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This is a bit of a silly question, but I've been puzzled on how this can be done, so I ask here.

You are given an implicit Cartesian equation like

$$4 x^4-4 x^3+8 y^2 x^2-27 x^2+12 y^2 x+4 y^4-27 y^2+27=0$$

or

$$x^6+3 y^2 x^4-3 x^4+3 y^4 x^2+21 y^2 x^2+3 x^2+y^6-3 y^4+3 y^2-1=0$$

and are told that you are not allowed to graph the equation (which makes the determination of geometric properties a bit difficult), but are allowed to do manipulations to the equation. Is it possible to determine the symmetry of the graph of a curve only from its equation? Or to be more precise, can one determine $n$ in the sentence "the curve with implicit Cartesian equation so-and-so has n-fold symmetry" just from the implicit equation?

For some curves, one quick way of determining symmetry is to transform to polar coordinates: for instance,

$$(x^2+y^2)^2=x^2-y^2$$

transforms into the polar equation

$$r^2=\cos\;2\theta$$

and due to the periodicity of the cosine, it is seen that it has two-fold symmetry (i.e., invariant under rotations of $\frac{2\pi}{2}=\pi$ radians.) However, we all know that not all algebraic curves can undergo such a neat conversion, and that this method presumes that the origin is "in the vicinity" of the curve. (On the other hand, if you determine through the usual methods that the curve has the origin as a singular point, conversion to polar coordinates is especially attractive!)

Thus, I would like to see if a more general method is available.

Tangentially, we know that only the circle has infinitely many axes of symmetry, so if the method can also demonstrate that only the circle has this property, that would be a plus.

Last but not the least (and since I feel it does not deserve to be asked as a separate question): is there a bounded algebraic curve that has no axes of symmetry whatsoever? Most of the examples I see have at least two-fold symmetry.

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A "generic" algebraic curve should have no rotational or reflectional symmetries; an example like y^4 + y + x^6 + x = 5 should work. (In general, what kind of symmetries are we talking about?) –  Qiaochu Yuan Nov 8 '10 at 0:59
    
@Qiaochu: Yes, that's why I'm interested if there's a procedure to determine if a curve does have symmetry of some sort, since a curve having symmetry is exceptional. I suppose considering just rotational symmetry would be sufficient for now (like in the example I transformed to polar coordinates). Your example settles my last question, so thanks for that! –  J. M. Nov 8 '10 at 1:10
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1 Answer

I'll assume we're talking about Euclidean symmetries. Then we're looking at transformations of the form $x \mapsto ax + by + c, y = -bx + ay + d$ where $a^2 + b^2 = 1$ (or the analogous set of equations for orientation-reversing symmetries). The condition $k f(x, y) = f(ax + by + c, -bx + ay + d)$ for some nonzero $k$ (which can be determined in terms of $a, b, c, d$ by evaluating both sides at a point where $f(x, y) \neq 0$), when equating coefficients, into a system of polynomial equations in $a, b, c, d, k$. Put this together with the equation $a^2 + b^2 = 1$ and you have a system you can plug into a CAS which will, using Grobner basis methods, tell you what the solutions look like.

(I am assuming that $f$ has a certain nondegeneracy property here, namely that $f$ generates the ideal of polynomial functions vanishing on $f(x, y) = 0$ in $\mathbb{R}[x, y]$.)

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So in essence, I use Gröbner to determine sines and cosines? Looks good, thanks! I'll wait for other answers and accept this if I don't get anything else. –  J. M. Nov 8 '10 at 1:21
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