Timeline for Is it possible to check two curves on birational equivalence by some computer algebra system?
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
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| Nov 9, 2011 at 14:08 | comment | added | Jack Huizenga | Yes, sorry if my quantifiers weren't clear. | |
| Nov 9, 2011 at 12:52 | comment | added | David E Speyer | Re Jack Huizenga's first suggestion: You have to check all triples of roots of the second branch locus, you can't just choose $3$ roots of $f$ and $3$ roots of $g$. | |
| Nov 9, 2011 at 12:22 | history | edited | Jack Huizenga | CC BY-SA 3.0 |
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| Nov 9, 2011 at 9:44 | comment | added | Felipe Voloch | For genus two (as in the OP's example), there are the Igusa invariants. | |
| Nov 9, 2011 at 5:00 | comment | added | Will Sawin | The problem is just the classification of binary quadratic/cubic/quintic/etc. forms, up to $GL(n)$ action. (We view a function on $\mathbb P^1$ as a function on $\mathbb A^2$, and see what kind of function it is). This is a fairly standard representation theory problem, but I don't know the answer. Probably, one can show that the ring of polynomial invariants completely classifies the orbit. For elliptic curves, the $j$-invariant, which is a rational function of the coefficients, fits the bill. But the formulas for these in higher dimensions might be even nastier. | |
| Nov 9, 2011 at 3:48 | comment | added | Jack Huizenga | Perhaps something smarter would be to start with $y^2=f(x)$ and $y^2=g(x)$ and check if $f$ and $g$ are in the same orbit under $PGL(2)$, as this skips the root-finding process. I'm not sure how to do this directly though. | |
| Nov 9, 2011 at 3:47 | comment | added | Jack Huizenga | I don't have a "smart" way, but if the curve is $y^2 = f(x)$ then the branch points are the roots of $f$. Normalize the branch points of one of the curves by sending 3 points to $0,1,\infty$. Given any 3 branch points of the second curve, map them to $0,1,\infty$ and check if the images of the other branch points coincide with the branch points of the original curve. | |
| Nov 9, 2011 at 2:59 | comment | added | David E Speyer | So, how do you check whether the branch divisors are projectively equivalent? My gut says that this should be easy, but I don't actually know how. | |
| Nov 9, 2011 at 0:33 | history | answered | Jack Huizenga | CC BY-SA 3.0 |