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Bring's curve or Bring's surface with genus 4 and $5!=120$ automorphisms can be given by the homogeneous equations,

$$x_1+x_2+x_3+x_4+x_5 = x_1^2+x_2^2+x_3^2+x_4^2+x_5^2 = \\x_1^3+x_2^3+x_3^3+x_4^3+x_5^3 = 0$$

This is also a property of the Bring quintic $x^5+ax+b = 0$ since its roots $x_i$ obey,

$$\sum_{i=1}^5 x_i^k = 0,\quad k = 1,2,3$$

But there is also a surface with six coordinates such that,

$$y_1 + y_2 + y_3 + y_4 + y_5 + y_6 = y_1^2 + y_2^2 + y_3^2 + y_4^2 + y_5^2 + y_6^2 = \\ y_1^4 + y_2^4 + y_3^4 + y_4^4 + y_5^4 + y_6^4 = y_1^7 + y_2^7 + y_3^7 + y_4^7 + y_5^7 + y_6^7 = 0$$

This seems to be a property of the sextic $y^6+ay^3+by+c = 0$ since its roots $y_i$ obey,

$$\quad\sum_{i=1}^6 y_i^k = 0,\quad k = 1,2,4,\color{red}7$$

Questions:

  1. The general quintic and sextic can be transformed in radicals into $x^5+ax+b = 0$ and $y^6+ay^3+by+c = 0$, respectively. So does it follow if the first surface has $5!=120$ automorphisms, then the second has $6!=720$? (In this table, no genus $< 12$ has more than 720.)
  2. If the surface in $x$ has genus 4, what is the genus of the surface in $y$?
  3. Any other special property, such as why is it valid for the high exponent $k=7$?
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  • $\begingroup$ More on the Bring surface in this MO post. $\endgroup$ Commented May 3, 2023 at 6:27
  • $\begingroup$ 1. It has 720 automorphisms. The group is $\mathfrak{S}_6$. $\qquad\qquad$ 2. What do you call the genus of a surface? If it is the geometric genus, it is easily calculable, but not very interesting — certainly a 3-digit number. $\endgroup$
    – abx
    Commented May 3, 2023 at 18:55
  • $\begingroup$ @abx I realized it was a property of the general sextic (in disguise), so figured it had $6!=720$ automorphisms. If the Bring surface in $x$ has genus 4, then by analogy what is the genus of the surface in $y$? Definitely $>12$, but I'm curious how high it is. $\endgroup$ Commented May 3, 2023 at 19:07
  • $\begingroup$ @abx Can you convert your comment about 720 into an answer? If you have an estimate for the size of the genus, kindly include that as well. And where does $k=7$ fit in? $\endgroup$ Commented May 3, 2023 at 19:14

1 Answer 1

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The equations $\sum_{i=1}^6 y_i^k = 0$ for $k=1,2,4,7$ do not cut out a curve because the $k=1,2,4$ equations imply the one for $k=7$. (The 7th power sum is a polynomial in the power sums of degree 1 through 6, and there is no way to get 7 as a sum of numbers in $\{3,5,6\}$.) So you have some surface with 720 automorphisms.

If some family of sextics $y^6 + a y^3 + b y + c = 0$ is a sextic resolvent of $x^5 + Ax + B = 0$, then the corresponding curve -- call it $C$ -- is isomorphic to the Bring curve, and thus also has genus 4 and automorphisms by $S_5$, acting by permutations of the $y_i$ by a transitive subgroup of $S_6$. If you find enough $S_6$-invariant equations in $y_1,\ldots,y_6$ to cut out a 1-dimensional variety containing $C$, then that variety is the union of 6 curves each of which is isomorphic with the Bring curve and has automorphism group one of the six transitive subgroups of $S_6$ that's isomorphic with $S_5$.

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