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I am new to the subject of invariant theory, but the Reynolds operator popped up so I tried to calculated some examples for myself. I computed the invariant polynomials under the cylic group $C_3$ of order $3$, given by $120°$ rotations in the plane.

These polynomials, like

$$\frac 1 4 (x^3-3xy^2), \frac 1 4 (3x^2y-y^3), x^2+y^2, ...$$

and their combinations give rise to nice 120°-symmetric plane curves. However, from what I can judge by looking at the plotted images in Wolframalpha, there always seems to be a reflection symmetry along some axis. enter image description here

Can we get rid of that? Is there an algebraic curve with just 120°-rotational symmetry but without additional reflection axes?

Of course we can take a union of e.g. appropriately arranged circles expressed in a single equation, so we should just talk about irreducible curves. Is my observation even correct? Maybe there is some obvious counterexample. Thank you

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1 Answer 1

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Posting this just to get things started. I haven't checked that this is absolutely irreducible, but it looks promising:

$$x^3 - 3 x y^2 = C + (x^3 - 3 x y^2)^2 + y^3 - 3 y x^2. $$

Built from the invariants that you listed. If we have $C=0$ then there will be a singularity at the origin, and Mathematica has problems drawing it smoothly. Below there is a picture of the real points (I got the impression that you were only interested in real plane curves) with $C=1/24$.

enter image description here

Adding a positive definite term of a high enough degree should give us a compact variant: $$ x^3 - 3 x y^2 + (x^2 + y^2)^4 = \frac1{24} + (x^3 - 3 x y^2)^2 + y^3 - 3 y x^2 $$ enter image description here

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  • $\begingroup$ So just an inhomogeneous equation involving the real and imaginary parts of $(x+iy)^3$. Using $C<0$ in the latter example gives a real curve with three connectivity components. $\endgroup$ Nov 16, 2015 at 18:48
  • $\begingroup$ My first idea was to build a cyclic Galois extension of function fields, but I'm not sure that works well, because the elements of the Galois group do not necessarily come from rotations. $\endgroup$ Nov 16, 2015 at 19:13
  • $\begingroup$ Thanks. The trick for achieving compactness is neat, forgot about that. It seems that if one adds high enough powers, a sufficiently general polynomial will break the additional symmetry. Any more systematic approach would be appreciated though. Galois theory sounds interesting $\endgroup$
    – Dario
    Nov 17, 2015 at 17:26
  • $\begingroup$ What does 3-fold or n-fold symmetry say about the variety? I am facing some problem where n-fold symmetry is known, but the actual variety is not directly given (but rather requires complicated symbolic reduction). So far, Mathematica has failed me (3 days straight and the software won't budge)... $\endgroup$
    – Troy Woo
    Dec 6, 2016 at 14:02

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