2019
DOI: 10.1007/s10801-019-00871-0
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Unexpected curves arising from special line arrangements

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Cited by 23 publications
(23 citation statements)
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“…A line arrangement in is supersolvable if it has a so-called modular point , that is, a point with the property that if and if is the intersection of and then the line joining and is a line of . (See [DMO18] for examples of supersolvable line arrangements giving rise to unexpected curves.) A standard fact is that if is a supersolvable line arrangement consisting of lines, of which pass through the modular point , then is free, and the splitting type is .…”
Section: Examplesmentioning
confidence: 99%
“…A line arrangement in is supersolvable if it has a so-called modular point , that is, a point with the property that if and if is the intersection of and then the line joining and is a line of . (See [DMO18] for examples of supersolvable line arrangements giving rise to unexpected curves.) A standard fact is that if is a supersolvable line arrangement consisting of lines, of which pass through the modular point , then is free, and the splitting type is .…”
Section: Examplesmentioning
confidence: 99%
“…Recently, unexpected curves and hypersurfaces have been intensively studied. In the paper [6], Di Marca, Malara and Oneto present a way to produce families of unexpected curves using supersolvable arrangements of lines. In [2], Bauer, Malara, Szemberg and Szpond consider the existence of special linear systems in P 3 and exhibit there a quartic surface with unexpected postulation properties.…”
Section: Mathematicians Have Been Working On Interpolation Problems Fmentioning
confidence: 99%
“…, a n ] is a form (possibly a constant) describing the codimension 1 part of the locus of points B such that F (B, ·) ≡ 0. In the case of the unexpected plane quartics discussed above, we have n = 2, d = 4 and m = 3 so F has bidegree ( m+n−1 n (d−m+1), d) = (12, 4), but H = ab 3 c 3 (a+b−c)(a−b+c) (see Example 13), so in this case H has bidegree (9, 0) and G has bidegree (3,4) which gives the bidegree of F to be (12, 4), as asserted. While we do not have an explicit expression for the bidegree of H (or for G), our results show that BMSS duality holds for F (and hence for G, whatever its bidegree is).…”
Section: Introductionmentioning
confidence: 99%
“…The form G has bidegree (3,4) and defines a hypersurface in P 2 × P 2 . The fibers with respect to the first projection are quartics G(B, ·) = Q B (·) = 0 with a triple point at B, while the fibers with respect to the second projection are cubics G(·, S) = Q S (·) = 0 with a triple point at S. The connection between G(·, S) and G(B, ·) is that they have the same tangent cone at B = S.…”
Section: Introductionmentioning
confidence: 99%