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I stumbled upon the fact that the Bolza surface can be obtained as the locus of the equation,

$$y^2 = \color{blue}{x^5-x}.$$

Its automorphism group has the highest order for genus $2$, namely $48$. I recognized $x^5-x$ as a polynomial invariant of the octahedron. In fact, the Bolza surface is connected to the octahedron.

(Edit, Apr 27, 2023:) Since someone has bumped this to the front page, the connection to the octahedron can be illustrated by the j-function formula,

$$j(\tau)-1728 = \frac{16(x^8 + 14x^4 + 1)^3}{(\color{blue}{x^5 - x})^4}-1728 = \frac{16(x^{12} - 33x^8 - 33x^4 + 1)^2}{(\color{blue}{x^5 - x})^4}$$

where the numerator of the RHS square is also a polynomial invariant of the octahedral group $H_{48}$. (Dickson, p. 230, Algebraic Theories, 1959.) And perhaps not surprising, the central numerator can be expressed as powers of 8,

$$(x^2-1)^8+(2x)^8+(x^2+1)^8 = 2(x^8+14x^4+1)^2$$

Question:

If we use the analogous polynomial invariant of the icosahedron, then does the genus 5 surface,

$$y^2 = x(x^{10}+11x^5-1)$$

have special properties? How close does the order of its automorphism group get to the bound $84(g-1)$? (For $g = 5$, this would be $336$.)

POSTSCRIPT:

My thanks to Noam Elkies for the highly detailed answer below. The background to this question is an identity I found involving $x^{10}+11x^5-1$. Define,

$$a = \frac{r^5(r^{10}+11r^5-1)^5}{(r^{30}+522r^{25}-10005r^{20}-10005r^{10}-522r^5+1)^2}$$

and,

$$w = \frac{r^2(r^{10}+11r^5-1)^2(r^6+2r^5-5r^4-5r^2-2r+1)}{r^{30}+522r^{25}-10005r^{20}-10005r^{10}-522r^5+1}$$

then they can be neatly stitched together as,

$$w^5-10aw^3+45a^2w-a^2 = 0$$

for arbitrary $r$. This in fact is the Brioschi quintic form which the general quintic can be reduced into. Two of the polynomials are easily recognizable as icosahedral invariants, while $r^6+2r^5-5r^4-5r^2-2r+1$ is a factor of $r^{30}+522r^{25}-10005r^{20}-10005r^{10}-522r^5+1$.

So other than in formulas using Ramanujan's continued fractions, I wondered where else those polynomials appear. Since the Bolza surface involved an invariant of the octahedron, it was reasonable to consider if using the corresponding one for the icosahedron would also be special. As Elkies wonderfully showed, it turns out that it is.

I stumbled upon the fact that the Bolza surface can be obtained as the locus of the equation,

$$y^2 = \color{blue}{x^5-x}.$$

Its automorphism group has the highest order for genus $2$, namely $48$. I recognized $x^5-x$ as a polynomial invariant of the octahedron. In fact, the Bolza surface is connected to the octahedron.

(Edit, Apr 27, 2023:) Since someone has bumped this to the front page, the connection to the octahedron can be illustrated by the j-function formula,

$$j(\tau)-1728 = \frac{16(x^8 + 14x^4 + 1)^3}{(\color{blue}{x^5 - x})^4}-1728 = \frac{16(x^{12} - 33x^8 - 33x^4 + 1)^2}{(\color{blue}{x^5 - x})^4}$$

where the numerator of the RHS square is an invariant of the octahedral group $H_{48}$. (Dickson, p. 230, Algebraic Theories, 1959.) The factorization $\color{blue}{x^5-x} = x(x^2-1)(x^2+1)$ involving Pythagorean triples is of course familiar, but not so familiar is it can express the other numerator,

$$(x^2-1)^4+(2x)^4+(x^2+1)^4 = 2(x^8+14x^4+1)$$ $$\;(x^2-1)^8+(2x)^8+(x^2+1)^8 = 2(x^8+14x^4+1)^2$$

Question:

If we use the analogous polynomial invariant of the icosahedron, then does the genus 5 surface,

$$y^2 = x(x^{10}+11x^5-1)$$

have special properties? How close does the order of its automorphism group get to the bound $84(g-1)$? (For $g = 5$, this would be $336$.)

POSTSCRIPT:

My thanks to Noam Elkies for the highly detailed answer below. The background to this question is an identity I found involving $x^{10}+11x^5-1$. Define,

$$a = \frac{r^5(r^{10}+11r^5-1)^5}{(r^{30}+522r^{25}-10005r^{20}-10005r^{10}-522r^5+1)^2}$$

and,

$$w = \frac{r^2(r^{10}+11r^5-1)^2(r^6+2r^5-5r^4-5r^2-2r+1)}{r^{30}+522r^{25}-10005r^{20}-10005r^{10}-522r^5+1}$$

then they can be neatly stitched together as,

$$w^5-10aw^3+45a^2w-a^2 = 0$$

for arbitrary $r$. This in fact is the Brioschi quintic form which the general quintic can be reduced into. Two of the polynomials are easily recognizable as icosahedral invariants, while $r^6+2r^5-5r^4-5r^2-2r+1$ is a factor of $r^{30}+522r^{25}-10005r^{20}-10005r^{10}-522r^5+1$.

So other than in formulas using Ramanujan's continued fractions, I wondered where else those polynomials appear. Since the Bolza surface involved an invariant of the octahedron, it was reasonable to consider if using the corresponding one for the icosahedron would also be special. As Elkies wonderfully showed, it turns out that it is.

I stumbled upon the fact that the Bolza surface can be obtained as the locus of the equation,

$$y^2 = \color{blue}{x^5-x}.$$

Its automorphism group has the highest order for genus $2$, namely $48$. I recognized $x^5-x$ as a polynomial invariant of the octahedron. In fact, the Bolza surface is connected to the octahedron.

(Edit, Apr 27, 2023:) Since someone has bumped this to the front page, the connection to the octahedron can be illustrated by the j-function formula,

$$j(\tau)-1728 = \frac{16(x^8 + 14x^4 + 1)^3}{(\color{blue}{x^5 - x})^4}-1728 = \frac{16(x^{12} - 33x^8 - 33x^4 + 1)^2}{(\color{blue}{x^5 - x})^4}$$

where the numerator of the RHS square is also a polynomial invariant of the octahedral group $H_{48}$. (Dickson, p. 230, Algebraic Theories, 1959.) And perhaps not surprising, the central numerator can be expressed as powers of 8,

$$(x^2-1)^8+(2x)^8+(x^2+1)^8 = 2(x^8+14x^4+1)^2$$

Question:

If we use the analogous polynomial invariant of the icosahedron, then does the genus 5 surface,

$$y^2 = x(x^{10}+11x^5-1)$$

have special properties? How close does the order of its automorphism group get to the bound $84(g-1)$? (For $g = 5$, this would be $336$.)

POSTSCRIPT:

My thanks to Noam Elkies for the highly detailed answer below. The background to this question is an identity I found involving $x^{10}+11x^5-1$. Define,

$$a = \frac{r^5(r^{10}+11r^5-1)^5}{(r^{30}+522r^{25}-10005r^{20}-10005r^{10}-522r^5+1)^2}$$

and,

$$w = \frac{r^2(r^{10}+11r^5-1)^2(r^6+2r^5-5r^4-5r^2-2r+1)}{r^{30}+522r^{25}-10005r^{20}-10005r^{10}-522r^5+1}$$

then,

$$w^5-10aw^3+45a^2w-a^2 = 0$$

for arbitrary $r$. This in fact is the Brioschi quintic form which the general quintic can be reduced into. Two of the polynomials are easily recognizable as icosahedral invariants, while $r^6+2r^5-5r^4-5r^2-2r+1$ is a polynomial invariant for the octahedron.

Other than in formulas using Ramanujan's continued fractions, I wondered where else those polynomials appear. Since the Bolza surface involved an invariant of the octahedron, it was reasonable to consider if using the corresponding one for the icosahedron would also be special. As Elkies wonderfully showed, it turns out that it is.

I stumbled upon the fact that the Bolza surface can be obtained as the locus of the equation,

$$y^2 = \color{blue}{x^5-x}.$$

Its automorphism group has the highest order for genus $2$, namely $48$. I recognized $x^5-x$ as a polynomial invariant of the octahedron. In fact, the Bolza surface is connected to the octahedron.

(Edit, Apr 27, 2023:) Since someone has bumped this to the front page, the connection to the octahedron can be illustrated by the j-function formula,

$$j(\tau)-1728 = \frac{16(x^8 + 14x^4 + 1)^3}{(\color{blue}{x^5 - x})^4}-1728 = \frac{16(x^{12} - 33x^8 - 33x^4 + 1)^2}{(\color{blue}{x^5 - x})^4}$$

where the numerator of the RHS square is also a polynomial invariant of the octahedral group $H_{48}$. (Dickson, p. 230, Algebraic Theories, 1959.) And perhaps not surprising, the central numerator can be expressed as powers of 8,

$$(x^2-1)^8+(2x)^8+(x^2+1)^8 = 2(x^8+14x^4+1)^2$$

Question:

If we use the analogous polynomial invariant of the icosahedron, then does the genus 5 surface,

$$y^2 = x(x^{10}+11x^5-1)$$

have special properties? How close does the order of its automorphism group get to the bound $84(g-1)$? (For $g = 5$, this would be $336$.)

POSTSCRIPT:

My thanks to Noam Elkies for the highly detailed answer below. The background to this question is an identity I found involving $x^{10}+11x^5-1$. Define,

$$a = \frac{r^5(r^{10}+11r^5-1)^5}{(r^{30}+522r^{25}-10005r^{20}-10005r^{10}-522r^5+1)^2}$$

and,

$$w = \frac{r^2(r^{10}+11r^5-1)^2(r^6+2r^5-5r^4-5r^2-2r+1)}{r^{30}+522r^{25}-10005r^{20}-10005r^{10}-522r^5+1}$$

then they can be neatly stitched together as,

$$w^5-10aw^3+45a^2w-a^2 = 0$$

for arbitrary $r$. This in fact is the Brioschi quintic form which the general quintic can be reduced into. Two of the polynomials are easily recognizable as icosahedral invariants, while $r^6+2r^5-5r^4-5r^2-2r+1$ is a factor of $r^{30}+522r^{25}-10005r^{20}-10005r^{10}-522r^5+1$.

So other than in formulas using Ramanujan's continued fractions, I wondered where else those polynomials appear. Since the Bolza surface involved an invariant of the octahedron, it was reasonable to consider if using the corresponding one for the icosahedron would also be special. As Elkies wonderfully showed, it turns out that it is.

I stumbled upon the fact that the Bolza surface can be obtained as the locus of the equation,

$$y^2 = \color{blue}{x^5-x}.$$

Its automorphism group has the highest order for genus $2$, namely $48$. I recognized $x^5-x$ as a polynomial invariant of the octahedron. In fact, the Bolza surface is connected to the octahedron.

(Edit, Apr 27, 2023:) Since someone has bumped this to the front page, the connection to the octahedron can be illustrated by the j-function formula,

$$j(\tau)-1728 = \frac{16(x^8 + 14x^4 + 1)^3}{(\color{blue}{x^5 - x})^4}-1728 = \frac{16(x^{12} - 33x^8 - 33x^4 + 1)^2}{(\color{blue}{x^5 - x})^4}$$

where the numerator of the RHS square is also a polynomial invariant of the icosahedral group $H_{48}$. (Dickson, p. 230, Algebraic Theories, 1959.) And perhaps not surprising, the central numerator can be expressed as powers of 8,

$$(x^2-1)^8+(2x)^8+(x^2+1)^8 = 2(x^8+14x^4+1)^2$$

Question:

If we use the analogous polynomial invariant of the icosahedron, then does the genus 5 surface,

$$y^2 = x(x^{10}+11x^5-1)$$

have special properties? How close does the order of its automorphism group get to the bound $84(g-1)$? (For $g = 5$, this would be $336$.)

POSTSCRIPT:

My thanks to Noam Elkies for the highly detailed answer below. The background to this question is an identity I found involving $x^{10}+11x^5-1$. Define,

$$a = \frac{r^5(r^{10}+11r^5-1)^5}{(r^{30}+522r^{25}-10005r^{20}-10005r^{10}-522r^5+1)^2}$$

and,

$$w = \frac{r^2(r^{10}+11r^5-1)^2(r^6+2r^5-5r^4-5r^2-2r+1)}{r^{30}+522r^{25}-10005r^{20}-10005r^{10}-522r^5+1}$$

then,

$$w^5-10aw^3+45a^2w-a^2 = 0$$

for arbitrary $r$. This in fact is the Brioschi quintic form which the general quintic can be reduced into. Two of the polynomials are easily recognizable as icosahedral invariants, while $r^6+2r^5-5r^4-5r^2-2r+1$ is a polynomial invariant for the octahedron.

Other than in formulas using Ramanujan's continued fractions, I wondered where else those polynomials appear. Since the Bolza surface involved an invariant of the octahedron, it was reasonable to consider if using the corresponding one for the icosahedron would also be special. As Elkies wonderfully showed, it turns out that it is.

I stumbled upon the fact that the Bolza surface can be obtained as the locus of the equation,

$$y^2 = \color{blue}{x^5-x}.$$

Its automorphism group has the highest order for genus $2$, namely $48$. I recognized $x^5-x$ as a polynomial invariant of the octahedron. In fact, the Bolza surface is connected to the octahedron.

(Edit, Apr 27, 2023:) Since someone has bumped this to the front page, the connection to the octahedron can be illustrated by the j-function formula,

$$j(\tau)-1728 = \frac{16(x^8 + 14x^4 + 1)^3}{(\color{blue}{x^5 - x})^4}-1728 = \frac{16(x^{12} - 33x^8 - 33x^4 + 1)^2}{(\color{blue}{x^5 - x})^4}$$

where the numerator of the RHS square is also a polynomial invariant of the octahedral group $H_{48}$. (Dickson, p. 230, Algebraic Theories, 1959.) And perhaps not surprising, the central numerator can be expressed as powers of 8,

$$(x^2-1)^8+(2x)^8+(x^2+1)^8 = 2(x^8+14x^4+1)^2$$

Question:

If we use the analogous polynomial invariant of the icosahedron, then does the genus 5 surface,

$$y^2 = x(x^{10}+11x^5-1)$$

have special properties? How close does the order of its automorphism group get to the bound $84(g-1)$? (For $g = 5$, this would be $336$.)

POSTSCRIPT:

My thanks to Noam Elkies for the highly detailed answer below. The background to this question is an identity I found involving $x^{10}+11x^5-1$. Define,

$$a = \frac{r^5(r^{10}+11r^5-1)^5}{(r^{30}+522r^{25}-10005r^{20}-10005r^{10}-522r^5+1)^2}$$

and,

$$w = \frac{r^2(r^{10}+11r^5-1)^2(r^6+2r^5-5r^4-5r^2-2r+1)}{r^{30}+522r^{25}-10005r^{20}-10005r^{10}-522r^5+1}$$

then,

$$w^5-10aw^3+45a^2w-a^2 = 0$$

for arbitrary $r$. This in fact is the Brioschi quintic form which the general quintic can be reduced into. Two of the polynomials are easily recognizable as icosahedral invariants, while $r^6+2r^5-5r^4-5r^2-2r+1$ is a polynomial invariant for the octahedron.

Other than in formulas using Ramanujan's continued fractions, I wondered where else those polynomials appear. Since the Bolza surface involved an invariant of the octahedron, it was reasonable to consider if using the corresponding one for the icosahedron would also be special. As Elkies wonderfully showed, it turns out that it is.