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Sep 14, 2020 at 13:17 vote accept Ben C
Sep 13, 2020 at 15:07 answer added David E Speyer timeline score: 13
Sep 13, 2020 at 13:31 history edited David E Speyer
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Sep 13, 2020 at 2:11 comment added David E Speyer To be clear, by the width of a lattice polytope $P$, I mean the following: the minimum, over nonzero lattice vectors $v$, of $\max_{p \in P} \langle p,v \rangle - \min_{q \in P} \langle q,v \rangle$.
Sep 12, 2020 at 23:50 comment added David E Speyer Some observations: The divisor $C$ on $S$ corresponds to a lattice polygon, call it $P$. The number of interior lattice points of this polygon is $g$. If the polygon has width $k$ in any lattice direction, then the curve has gonality $\leq k$. Thus, if we are to get a generic curve of genus $g$, the width must be $\geq g/2+1$ in every direction. Also, if $N$ is the number of lattice points and we want to get a generic curve, we must have $N-3 \geq 3g-3$, since the moduli space of curves is $3g-3$ dimensional, and we lose $3$ dimensions rescaling coordinates.
Sep 11, 2020 at 16:46 comment added Ben C @Jef thanks for the comment. I am familiar with Castryck's work and I think you are correct nondegenerate is a stronger condition it corresponds to an embedding that intersects the toric divisor transversally. There is a related notion of weakly nondegenerate which I do believe corresponds to what I am looking for. However, it appears less is know about when a curve admits a weakly nondegenerate equation.
Sep 11, 2020 at 10:56 comment added Jef You might already know this, but there is the notion of a nondegenerate curve which is related but (I think) stronger than your condition, for a definition see Castryck and Voight's article. In that paper they prove that every curve over an algebraically closed field is nondegenerate if its genus is $\leq 4$, so your example will necessarily have genus at least $5$.
Sep 10, 2020 at 20:22 history edited Ben C CC BY-SA 4.0
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Sep 10, 2020 at 20:10 history asked Ben C CC BY-SA 4.0