the icosahedron.

I verified the correctness of these identities, and I am sure there is an easy algebraic proof of them. However, this does not mean that nothing deep is laying behind them; after all, historical identities like Diophantus's two squares identity (which can be explained by equating norms of complex numbers) and Euler's four squares identity (which can be explained by equating norms of quaternions) can be shown to be true by simply expanding the terms, but nevertheless deep and fruitfull ideas are contained in them.

So maybe these trigonometric identities can be somehow derived by Icosian algebra, and this line of thought will also lead directly to Gauss's combinations of fifth roots of unity.

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  • $\begingroup$ I guess most likely starting point for these calculations is an icosahedron inscribed in a cube (or a cube in a dodecahedron) - after the combinatorial type is clear one simply solves for all edge lengths to be equal. $\endgroup$ Commented May 13 at 19:44
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    $\begingroup$ I am sure there is a not very difficult geometric derivation of the coordinates, maybe with methods dating back to the time of Euclid, but since Gauss uses sums of fifth roots of unity to express them I suspect something more interesting is hiding here. Maybe he really discovered an algebraic structure here. $\endgroup$
    – user2554
    Commented May 13 at 20:40
  • $\begingroup$ There is a lot of discussion here. Can you please highlight your specific question? $\endgroup$ Commented May 13 at 20:52
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    $\begingroup$ @SamHopkins - I highlighted the main questions. I am looking for a point of view according to which the expressions of Gauss appear naturally as the outcome of the calculation, if there is such a way. For example, Gauss writes: $\pm\sqrt{5}k = \pm(\rho-\rho^2-\rho^3+\rho^4)k$ ; if he just wanted to indicate the coordinates he did not have to write such a sum of powers of $\rho$ for writing $\sqrt{5}$. This made me suspect that his final expression are not only a mode of presenting the coordinates but a result of a certain algebraic method. $\endgroup$
    – user2554
    Commented May 13 at 21:58

1 Answer 1

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Straightening out the nomenclature here.

The apparent second imaginary unit $k$ that Gauss introduces should not be interpreted as either an imaginary unit or a square root of $-1$. It's just a unit in the $z$ durection tacked on perpendicularly to the complex plane that is the $xy$ plane. We might better label the ten lateral vertices as $(\pm_12\rho^n,\pm_11)$ and the two axial vertices as $(0,\pm_2\sqrt5)$ where:

  • The first coordinate represents $(x,y)$ represented as a complex number $x+yi$

  • The second coordinate represents the $z$ coordinate, which Gauss renders as a coefficient of the "tacked-on" unit $k$

  • $\pm$ signs with like subscripts are to be chosen identically; those with different subscripts may be chosen independently.

With this nomenclature, $\rho$ may be taken as any primitive complex fifth root of unity; different choices for $\rho$ may change the sign of Gauss's quartic expression $\rho-\rho^2-\rho^3+\rho^4$ (either $+\sqrt5$ or $-\sqrt5$), but this does not change the set of axial vertices which always includes both signs on $\sqrt5$.

The circumscribed radius is, in fact, $\sqrt5$. The axial vertices are perforce $\sqrt5$ units from the origin, while the ten lateral vertices are $\sqrt{2^2+1^2}=\sqrt5$ units from the origin, where the $2$ comes from the absolute value of $2\rho^n$ representing the complex-number $(x,y)$ coordinates, and the $1$ is from the $z$ coordinate.

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  • $\begingroup$ Thanks! but I already understood the nomenclature. Anyway, is is good that you clarified it for anyone who did not understand it yet. [+1] (but I do not consider this as answer). $\endgroup$
    – user2554
    Commented May 14 at 10:18

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